contest_id
stringclasses 33
values | problem_id
stringclasses 14
values | statement
stringclasses 181
values | tags
sequencelengths 1
8
| code
stringlengths 21
64.5k
| language
stringclasses 3
values |
|---|---|---|---|---|---|
1324 | E | E. Sleeping Scheduletime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputVova had a pretty weird sleeping schedule. There are hh hours in a day. Vova will sleep exactly nn times. The ii-th time he will sleep exactly after aiai hours from the time he woke up. You can assume that Vova woke up exactly at the beginning of this story (the initial time is 00). Each time Vova sleeps exactly one day (in other words, hh hours).Vova thinks that the ii-th sleeping time is good if he starts to sleep between hours ll and rr inclusive.Vova can control himself and before the ii-th time can choose between two options: go to sleep after aiai hours or after ai−1ai−1 hours.Your task is to say the maximum number of good sleeping times Vova can obtain if he acts optimally.InputThe first line of the input contains four integers n,h,ln,h,l and rr (1≤n≤2000,3≤h≤2000,0≤l≤r<h1≤n≤2000,3≤h≤2000,0≤l≤r<h) — the number of times Vova goes to sleep, the number of hours in a day and the segment of the good sleeping time.The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai<h1≤ai<h), where aiai is the number of hours after which Vova goes to sleep the ii-th time.OutputPrint one integer — the maximum number of good sleeping times Vova can obtain if he acts optimally.ExampleInputCopy7 24 21 23
16 17 14 20 20 11 22
OutputCopy3
NoteThe maximum number of good times in the example is 33.The story starts from t=0t=0. Then Vova goes to sleep after a1−1a1−1 hours, now the time is 1515. This time is not good. Then Vova goes to sleep after a2−1a2−1 hours, now the time is 15+16=715+16=7. This time is also not good. Then Vova goes to sleep after a3a3 hours, now the time is 7+14=217+14=21. This time is good. Then Vova goes to sleep after a4−1a4−1 hours, now the time is 21+19=1621+19=16. This time is not good. Then Vova goes to sleep after a5a5 hours, now the time is 16+20=1216+20=12. This time is not good. Then Vova goes to sleep after a6a6 hours, now the time is 12+11=2312+11=23. This time is good. Then Vova goes to sleep after a7a7 hours, now the time is 23+22=2123+22=21. This time is also good. | [
"dp",
"implementation"
] | from heapq import heappush, heappop
from collections import defaultdict, Counter, deque
import threading
import sys
import bisect
# input = sys.stdin.readline
def ri(): return int(input())
def rs(): return input()
def rl(): return list(map(int, input().split()))
def rls(): return list(input().split())
# threading.stack_size(10**8)
# sys.setrecursionlimit(10**6)
def main():
n, h, l, r = rl()
a = rl()
dp = [[-1 << 33 for i in range(n+1)] for i in range(n+1)]
def ins(x):
return (l <= x and x <= r)
dp[0][0] = 0
cs = 0
for i in range(n):
cs += a[i]
for j in range(n+1):
dp[i+1][j] = max(dp[i+1][j], dp[i][j]+ins((cs-j) % h))
if j < n:
dp[i+1][j+1] = max(dp[i+1][j+1], dp[i][j]+ins((cs-j-1) % h))
print(max(dp[-1]))
pass
main()
# threading.Thread(target=main).start()
| py |
1285 | B | B. Just Eat It!time limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputToday; Yasser and Adel are at the shop buying cupcakes. There are nn cupcake types, arranged from 11 to nn on the shelf, and there are infinitely many of each type. The tastiness of a cupcake of type ii is an integer aiai. There are both tasty and nasty cupcakes, so the tastiness can be positive, zero or negative.Yasser, of course, wants to try them all, so he will buy exactly one cupcake of each type.On the other hand, Adel will choose some segment [l,r][l,r] (1≤l≤r≤n)(1≤l≤r≤n) that does not include all of cupcakes (he can't choose [l,r]=[1,n][l,r]=[1,n]) and buy exactly one cupcake of each of types l,l+1,…,rl,l+1,…,r.After that they will compare the total tastiness of the cupcakes each of them have bought. Yasser will be happy if the total tastiness of cupcakes he buys is strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice.For example, let the tastinesses of the cupcakes be [7,4,−1][7,4,−1]. Yasser will buy all of them, the total tastiness will be 7+4−1=107+4−1=10. Adel can choose segments [7],[4],[−1],[7,4][7],[4],[−1],[7,4] or [4,−1][4,−1], their total tastinesses are 7,4,−1,117,4,−1,11 and 33, respectively. Adel can choose segment with tastiness 1111, and as 1010 is not strictly greater than 1111, Yasser won't be happy :(Find out if Yasser will be happy after visiting the shop.InputEach test contains multiple test cases. The first line contains the number of test cases tt (1≤t≤1041≤t≤104). The description of the test cases follows.The first line of each test case contains nn (2≤n≤1052≤n≤105).The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (−109≤ai≤109−109≤ai≤109), where aiai represents the tastiness of the ii-th type of cupcake.It is guaranteed that the sum of nn over all test cases doesn't exceed 105105.OutputFor each test case, print "YES", if the total tastiness of cupcakes Yasser buys will always be strictly greater than the total tastiness of cupcakes Adel buys regardless of Adel's choice. Otherwise, print "NO".ExampleInputCopy3
4
1 2 3 4
3
7 4 -1
3
5 -5 5
OutputCopyYES
NO
NO
NoteIn the first example; the total tastiness of any segment Adel can choose is less than the total tastiness of all cupcakes.In the second example; Adel will choose the segment [1,2][1,2] with total tastiness 1111, which is not less than the total tastiness of all cupcakes, which is 1010.In the third example, Adel can choose the segment [3,3][3,3] with total tastiness of 55. Note that Yasser's cupcakes' total tastiness is also 55, so in that case, the total tastiness of Yasser's cupcakes isn't strictly greater than the total tastiness of Adel's cupcakes. | [
"dp",
"greedy",
"implementation"
] | n = int(input())
for t in range(n):
m = int(input())
a = list(map(int, input().split()))
ans1 = sum(a)
ans2 = 0
dlina = 0
maxim = 0
for i in range(m):
if dlina != m - 1:
ans2 += a[i]
if ans2 > 0:
dlina += 1
else:
ans2 = 0
dlina = 0
maxim = max(maxim ,ans2)
else:
maxim = max(maxim, sum(a[0:m-1]), sum(a[1:]))
break
if maxim >= ans1:
print("NO")
else:
print("YES")
| py |
1292 | B | B. Aroma's Searchtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputTHE SxPLAY & KIVΛ - 漂流 KIVΛ & Nikki Simmons - PerspectivesWith a new body; our idol Aroma White (or should we call her Kaori Minamiya?) begins to uncover her lost past through the OS space.The space can be considered a 2D plane, with an infinite number of data nodes, indexed from 00, with their coordinates defined as follows: The coordinates of the 00-th node is (x0,y0)(x0,y0) For i>0i>0, the coordinates of ii-th node is (ax⋅xi−1+bx,ay⋅yi−1+by)(ax⋅xi−1+bx,ay⋅yi−1+by) Initially Aroma stands at the point (xs,ys)(xs,ys). She can stay in OS space for at most tt seconds, because after this time she has to warp back to the real world. She doesn't need to return to the entry point (xs,ys)(xs,ys) to warp home.While within the OS space, Aroma can do the following actions: From the point (x,y)(x,y), Aroma can move to one of the following points: (x−1,y)(x−1,y), (x+1,y)(x+1,y), (x,y−1)(x,y−1) or (x,y+1)(x,y+1). This action requires 11 second. If there is a data node at where Aroma is staying, she can collect it. We can assume this action costs 00 seconds. Of course, each data node can be collected at most once. Aroma wants to collect as many data as possible before warping back. Can you help her in calculating the maximum number of data nodes she could collect within tt seconds?InputThe first line contains integers x0x0, y0y0, axax, ayay, bxbx, byby (1≤x0,y0≤10161≤x0,y0≤1016, 2≤ax,ay≤1002≤ax,ay≤100, 0≤bx,by≤10160≤bx,by≤1016), which define the coordinates of the data nodes.The second line contains integers xsxs, ysys, tt (1≤xs,ys,t≤10161≤xs,ys,t≤1016) – the initial Aroma's coordinates and the amount of time available.OutputPrint a single integer — the maximum number of data nodes Aroma can collect within tt seconds.ExamplesInputCopy1 1 2 3 1 0
2 4 20
OutputCopy3InputCopy1 1 2 3 1 0
15 27 26
OutputCopy2InputCopy1 1 2 3 1 0
2 2 1
OutputCopy0NoteIn all three examples; the coordinates of the first 55 data nodes are (1,1)(1,1), (3,3)(3,3), (7,9)(7,9), (15,27)(15,27) and (31,81)(31,81) (remember that nodes are numbered from 00).In the first example, the optimal route to collect 33 nodes is as follows: Go to the coordinates (3,3)(3,3) and collect the 11-st node. This takes |3−2|+|3−4|=2|3−2|+|3−4|=2 seconds. Go to the coordinates (1,1)(1,1) and collect the 00-th node. This takes |1−3|+|1−3|=4|1−3|+|1−3|=4 seconds. Go to the coordinates (7,9)(7,9) and collect the 22-nd node. This takes |7−1|+|9−1|=14|7−1|+|9−1|=14 seconds. In the second example, the optimal route to collect 22 nodes is as follows: Collect the 33-rd node. This requires no seconds. Go to the coordinates (7,9)(7,9) and collect the 22-th node. This takes |15−7|+|27−9|=26|15−7|+|27−9|=26 seconds. In the third example, Aroma can't collect any nodes. She should have taken proper rest instead of rushing into the OS space like that. | [
"brute force",
"constructive algorithms",
"geometry",
"greedy",
"implementation"
] | T = 1
for test_no in range(T):
x0, y0, ax, ay, bx, by = map(int, input().split())
xs, ys, t = map(int, input().split())
LIMIT = 2 ** 62 - 1
x, y = [x0], [y0]
while ((LIMIT - bx) / ax >= x[-1] and (LIMIT - by) / ay >= y[-1]):
x.append(ax * x[-1] + bx)
y.append(ay * y[-1] + by)
n = len(x)
ans = 0
for i in range(n):
for j in range(i, n):
length = x[j] - x[i] + y[j] - y[i]
dist2Left = abs(xs - x[i]) + abs(ys - y[i])
dist2Right = abs(xs - x[j]) + abs(ys - y[j])
if (length <= t - dist2Left or length <= t - dist2Right): ans = max(ans, j-i+1)
print(ans) | py |
1321 | A | A. Contest for Robotstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputPolycarp is preparing the first programming contest for robots. There are nn problems in it, and a lot of robots are going to participate in it. Each robot solving the problem ii gets pipi points, and the score of each robot in the competition is calculated as the sum of pipi over all problems ii solved by it. For each problem, pipi is an integer not less than 11.Two corporations specializing in problem-solving robot manufacturing, "Robo-Coder Inc." and "BionicSolver Industries", are going to register two robots (one for each corporation) for participation as well. Polycarp knows the advantages and flaws of robots produced by these companies, so, for each problem, he knows precisely whether each robot will solve it during the competition. Knowing this, he can try predicting the results — or manipulating them. For some reason (which absolutely cannot involve bribing), Polycarp wants the "Robo-Coder Inc." robot to outperform the "BionicSolver Industries" robot in the competition. Polycarp wants to set the values of pipi in such a way that the "Robo-Coder Inc." robot gets strictly more points than the "BionicSolver Industries" robot. However, if the values of pipi will be large, it may look very suspicious — so Polycarp wants to minimize the maximum value of pipi over all problems. Can you help Polycarp to determine the minimum possible upper bound on the number of points given for solving the problems?InputThe first line contains one integer nn (1≤n≤1001≤n≤100) — the number of problems.The second line contains nn integers r1r1, r2r2, ..., rnrn (0≤ri≤10≤ri≤1). ri=1ri=1 means that the "Robo-Coder Inc." robot will solve the ii-th problem, ri=0ri=0 means that it won't solve the ii-th problem.The third line contains nn integers b1b1, b2b2, ..., bnbn (0≤bi≤10≤bi≤1). bi=1bi=1 means that the "BionicSolver Industries" robot will solve the ii-th problem, bi=0bi=0 means that it won't solve the ii-th problem.OutputIf "Robo-Coder Inc." robot cannot outperform the "BionicSolver Industries" robot by any means, print one integer −1−1.Otherwise, print the minimum possible value of maxi=1npimaxi=1npi, if all values of pipi are set in such a way that the "Robo-Coder Inc." robot gets strictly more points than the "BionicSolver Industries" robot.ExamplesInputCopy5
1 1 1 0 0
0 1 1 1 1
OutputCopy3
InputCopy3
0 0 0
0 0 0
OutputCopy-1
InputCopy4
1 1 1 1
1 1 1 1
OutputCopy-1
InputCopy9
1 0 0 0 0 0 0 0 1
0 1 1 0 1 1 1 1 0
OutputCopy4
NoteIn the first example; one of the valid score assignments is p=[3,1,3,1,1]p=[3,1,3,1,1]. Then the "Robo-Coder" gets 77 points, the "BionicSolver" — 66 points.In the second example, both robots get 00 points, and the score distribution does not matter.In the third example, both robots solve all problems, so their points are equal. | [
"greedy"
] |
n = int(input())
l1 = list(map(int,input().split()))
l2 = list(map(int,input().split()))
s = 0
r = 0
for i in range(n):
if(l1[i]==1 and l2[i]==0):
s = s +1
elif(l1[i]==0 and l2[i]==1):
r = r +1
if(s==r):
if(s==0):
print(-1)
else:
print(2)
elif(s>r):
print(1)
else:
if(s!=0):
c = r//s +1
print(c)
else:
print(-1) | py |
1288 | A | A. Deadlinetime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputAdilbek was assigned to a special project. For Adilbek it means that he has nn days to run a special program and provide its results. But there is a problem: the program needs to run for dd days to calculate the results.Fortunately, Adilbek can optimize the program. If he spends xx (xx is a non-negative integer) days optimizing the program, he will make the program run in ⌈dx+1⌉⌈dx+1⌉ days (⌈a⌉⌈a⌉ is the ceiling function: ⌈2.4⌉=3⌈2.4⌉=3, ⌈2⌉=2⌈2⌉=2). The program cannot be run and optimized simultaneously, so the total number of days he will spend is equal to x+⌈dx+1⌉x+⌈dx+1⌉.Will Adilbek be able to provide the generated results in no more than nn days?InputThe first line contains a single integer TT (1≤T≤501≤T≤50) — the number of test cases.The next TT lines contain test cases – one per line. Each line contains two integers nn and dd (1≤n≤1091≤n≤109, 1≤d≤1091≤d≤109) — the number of days before the deadline and the number of days the program runs.OutputPrint TT answers — one per test case. For each test case print YES (case insensitive) if Adilbek can fit in nn days or NO (case insensitive) otherwise.ExampleInputCopy3
1 1
4 5
5 11
OutputCopyYES
YES
NO
NoteIn the first test case; Adilbek decides not to optimize the program at all; since d≤nd≤n.In the second test case, Adilbek can spend 11 day optimizing the program and it will run ⌈52⌉=3⌈52⌉=3 days. In total, he will spend 44 days and will fit in the limit.In the third test case, it's impossible to fit in the limit. For example, if Adilbek will optimize the program 22 days, it'll still work ⌈112+1⌉=4⌈112+1⌉=4 days. | [
"binary search",
"brute force",
"math",
"ternary search"
] | from math import ceil
for _ in range(int(input())):
n, d = map(int, input().split())
if d <= n:
print("YES")
else:
low = 0
high = n
flag = False
while low <= high:
mid = (low + high) // 2
time = mid + ceil(d / (mid + 1))
if time <= n:
flag = True
break
else:
high = mid - 1
if flag:
print("YES")
else:
print("NO")
| py |
1290 | B | B. Irreducible Anagramstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputLet's call two strings ss and tt anagrams of each other if it is possible to rearrange symbols in the string ss to get a string, equal to tt.Let's consider two strings ss and tt which are anagrams of each other. We say that tt is a reducible anagram of ss if there exists an integer k≥2k≥2 and 2k2k non-empty strings s1,t1,s2,t2,…,sk,tks1,t1,s2,t2,…,sk,tk that satisfy the following conditions: If we write the strings s1,s2,…,sks1,s2,…,sk in order, the resulting string will be equal to ss; If we write the strings t1,t2,…,tkt1,t2,…,tk in order, the resulting string will be equal to tt; For all integers ii between 11 and kk inclusive, sisi and titi are anagrams of each other. If such strings don't exist, then tt is said to be an irreducible anagram of ss. Note that these notions are only defined when ss and tt are anagrams of each other.For example, consider the string s=s= "gamegame". Then the string t=t= "megamage" is a reducible anagram of ss, we may choose for example s1=s1= "game", s2=s2= "gam", s3=s3= "e" and t1=t1= "mega", t2=t2= "mag", t3=t3= "e": On the other hand, we can prove that t=t= "memegaga" is an irreducible anagram of ss.You will be given a string ss and qq queries, represented by two integers 1≤l≤r≤|s|1≤l≤r≤|s| (where |s||s| is equal to the length of the string ss). For each query, you should find if the substring of ss formed by characters from the ll-th to the rr-th has at least one irreducible anagram.InputThe first line contains a string ss, consisting of lowercase English characters (1≤|s|≤2⋅1051≤|s|≤2⋅105).The second line contains a single integer qq (1≤q≤1051≤q≤105) — the number of queries.Each of the following qq lines contain two integers ll and rr (1≤l≤r≤|s|1≤l≤r≤|s|), representing a query for the substring of ss formed by characters from the ll-th to the rr-th.OutputFor each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.ExamplesInputCopyaaaaa
3
1 1
2 4
5 5
OutputCopyYes
No
Yes
InputCopyaabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
OutputCopyNo
Yes
Yes
Yes
No
No
NoteIn the first sample; in the first and third queries; the substring is "a"; which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand; in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s1=s1= "a", s2=s2= "aa", t1=t1= "a", t2=t2= "aa" to show that it is a reducible anagram.In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram. | [
"binary search",
"constructive algorithms",
"data structures",
"strings",
"two pointers"
] | import sys
input = sys.stdin.readline
s = [ord(i)-97 for i in input().strip()]
p = [[0] for i in range(26)]
for i in range(len(s)):
for j in range(26):
p[j].append(p[j][-1])
p[s[i]][i+1] += 1
for z in range(int(input())):
l, r = map(int, input().split())
if l == r:
print('Yes')
continue
x = 0
for i in range(26):
cur = p[i][r]-p[i][l-1]
if cur:
x += 1
if x == 1 or (x == 2 and s[l-1] == s[r-1]):
print('No')
else:
print('Yes')
| py |
1286 | A | A. Garlandtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputVadim loves decorating the Christmas tree; so he got a beautiful garland as a present. It consists of nn light bulbs in a single row. Each bulb has a number from 11 to nn (in arbitrary order), such that all the numbers are distinct. While Vadim was solving problems, his home Carp removed some light bulbs from the garland. Now Vadim wants to put them back on.Vadim wants to put all bulb back on the garland. Vadim defines complexity of a garland to be the number of pairs of adjacent bulbs with numbers with different parity (remainder of the division by 22). For example, the complexity of 1 4 2 3 5 is 22 and the complexity of 1 3 5 7 6 4 2 is 11.No one likes complexity, so Vadim wants to minimize the number of such pairs. Find the way to put all bulbs back on the garland, such that the complexity is as small as possible.InputThe first line contains a single integer nn (1≤n≤1001≤n≤100) — the number of light bulbs on the garland.The second line contains nn integers p1, p2, …, pnp1, p2, …, pn (0≤pi≤n0≤pi≤n) — the number on the ii-th bulb, or 00 if it was removed.OutputOutput a single number — the minimum complexity of the garland.ExamplesInputCopy5
0 5 0 2 3
OutputCopy2
InputCopy7
1 0 0 5 0 0 2
OutputCopy1
NoteIn the first example; one should place light bulbs as 1 5 4 2 3. In that case; the complexity would be equal to 2; because only (5,4)(5,4) and (2,3)(2,3) are the pairs of adjacent bulbs that have different parity.In the second case, one of the correct answers is 1 7 3 5 6 4 2. | [
"dp",
"greedy",
"sortings"
] | import sys
from collections import *
from itertools import *
from math import *
from array import *
from functools import lru_cache
import heapq
import bisect
import random
import io, os
from bisect import *
if sys.hexversion == 50924784:
sys.stdin = open('cfinput.txt')
RI = lambda: map(int, sys.stdin.buffer.readline().split())
RS = lambda: map(bytes.decode, sys.stdin.buffer.readline().strip().split())
RILST = lambda: list(RI())
MOD = 10 ** 9 + 7
"""https://codeforces.com/problemset/problem/1286/A
输入 n(≤100) 和一个长为 n 的数组 p，p 原本是一个 1~n 的排列，但是有些数字丢失了，丢失的数字用 0 表示。
你需要还原 p，使得 p 中相邻元素奇偶性不同的对数最少。输出这个最小值。
输入
5
0 5 0 2 3
输出 2
输入
7
1 0 0 5 0 0 2
输出 1
"""
# 264 ms
def solve1(n, a):
if n == 1:
return print(0)
even = n // 2 # 偶数
ood = n - even # 奇数
for v in a:
if not v:
continue
if v & 1:
ood -= 1
else:
even -= 1
# f[i][j][k][0] 前i个数，填了j个奇数，l个偶数时，且末位是偶数的情况
# f[i][j][k][1] 前i个数，填了j个奇数，l个偶数时，且末位是奇数的情况
f = [[[[inf] * 2 for _ in range(even + 1)] for _ in range(ood + 1)] for _ in range(n)]
# f[0][0][0] = 0
if a[0] == 0:
f[0][1][0][1] = 0 # a[0]放奇数
f[0][0][1][0] = 0 # a[0]放偶数
else:
if a[0] & 1: # 奇数
f[0][0][0][1] = 0
else:
f[0][0][0][0] = 0
# print(ood,even)
for i in range(1, n):
if a[i] == 0:
for j in range(ood + 1):
for k in range(even + 1):
if k:
# a[i]放偶数，前一个是奇数就+1否则不+
f[i][j][k][0] = min(f[i][j][k][0], f[i - 1][j][k - 1][0])
f[i][j][k][0] = min(f[i][j][k][0], f[i - 1][j][k - 1][1] + 1)
if j:
# a[i]放奇数
f[i][j][k][1] = min(f[i][j][k][1], f[i - 1][j - 1][k][0] + 1)
f[i][j][k][1] = min(f[i][j][k][1], f[i - 1][j - 1][k][1])
else:
if a[i] & 1:
for j in range(ood + 1):
for k in range(even + 1):
f[i][j][k][1] = min(f[i - 1][j][k][1], f[i - 1][j][k][0] + 1)
else:
for j in range(ood + 1):
for k in range(even + 1):
f[i][j][k][0] = min(f[i - 1][j][k][1] + 1, f[i - 1][j][k][0])
# print(f)
print(min(f[-1][-1][-1]))
# 155 ms
def solve2(n, a):
if n == 1:
return print(0)
even = n // 2 # 偶数
ood = n - even # 奇数
for v in a:
if not v:
continue
if v & 1:
ood -= 1
else:
even -= 1
f = [[[inf] * 2 for _ in range(even + 1)] for _ in range(ood + 1)]
# f[0][0][0] = 0
if a[0] == 0:
f[1][0][1] = 0 # a[0]放奇数
f[0][1][0] = 0 # a[0]放偶数
else:
if a[0] & 1: # 奇数
f[0][0][1] = 0
else:
f[0][0][0] = 0
# print(ood,even)
for i in range(1, n):
g = f
f = [[[inf] * 2 for _ in range(even + 1)] for _ in range(ood + 1)]
if a[i] == 0:
for j in range(ood + 1):
for k in range(even + 1):
if k:
# a[i]放偶数，前一个是奇数就+1否则不+
f[j][k][0] = min(f[j][k][0], g[j][k - 1][0])
f[j][k][0] = min(f[j][k][0], g[j][k - 1][1] + 1)
if j:
# a[i]放奇数
f[j][k][1] = min(f[j][k][1], g[j - 1][k][0] + 1)
f[j][k][1] = min(f[j][k][1], g[j - 1][k][1])
else:
if a[i] & 1:
for j in range(ood + 1):
for k in range(even + 1):
f[j][k][1] = min(g[j][k][1], g[j][k][0] + 1)
else:
for j in range(ood + 1):
for k in range(even + 1):
f[j][k][0] = min(g[j][k][1] + 1, g[j][k][0])
# print(f)
print(min(f[-1][-1]))
# ms
def solve(n, a):
if n == 1:
return print(0)
even = n // 2 # 偶数
ood = n - even # 奇数
for v in a:
if not v:
continue
if v & 1:
ood -= 1
else:
even -= 1
# f[i][j][0] 前i个数，填了j个偶数且末尾是偶数的情况
f = [[inf] * 2 for _ in range(even + 1)]
if a[0] == 0:
f[0][1] = 0 # a[0]放奇数
f[1][0] = 0 # a[0]放偶数
else:
if a[0] & 1: # 奇数
f[0][1] = 0
else:
f[0][0] = 0
# print(ood,even)
pos = 0 + (a[0] == 0)
for i in range(1, n):
g = f
f = [[inf] * 2 for _ in range(even + 1)]
if a[i] == 0:
pos += 1
for j in range(min(pos + 1, even + 1)):
if j:
f[j][0] = min(g[j - 1][0], g[j - 1][1] + 1)
f[j][1] = min(g[j][1], g[j][0] + 1)
else:
p = a[i] & 1
for j in range(min(pos + 1, even + 1)):
f[j][p] = min(g[j][p], g[j][p ^ 1]+1)
# if a[i] & 1:
# for j in range(min(pos + 1, ood + 1)):
# f[j][1] = min(g[j][1], g[j][0] + 1)
# else:
# for j in range(min(pos + 1, ood + 1)):
# f[j][0] = min(g[j][1] + 1, g[j][0])
# print(f)
print(min(f[-1]))
if __name__ == '__main__':
n, = RI()
a = RILST()
solve(n, a)
| py |
1299 | B | B. Aerodynamictime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputGuy-Manuel and Thomas are going to build a polygon spaceship. You're given a strictly convex (i. e. no three points are collinear) polygon PP which is defined by coordinates of its vertices. Define P(x,y)P(x,y) as a polygon obtained by translating PP by vector (x,y)−→−−(x,y)→. The picture below depicts an example of the translation:Define TT as a set of points which is the union of all P(x,y)P(x,y) such that the origin (0,0)(0,0) lies in P(x,y)P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y)(x,y) lies in TT only if there are two points A,BA,B in PP such that AB−→−=(x,y)−→−−AB→=(x,y)→. One can prove TT is a polygon too. For example, if PP is a regular triangle then TT is a regular hexagon. At the picture below PP is drawn in black and some P(x,y)P(x,y) which contain the origin are drawn in colored: The spaceship has the best aerodynamic performance if PP and TT are similar. Your task is to check whether the polygons PP and TT are similar.InputThe first line of input will contain a single integer nn (3≤n≤1053≤n≤105) — the number of points.The ii-th of the next nn lines contains two integers xi,yixi,yi (|xi|,|yi|≤109|xi|,|yi|≤109), denoting the coordinates of the ii-th vertex.It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.OutputOutput "YES" in a separate line, if PP and TT are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).ExamplesInputCopy4
1 0
4 1
3 4
0 3
OutputCopyYESInputCopy3
100 86
50 0
150 0
OutputCopynOInputCopy8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
OutputCopyYESNoteThe following image shows the first sample: both PP and TT are squares. The second sample was shown in the statements. | [
"geometry"
] | import sys
from array import array
def center_symmetry(xs, ys):
if len(xs) & 1:
return 0
curx, cury = 1e9 + 1, 1e9 + 1
for i in range(n // 2):
l, r = i, i + n // 2
mx = abs(xs[l] - xs[r]) + 2 * min(xs[l], xs[r])
my = abs(ys[l] - ys[r]) + 2 * min(ys[l], ys[r])
if (curx == cury == 1e9 + 1) or curx == mx and cury == my:
curx, cury = mx, my
else:
return 0
return curx, cury
input = lambda: sys.stdin.buffer.readline().decode().strip()
n = int(input())
ys, xs = array('i'), array('i')
for _ in range(n):
x, y = map(int, input().split())
ys.append(y)
xs.append(x)
ret = center_symmetry(xs, ys)
print('no' if not ret else 'yes')
| py |
1284 | B | B. New Year and Ascent Sequencetime limit per test2 secondsmemory limit per test1024 megabytesinputstandard inputoutputstandard outputA sequence a=[a1,a2,…,al]a=[a1,a2,…,al] of length ll has an ascent if there exists a pair of indices (i,j)(i,j) such that 1≤i<j≤l1≤i<j≤l and ai<ajai<aj. For example, the sequence [0,2,0,2,0][0,2,0,2,0] has an ascent because of the pair (1,4)(1,4), but the sequence [4,3,3,3,1][4,3,3,3,1] doesn't have an ascent.Let's call a concatenation of sequences pp and qq the sequence that is obtained by writing down sequences pp and qq one right after another without changing the order. For example, the concatenation of the [0,2,0,2,0][0,2,0,2,0] and [4,3,3,3,1][4,3,3,3,1] is the sequence [0,2,0,2,0,4,3,3,3,1][0,2,0,2,0,4,3,3,3,1]. The concatenation of sequences pp and qq is denoted as p+qp+q.Gyeonggeun thinks that sequences with ascents bring luck. Therefore, he wants to make many such sequences for the new year. Gyeonggeun has nn sequences s1,s2,…,sns1,s2,…,sn which may have different lengths. Gyeonggeun will consider all n2n2 pairs of sequences sxsx and sysy (1≤x,y≤n1≤x,y≤n), and will check if its concatenation sx+sysx+sy has an ascent. Note that he may select the same sequence twice, and the order of selection matters.Please count the number of pairs (x,yx,y) of sequences s1,s2,…,sns1,s2,…,sn whose concatenation sx+sysx+sy contains an ascent.InputThe first line contains the number nn (1≤n≤1000001≤n≤100000) denoting the number of sequences.The next nn lines contain the number lili (1≤li1≤li) denoting the length of sisi, followed by lili integers si,1,si,2,…,si,lisi,1,si,2,…,si,li (0≤si,j≤1060≤si,j≤106) denoting the sequence sisi. It is guaranteed that the sum of all lili does not exceed 100000100000.OutputPrint a single integer, the number of pairs of sequences whose concatenation has an ascent.ExamplesInputCopy5
1 1
1 1
1 2
1 4
1 3
OutputCopy9
InputCopy3
4 2 0 2 0
6 9 9 8 8 7 7
1 6
OutputCopy7
InputCopy10
3 62 24 39
1 17
1 99
1 60
1 64
1 30
2 79 29
2 20 73
2 85 37
1 100
OutputCopy72
NoteFor the first example; the following 99 arrays have an ascent: [1,2],[1,2],[1,3],[1,3],[1,4],[1,4],[2,3],[2,4],[3,4][1,2],[1,2],[1,3],[1,3],[1,4],[1,4],[2,3],[2,4],[3,4]. Arrays with the same contents are counted as their occurences. | [
"binary search",
"combinatorics",
"data structures",
"dp",
"implementation",
"sortings"
] | n = int(input())
minimos, maximos = [], []
asc, nasc = 0, 0
for i in range(n):
sequencia = list(map(int, input().split(" ")))[1:]
for j in range(len(sequencia)):
if j > 0 and sequencia[j] > sequencia[j - 1]:
asc += 1
break
elif j == len(sequencia) - 1:
nasc += 1
minimos.append(min(sequencia))
maximos.append((max(sequencia)))
total = asc * asc + 2 * asc * nasc
minimos.sort()
maximos.sort()
i = 0
for k in maximos:
while i < nasc and minimos[i] < k:
i += 1
total += i
print(total)
| py |
1292 | D | D. Chaotic V.time limit per test2 secondsmemory limit per test512 megabytesinputstandard inputoutputstandard outputÆsir - CHAOS Æsir - V."Everything has been planned out. No more hidden concerns. The condition of Cytus is also perfect.The time right now...... 00:01:12......It's time."The emotion samples are now sufficient. After almost 3 years; it's time for Ivy to awake her bonded sister, Vanessa.The system inside A.R.C.'s Library core can be considered as an undirected graph with infinite number of processing nodes, numbered with all positive integers (1,2,3,…1,2,3,…). The node with a number xx (x>1x>1), is directly connected with a node with number xf(x)xf(x), with f(x)f(x) being the lowest prime divisor of xx.Vanessa's mind is divided into nn fragments. Due to more than 500 years of coma, the fragments have been scattered: the ii-th fragment is now located at the node with a number ki!ki! (a factorial of kiki).To maximize the chance of successful awakening, Ivy decides to place the samples in a node PP, so that the total length of paths from each fragment to PP is smallest possible. If there are multiple fragments located at the same node, the path from that node to PP needs to be counted multiple times.In the world of zeros and ones, such a requirement is very simple for Ivy. Not longer than a second later, she has already figured out such a node.But for a mere human like you, is this still possible?For simplicity, please answer the minimal sum of paths' lengths from every fragment to the emotion samples' assembly node PP.InputThe first line contains an integer nn (1≤n≤1061≤n≤106) — number of fragments of Vanessa's mind.The second line contains nn integers: k1,k2,…,knk1,k2,…,kn (0≤ki≤50000≤ki≤5000), denoting the nodes where fragments of Vanessa's mind are located: the ii-th fragment is at the node with a number ki!ki!.OutputPrint a single integer, denoting the minimal sum of path from every fragment to the node with the emotion samples (a.k.a. node PP).As a reminder, if there are multiple fragments at the same node, the distance from that node to PP needs to be counted multiple times as well.ExamplesInputCopy3
2 1 4
OutputCopy5
InputCopy4
3 1 4 4
OutputCopy6
InputCopy4
3 1 4 1
OutputCopy6
InputCopy5
3 1 4 1 5
OutputCopy11
NoteConsidering the first 2424 nodes of the system; the node network will look as follows (the nodes 1!1!, 2!2!, 3!3!, 4!4! are drawn bold):For the first example, Ivy will place the emotion samples at the node 11. From here: The distance from Vanessa's first fragment to the node 11 is 11. The distance from Vanessa's second fragment to the node 11 is 00. The distance from Vanessa's third fragment to the node 11 is 44. The total length is 55.For the second example, the assembly node will be 66. From here: The distance from Vanessa's first fragment to the node 66 is 00. The distance from Vanessa's second fragment to the node 66 is 22. The distance from Vanessa's third fragment to the node 66 is 22. The distance from Vanessa's fourth fragment to the node 66 is again 22. The total path length is 66. | [
"dp",
"graphs",
"greedy",
"math",
"number theory",
"trees"
] |
import copy
M = 5003
Primes = []
Rob = [0] * M
for x in range(2,M):
if Rob[x]==0:
Primes.append(x)
y = x+x
while y<M:
Rob[y]+=1
y+=x
P = len(Primes)
n = int(input())
Cnt = [0] * M
for a in input().split():
Cnt[int(a)] +=1
PF = []
PF.append( [0] * P)
s = 0
res = 0
for k in range(1, M):
PF.append(copy.copy(PF[-1]))
x = k
for p in range(P):
while x % Primes[p] ==0:
PF[k][p]+=1
x/=Primes[p]
s+=1
res += Cnt[k] * s
for i in range(M):
while (len(PF[i])>0 and PF[i][-1]==0):
PF[i].pop()
BPD = [ len(PF[i]) for i in range(M) ]
branched = 0
frequency = [0] * (P+1)
while ( max(BPD) > 0):
# Find the best prime divisors:
for i in range(P+1): frequency[i] = 0
for i in range(M): frequency[BPD[i]] += Cnt[i]
m = max(frequency)
pr = frequency.index(m)
if pr==0:
break
for i in range(M):
if BPD[i]!=pr:
branched += Cnt[i]
Cnt[i] = 0
BPD[i] = 0
if branched * 2 >= n:
break
res += 2* branched - n
for i in range(M):
if (BPD[i]==0): continue
if (len(PF[i])>0):
PF[i][-1]-=1
while (len(PF[i])>0 and PF[i][-1]==0):
PF[i].pop()
BPD[i] = len(PF[i])
print(res)
| py |
1324 | C | C. Frog Jumpstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThere is a frog staying to the left of the string s=s1s2…sns=s1s2…sn consisting of nn characters (to be more precise, the frog initially stays at the cell 00). Each character of ss is either 'L' or 'R'. It means that if the frog is staying at the ii-th cell and the ii-th character is 'L', the frog can jump only to the left. If the frog is staying at the ii-th cell and the ii-th character is 'R', the frog can jump only to the right. The frog can jump only to the right from the cell 00.Note that the frog can jump into the same cell twice and can perform as many jumps as it needs.The frog wants to reach the n+1n+1-th cell. The frog chooses some positive integer value dd before the first jump (and cannot change it later) and jumps by no more than dd cells at once. I.e. if the ii-th character is 'L' then the frog can jump to any cell in a range [max(0,i−d);i−1][max(0,i−d);i−1], and if the ii-th character is 'R' then the frog can jump to any cell in a range [i+1;min(n+1;i+d)][i+1;min(n+1;i+d)].The frog doesn't want to jump far, so your task is to find the minimum possible value of dd such that the frog can reach the cell n+1n+1 from the cell 00 if it can jump by no more than dd cells at once. It is guaranteed that it is always possible to reach n+1n+1 from 00.You have to answer tt independent test cases.InputThe first line of the input contains one integer tt (1≤t≤1041≤t≤104) — the number of test cases.The next tt lines describe test cases. The ii-th test case is described as a string ss consisting of at least 11 and at most 2⋅1052⋅105 characters 'L' and 'R'.It is guaranteed that the sum of lengths of strings over all test cases does not exceed 2⋅1052⋅105 (∑|s|≤2⋅105∑|s|≤2⋅105).OutputFor each test case, print the answer — the minimum possible value of dd such that the frog can reach the cell n+1n+1 from the cell 00 if it jumps by no more than dd at once.ExampleInputCopy6
LRLRRLL
L
LLR
RRRR
LLLLLL
R
OutputCopy3
2
3
1
7
1
NoteThe picture describing the first test case of the example and one of the possible answers:In the second test case of the example; the frog can only jump directly from 00 to n+1n+1.In the third test case of the example, the frog can choose d=3d=3, jump to the cell 33 from the cell 00 and then to the cell 44 from the cell 33.In the fourth test case of the example, the frog can choose d=1d=1 and jump 55 times to the right.In the fifth test case of the example, the frog can only jump directly from 00 to n+1n+1.In the sixth test case of the example, the frog can choose d=1d=1 and jump 22 times to the right. | [
"binary search",
"data structures",
"dfs and similar",
"greedy",
"implementation"
] | for _ in range(int(input())):
string = input()
len_max = 0
len = 0
for i in string:
if i == 'R':
len = 0
else:
len += 1
if len > len_max:
len_max = len
print(len_max + 1)
| py |
1141 | F2 | F2. Same Sum Blocks (Hard)time limit per test3 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThis problem is given in two editions; which differ exclusively in the constraints on the number nn.You are given an array of integers a[1],a[2],…,a[n].a[1],a[2],…,a[n]. A block is a sequence of contiguous (consecutive) elements a[l],a[l+1],…,a[r]a[l],a[l+1],…,a[r] (1≤l≤r≤n1≤l≤r≤n). Thus, a block is defined by a pair of indices (l,r)(l,r).Find a set of blocks (l1,r1),(l2,r2),…,(lk,rk)(l1,r1),(l2,r2),…,(lk,rk) such that: They do not intersect (i.e. they are disjoint). Formally, for each pair of blocks (li,ri)(li,ri) and (lj,rj(lj,rj) where i≠ji≠j either ri<ljri<lj or rj<lirj<li. For each block the sum of its elements is the same. Formally, a[l1]+a[l1+1]+⋯+a[r1]=a[l2]+a[l2+1]+⋯+a[r2]=a[l1]+a[l1+1]+⋯+a[r1]=a[l2]+a[l2+1]+⋯+a[r2]= ⋯=⋯= a[lk]+a[lk+1]+⋯+a[rk].a[lk]+a[lk+1]+⋯+a[rk]. The number of the blocks in the set is maximum. Formally, there does not exist a set of blocks (l′1,r′1),(l′2,r′2),…,(l′k′,r′k′)(l1′,r1′),(l2′,r2′),…,(lk′′,rk′′) satisfying the above two requirements with k′>kk′>k. The picture corresponds to the first example. Blue boxes illustrate blocks. Write a program to find such a set of blocks.InputThe first line contains integer nn (1≤n≤15001≤n≤1500) — the length of the given array. The second line contains the sequence of elements a[1],a[2],…,a[n]a[1],a[2],…,a[n] (−105≤ai≤105−105≤ai≤105).OutputIn the first line print the integer kk (1≤k≤n1≤k≤n). The following kk lines should contain blocks, one per line. In each line print a pair of indices li,rili,ri (1≤li≤ri≤n1≤li≤ri≤n) — the bounds of the ii-th block. You can print blocks in any order. If there are multiple answers, print any of them.ExamplesInputCopy7
4 1 2 2 1 5 3
OutputCopy3
7 7
2 3
4 5
InputCopy11
-5 -4 -3 -2 -1 0 1 2 3 4 5
OutputCopy2
3 4
1 1
InputCopy4
1 1 1 1
OutputCopy4
4 4
1 1
2 2
3 3
| [
"data structures",
"greedy"
] | #!/usr/bin/env python3
# -*- encoding: utf-8 -*-
'''
@File : F_2_Same_Sum_Blocks_Hard.py
@Time : 2023/02/08 23:31:26
@Author : @bvf
'''
import sys
import os
from io import BytesIO, IOBase
from types import GeneratorType
BUFSIZE = 4096
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = 'x' in file.mode or 'r' not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self, **kwargs):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b'\n') + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode('ascii'))
self.read = lambda: self.buffer.read().decode('ascii')
self.readline = lambda: self.buffer.readline().decode('ascii')
def debug(func):
def wrapper(*args, **kwargs):
res = func(*args, **kwargs)
print('----------------')
return res
return wrapper
# 'return->yield','dfs()->yield dfs()'
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
def I():
return input()
def II():
return int(input())
def MI():
return map(int, input().split())
def LI():
return list(input().split())
def LII():
return list(map(int, input().split()))
def GMI():
return map(lambda x: int(x) - 1, input().split())
def LGMI():
return list(map(lambda x: int(x) - 1, input().split()))
# ------------------------------FastIO---------------------------------
# from typing import *
# from bisect import bisect_left, bisect_right
# from copy import deepcopy, copy
# from math import comb, gcd, factorial, log, log2
from collections import Counter, defaultdict, deque
# from itertools import accumulate, combinations, permutations, groupby
# from heapq import nsmallest, nlargest, heapify, heappop, heappush, heapreplace
# from functools import lru_cache, reduce
# from sortedcontainers import SortedList, SortedSet
MOD = int(1e9 + 7)
M9D = int(998244353)
INF = int(1e20)
# -------------------------------@bvf----------------------------------
n = II()
a = LII()
d = defaultdict(list)
for i in range(n):
tt = 0
for j in range(i,n):
tt += a[j]
d[tt].append((i+1,j+1))
res = 0
ret = []
for v in d.values():
r = -1
cnt = 0
o = []
v.sort(key= lambda x: x[1])
for s,t in v:
if s>r:
r = t
cnt += 1
o.append((s,t))
if cnt>res:
res = cnt
ret = o.copy()
print(res)
for x,y in ret:
print(x,y)
| py |
1285 | C | C. Fadi and LCMtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputToday; Osama gave Fadi an integer XX, and Fadi was wondering about the minimum possible value of max(a,b)max(a,b) such that LCM(a,b)LCM(a,b) equals XX. Both aa and bb should be positive integers.LCM(a,b)LCM(a,b) is the smallest positive integer that is divisible by both aa and bb. For example, LCM(6,8)=24LCM(6,8)=24, LCM(4,12)=12LCM(4,12)=12, LCM(2,3)=6LCM(2,3)=6.Of course, Fadi immediately knew the answer. Can you be just like Fadi and find any such pair?InputThe first and only line contains an integer XX (1≤X≤10121≤X≤1012).OutputPrint two positive integers, aa and bb, such that the value of max(a,b)max(a,b) is minimum possible and LCM(a,b)LCM(a,b) equals XX. If there are several possible such pairs, you can print any.ExamplesInputCopy2
OutputCopy1 2
InputCopy6
OutputCopy2 3
InputCopy4
OutputCopy1 4
InputCopy1
OutputCopy1 1
| [
"brute force",
"math",
"number theory"
] | from math import isqrt, lcm
import sys
x = int(input())
for i in range(isqrt(x), 0, -1):
j, r = divmod(x, i)
if r == 0 and lcm(i, j) == x:
print(i, j)
sys.exit(0)
| py |
1311 | D | D. Three Integerstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given three integers a≤b≤ca≤b≤c.In one move, you can add +1+1 or −1−1 to any of these integers (i.e. increase or decrease any number by one). You can perform such operation any (possibly, zero) number of times, you can even perform this operation several times with one number. Note that you cannot make non-positive numbers using such operations.You have to perform the minimum number of such operations in order to obtain three integers A≤B≤CA≤B≤C such that BB is divisible by AA and CC is divisible by BB.You have to answer tt independent test cases. InputThe first line of the input contains one integer tt (1≤t≤1001≤t≤100) — the number of test cases.The next tt lines describe test cases. Each test case is given on a separate line as three space-separated integers a,ba,b and cc (1≤a≤b≤c≤1041≤a≤b≤c≤104).OutputFor each test case, print the answer. In the first line print resres — the minimum number of operations you have to perform to obtain three integers A≤B≤CA≤B≤C such that BB is divisible by AA and CC is divisible by BB. On the second line print any suitable triple A,BA,B and CC.ExampleInputCopy8
1 2 3
123 321 456
5 10 15
15 18 21
100 100 101
1 22 29
3 19 38
6 30 46
OutputCopy1
1 1 3
102
114 228 456
4
4 8 16
6
18 18 18
1
100 100 100
7
1 22 22
2
1 19 38
8
6 24 48
| [
"brute force",
"math"
] | from math import inf,floor,ceil
answers=[]
smallest_prime_factor=[0]*(2*10**4+1)
primes=[]
for i in range(2,2*10**4+1):
if smallest_prime_factor[i]==0:
smallest_prime_factor[i]=i
primes.append(i)
j=0
while i*primes[j]<=2*10**4:
smallest_prime_factor[i*primes[j]]=primes[j]
if primes[j]==smallest_prime_factor[i]:break
j+=1
T=int(input())
for test_i in range(T):
A,B,C=map(int,input().split())
ans_moves=inf
ans_vals=None
for b_val in range(1,11000+1):
c_val=min([b_val*(floor(C/b_val)+add) for add in [-1,0,1]],key=lambda m:abs(m-C))
prime_factorization=[]
b_val2=b_val
while b_val2!=1:
prime_factor=smallest_prime_factor[b_val2]
exponent=0
while b_val2%prime_factor==0:
b_val2//=prime_factor
exponent+=1
prime_factorization.append([prime_factor,exponent])
factors=[1]
for prime,exponent in prime_factorization:
for i in range(len(factors)):
for count in range(1,exponent+1):
factors.append(factors[i]*prime**count)
a_val=min(factors,key=lambda factor:abs(factor-A))
vals=[a_val,b_val,c_val]
moves=sum(abs(new-old) for new,old in zip(vals,[A,B,C]))
if moves<ans_moves:
ans_moves=moves
ans_vals=vals
answers.append([ans_moves,ans_vals])
for moves,vals in answers:
print(moves)
print(*vals) | py |
1296 | E2 | E2. String Coloring (hard version)time limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputThis is a hard version of the problem. The actual problems are different; but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.You are given a string ss consisting of nn lowercase Latin letters.You have to color all its characters the minimum number of colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in ss).After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.The goal is to make the string sorted, i.e. all characters should be in alphabetical order.Your task is to find the minimum number of colors which you have to color the given string in so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.InputThe first line of the input contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the length of ss.The second line of the input contains the string ss consisting of exactly nn lowercase Latin letters.OutputIn the first line print one integer resres (1≤res≤n1≤res≤n) — the minimum number of colors in which you have to color the given string so that after coloring it can become sorted by some sequence of swaps.In the second line print any possible coloring that can be used to sort the string using some sequence of swaps described in the problem statement. The coloring is the array cc of length nn, where 1≤ci≤res1≤ci≤res and cici means the color of the ii-th character.ExamplesInputCopy9
abacbecfd
OutputCopy2
1 1 2 1 2 1 2 1 2
InputCopy8
aaabbcbb
OutputCopy2
1 2 1 2 1 2 1 1
InputCopy7
abcdedc
OutputCopy3
1 1 1 1 1 2 3
InputCopy5
abcde
OutputCopy1
1 1 1 1 1
| [
"data structures",
"dp"
] | from collections import defaultdict, Counter,deque
from math import sqrt, log10, log, floor, factorial,gcd
from bisect import bisect_left, bisect_right
from itertools import permutations,combinations
import sys, io, os
input = sys.stdin.readline
# input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# sys.setrecursionlimit(10000)
inf = float('inf')
mod = 10 ** 9 + 7
def get_list(): return [int(i) for i in input().split()]
def yn(a): print("YES" if a else "NO")
ceil = lambda a, b: (a + b - 1) // b
def lis_length(nums, cmp=lambda x, y: x > y):
P = [0] * len(nums)
M = [0] * (len(nums) + 1)
L = 0
for i in range(len(nums)):
lo, hi = 1, L
while lo <= hi:
mid = (lo + hi) // 2
if cmp(nums[M[mid]], nums[i]):
lo = mid + 1
else:
hi = mid - 1
newL = lo
P[i] = M[newL - 1]
M[newL] = i
L = max(L, newL)
lislen=[1 for i in P]
for i in range(1,len(nums)):
if P[i]==0:
if cmp(nums[P[i]],nums[i]):
lislen[i]=lislen[P[i]]+1
else:
lislen[i]=lislen[P[i]]
else:
lislen[i] = lislen[P[i]] + 1
return lislen
t=1
for i in range(t):
n=int(input())
s=input().strip()
lislen=lis_length(s)
print(max(lislen))
print(*lislen)
| py |
1307 | B | B. Cow and Friendtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputBessie has way too many friends because she is everyone's favorite cow! Her new friend Rabbit is trying to hop over so they can play! More specifically; he wants to get from (0,0)(0,0) to (x,0)(x,0) by making multiple hops. He is only willing to hop from one point to another point on the 2D plane if the Euclidean distance between the endpoints of a hop is one of its nn favorite numbers: a1,a2,…,ana1,a2,…,an. What is the minimum number of hops Rabbit needs to get from (0,0)(0,0) to (x,0)(x,0)? Rabbit may land on points with non-integer coordinates. It can be proved that Rabbit can always reach his destination.Recall that the Euclidean distance between points (xi,yi)(xi,yi) and (xj,yj)(xj,yj) is (xi−xj)2+(yi−yj)2−−−−−−−−−−−−−−−−−−√(xi−xj)2+(yi−yj)2.For example, if Rabbit has favorite numbers 11 and 33 he could hop from (0,0)(0,0) to (4,0)(4,0) in two hops as shown below. Note that there also exists other valid ways to hop to (4,0)(4,0) in 22 hops (e.g. (0,0)(0,0) →→ (2,−5–√)(2,−5) →→ (4,0)(4,0)). Here is a graphic for the first example. Both hops have distance 33, one of Rabbit's favorite numbers. In other words, each time Rabbit chooses some number aiai and hops with distance equal to aiai in any direction he wants. The same number can be used multiple times.InputThe input consists of multiple test cases. The first line contains an integer tt (1≤t≤10001≤t≤1000) — the number of test cases. Next 2t2t lines contain test cases — two lines per test case.The first line of each test case contains two integers nn and xx (1≤n≤1051≤n≤105, 1≤x≤1091≤x≤109) — the number of favorite numbers and the distance Rabbit wants to travel, respectively.The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) — Rabbit's favorite numbers. It is guaranteed that the favorite numbers are distinct.It is guaranteed that the sum of nn over all the test cases will not exceed 105105.OutputFor each test case, print a single integer — the minimum number of hops needed.ExampleInputCopy42 41 33 123 4 51 552 1015 4OutputCopy2
3
1
2
NoteThe first test case of the sample is shown in the picture above. Rabbit can hop to (2,5–√)(2,5), then to (4,0)(4,0) for a total of two hops. Each hop has a distance of 33, which is one of his favorite numbers.In the second test case of the sample, one way for Rabbit to hop 33 times is: (0,0)(0,0) →→ (4,0)(4,0) →→ (8,0)(8,0) →→ (12,0)(12,0).In the third test case of the sample, Rabbit can hop from (0,0)(0,0) to (5,0)(5,0).In the fourth test case of the sample, Rabbit can hop: (0,0)(0,0) →→ (5,102–√)(5,102) →→ (10,0)(10,0). | [
"geometry",
"greedy",
"math"
] | p = int(input())
for j in range (p):
n, x = map(int, input().split())
s = set()
num = 0
for i in input().split():
s.add(int(i))
m = max(s)
if x in s:
print(1)
else:
if x % m == 0:
num += x // m
else:
if x//m< 1:
k = 1
else:
k = x//m
num += k+1
print(num)
| py |
1311 | B | B. WeirdSorttime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given an array aa of length nn.You are also given a set of distinct positions p1,p2,…,pmp1,p2,…,pm, where 1≤pi<n1≤pi<n. The position pipi means that you can swap elements a[pi]a[pi] and a[pi+1]a[pi+1]. You can apply this operation any number of times for each of the given positions.Your task is to determine if it is possible to sort the initial array in non-decreasing order (a1≤a2≤⋯≤ana1≤a2≤⋯≤an) using only allowed swaps.For example, if a=[3,2,1]a=[3,2,1] and p=[1,2]p=[1,2], then we can first swap elements a[2]a[2] and a[3]a[3] (because position 22 is contained in the given set pp). We get the array a=[3,1,2]a=[3,1,2]. Then we swap a[1]a[1] and a[2]a[2] (position 11 is also contained in pp). We get the array a=[1,3,2]a=[1,3,2]. Finally, we swap a[2]a[2] and a[3]a[3] again and get the array a=[1,2,3]a=[1,2,3], sorted in non-decreasing order.You can see that if a=[4,1,2,3]a=[4,1,2,3] and p=[3,2]p=[3,2] then you cannot sort the array.You have to answer tt independent test cases.InputThe first line of the input contains one integer tt (1≤t≤1001≤t≤100) — the number of test cases.Then tt test cases follow. The first line of each test case contains two integers nn and mm (1≤m<n≤1001≤m<n≤100) — the number of elements in aa and the number of elements in pp. The second line of the test case contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1001≤ai≤100). The third line of the test case contains mm integers p1,p2,…,pmp1,p2,…,pm (1≤pi<n1≤pi<n, all pipi are distinct) — the set of positions described in the problem statement.OutputFor each test case, print the answer — "YES" (without quotes) if you can sort the initial array in non-decreasing order (a1≤a2≤⋯≤ana1≤a2≤⋯≤an) using only allowed swaps. Otherwise, print "NO".ExampleInputCopy6
3 2
3 2 1
1 2
4 2
4 1 2 3
3 2
5 1
1 2 3 4 5
1
4 2
2 1 4 3
1 3
4 2
4 3 2 1
1 3
5 2
2 1 2 3 3
1 4
OutputCopyYES
NO
YES
YES
NO
YES
| [
"dfs and similar",
"sortings"
] | #!/usr/bin/env python3
import math
import sys
input = lambda: sys.stdin.readline().rstrip("\r\n")
from bisect import bisect_left as bs
def test_case():
n, m = map(int, input().split())
a = list(map(int, input().split()))
p = sorted(list(map(lambda x: int(x)-1, input().split())))
i = 0
while i < m:
j = i+1
while j < m and p[j] == p[j-1]+1:
j += 1
x = p[i]
y = p[j-1]+1
a = a[:x] + sorted(a[x:y+1]) + a[y+1:]
i = j
done = True
for i in range(n-1):
if a[i] > a[i+1]:
done = False
break
print("YES" if done else "NO")
def main():
t = 1
t = int(input())
for _ in range(t):
test_case()
if __name__ == '__main__':
main()
| py |
1284 | D | D. New Year and Conferencetime limit per test2 secondsmemory limit per test1024 megabytesinputstandard inputoutputstandard outputFilled with optimism; Hyunuk will host a conference about how great this new year will be!The conference will have nn lectures. Hyunuk has two candidate venues aa and bb. For each of the nn lectures, the speaker specified two time intervals [sai,eai][sai,eai] (sai≤eaisai≤eai) and [sbi,ebi][sbi,ebi] (sbi≤ebisbi≤ebi). If the conference is situated in venue aa, the lecture will be held from saisai to eaieai, and if the conference is situated in venue bb, the lecture will be held from sbisbi to ebiebi. Hyunuk will choose one of these venues and all lectures will be held at that venue.Two lectures are said to overlap if they share any point in time in common. Formally, a lecture held in interval [x,y][x,y] overlaps with a lecture held in interval [u,v][u,v] if and only if max(x,u)≤min(y,v)max(x,u)≤min(y,v).We say that a participant can attend a subset ss of the lectures if the lectures in ss do not pairwise overlap (i.e. no two lectures overlap). Note that the possibility of attending may depend on whether Hyunuk selected venue aa or venue bb to hold the conference.A subset of lectures ss is said to be venue-sensitive if, for one of the venues, the participant can attend ss, but for the other venue, the participant cannot attend ss.A venue-sensitive set is problematic for a participant who is interested in attending the lectures in ss because the participant cannot be sure whether the lecture times will overlap. Hyunuk will be happy if and only if there are no venue-sensitive sets. Determine whether Hyunuk will be happy.InputThe first line contains an integer nn (1≤n≤1000001≤n≤100000), the number of lectures held in the conference.Each of the next nn lines contains four integers saisai, eaieai, sbisbi, ebiebi (1≤sai,eai,sbi,ebi≤1091≤sai,eai,sbi,ebi≤109, sai≤eai,sbi≤ebisai≤eai,sbi≤ebi).OutputPrint "YES" if Hyunuk will be happy. Print "NO" otherwise.You can print each letter in any case (upper or lower).ExamplesInputCopy2
1 2 3 6
3 4 7 8
OutputCopyYES
InputCopy3
1 3 2 4
4 5 6 7
3 4 5 5
OutputCopyNO
InputCopy6
1 5 2 9
2 4 5 8
3 6 7 11
7 10 12 16
8 11 13 17
9 12 14 18
OutputCopyYES
NoteIn second example; lecture set {1,3}{1,3} is venue-sensitive. Because participant can't attend this lectures in venue aa, but can attend in venue bb.In first and third example, venue-sensitive set does not exist. | [
"binary search",
"data structures",
"hashing",
"sortings"
] | import sys
input = sys.stdin.readline
import heapq
n=int(input())
C=[tuple(map(int,input().split())) for i in range(n)]
CA=[]
CB=[]
for ind,(a,b,c,d) in enumerate(C):
CA.append((a,0,ind))
CA.append((b,1,ind))
CB.append((c,0,ind))
CB.append((d,1,ind))
CA.sort()
CB.sort()
SMAX=[]
EMIN=[]
FINISHED=[0]*n
for time,flag,ind in CA:
if flag==0:
heapq.heappush(SMAX,(-C[ind][2],ind))
heapq.heappush(EMIN,(C[ind][3],ind))
while FINISHED[SMAX[0][1]]:
heapq.heappop(SMAX)
while FINISHED[EMIN[0][1]]:
heapq.heappop(EMIN)
if -SMAX[0][0]>EMIN[0][0]:
print("NO")
sys.exit()
else:
FINISHED[ind]=1
SMAX=[]
EMIN=[]
FINISHED=[0]*n
for time,flag,ind in CB:
if flag==0:
heapq.heappush(SMAX,(-C[ind][0],ind))
heapq.heappush(EMIN,(C[ind][1],ind))
while FINISHED[SMAX[0][1]]:
heapq.heappop(SMAX)
while FINISHED[EMIN[0][1]]:
heapq.heappop(EMIN)
if -SMAX[0][0]>EMIN[0][0]:
print("NO")
sys.exit()
else:
FINISHED[ind]=1
print("YES")
| py |
1315 | C | C. Restoring Permutationtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given a sequence b1,b2,…,bnb1,b2,…,bn. Find the lexicographically minimal permutation a1,a2,…,a2na1,a2,…,a2n such that bi=min(a2i−1,a2i)bi=min(a2i−1,a2i), or determine that it is impossible.InputEach test contains one or more test cases. The first line contains the number of test cases tt (1≤t≤1001≤t≤100).The first line of each test case consists of one integer nn — the number of elements in the sequence bb (1≤n≤1001≤n≤100).The second line of each test case consists of nn different integers b1,…,bnb1,…,bn — elements of the sequence bb (1≤bi≤2n1≤bi≤2n).It is guaranteed that the sum of nn by all test cases doesn't exceed 100100.OutputFor each test case, if there is no appropriate permutation, print one number −1−1.Otherwise, print 2n2n integers a1,…,a2na1,…,a2n — required lexicographically minimal permutation of numbers from 11 to 2n2n.ExampleInputCopy5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
OutputCopy1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10
| [
"greedy"
] | import itertools
import math
import threading
import time
from builtins import input, range
from math import gcd as gcd
import sys
from io import BytesIO, IOBase
import queue
import itertools
import collections
from heapq import heappop, heappush
import random
import os
from random import randint
import decimal
from math import factorial as fac
def solve():
n = int(input())
b = list(map(int,input().split()))
a = [0 for i in range(2* n)]
can_use = {i+1 for i in range(2*n)}
for i in range(n):
a[2*i] = b[i]
can_use.discard(b[i])
can_use = list(can_use)
if len(can_use) != n:
print(-1)
return
can_use = set(can_use)
for i in range(n):
k = a[2*i]
mn = 1<<60
for j in can_use:
if j > k and j < mn:
mn = j
if mn == 1<<60:
print(-1)
return
a[2*i + 1] = mn
can_use.discard(mn)
print(*a)
if __name__ == '__main__':
multytest = True
if multytest:
t = int(input())
for i in range(t):
solve()
else:
solve()
| py |
1287 | A | A. Angry Studentstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputIt's a walking tour day in SIS.Winter; so tt groups of students are visiting Torzhok. Streets of Torzhok are so narrow that students have to go in a row one after another.Initially, some students are angry. Let's describe a group of students by a string of capital letters "A" and "P": "A" corresponds to an angry student "P" corresponds to a patient student Such string describes the row from the last to the first student.Every minute every angry student throws a snowball at the next student. Formally, if an angry student corresponds to the character with index ii in the string describing a group then they will throw a snowball at the student that corresponds to the character with index i+1i+1 (students are given from the last to the first student). If the target student was not angry yet, they become angry. Even if the first (the rightmost in the string) student is angry, they don't throw a snowball since there is no one in front of them.Let's look at the first example test. The row initially looks like this: PPAP. Then, after a minute the only single angry student will throw a snowball at the student in front of them, and they also become angry: PPAA. After that, no more students will become angry.Your task is to help SIS.Winter teachers to determine the last moment a student becomes angry for every group.InputThe first line contains a single integer tt — the number of groups of students (1≤t≤1001≤t≤100). The following 2t2t lines contain descriptions of groups of students.The description of the group starts with an integer kiki (1≤ki≤1001≤ki≤100) — the number of students in the group, followed by a string sisi, consisting of kiki letters "A" and "P", which describes the ii-th group of students.OutputFor every group output single integer — the last moment a student becomes angry.ExamplesInputCopy1
4
PPAP
OutputCopy1
InputCopy3
12
APPAPPPAPPPP
3
AAP
3
PPA
OutputCopy4
1
0
NoteIn the first test; after 11 minute the state of students becomes PPAA. After that, no new angry students will appear.In the second tets, state of students in the first group is: after 11 minute — AAPAAPPAAPPP after 22 minutes — AAAAAAPAAAPP after 33 minutes — AAAAAAAAAAAP after 44 minutes all 1212 students are angry In the second group after 11 minute, all students are angry. | [
"greedy",
"implementation"
] | a = int(input())
for i in range(a):
q = int(input())
s = input()
if "A" not in s:
print(0)
else:
w = s.index("A")
d = []
e = 0
for j in range(w, q):
if s[j] == "A":
d.append(e)
e = 0
else:
e += 1
d.append(e)
print(max(d))
| py |
1321 | A | A. Contest for Robotstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputPolycarp is preparing the first programming contest for robots. There are nn problems in it, and a lot of robots are going to participate in it. Each robot solving the problem ii gets pipi points, and the score of each robot in the competition is calculated as the sum of pipi over all problems ii solved by it. For each problem, pipi is an integer not less than 11.Two corporations specializing in problem-solving robot manufacturing, "Robo-Coder Inc." and "BionicSolver Industries", are going to register two robots (one for each corporation) for participation as well. Polycarp knows the advantages and flaws of robots produced by these companies, so, for each problem, he knows precisely whether each robot will solve it during the competition. Knowing this, he can try predicting the results — or manipulating them. For some reason (which absolutely cannot involve bribing), Polycarp wants the "Robo-Coder Inc." robot to outperform the "BionicSolver Industries" robot in the competition. Polycarp wants to set the values of pipi in such a way that the "Robo-Coder Inc." robot gets strictly more points than the "BionicSolver Industries" robot. However, if the values of pipi will be large, it may look very suspicious — so Polycarp wants to minimize the maximum value of pipi over all problems. Can you help Polycarp to determine the minimum possible upper bound on the number of points given for solving the problems?InputThe first line contains one integer nn (1≤n≤1001≤n≤100) — the number of problems.The second line contains nn integers r1r1, r2r2, ..., rnrn (0≤ri≤10≤ri≤1). ri=1ri=1 means that the "Robo-Coder Inc." robot will solve the ii-th problem, ri=0ri=0 means that it won't solve the ii-th problem.The third line contains nn integers b1b1, b2b2, ..., bnbn (0≤bi≤10≤bi≤1). bi=1bi=1 means that the "BionicSolver Industries" robot will solve the ii-th problem, bi=0bi=0 means that it won't solve the ii-th problem.OutputIf "Robo-Coder Inc." robot cannot outperform the "BionicSolver Industries" robot by any means, print one integer −1−1.Otherwise, print the minimum possible value of maxi=1npimaxi=1npi, if all values of pipi are set in such a way that the "Robo-Coder Inc." robot gets strictly more points than the "BionicSolver Industries" robot.ExamplesInputCopy5
1 1 1 0 0
0 1 1 1 1
OutputCopy3
InputCopy3
0 0 0
0 0 0
OutputCopy-1
InputCopy4
1 1 1 1
1 1 1 1
OutputCopy-1
InputCopy9
1 0 0 0 0 0 0 0 1
0 1 1 0 1 1 1 1 0
OutputCopy4
NoteIn the first example; one of the valid score assignments is p=[3,1,3,1,1]p=[3,1,3,1,1]. Then the "Robo-Coder" gets 77 points, the "BionicSolver" — 66 points.In the second example, both robots get 00 points, and the score distribution does not matter.In the third example, both robots solve all problems, so their points are equal. | [
"greedy"
] |
n = int(input())
a = [int(l) for l in input().split()]
b = [int(l) for l in input().split()]
unfav = 0
fav = 0
imp = 1
for i in range(n):
d = a[i] - b[i]
if d == -1:
unfav +=1
elif d==1:
fav +=1
imp = 0
if imp == 0:
if (unfav+1)%fav == 0:
print((unfav+1)//fav)
elif (unfav+1)%fav >0:
print(((unfav + 1) // fav)+1)
else:
print("-1") | py |
1141 | D | D. Colored Bootstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThere are nn left boots and nn right boots. Each boot has a color which is denoted as a lowercase Latin letter or a question mark ('?'). Thus, you are given two strings ll and rr, both of length nn. The character lili stands for the color of the ii-th left boot and the character riri stands for the color of the ii-th right boot.A lowercase Latin letter denotes a specific color, but the question mark ('?') denotes an indefinite color. Two specific colors are compatible if they are exactly the same. An indefinite color is compatible with any (specific or indefinite) color.For example, the following pairs of colors are compatible: ('f', 'f'), ('?', 'z'), ('a', '?') and ('?', '?'). The following pairs of colors are not compatible: ('f', 'g') and ('a', 'z').Compute the maximum number of pairs of boots such that there is one left and one right boot in a pair and their colors are compatible.Print the maximum number of such pairs and the pairs themselves. A boot can be part of at most one pair.InputThe first line contains nn (1≤n≤1500001≤n≤150000), denoting the number of boots for each leg (i.e. the number of left boots and the number of right boots).The second line contains the string ll of length nn. It contains only lowercase Latin letters or question marks. The ii-th character stands for the color of the ii-th left boot.The third line contains the string rr of length nn. It contains only lowercase Latin letters or question marks. The ii-th character stands for the color of the ii-th right boot.OutputPrint kk — the maximum number of compatible left-right pairs of boots, i.e. pairs consisting of one left and one right boot which have compatible colors.The following kk lines should contain pairs aj,bjaj,bj (1≤aj,bj≤n1≤aj,bj≤n). The jj-th of these lines should contain the index ajaj of the left boot in the jj-th pair and index bjbj of the right boot in the jj-th pair. All the numbers ajaj should be distinct (unique), all the numbers bjbj should be distinct (unique).If there are many optimal answers, print any of them.ExamplesInputCopy10
codeforces
dodivthree
OutputCopy5
7 8
4 9
2 2
9 10
3 1
InputCopy7
abaca?b
zabbbcc
OutputCopy5
6 5
2 3
4 6
7 4
1 2
InputCopy9
bambarbia
hellocode
OutputCopy0
InputCopy10
code??????
??????test
OutputCopy10
6 2
1 6
7 3
3 5
4 8
9 7
5 1
2 4
10 9
8 10
| [
"greedy",
"implementation"
] | from heapq import heapify, heappop, heappush
from itertools import cycle
from math import sqrt,ceil
import os
import sys
from collections import defaultdict,deque
from io import BytesIO, IOBase
# prime = [True for i in range(5*10**7 + 1)]
# def SieveOfEratosthenes(n):
# p = 2
# while (p * p <= n):
# # If prime[p] is not changed, then it is a prime
# if (prime[p] == True):
# # Update all multiples of p
# for i in range(p ** 2, n + 1, p):
# prime[i] = False
# p += 1
# prime[0]= False
# prime[1]= False
# SieveOfEratosthenes(5*10**7)
# res=[]
# for i,el in prime:
# if i>=2 and el:
# res.append(el)
# MOD = 998244353
# nmax = (2*(10**5))+2
# fact = [1] * (nmax+1)
# for i in range(2, nmax+1):
# fact[i] = fact[i-1] * i % MOD
# inv = [1] * (nmax+1)
# for i in range(2, nmax+1):
# inv[i] = pow(fact[i], MOD-2, MOD)
# def C(n, m):
# return fact[n] * inv[m] % MOD * inv[n-m] % MOD if 0 <= m <= n else 0
# import sys
# import math
# mod = 10**7+1
# LI=lambda:[int(k) for k in input().split()]
# input = lambda: sys.stdin.readline().rstrip()
# IN=lambda:int(input())
# S=lambda:input()
# r=range
# fact=[i for i in r(mod)]
# for i in reversed(r(2,int(mod**0.5))):
# i=i**2
# for j in range(i,mod,i):
# if fact[j]%i==0:
# fact[j]//=i
from collections import Counter
from functools import lru_cache
from collections import deque
def main():
from heapq import heappush,heappop,heapify
# class SegmentTree:
# def __init__(self, data, default=0, func=max):
# self._default = default
# self._func = func
# self._len = len(data)
# self._size = _size = 1 << (self._len - 1).bit_length()
# self.data = [default] * (2 * _size)
# self.data[_size:_size + self._len] = data
# for i in reversed(range(_size)):
# self.data[i] = func(self.data[i + i], self.data[i + i + 1])
# def __delitem__(self, idx):
# self[idx] = self._default
# def __getitem__(self, idx):
# return self.data[idx + self._size]
# def __setitem__(self, idx, value):
# idx += self._size
# self.data[idx] = value
# idx >>= 1
# while idx:
# self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
# idx >>= 1
# def __len__(self):
# return self._len
# def query(self, start, stop):
# """func of data[start, stop)"""
# start += self._size
# stop += self._size
# if start==stop:
# return self._default
# res_left = res_right = self._default
# while start < stop:
# if start & 1:
# res_left = self._func(res_left, self.data[start])
# start += 1
# if stop & 1:
# stop -= 1
# res_right = self._func(self.data[stop], res_right)
# start >>= 1
# stop >>= 1
# return self._func(res_left, res_right)
# def __repr__(self):
# return "SegmentTree({0})".format(self.data)
import bisect,math
# ---------------------------------------------------------
class SortedList():
BUCKET_RATIO = 50
REBUILD_RATIO = 170
def __init__(self,buckets):
buckets = list(buckets)
buckets = sorted(buckets)
self._build(buckets)
def __iter__(self):
for i in self.buckets:
for j in i: yield j
def __reversed__(self):
for i in reversed(self.buckets):
for j in reversed(i): yield j
def __len__(self):
return self.size
def __contains__(self,x):
if self.size == 0: return False
bucket = self._find_bucket(x)
i = bisect.bisect_left(bucket,x)
return i != len(bucket) and bucket[i] == x
def __getitem__(self,x):
if x < 0: x += self.size
if x < 0: raise IndexError
for i in self.buckets:
if x < len(i): return i[x]
x -= len(i)
raise IndexError
def _build(self,buckets=None):
if buckets is None: buckets = list(self)
self.size = len(buckets)
bucket_size = int(math.ceil(math.sqrt(self.size/self.BUCKET_RATIO)))
tmp = []
for i in range(bucket_size):
t = buckets[(self.size*i)//bucket_size:(self.size*(i+1))//bucket_size]
tmp.append(t)
self.buckets = tmp
def _find_bucket(self,x):
for i in self.buckets:
if x <= i[-1]:
return i
return i
def add(self,x):
# O(√N)
if self.size == 0:
self.buckets = [[x]]
self.size = 1
return True
bucket = self._find_bucket(x)
bisect.insort(bucket,x)
self.size += 1
if len(bucket) > len(self.buckets) * self.REBUILD_RATIO:
self._build()
return True
def remove(self,x):
# O(√N)
if self.size == 0: return False
bucket = self._find_bucket(x)
i = bisect.bisect_left(bucket,x)
if i == len(bucket) or bucket[i] != x: return False
bucket.pop(i)
self.size -= 1
if len(bucket) == 0: self._build()
return True
def lt(self,x):
# less than < x
for i in reversed(self.buckets):
if i[0] < x:
return i[bisect.bisect_left(i,x) - 1]
def le(self,x):
# less than or equal to <= x
for i in reversed(self.buckets):
if i[0] <= x:
return i[bisect.bisect_right(i,x) - 1]
def gt(self,x):
# greater than > x
for i in self.buckets:
if i[-1] > x:
return i[bisect.bisect_right(i,x)]
def ge(self,x):
# greater than or equal to >= x
for i in self.buckets:
if i[-1] >= x:
return i[bisect.bisect_left(i,x)]
def index(self,x):
# the number of elements < x
ans = 0
for i in self.buckets:
if i[-1] >= x:
return ans + bisect.bisect_left(i,x)
ans += len(i)
return ans
def index_right(self,x):
# the number of elements < x
ans = 0
for i in self.buckets:
if i[-1] > x:
return ans + bisect.bisect_right(i,x)
ans += len(i)
return ans
#-------------------------------------------------------------
from collections import defaultdict
n=int(input())
s=input()
t=input()
d=defaultdict(lambda :[[],[]])
res=[]
a=[]
b=[]
for i in range(n):
if s[i]==t[i] and t[i]!="?":
res.append((i+1,i+1))
else :
if s[i]!="?":
d[ord(s[i])-ord("a")][0].append(i+1)
else :
a.append(i+1)
if t[i]!="?":
d[ord(t[i])-ord("a")][1].append(i+1)
else :
b.append(i+1)
for i in range(26):
am=min(len(d[i][0]),len(d[i][1]))
for el in range(am):
p=d[i][0].pop()
q=d[i][1].pop()
res.append((p,q))
for j in range(len(d[i][0])):
if b:
res.append((d[i][0][j],b.pop()))
for j in range(len(d[i][1])):
if a:
res.append((a.pop(),d[i][1][j]))
while a and b:
res.append((a.pop(),b.pop()))
print(len(res))
for a,b in res:
print(a,b)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = 'x' in file.mode or 'r' not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b'\n') + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode('ascii'))
self.read = lambda: self.buffer.read().decode('ascii')
self.readline = lambda: self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
# endregion
if __name__ == '__main__':
main() | py |
1294 | E | E. Obtain a Permutationtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given a rectangular matrix of size n×mn×m consisting of integers from 11 to 2⋅1052⋅105.In one move, you can: choose any element of the matrix and change its value to any integer between 11 and n⋅mn⋅m, inclusive; take any column and shift it one cell up cyclically (see the example of such cyclic shift below). A cyclic shift is an operation such that you choose some jj (1≤j≤m1≤j≤m) and set a1,j:=a2,j,a2,j:=a3,j,…,an,j:=a1,ja1,j:=a2,j,a2,j:=a3,j,…,an,j:=a1,j simultaneously. Example of cyclic shift of the first column You want to perform the minimum number of moves to make this matrix look like this: In other words, the goal is to obtain the matrix, where a1,1=1,a1,2=2,…,a1,m=m,a2,1=m+1,a2,2=m+2,…,an,m=n⋅ma1,1=1,a1,2=2,…,a1,m=m,a2,1=m+1,a2,2=m+2,…,an,m=n⋅m (i.e. ai,j=(i−1)⋅m+jai,j=(i−1)⋅m+j) with the minimum number of moves performed.InputThe first line of the input contains two integers nn and mm (1≤n,m≤2⋅105,n⋅m≤2⋅1051≤n,m≤2⋅105,n⋅m≤2⋅105) — the size of the matrix.The next nn lines contain mm integers each. The number at the line ii and position jj is ai,jai,j (1≤ai,j≤2⋅1051≤ai,j≤2⋅105).OutputPrint one integer — the minimum number of moves required to obtain the matrix, where a1,1=1,a1,2=2,…,a1,m=m,a2,1=m+1,a2,2=m+2,…,an,m=n⋅ma1,1=1,a1,2=2,…,a1,m=m,a2,1=m+1,a2,2=m+2,…,an,m=n⋅m (ai,j=(i−1)m+jai,j=(i−1)m+j).ExamplesInputCopy3 3
3 2 1
1 2 3
4 5 6
OutputCopy6
InputCopy4 3
1 2 3
4 5 6
7 8 9
10 11 12
OutputCopy0
InputCopy3 4
1 6 3 4
5 10 7 8
9 2 11 12
OutputCopy2
NoteIn the first example; you can set a1,1:=7,a1,2:=8a1,1:=7,a1,2:=8 and a1,3:=9a1,3:=9 then shift the first, the second and the third columns cyclically, so the answer is 66. It can be shown that you cannot achieve a better answer.In the second example, the matrix is already good so the answer is 00.In the third example, it is enough to shift the second column cyclically twice to obtain a good matrix, so the answer is 22. | [
"greedy",
"implementation",
"math"
] | # Thank God that I'm not you.
from collections import Counter, deque
import heapq
import itertools
import math
import random
import sys
from types import GeneratorType;
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
def primeFactors(n):
counter = Counter();
while n % 2 == 0:
counter[2] += 1;
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
counter[i] += 1;
n = n / i
if (n > 2):
counter[n] += 1;
return counter;
input = sys.stdin.readline;
m = (10**9 + 7)
def mod_inv(n,m):
n=n%m
return pow(n,m-2,m)
def nCr(n,r,m):
numerator=1
for i in range(r):
numerator=(numerator*(n-i))%m
denomenator=1
for i in range(2,r+1):
denomenator=(denomenator*i)%m
return (numerator*mod_inv(denomenator,m))%m
n, m = map(int, input().split())
matrix = [];
for i in range(n):
matrix.append(list(map(int, input().split())));
identityMatrix = [];
num = 1;
for i in range(n):
col = [];
for j in range(m):
col.append(num)
num += 1;
identityMatrix.append(col);
count = 0;
def solve(colOne, colTwo):
position = {}
for i, num in enumerate(colTwo):
position[num] = i;
counter = Counter()
currentMin = float('inf')
for i, num in enumerate(colOne):
if num not in position:
continue
pos = position[num]
currDif = None
if i >= pos:
currDiff = i - pos;
else:
newPos = len(colOne) - 1
currDiff = (i + 1) + (newPos - pos)
counter[currDiff] += 1;
for diff in counter:
currentMin = min(currentMin, diff + (len(colOne) - counter[diff]))
return min(currentMin, len(colOne))
count = 0
for col in range(m):
colOne, colTwo = [], []
for row in range(n):
colOne.append(matrix[row][col])
colTwo.append(identityMatrix[row][col])
count += solve(colOne, colTwo)
print(count)
| py |
13 | A | A. Numberstime limit per test1 secondmemory limit per test64 megabytesinputstdinoutputstdoutLittle Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.Now he wonders what is an average value of sum of digits of the number A written in all bases from 2 to A - 1.Note that all computations should be done in base 10. You should find the result as an irreducible fraction; written in base 10.InputInput contains one integer number A (3 ≤ A ≤ 1000).OutputOutput should contain required average value in format «X/Y», where X is the numerator and Y is the denominator.ExamplesInputCopy5OutputCopy7/3InputCopy3OutputCopy2/1NoteIn the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively. | [
"implementation",
"math"
] | from math import gcd
def sanoq(n,x):
s=0
while n:
s, n = s+n%x, n//x
return s
a=int(input())
j=0
for x in range(2,a):
j+=sanoq(a,x)
g=gcd(j,a-2)
print(f"{j//g}/{(a-2)//g}") | py |
1301 | A | A. Three Stringstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given three strings aa, bb and cc of the same length nn. The strings consist of lowercase English letters only. The ii-th letter of aa is aiai, the ii-th letter of bb is bibi, the ii-th letter of cc is cici.For every ii (1≤i≤n1≤i≤n) you must swap (i.e. exchange) cici with either aiai or bibi. So in total you'll perform exactly nn swap operations, each of them either ci↔aici↔ai or ci↔bici↔bi (ii iterates over all integers between 11 and nn, inclusive).For example, if aa is "code", bb is "true", and cc is "help", you can make cc equal to "crue" taking the 11-st and the 44-th letters from aa and the others from bb. In this way aa becomes "hodp" and bb becomes "tele".Is it possible that after these swaps the string aa becomes exactly the same as the string bb?InputThe input consists of multiple test cases. The first line contains a single integer tt (1≤t≤1001≤t≤100) — the number of test cases. The description of the test cases follows.The first line of each test case contains a string of lowercase English letters aa.The second line of each test case contains a string of lowercase English letters bb.The third line of each test case contains a string of lowercase English letters cc.It is guaranteed that in each test case these three strings are non-empty and have the same length, which is not exceeding 100100.OutputPrint tt lines with answers for all test cases. For each test case:If it is possible to make string aa equal to string bb print "YES" (without quotes), otherwise print "NO" (without quotes).You can print either lowercase or uppercase letters in the answers.ExampleInputCopy4
aaa
bbb
ccc
abc
bca
bca
aabb
bbaa
baba
imi
mii
iim
OutputCopyNO
YES
YES
NO
NoteIn the first test case; it is impossible to do the swaps so that string aa becomes exactly the same as string bb.In the second test case, you should swap cici with aiai for all possible ii. After the swaps aa becomes "bca", bb becomes "bca" and cc becomes "abc". Here the strings aa and bb are equal.In the third test case, you should swap c1c1 with a1a1, c2c2 with b2b2, c3c3 with b3b3 and c4c4 with a4a4. Then string aa becomes "baba", string bb becomes "baba" and string cc becomes "abab". Here the strings aa and bb are equal.In the fourth test case, it is impossible to do the swaps so that string aa becomes exactly the same as string bb. | [
"implementation",
"strings"
] | def solve():
a = input()
b = input()
c = input()
for i in range(len(a)):
if c[i] not in [a[i],b[i]]:
return 'NO'
return 'YES'
for i in range(int(input())):
print(solve()) | py |
1305 | E | E. Kuroni and the Score Distributiontime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputKuroni is the coordinator of the next Mathforces round written by the "Proof by AC" team. All the preparation has been done; and he is discussing with the team about the score distribution for the round.The round consists of nn problems, numbered from 11 to nn. The problems are ordered in increasing order of difficulty, no two problems have the same difficulty. A score distribution for the round can be denoted by an array a1,a2,…,ana1,a2,…,an, where aiai is the score of ii-th problem. Kuroni thinks that the score distribution should satisfy the following requirements: The score of each problem should be a positive integer not exceeding 109109. A harder problem should grant a strictly higher score than an easier problem. In other words, 1≤a1<a2<⋯<an≤1091≤a1<a2<⋯<an≤109. The balance of the score distribution, defined as the number of triples (i,j,k)(i,j,k) such that 1≤i<j<k≤n1≤i<j<k≤n and ai+aj=akai+aj=ak, should be exactly mm. Help the team find a score distribution that satisfies Kuroni's requirement. In case such a score distribution does not exist, output −1−1.InputThe first and single line contains two integers nn and mm (1≤n≤50001≤n≤5000, 0≤m≤1090≤m≤109) — the number of problems and the required balance.OutputIf there is no solution, print a single integer −1−1.Otherwise, print a line containing nn integers a1,a2,…,ana1,a2,…,an, representing a score distribution that satisfies all the requirements. If there are multiple answers, print any of them.ExamplesInputCopy5 3
OutputCopy4 5 9 13 18InputCopy8 0
OutputCopy10 11 12 13 14 15 16 17
InputCopy4 10
OutputCopy-1
NoteIn the first example; there are 33 triples (i,j,k)(i,j,k) that contribute to the balance of the score distribution. (1,2,3)(1,2,3) (1,3,4)(1,3,4) (2,4,5)(2,4,5) | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | n, m = map(int, input().split())
numList = [x+1 for x in range(n)]
backdoor = []
count = sum([(i-1) // 2 for i in range(1, n+1)])
if count < m: exit(print(-1))
while count > m:
lastpop = numList.pop()
count -= (lastpop - 1) // 2
if count >= m:
if len(backdoor) == 0: backdoor.append(10 ** 9)
else: backdoor.append(backdoor[-1] - 2 ** 16)
else:
gap = m - count
backdoor.append(2 * (lastpop - gap) - 1)
count += gap
while len(backdoor) > 0: numList.append(backdoor.pop())
print(' '.join([str(x) for x in numList])) | py |
1299 | A | A. Anu Has a Functiontime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputAnu has created her own function ff: f(x,y)=(x|y)−yf(x,y)=(x|y)−y where || denotes the bitwise OR operation. For example, f(11,6)=(11|6)−6=15−6=9f(11,6)=(11|6)−6=15−6=9. It can be proved that for any nonnegative numbers xx and yy value of f(x,y)f(x,y) is also nonnegative. She would like to research more about this function and has created multiple problems for herself. But she isn't able to solve all of them and needs your help. Here is one of these problems.A value of an array [a1,a2,…,an][a1,a2,…,an] is defined as f(f(…f(f(a1,a2),a3),…an−1),an)f(f(…f(f(a1,a2),a3),…an−1),an) (see notes). You are given an array with not necessarily distinct elements. How should you reorder its elements so that the value of the array is maximal possible?InputThe first line contains a single integer nn (1≤n≤1051≤n≤105).The second line contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤1090≤ai≤109). Elements of the array are not guaranteed to be different.OutputOutput nn integers, the reordering of the array with maximum value. If there are multiple answers, print any.ExamplesInputCopy4
4 0 11 6
OutputCopy11 6 4 0InputCopy1
13
OutputCopy13 NoteIn the first testcase; value of the array [11,6,4,0][11,6,4,0] is f(f(f(11,6),4),0)=f(f(9,4),0)=f(9,0)=9f(f(f(11,6),4),0)=f(f(9,4),0)=f(9,0)=9.[11,4,0,6][11,4,0,6] is also a valid answer. | [
"brute force",
"greedy",
"math"
] | import sys, random
input = lambda : sys.stdin.readline().rstrip()
write = lambda x: sys.stdout.write(x+"\n"); writef = lambda x: print("{:.12f}".format(x))
debug = lambda x: sys.stderr.write(x+"\n")
YES="Yes"; NO="No"; pans = lambda v: print(YES if v else NO); INF=10**18
LI = lambda : list(map(int, input().split())); II=lambda : int(input())
def debug(_l_):
for s in _l_.split():
print(f"{s}={eval(s)}", end=" ")
print()
def dlist(*l, fill=0):
if len(l)==1:
return [fill]*l[0]
ll = l[1:]
return [dlist(*ll, fill=fill) for _ in range(l[0])]
n = int(input())
a = list(map(int, input().split()))
for b in range(32)[::-1]:
c = [[],[]]
for v in a:
c[v>>b&1].append(v)
if len(c[1])==1:
val = c[1][0]
ans = [val]
for v in a:
if val==v:
continue
ans.append(v)
break
else:
ans = a
write(" ".join(map(str, ans))) | py |
1285 | D | D. Dr. Evil Underscorestime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputToday; as a friendship gift, Bakry gave Badawy nn integers a1,a2,…,ana1,a2,…,an and challenged him to choose an integer XX such that the value max1≤i≤n(ai⊕X)max1≤i≤n(ai⊕X) is minimum possible, where ⊕⊕ denotes the bitwise XOR operation.As always, Badawy is too lazy, so you decided to help him and find the minimum possible value of max1≤i≤n(ai⊕X)max1≤i≤n(ai⊕X).InputThe first line contains integer nn (1≤n≤1051≤n≤105).The second line contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤230−10≤ai≤230−1).OutputPrint one integer — the minimum possible value of max1≤i≤n(ai⊕X)max1≤i≤n(ai⊕X).ExamplesInputCopy3
1 2 3
OutputCopy2
InputCopy2
1 5
OutputCopy4
NoteIn the first sample; we can choose X=3X=3.In the second sample, we can choose X=5X=5. | [
"bitmasks",
"brute force",
"dfs and similar",
"divide and conquer",
"dp",
"greedy",
"strings",
"trees"
] | import gc
import heapq
import itertools
import math
from collections import Counter, deque, defaultdict
from sys import stdout
import time
from math import factorial, log, gcd
import sys
from decimal import Decimal
import threading
from heapq import *
from fractions import Fraction
import bisect
def S():
return sys.stdin.readline().split()
def I():
return [int(i) for i in sys.stdin.readline().split()]
def II():
return int(sys.stdin.readline())
def IS():
return sys.stdin.readline().replace('\n', '')
def main():
n = II()
a = I()
two = []
twotwo = 1
for i in range(30):
two.append(twotwo)
twotwo *= 2
two = two[::-1]
queue = deque([(a, set(range(30)), 0)])
ans = float('inf')
while queue:
m, s, h = queue.pop()
c, idx, new_s = 0, -1, set()
for i in s:
first = (m[0] >> (29 - i)) & 1
for j in m:
if (j >> (29 - i)) & 1 != first:
new_s.add(i)
c += 1
if idx == -1:
idx = i
break
if c == 0:
ans = min(ans, h)
else:
m1, m2 = [], []
for i in range(len(m)):
if (m[i] >> (29 - idx)) & 1 == 0:
m1.append(m[i])
else:
m2.append(m[i])
h += two[idx]
new_s.remove(idx)
if h <= ans:
queue.append((m1, new_s.copy(), h))
queue.append((m2, new_s.copy(), h))
print(ans)
if __name__ == '__main__':
# for _ in range(II()):
# main()
main()
| py |
1141 | F1 | F1. Same Sum Blocks (Easy)time limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThis problem is given in two editions; which differ exclusively in the constraints on the number nn.You are given an array of integers a[1],a[2],…,a[n].a[1],a[2],…,a[n]. A block is a sequence of contiguous (consecutive) elements a[l],a[l+1],…,a[r]a[l],a[l+1],…,a[r] (1≤l≤r≤n1≤l≤r≤n). Thus, a block is defined by a pair of indices (l,r)(l,r).Find a set of blocks (l1,r1),(l2,r2),…,(lk,rk)(l1,r1),(l2,r2),…,(lk,rk) such that: They do not intersect (i.e. they are disjoint). Formally, for each pair of blocks (li,ri)(li,ri) and (lj,rj(lj,rj) where i≠ji≠j either ri<ljri<lj or rj<lirj<li. For each block the sum of its elements is the same. Formally, a[l1]+a[l1+1]+⋯+a[r1]=a[l2]+a[l2+1]+⋯+a[r2]=a[l1]+a[l1+1]+⋯+a[r1]=a[l2]+a[l2+1]+⋯+a[r2]= ⋯=⋯= a[lk]+a[lk+1]+⋯+a[rk].a[lk]+a[lk+1]+⋯+a[rk]. The number of the blocks in the set is maximum. Formally, there does not exist a set of blocks (l′1,r′1),(l′2,r′2),…,(l′k′,r′k′)(l1′,r1′),(l2′,r2′),…,(lk′′,rk′′) satisfying the above two requirements with k′>kk′>k. The picture corresponds to the first example. Blue boxes illustrate blocks. Write a program to find such a set of blocks.InputThe first line contains integer nn (1≤n≤501≤n≤50) — the length of the given array. The second line contains the sequence of elements a[1],a[2],…,a[n]a[1],a[2],…,a[n] (−105≤ai≤105−105≤ai≤105).OutputIn the first line print the integer kk (1≤k≤n1≤k≤n). The following kk lines should contain blocks, one per line. In each line print a pair of indices li,rili,ri (1≤li≤ri≤n1≤li≤ri≤n) — the bounds of the ii-th block. You can print blocks in any order. If there are multiple answers, print any of them.ExamplesInputCopy7
4 1 2 2 1 5 3
OutputCopy3
7 7
2 3
4 5
InputCopy11
-5 -4 -3 -2 -1 0 1 2 3 4 5
OutputCopy2
3 4
1 1
InputCopy4
1 1 1 1
OutputCopy4
4 4
1 1
2 2
3 3
| [
"greedy"
] | n = int(input())
A = list(map(int, input().split()))
from collections import defaultdict
D = defaultdict(lambda: [])
for r in range(n):
s = 0
for l in range(r, -1, -1):
s += A[l]
D[s].append((l, r))
k = 0
ans = -1
for s, LR in D.items():
cand_k = 0
cand_ans = []
cur = -1
for l, r in LR:
if l <= cur:
continue
cur = r
cand_ans.append((l+1, r+1))
cand_k += 1
if cand_k > k:
k = cand_k
ans = cand_ans
print(k)
for i in range(len(ans)):
print(*ans[i])
| py |
1307 | A | A. Cow and Haybalestime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThe USA Construction Operation (USACO) recently ordered Farmer John to arrange a row of nn haybale piles on the farm. The ii-th pile contains aiai haybales. However, Farmer John has just left for vacation, leaving Bessie all on her own. Every day, Bessie the naughty cow can choose to move one haybale in any pile to an adjacent pile. Formally, in one day she can choose any two indices ii and jj (1≤i,j≤n1≤i,j≤n) such that |i−j|=1|i−j|=1 and ai>0ai>0 and apply ai=ai−1ai=ai−1, aj=aj+1aj=aj+1. She may also decide to not do anything on some days because she is lazy.Bessie wants to maximize the number of haybales in pile 11 (i.e. to maximize a1a1), and she only has dd days to do so before Farmer John returns. Help her find the maximum number of haybales that may be in pile 11 if she acts optimally!InputThe input consists of multiple test cases. The first line contains an integer tt (1≤t≤1001≤t≤100) — the number of test cases. Next 2t2t lines contain a description of test cases — two lines per test case.The first line of each test case contains integers nn and dd (1≤n,d≤1001≤n,d≤100) — the number of haybale piles and the number of days, respectively. The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤1000≤ai≤100) — the number of haybales in each pile.OutputFor each test case, output one integer: the maximum number of haybales that may be in pile 11 after dd days if Bessie acts optimally.ExampleInputCopy3
4 5
1 0 3 2
2 2
100 1
1 8
0
OutputCopy3
101
0
NoteIn the first test case of the sample; this is one possible way Bessie can end up with 33 haybales in pile 11: On day one, move a haybale from pile 33 to pile 22 On day two, move a haybale from pile 33 to pile 22 On day three, move a haybale from pile 22 to pile 11 On day four, move a haybale from pile 22 to pile 11 On day five, do nothing In the second test case of the sample, Bessie can do nothing on the first day and move a haybale from pile 22 to pile 11 on the second day. | [
"greedy",
"implementation"
] | def tst(n, d, l):
for i in range(1, n):
if d >= i * l[i]:
d -= l[i] * i
l[0] += l[i]
else:
l[0] += d // i
break
print(l[0])
pass
s = int(input())
for i in range(s):
n, d = map(int, input().split())
l = list(map(int, input().split()))
tst(n, d, l)
| py |
1287 | B | B. Hypersettime limit per test3 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputBees Alice and Alesya gave beekeeper Polina famous card game "Set" as a Christmas present. The deck consists of cards that vary in four features across three options for each kind of feature: number of shapes; shape, shading, and color. In this game, some combinations of three cards are said to make up a set. For every feature — color, number, shape, and shading — the three cards must display that feature as either all the same, or pairwise different. The picture below shows how sets look.Polina came up with a new game called "Hyperset". In her game, there are nn cards with kk features, each feature has three possible values: "S", "E", or "T". The original "Set" game can be viewed as "Hyperset" with k=4k=4.Similarly to the original game, three cards form a set, if all features are the same for all cards or are pairwise different. The goal of the game is to compute the number of ways to choose three cards that form a set.Unfortunately, winter holidays have come to an end, and it's time for Polina to go to school. Help Polina find the number of sets among the cards lying on the table.InputThe first line of each test contains two integers nn and kk (1≤n≤15001≤n≤1500, 1≤k≤301≤k≤30) — number of cards and number of features.Each of the following nn lines contains a card description: a string consisting of kk letters "S", "E", "T". The ii-th character of this string decribes the ii-th feature of that card. All cards are distinct.OutputOutput a single integer — the number of ways to choose three cards that form a set.ExamplesInputCopy3 3
SET
ETS
TSE
OutputCopy1InputCopy3 4
SETE
ETSE
TSES
OutputCopy0InputCopy5 4
SETT
TEST
EEET
ESTE
STES
OutputCopy2NoteIn the third example test; these two triples of cards are sets: "SETT"; "TEST"; "EEET" "TEST"; "ESTE", "STES" | [
"brute force",
"data structures",
"implementation"
] | import os
import sys
from io import BytesIO, IOBase
from collections import Counter, defaultdict
import math
import heapq
import bisect
import collections
def ceil(a, b):
return (a + b - 1) // b
BUFSIZE = 8192
inf = float('inf')
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
get = lambda: sys.stdin.readline().rstrip("\r\n")
MOD = 1000000007
def v():
return list()
# for _ in range(int(get())):
# n=int(get())
# l=list(map(int,get().split()))
# = map(int,get().split())
###########################################################################
###########################################################################
n,k = map(int,get().split())
l=[]
for i in range(n):
s=get()
l.append(s)
temp=set(l)
ans=0
for i in range(n):
for j in range(i+1,n):
c=""
for t in range(k):
if l[i][t]==l[j][t]:
c+=l[i][t]
else:
s1=["S","T","E"]
for f in s1:
if f!=l[i][t] and f!=l[j][t]:
c+=f
if c in temp:
ans+=1
print(ans//3)
| py |
1141 | E | E. Superhero Battletime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputA superhero fights with a monster. The battle consists of rounds; each of which lasts exactly nn minutes. After a round ends, the next round starts immediately. This is repeated over and over again.Each round has the same scenario. It is described by a sequence of nn numbers: d1,d2,…,dnd1,d2,…,dn (−106≤di≤106−106≤di≤106). The ii-th element means that monster's hp (hit points) changes by the value didi during the ii-th minute of each round. Formally, if before the ii-th minute of a round the monster's hp is hh, then after the ii-th minute it changes to h:=h+dih:=h+di.The monster's initial hp is HH. It means that before the battle the monster has HH hit points. Print the first minute after which the monster dies. The monster dies if its hp is less than or equal to 00. Print -1 if the battle continues infinitely.InputThe first line contains two integers HH and nn (1≤H≤10121≤H≤1012, 1≤n≤2⋅1051≤n≤2⋅105). The second line contains the sequence of integers d1,d2,…,dnd1,d2,…,dn (−106≤di≤106−106≤di≤106), where didi is the value to change monster's hp in the ii-th minute of a round.OutputPrint -1 if the superhero can't kill the monster and the battle will last infinitely. Otherwise, print the positive integer kk such that kk is the first minute after which the monster is dead.ExamplesInputCopy1000 6
-100 -200 -300 125 77 -4
OutputCopy9
InputCopy1000000000000 5
-1 0 0 0 0
OutputCopy4999999999996
InputCopy10 4
-3 -6 5 4
OutputCopy-1
| [
"math"
] | # Thank God that I'm not you.
import bisect
from collections import Counter, deque
import heapq
import itertools
import math
import random
import sys
from types import GeneratorType;
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
def primeFactors(n):
counter = Counter();
while n % 2 == 0:
counter[2] += 1;
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
counter[i] += 1;
n = n / i
if (n > 2):
counter[n] += 1;
return counter;
h, n = map(int, input().split())
array = list(map(int, input().split()))
def solve():
sumElems = sum(array)
currHp = h;
for i, num in enumerate(array):
currHp += num;
if currHp <= 0:
return i + 1
if sumElems >= 0:
return -1;
leftPointer, rightPointer = 0, 10**15 + 5;
ans = None;
def can(rnds):
currHp = h;
currHp += (rnds * sumElems)
if currHp <= 0:
return True;
for num in array:
currHp += num;
if currHp <= 0:
return True;
return False;
while(rightPointer >= leftPointer):
middlePointer = (leftPointer + rightPointer) // 2;
if can(middlePointer):
ans = middlePointer;
rightPointer = middlePointer - 1;
else:
leftPointer = middlePointer + 1
newAn = ans * n;
currHp = h;
currHp += (ans * sumElems);
if currHp <= 0:
return newAn;
for num in array:
currHp += num;
newAn += 1
if currHp <= 0:
return newAn;
return -1
print(solve()) | py |
1301 | C | C. Ayoub's functiontime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputAyoub thinks that he is a very smart person; so he created a function f(s)f(s), where ss is a binary string (a string which contains only symbols "0" and "1"). The function f(s)f(s) is equal to the number of substrings in the string ss that contains at least one symbol, that is equal to "1".More formally, f(s)f(s) is equal to the number of pairs of integers (l,r)(l,r), such that 1≤l≤r≤|s|1≤l≤r≤|s| (where |s||s| is equal to the length of string ss), such that at least one of the symbols sl,sl+1,…,srsl,sl+1,…,sr is equal to "1". For example, if s=s="01010" then f(s)=12f(s)=12, because there are 1212 such pairs (l,r)(l,r): (1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4),(2,5),(3,4),(3,5),(4,4),(4,5)(1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4),(2,5),(3,4),(3,5),(4,4),(4,5).Ayoub also thinks that he is smarter than Mahmoud so he gave him two integers nn and mm and asked him this problem. For all binary strings ss of length nn which contains exactly mm symbols equal to "1", find the maximum value of f(s)f(s).Mahmoud couldn't solve the problem so he asked you for help. Can you help him? InputThe input consists of multiple test cases. The first line contains a single integer tt (1≤t≤1051≤t≤105) — the number of test cases. The description of the test cases follows.The only line for each test case contains two integers nn, mm (1≤n≤1091≤n≤109, 0≤m≤n0≤m≤n) — the length of the string and the number of symbols equal to "1" in it.OutputFor every test case print one integer number — the maximum value of f(s)f(s) over all strings ss of length nn, which has exactly mm symbols, equal to "1".ExampleInputCopy5
3 1
3 2
3 3
4 0
5 2
OutputCopy4
5
6
0
12
NoteIn the first test case; there exists only 33 strings of length 33, which has exactly 11 symbol, equal to "1". These strings are: s1=s1="100", s2=s2="010", s3=s3="001". The values of ff for them are: f(s1)=3,f(s2)=4,f(s3)=3f(s1)=3,f(s2)=4,f(s3)=3, so the maximum value is 44 and the answer is 44.In the second test case, the string ss with the maximum value is "101".In the third test case, the string ss with the maximum value is "111".In the fourth test case, the only string ss of length 44, which has exactly 00 symbols, equal to "1" is "0000" and the value of ff for that string is 00, so the answer is 00.In the fifth test case, the string ss with the maximum value is "01010" and it is described as an example in the problem statement. | [
"binary search",
"combinatorics",
"greedy",
"math",
"strings"
] | ######################################################################################
#------------------------------------Template---------------------------------------#
######################################################################################
import collections
import heapq
import sys
import math
import itertools
import bisect
from io import BytesIO, IOBase
import os
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def input(): return sys.stdin.readline().rstrip("\r\n")
######################################################################################
#--------------------------------------Input-----------------------------------------#
######################################################################################
def value(): return tuple(map(int, input().split()))
def inlt(): return [int(i) for i in input().split()]
def inp(): return int(input())
def insr(): return input()
def stlt(): return [i for i in input().split()]
######################################################################################
#------------------------------------Functions---------------------------------------#
######################################################################################
def SieveOfEratosthenes(n):
prime = [True for i in range(n+1)]
p = 2
l = []
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
for p in range(2, n+1):
if prime[p]:
l.append(p)
# print(p)
else:
continue
return l
def isPrime(n):
prime_flag = 0
if n > 1:
for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0:
prime_flag = 1
break
if prime_flag == 0:
return True
else:
return False
else:
return False
def gcdofarray(a):
x = 0
for p in a:
x = math.gcd(x, p)
return x
def printDivisors(n):
i = 1
ans = []
while i <= math.sqrt(n):
if (n % i == 0):
if (n / i == i):
ans.append(i)
else:
ans.append(i)
ans.append(n // i)
i = i + 1
ans.sort()
return ans
def binaryToDecimal(n):
return int(n, 2)
def countTriplets(a, n):
s = set()
for i in range(n):
s.add(a[i])
count = 0
for i in range(n):
for j in range(i + 1, n, 1):
xr = a[i] ^ a[j]
if xr in s and xr != a[i] and xr != a[j]:
count += 1
return int(count // 3)
def countOdd(L, R):
N = (R - L) // 2
if (R % 2 != 0 or L % 2 != 0):
N += 1
return N
def isPalindrome(s):
return s == s[::-1]
def sufsum(a):
test_list = a.copy()
test_list.reverse()
n = len(test_list)
for i in range(1, n):
test_list[i] = test_list[i] + test_list[i - 1]
return test_list
def prsum(b):
a = b.copy()
for i in range(1, len(a)):
a[i] += a[i - 1]
return a
def badachotabadachota(nums):
nums.sort()
i = 0
j = len(nums) - 1
ans = []
cc = 0
while len(ans) != len(nums):
if cc % 2 == 0:
ans.append(nums[j])
j -= 1
else:
ans.append(nums[i])
i += 1
cc += 1
return ans
def chotabadachotabada(nums):
nums.sort()
i = 0
j = len(nums) - 1
ans = []
cc = 0
while len(ans) != len(nums):
if cc % 2 == 1:
ans.append(nums[j])
j -= 1
else:
ans.append(nums[i])
i += 1
cc += 1
return ans
def primeFactors(n):
ans = []
while n % 2 == 0:
ans.append(2)
n = n // 2
for i in range(3, int(math.sqrt(n))+1, 2):
while n % i == 0:
ans.append(i)
n = n / i
if n > 2:
ans.append(n)
return ans
def closestMultiple(n, x):
if x > n:
return x
z = (int)(x / 2)
n = n + z
n = n - (n % x)
return n
def getPairsCount(arr, n, sum):
m = [0] * 1000
for i in range(0, n):
m[arr[i]] += 1
twice_count = 0
for i in range(0, n):
twice_count += m[int(sum - arr[i])]
if (int(sum - arr[i]) == arr[i]):
twice_count -= 1
return int(twice_count / 2)
def remove_consec_duplicates(test_list):
res = [i[0] for i in itertools.groupby(test_list)]
return res
def BigPower(a, b, mod):
if b == 0:
return 1
ans = BigPower(a, b//2, mod)
ans *= ans
ans %= mod
if b % 2:
ans *= a
return ans % mod
def nextPowerOf2(n):
count = 0
if (n and not(n & (n - 1))):
return n
while(n != 0):
n >>= 1
count += 1
return 1 << count
# This function multiplies x with the number
# represented by res[]. res_size is size of res[]
# or number of digits in the number represented
# by res[]. This function uses simple school
# mathematics for multiplication. This function
# may value of res_size and returns the new value
# of res_size
def multiply(x, res, res_size):
carry = 0 # Initialize carry
# One by one multiply n with individual
# digits of res[]
i = 0
while i < res_size:
prod = res[i] * x + carry
res[i] = prod % 10 # Store last digit of
# 'prod' in res[]
# make sure floor division is used
carry = prod//10 # Put rest in carry
i = i + 1
# Put carry in res and increase result size
while (carry):
res[res_size] = carry % 10
# make sure floor division is used
# to avoid floating value
carry = carry // 10
res_size = res_size + 1
return res_size
def Kadane(a, size):
max_so_far = -1e18 - 1
max_ending_here = 0
for i in range(0, size):
max_ending_here = max_ending_here + a[i]
if (max_so_far < max_ending_here):
max_so_far = max_ending_here
if max_ending_here < 0:
max_ending_here = 0
return max_so_far
def highestPowerof2(n):
res = 0
for i in range(n, 0, -1):
if ((i & (i - 1)) == 0):
res = i
break
return res
def lower_bound(numbers, length, searchnum):
answer = -1
start = 0
end = length - 1
while start <= end:
middle = (start + end)//2
if numbers[middle] == searchnum:
answer = middle
end = middle - 1
elif numbers[middle] > searchnum:
end = middle - 1
else:
start = middle + 1
return answer
def MEX(nList, start):
nList = set(nList)
mex = start
while mex in nList:
mex += 1
return mex
def MinValue(N, X):
N = list(N)
ln = len(N)
position = ln + 1
# # If the given string N represent
# # a negative value
# if (N[0] == '-'):
# # X must be place at the last
# # index where is greater than N[i]
# for i in range(ln - 1, 0, -1):
# if ((ord(N[i]) - ord('0')) < X):
# position = i
# else:
# # For positive numbers, X must be
# # placed at the last index where
# # it is smaller than N[i]
for i in range(ln - 1, -1, -1):
if ((ord(N[i]) - ord('0')) > X):
position = i
# Insert X at that position
c = chr(X + ord('0'))
str = N.insert(position, c)
# Return the string
return ''.join(N)
def findClosest(arr, n, target):
if (target <= arr[0]):
return arr[0]
if (target >= arr[n - 1]):
return arr[n - 1]
i = 0
j = n
mid = 0
while (i < j):
mid = (i + j) // 2
if (arr[mid] == target):
return arr[mid]
if (target < arr[mid]):
if (mid > 0 and target > arr[mid - 1]):
return getClosest(arr[mid - 1], arr[mid], target)
j = mid
else:
if (mid < n - 1 and target < arr[mid + 1]):
return getClosest(arr[mid], arr[mid + 1], target)
i = mid + 1
return arr[mid]
def getClosest(val1, val2, target):
if (target - val1 >= val2 - target):
return val2
else:
return val1
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den, p - 2, p)) % p
# #####################################################################################################
# #------------------------------------------GRAPHS---------------------------------------------------#
# #####################################################################################################
class Graph:
def __init__(self, edges, n):
self.adjList = [[] for _ in range(n)]
for (src, dest) in edges:
self.adjList[src].append(dest)
self.adjList[dest].append(src)
def BFS(graph, v, discovered):
q = collections.deque()
discovered[v] = True
q.append(v)
ans = []
while q:
v = q.popleft()
ans.append(v)
# print(v, end=' ')
for u in graph.adjList[v]:
if not discovered[u]:
discovered[u] = True
q.append(u)
return ans
#############################################################################################################
#--------------------------------------------Fenwick Tree---------------------------------------------------#
#############################################################################################################
def getsumFT(BITTree, i):
s = 0 # initialize result
# index in BITree[] is 1 more than the index in arr[]
i = i+1
# Traverse ancestors of BITree[index]
while i > 0:
# Add current element of BITree to sum
s += BITTree[i]
# Move index to parent node in getSum View
i -= i & (-i)
return s
# Updates a node in Binary Index Tree (BITree) at given index
# in BITree. The given value 'val' is added to BITree[i] and
# all of its ancestors in tree.
def updatebit(BITTree, n, i, v):
# index in BITree[] is 1 more than the index in arr[]
i += 1
# Traverse all ancestors and add 'val'
while i <= n:
# Add 'val' to current node of BI Tree
BITTree[i] += v
# Update index to that of parent in update View
i += i & (-i)
# Constructs and returns a Binary Indexed Tree for given
# array of size n.
def construct(arr, n):
# Create and initialize BITree[] as 0
BITTree = [0]*(n+1)
# Store the actual values in BITree[] using update()
for i in range(n):
updatebit(BITTree, n, i, arr[i])
# Uncomment below lines to see contents of BITree[]
# for i in range(1,n+1):
# print BITTree[i],
return BITTree
#############################################################################################################
#--------------------------------------------Segment Tree---------------------------------------------------#
#############################################################################################################
def getMid(s, e):
return s + (e - s) // 2
""" A recursive function to get the sum of values
in the given range of the array. The following
are parameters for this function.
st --> Pointer to segment tree
si --> Index of current node in the segment tree.
Initially 0 is passed as root is always at index 0
ss & se --> Starting and ending indexes of the segment
represented by current node, i.e., st[si]
qs & qe --> Starting and ending indexes of query range """
def getSumUtil(st, ss, se, qs, qe, si):
# If segment of this node is a part of given range,
# then return the sum of the segment
if (qs <= ss and qe >= se):
return st[si]
# If segment of this node is
# outside the given range
if (se < qs or ss > qe):
return 0
# If a part of this segment overlaps
# with the given range
mid = getMid(ss, se)
return getSumUtil(st, ss, mid, qs, qe, 2 * si + 1) + getSumUtil(st, mid + 1, se, qs, qe, 2 * si + 2)
def updateValueUtil(st, ss, se, i, diff, si):
# Base Case: If the input index lies
# outside the range of this segment
if (i < ss or i > se):
return
# If the input index is in range of this node,
# then update the value of the node and its children
st[si] = st[si] + diff
if (se != ss):
mid = getMid(ss, se)
updateValueUtil(st, ss, mid, i,
diff, 2 * si + 1)
updateValueUtil(st, mid + 1, se, i,
diff, 2 * si + 2)
# The function to update a value in input array
# and segment tree. It uses updateValueUtil()
# to update the value in segment tree
def updateValue(arr, st, n, i, new_val):
# Check for erroneous input index
if (i < 0 or i > n - 1):
print("Invalid Input", end="")
return
# Get the difference between
# new value and old value
diff = new_val - arr[i]
# Update the value in array
arr[i] = new_val
# Update the values of nodes in segment tree
updateValueUtil(st, 0, n - 1, i, diff, 0)
# Return sum of elements in range from
# index qs (query start) to qe (query end).
# It mainly uses getSumUtil()
def getSum(st, n, qs, qe):
# Check for erroneous input values
if (qs < 0 or qe > n - 1 or qs > qe):
print("Invalid Input", end="")
return -1
return getSumUtil(st, 0, n - 1, qs, qe, 0)
# A recursive function that constructs
# Segment Tree for array[ss..se].
# si is index of current node in segment tree st
def constructSTUtil(arr, ss, se, st, si):
# If there is one element in array,
# store it in current node of
# segment tree and return
if (ss == se):
st[si] = arr[ss]
return arr[ss]
# If there are more than one elements,
# then recur for left and right subtrees
# and store the sum of values in this node
mid = getMid(ss, se)
st[si] = constructSTUtil(arr, ss, mid, st, si * 2 + 1) + \
constructSTUtil(arr, mid + 1, se, st, si * 2 + 2)
return st[si]
""" Function to construct segment tree
from given array. This function allocates memory
for segment tree and calls constructSTUtil() to
fill the allocated memory """
def constructST(arr, n):
# Allocate memory for the segment tree
# Height of segment tree
x = (int)(math.ceil(math.log2(n)))
# Maximum size of segment tree
max_size = 2 * (int)(2**x) - 1
# Allocate memory
st = [0] * max_size
# Fill the allocated memory st
constructSTUtil(arr, 0, n - 1, st, 0)
# Return the constructed segment tree
return st
def tobinary(a): return bin(a).replace('0b', '')
def nextPerfectSquare(N):
nextN = math.floor(math.sqrt(N)) + 1
return nextN * nextN
# #####################################################################################################
alphabets = list("abcdefghijklmnopqrstuvwxyz")
vowels = list("aeiou")
MOD1 = int(1e9 + 7)
MOD2 = 998244353
INF = int(1e18)
# ########################################################################
# #-----------------------------Code Here--------------------------------#
# ########################################################################
# vsInput()
for _ in range(inp()):
n, m = inlt()
zero = n - m
k = zero // (m + 1)
ans = n * (n + 1) // 2
ans -= (m + 1) * ((k + 1) * k) // 2
ans -= (zero % (m + 1)) * (k + 1)
print(ans) | py |
1290 | A | A. Mind Controltime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou and your n−1n−1 friends have found an array of integers a1,a2,…,ana1,a2,…,an. You have decided to share it in the following way: All nn of you stand in a line in a particular order. Each minute, the person at the front of the line chooses either the first or the last element of the array, removes it, and keeps it for himself. He then gets out of line, and the next person in line continues the process.You are standing in the mm-th position in the line. Before the process starts, you may choose up to kk different people in the line, and persuade them to always take either the first or the last element in the array on their turn (for each person his own choice, not necessarily equal for all people), no matter what the elements themselves are. Once the process starts, you cannot persuade any more people, and you cannot change the choices for the people you already persuaded.Suppose that you're doing your choices optimally. What is the greatest integer xx such that, no matter what are the choices of the friends you didn't choose to control, the element you will take from the array will be greater than or equal to xx?Please note that the friends you don't control may do their choice arbitrarily, and they will not necessarily take the biggest element available.InputThe input consists of multiple test cases. The first line contains a single integer tt (1≤t≤10001≤t≤1000) — the number of test cases. The description of the test cases follows.The first line of each test case contains three space-separated integers nn, mm and kk (1≤m≤n≤35001≤m≤n≤3500, 0≤k≤n−10≤k≤n−1) — the number of elements in the array, your position in line and the number of people whose choices you can fix.The second line of each test case contains nn positive integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) — elements of the array.It is guaranteed that the sum of nn over all test cases does not exceed 35003500.OutputFor each test case, print the largest integer xx such that you can guarantee to obtain at least xx.ExampleInputCopy4
6 4 2
2 9 2 3 8 5
4 4 1
2 13 60 4
4 1 3
1 2 2 1
2 2 0
1 2
OutputCopy8
4
1
1
NoteIn the first test case; an optimal strategy is to force the first person to take the last element and the second person to take the first element. the first person will take the last element (55) because he or she was forced by you to take the last element. After this turn the remaining array will be [2,9,2,3,8][2,9,2,3,8]; the second person will take the first element (22) because he or she was forced by you to take the first element. After this turn the remaining array will be [9,2,3,8][9,2,3,8]; if the third person will choose to take the first element (99), at your turn the remaining array will be [2,3,8][2,3,8] and you will take 88 (the last element); if the third person will choose to take the last element (88), at your turn the remaining array will be [9,2,3][9,2,3] and you will take 99 (the first element). Thus, this strategy guarantees to end up with at least 88. We can prove that there is no strategy that guarantees to end up with at least 99. Hence, the answer is 88.In the second test case, an optimal strategy is to force the first person to take the first element. Then, in the worst case, both the second and the third person will take the first element: you will end up with 44. | [
"brute force",
"data structures",
"implementation"
] | #!/usr/bin/env python3
def solution(n: int, m: int, k: int, soldier: list) -> int:
arr = soldier
max_val = None
pop_num_r0 = min(k, m-1)
for left_r0 in range(pop_num_r0+1):
right_r0 = n-pop_num_r0+left_r0
# print('l0, r0:', left_r0, right_r0)
len_arr_r0 = right_r0 - left_r0
pop_num_r1 = m - 1 - pop_num_r0
val = None
for left_r1 in range(left_r0, left_r0 + pop_num_r1+1):
right_r1 = len_arr_r0-pop_num_r1 + left_r1
# print('l1, r1:', left_r1, right_r1)
if val is None:
val = max(arr[left_r1], arr[right_r1-1])
val = min(max(arr[left_r1], arr[right_r1-1]), val)
if max_val is None:
max_val = val
max_val = max(val, max_val)
return max_val
if __name__ == '__main__':
n = int(input())
for _ in range(n):
n, m, k = [int(i) for i in input().split()]
arr = [int(i) for i in input().split()]
answer = solution(n, m, k, arr)
print(answer)
| py |
1288 | B | B. Yet Another Meme Problemtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given two integers AA and BB, calculate the number of pairs (a,b)(a,b) such that 1≤a≤A1≤a≤A, 1≤b≤B1≤b≤B, and the equation a⋅b+a+b=conc(a,b)a⋅b+a+b=conc(a,b) is true; conc(a,b)conc(a,b) is the concatenation of aa and bb (for example, conc(12,23)=1223conc(12,23)=1223, conc(100,11)=10011conc(100,11)=10011). aa and bb should not contain leading zeroes.InputThe first line contains tt (1≤t≤1001≤t≤100) — the number of test cases.Each test case contains two integers AA and BB (1≤A,B≤109)(1≤A,B≤109).OutputPrint one integer — the number of pairs (a,b)(a,b) such that 1≤a≤A1≤a≤A, 1≤b≤B1≤b≤B, and the equation a⋅b+a+b=conc(a,b)a⋅b+a+b=conc(a,b) is true.ExampleInputCopy31 114 2191 31415926OutputCopy1
0
1337
NoteThere is only one suitable pair in the first test case: a=1a=1; b=9b=9 (1+9+1⋅9=191+9+1⋅9=19). | [
"math"
] | for t in range(int(input())):
a, b = map(int, input().split())
print(a * (len(str(b + 1)) - 1)) | py |
1141 | D | D. Colored Bootstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThere are nn left boots and nn right boots. Each boot has a color which is denoted as a lowercase Latin letter or a question mark ('?'). Thus, you are given two strings ll and rr, both of length nn. The character lili stands for the color of the ii-th left boot and the character riri stands for the color of the ii-th right boot.A lowercase Latin letter denotes a specific color, but the question mark ('?') denotes an indefinite color. Two specific colors are compatible if they are exactly the same. An indefinite color is compatible with any (specific or indefinite) color.For example, the following pairs of colors are compatible: ('f', 'f'), ('?', 'z'), ('a', '?') and ('?', '?'). The following pairs of colors are not compatible: ('f', 'g') and ('a', 'z').Compute the maximum number of pairs of boots such that there is one left and one right boot in a pair and their colors are compatible.Print the maximum number of such pairs and the pairs themselves. A boot can be part of at most one pair.InputThe first line contains nn (1≤n≤1500001≤n≤150000), denoting the number of boots for each leg (i.e. the number of left boots and the number of right boots).The second line contains the string ll of length nn. It contains only lowercase Latin letters or question marks. The ii-th character stands for the color of the ii-th left boot.The third line contains the string rr of length nn. It contains only lowercase Latin letters or question marks. The ii-th character stands for the color of the ii-th right boot.OutputPrint kk — the maximum number of compatible left-right pairs of boots, i.e. pairs consisting of one left and one right boot which have compatible colors.The following kk lines should contain pairs aj,bjaj,bj (1≤aj,bj≤n1≤aj,bj≤n). The jj-th of these lines should contain the index ajaj of the left boot in the jj-th pair and index bjbj of the right boot in the jj-th pair. All the numbers ajaj should be distinct (unique), all the numbers bjbj should be distinct (unique).If there are many optimal answers, print any of them.ExamplesInputCopy10
codeforces
dodivthree
OutputCopy5
7 8
4 9
2 2
9 10
3 1
InputCopy7
abaca?b
zabbbcc
OutputCopy5
6 5
2 3
4 6
7 4
1 2
InputCopy9
bambarbia
hellocode
OutputCopy0
InputCopy10
code??????
??????test
OutputCopy10
6 2
1 6
7 3
3 5
4 8
9 7
5 1
2 4
10 9
8 10
| [
"greedy",
"implementation"
] | import sys
input = sys.stdin.readline
from collections import defaultdict
n = int(input())
l = set()
r = defaultdict(list)
a = input()[:-1]
b = input()[:-1]
for i in range(n):
r[b[i]].append(i)
if a[i] == '?':
l.add(i)
ew = []
x = [0]*n
for i in range(n):
if a[i] != '?':
if a[i] in r:
q = r[a[i]].pop()
ew.append((i+1, q+1))
x[q] = 1
if not r[a[i]]:
del r[a[i]]
else:
if '?' in r:
q = r['?'].pop()
ew.append((i + 1, q + 1))
x[q] = 1
if not r['?']:
del r['?']
j = 0
for i in l:
while x[j] == 1:
j += 1
ew.append((i+1, j+1))
j += 1
print(len(ew))
for i in ew:
print(' '.join(map(str, i))) | py |
1296 | D | D. Fight with Monsterstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputThere are nn monsters standing in a row numbered from 11 to nn. The ii-th monster has hihi health points (hp). You have your attack power equal to aa hp and your opponent has his attack power equal to bb hp.You and your opponent are fighting these monsters. Firstly, you and your opponent go to the first monster and fight it till his death, then you and your opponent go the second monster and fight it till his death, and so on. A monster is considered dead if its hp is less than or equal to 00.The fight with a monster happens in turns. You hit the monster by aa hp. If it is dead after your hit, you gain one point and you both proceed to the next monster. Your opponent hits the monster by bb hp. If it is dead after his hit, nobody gains a point and you both proceed to the next monster. You have some secret technique to force your opponent to skip his turn. You can use this technique at most kk times in total (for example, if there are two monsters and k=4k=4, then you can use the technique 22 times on the first monster and 11 time on the second monster, but not 22 times on the first monster and 33 times on the second monster).Your task is to determine the maximum number of points you can gain if you use the secret technique optimally.InputThe first line of the input contains four integers n,a,bn,a,b and kk (1≤n≤2⋅105,1≤a,b,k≤1091≤n≤2⋅105,1≤a,b,k≤109) — the number of monsters, your attack power, the opponent's attack power and the number of times you can use the secret technique.The second line of the input contains nn integers h1,h2,…,hnh1,h2,…,hn (1≤hi≤1091≤hi≤109), where hihi is the health points of the ii-th monster.OutputPrint one integer — the maximum number of points you can gain if you use the secret technique optimally.ExamplesInputCopy6 2 3 3
7 10 50 12 1 8
OutputCopy5
InputCopy1 1 100 99
100
OutputCopy1
InputCopy7 4 2 1
1 3 5 4 2 7 6
OutputCopy6
| [
"greedy",
"sortings"
] | ######################################################################################
#------------------------------------Template---------------------------------------#
######################################################################################
import collections
import heapq
import sys
import math
import itertools
import bisect
from io import BytesIO, IOBase
import os
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def input(): return sys.stdin.readline().rstrip("\r\n")
######################################################################################
#--------------------------------------Input-----------------------------------------#
######################################################################################
def value(): return tuple(map(int, input().split()))
def inlt(): return [int(i) for i in input().split()]
def inp(): return int(input())
def insr(): return input()
def stlt(): return [i for i in input().split()]
######################################################################################
#------------------------------------Functions---------------------------------------#
######################################################################################
def SieveOfEratosthenes(n):
prime = [True for i in range(n+1)]
p = 2
l = []
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
for p in range(2, n+1):
if prime[p]:
l.append(p)
# print(p)
else:
continue
return l
def isPrime(n):
prime_flag = 0
if n > 1:
for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0:
prime_flag = 1
break
if prime_flag == 0:
return True
else:
return False
else:
return False
def gcdofarray(a):
x = 0
for p in a:
x = math.gcd(x, p)
return x
def printDivisors(n):
i = 1
ans = []
while i <= math.sqrt(n):
if (n % i == 0):
if (n / i == i):
ans.append(i)
else:
ans.append(i)
ans.append(n // i)
i = i + 1
ans.sort()
return ans
def binaryToDecimal(n):
return int(n, 2)
def countTriplets(a, n):
s = set()
for i in range(n):
s.add(a[i])
count = 0
for i in range(n):
for j in range(i + 1, n, 1):
xr = a[i] ^ a[j]
if xr in s and xr != a[i] and xr != a[j]:
count += 1
return int(count // 3)
def countOdd(L, R):
N = (R - L) // 2
if (R % 2 != 0 or L % 2 != 0):
N += 1
return N
def isPalindrome(s):
return s == s[::-1]
def sufsum(a):
test_list = a.copy()
test_list.reverse()
n = len(test_list)
for i in range(1, n):
test_list[i] = test_list[i] + test_list[i - 1]
return test_list
def prsum(b):
a = b.copy()
for i in range(1, len(a)):
a[i] += a[i - 1]
return a
def badachotabadachota(nums):
nums.sort()
i = 0
j = len(nums) - 1
ans = []
cc = 0
while len(ans) != len(nums):
if cc % 2 == 0:
ans.append(nums[j])
j -= 1
else:
ans.append(nums[i])
i += 1
cc += 1
return ans
def chotabadachotabada(nums):
nums.sort()
i = 0
j = len(nums) - 1
ans = []
cc = 0
while len(ans) != len(nums):
if cc % 2 == 1:
ans.append(nums[j])
j -= 1
else:
ans.append(nums[i])
i += 1
cc += 1
return ans
def primeFactors(n):
ans = []
while n % 2 == 0:
ans.append(2)
n = n // 2
for i in range(3, int(math.sqrt(n))+1, 2):
while n % i == 0:
ans.append(i)
n = n / i
if n > 2:
ans.append(n)
return ans
def closestMultiple(n, x):
if x > n:
return x
z = (int)(x / 2)
n = n + z
n = n - (n % x)
return n
def getPairsCount(arr, n, sum):
m = [0] * 1000
for i in range(0, n):
m[arr[i]] += 1
twice_count = 0
for i in range(0, n):
twice_count += m[int(sum - arr[i])]
if (int(sum - arr[i]) == arr[i]):
twice_count -= 1
return int(twice_count / 2)
def remove_consec_duplicates(test_list):
res = [i[0] for i in itertools.groupby(test_list)]
return res
def BigPower(a, b, mod):
if b == 0:
return 1
ans = BigPower(a, b//2, mod)
ans *= ans
ans %= mod
if b % 2:
ans *= a
return ans % mod
def nextPowerOf2(n):
count = 0
if (n and not(n & (n - 1))):
return n
while(n != 0):
n >>= 1
count += 1
return 1 << count
# This function multiplies x with the number
# represented by res[]. res_size is size of res[]
# or number of digits in the number represented
# by res[]. This function uses simple school
# mathematics for multiplication. This function
# may value of res_size and returns the new value
# of res_size
def multiply(x, res, res_size):
carry = 0 # Initialize carry
# One by one multiply n with individual
# digits of res[]
i = 0
while i < res_size:
prod = res[i] * x + carry
res[i] = prod % 10 # Store last digit of
# 'prod' in res[]
# make sure floor division is used
carry = prod//10 # Put rest in carry
i = i + 1
# Put carry in res and increase result size
while (carry):
res[res_size] = carry % 10
# make sure floor division is used
# to avoid floating value
carry = carry // 10
res_size = res_size + 1
return res_size
def Kadane(a, size):
max_so_far = -1e18 - 1
max_ending_here = 0
for i in range(0, size):
max_ending_here = max_ending_here + a[i]
if (max_so_far < max_ending_here):
max_so_far = max_ending_here
if max_ending_here < 0:
max_ending_here = 0
return max_so_far
def highestPowerof2(n):
res = 0
for i in range(n, 0, -1):
if ((i & (i - 1)) == 0):
res = i
break
return res
def lower_bound(numbers, length, searchnum):
answer = -1
start = 0
end = length - 1
while start <= end:
middle = (start + end)//2
if numbers[middle] == searchnum:
answer = middle
end = middle - 1
elif numbers[middle] > searchnum:
end = middle - 1
else:
start = middle + 1
return answer
def MEX(nList, start):
nList = set(nList)
mex = start
while mex in nList:
mex += 1
return mex
def MinValue(N, X):
N = list(N)
ln = len(N)
position = ln + 1
# # If the given string N represent
# # a negative value
# if (N[0] == '-'):
# # X must be place at the last
# # index where is greater than N[i]
# for i in range(ln - 1, 0, -1):
# if ((ord(N[i]) - ord('0')) < X):
# position = i
# else:
# # For positive numbers, X must be
# # placed at the last index where
# # it is smaller than N[i]
for i in range(ln - 1, -1, -1):
if ((ord(N[i]) - ord('0')) > X):
position = i
# Insert X at that position
c = chr(X + ord('0'))
str = N.insert(position, c)
# Return the string
return ''.join(N)
def findClosest(arr, n, target):
if (target <= arr[0]):
return arr[0]
if (target >= arr[n - 1]):
return arr[n - 1]
i = 0
j = n
mid = 0
while (i < j):
mid = (i + j) // 2
if (arr[mid] == target):
return arr[mid]
if (target < arr[mid]):
if (mid > 0 and target > arr[mid - 1]):
return getClosest(arr[mid - 1], arr[mid], target)
j = mid
else:
if (mid < n - 1 and target < arr[mid + 1]):
return getClosest(arr[mid], arr[mid + 1], target)
i = mid + 1
return arr[mid]
def getClosest(val1, val2, target):
if (target - val1 >= val2 - target):
return val2
else:
return val1
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den, p - 2, p)) % p
# #####################################################################################################
# #------------------------------------------GRAPHS---------------------------------------------------#
# #####################################################################################################
class Graph:
def __init__(self, edges, n):
self.adjList = [[] for _ in range(n)]
for (src, dest) in edges:
self.adjList[src].append(dest)
self.adjList[dest].append(src)
def BFS(graph, v, discovered):
q = collections.deque()
discovered[v] = True
q.append(v)
ans = []
while q:
v = q.popleft()
ans.append(v)
# print(v, end=' ')
for u in graph.adjList[v]:
if not discovered[u]:
discovered[u] = True
q.append(u)
return ans
#############################################################################################################
#--------------------------------------------Fenwick Tree---------------------------------------------------#
#############################################################################################################
def getsumFT(BITTree, i):
s = 0 # initialize result
# index in BITree[] is 1 more than the index in arr[]
i = i+1
# Traverse ancestors of BITree[index]
while i > 0:
# Add current element of BITree to sum
s += BITTree[i]
# Move index to parent node in getSum View
i -= i & (-i)
return s
# Updates a node in Binary Index Tree (BITree) at given index
# in BITree. The given value 'val' is added to BITree[i] and
# all of its ancestors in tree.
def updatebit(BITTree, n, i, v):
# index in BITree[] is 1 more than the index in arr[]
i += 1
# Traverse all ancestors and add 'val'
while i <= n:
# Add 'val' to current node of BI Tree
BITTree[i] += v
# Update index to that of parent in update View
i += i & (-i)
# Constructs and returns a Binary Indexed Tree for given
# array of size n.
def construct(arr, n):
# Create and initialize BITree[] as 0
BITTree = [0]*(n+1)
# Store the actual values in BITree[] using update()
for i in range(n):
updatebit(BITTree, n, i, arr[i])
# Uncomment below lines to see contents of BITree[]
# for i in range(1,n+1):
# print BITTree[i],
return BITTree
#############################################################################################################
#--------------------------------------------Segment Tree---------------------------------------------------#
#############################################################################################################
def getMid(s, e):
return s + (e - s) // 2
""" A recursive function to get the sum of values
in the given range of the array. The following
are parameters for this function.
st --> Pointer to segment tree
si --> Index of current node in the segment tree.
Initially 0 is passed as root is always at index 0
ss & se --> Starting and ending indexes of the segment
represented by current node, i.e., st[si]
qs & qe --> Starting and ending indexes of query range """
def getSumUtil(st, ss, se, qs, qe, si):
# If segment of this node is a part of given range,
# then return the sum of the segment
if (qs <= ss and qe >= se):
return st[si]
# If segment of this node is
# outside the given range
if (se < qs or ss > qe):
return 0
# If a part of this segment overlaps
# with the given range
mid = getMid(ss, se)
return getSumUtil(st, ss, mid, qs, qe, 2 * si + 1) + getSumUtil(st, mid + 1, se, qs, qe, 2 * si + 2)
def updateValueUtil(st, ss, se, i, diff, si):
# Base Case: If the input index lies
# outside the range of this segment
if (i < ss or i > se):
return
# If the input index is in range of this node,
# then update the value of the node and its children
st[si] = st[si] + diff
if (se != ss):
mid = getMid(ss, se)
updateValueUtil(st, ss, mid, i,
diff, 2 * si + 1)
updateValueUtil(st, mid + 1, se, i,
diff, 2 * si + 2)
# The function to update a value in input array
# and segment tree. It uses updateValueUtil()
# to update the value in segment tree
def updateValue(arr, st, n, i, new_val):
# Check for erroneous input index
if (i < 0 or i > n - 1):
print("Invalid Input", end="")
return
# Get the difference between
# new value and old value
diff = new_val - arr[i]
# Update the value in array
arr[i] = new_val
# Update the values of nodes in segment tree
updateValueUtil(st, 0, n - 1, i, diff, 0)
# Return sum of elements in range from
# index qs (query start) to qe (query end).
# It mainly uses getSumUtil()
def getSum(st, n, qs, qe):
# Check for erroneous input values
if (qs < 0 or qe > n - 1 or qs > qe):
print("Invalid Input", end="")
return -1
return getSumUtil(st, 0, n - 1, qs, qe, 0)
# A recursive function that constructs
# Segment Tree for array[ss..se].
# si is index of current node in segment tree st
def constructSTUtil(arr, ss, se, st, si):
# If there is one element in array,
# store it in current node of
# segment tree and return
if (ss == se):
st[si] = arr[ss]
return arr[ss]
# If there are more than one elements,
# then recur for left and right subtrees
# and store the sum of values in this node
mid = getMid(ss, se)
st[si] = constructSTUtil(arr, ss, mid, st, si * 2 + 1) + \
constructSTUtil(arr, mid + 1, se, st, si * 2 + 2)
return st[si]
""" Function to construct segment tree
from given array. This function allocates memory
for segment tree and calls constructSTUtil() to
fill the allocated memory """
def constructST(arr, n):
# Allocate memory for the segment tree
# Height of segment tree
x = (int)(math.ceil(math.log2(n)))
# Maximum size of segment tree
max_size = 2 * (int)(2**x) - 1
# Allocate memory
st = [0] * max_size
# Fill the allocated memory st
constructSTUtil(arr, 0, n - 1, st, 0)
# Return the constructed segment tree
return st
def tobinary(a): return bin(a).replace('0b', '')
def nextPerfectSquare(N):
nextN = math.floor(math.sqrt(N)) + 1
return nextN * nextN
# #####################################################################################################
alphabets = list("abcdefghijklmnopqrstuvwxyz")
vowels = list("aeiou")
MOD1 = int(1e9 + 7)
MOD2 = 998244353
INF = int(1e18)
# ########################################################################
# #-----------------------------Code Here--------------------------------#
# ########################################################################
# vsInput()
# for _ in range(inp()):
n, a, b, k = inlt()
h = inlt()
loose = []
check = []
ans = 0
for i in range(n):
temp = a + b
mod = h[i] % temp
if mod - a <= 0 and mod:
ans += 1
elif mod == 0:
loose.append(a + b)
else:
check.append(mod)
new = sorted(loose + check)
for i in new:
temp = math.ceil(i / a) - 1
if k - temp >= 0:
k -= temp
ans += 1
else:
break
print(ans)
| py |
1292 | A | A. NEKO's Maze Gametime limit per test1.5 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output3R2 as DJ Mashiro - Happiness Breeze Ice - DJ Mashiro is dead or aliveNEKO#ΦωΦ has just got a new maze game on her PC!The game's main puzzle is a maze; in the forms of a 2×n2×n rectangle grid. NEKO's task is to lead a Nekomimi girl from cell (1,1)(1,1) to the gate at (2,n)(2,n) and escape the maze. The girl can only move between cells sharing a common side.However, at some moments during the game, some cells may change their state: either from normal ground to lava (which forbids movement into that cell), or vice versa (which makes that cell passable again). Initially all cells are of the ground type.After hours of streaming, NEKO finally figured out there are only qq such moments: the ii-th moment toggles the state of cell (ri,ci)(ri,ci) (either from ground to lava or vice versa).Knowing this, NEKO wonders, after each of the qq moments, whether it is still possible to move from cell (1,1)(1,1) to cell (2,n)(2,n) without going through any lava cells.Although NEKO is a great streamer and gamer, she still can't get through quizzes and problems requiring large amount of Brain Power. Can you help her?InputThe first line contains integers nn, qq (2≤n≤1052≤n≤105, 1≤q≤1051≤q≤105).The ii-th of qq following lines contains two integers riri, cici (1≤ri≤21≤ri≤2, 1≤ci≤n1≤ci≤n), denoting the coordinates of the cell to be flipped at the ii-th moment.It is guaranteed that cells (1,1)(1,1) and (2,n)(2,n) never appear in the query list.OutputFor each moment, if it is possible to travel from cell (1,1)(1,1) to cell (2,n)(2,n), print "Yes", otherwise print "No". There should be exactly qq answers, one after every update.You can print the words in any case (either lowercase, uppercase or mixed).ExampleInputCopy5 5
2 3
1 4
2 4
2 3
1 4
OutputCopyYes
No
No
No
Yes
NoteWe'll crack down the example test here: After the first query; the girl still able to reach the goal. One of the shortest path ways should be: (1,1)→(1,2)→(1,3)→(1,4)→(1,5)→(2,5)(1,1)→(1,2)→(1,3)→(1,4)→(1,5)→(2,5). After the second query, it's impossible to move to the goal, since the farthest cell she could reach is (1,3)(1,3). After the fourth query, the (2,3)(2,3) is not blocked, but now all the 44-th column is blocked, so she still can't reach the goal. After the fifth query, the column barrier has been lifted, thus she can go to the final goal again. | [
"data structures",
"dsu",
"implementation"
] | N,Q = map(int,input().split())
dat = [[0]*N for _ in range(2)]
num = 0
def check(x,y):
l = 0
for i in range(-1,2):
if 0<=i+x<N and dat[abs(y-1)][i+x]==1:
l += 1
return l
num = 0
for i in range(Q):
y,x = map(int,input().split())
l = check(x-1,y-1)
if dat[y-1][x-1]==0:
if l!=0:
num += l
else:
num -= l
dat[y-1][x-1] = abs(dat[y-1][x-1]-1)
if num==0:
print("Yes")
else:
print("No")
| py |
1313 | B | B. Different Rulestime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputNikolay has only recently started in competitive programming; but already qualified to the finals of one prestigious olympiad. There going to be nn participants, one of whom is Nikolay. Like any good olympiad, it consists of two rounds. Tired of the traditional rules, in which the participant who solved the largest number of problems wins, the organizers came up with different rules.Suppose in the first round participant A took xx-th place and in the second round — yy-th place. Then the total score of the participant A is sum x+yx+y. The overall place of the participant A is the number of participants (including A) having their total score less than or equal to the total score of A. Note, that some participants may end up having a common overall place. It is also important to note, that in both the first and the second round there were no two participants tying at a common place. In other words, for every ii from 11 to nn exactly one participant took ii-th place in first round and exactly one participant took ii-th place in second round.Right after the end of the Olympiad, Nikolay was informed that he got xx-th place in first round and yy-th place in the second round. Nikolay doesn't know the results of other participants, yet he wonders what is the minimum and maximum place he can take, if we consider the most favorable and unfavorable outcome for him. Please help Nikolay to find the answer to this question.InputThe first line contains an integer tt (1≤t≤1001≤t≤100) — the number of test cases to solve.Each of the following tt lines contains integers nn, xx, yy (1≤n≤1091≤n≤109, 1≤x,y≤n1≤x,y≤n) — the number of participants in the olympiad, the place that Nikolay took in the first round and the place that Nikolay took in the second round.OutputPrint two integers — the minimum and maximum possible overall place Nikolay could take.ExamplesInputCopy15 1 3OutputCopy1 3InputCopy16 3 4OutputCopy2 6NoteExplanation for the first example:Suppose there were 5 participants A-E. Let's denote Nikolay as A. The the most favorable results for Nikolay could look as follows: However, the results of the Olympiad could also look like this: In the first case Nikolay would have taken first place, and in the second — third place. | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | for _ in range(int(input())):
n,x,y=map(int,input().split())
print(max(1,min(n,x+y-n+1)),min(n,x+y-1)) | py |
1310 | A | A. Recommendationstime limit per test2 secondsmemory limit per test512 megabytesinputstandard inputoutputstandard outputVK news recommendation system daily selects interesting publications of one of nn disjoint categories for each user. Each publication belongs to exactly one category. For each category ii batch algorithm selects aiai publications.The latest A/B test suggests that users are reading recommended publications more actively if each category has a different number of publications within daily recommendations. The targeted algorithm can find a single interesting publication of ii-th category within titi seconds. What is the minimum total time necessary to add publications to the result of batch algorithm execution, so all categories have a different number of publications? You can't remove publications recommended by the batch algorithm.InputThe first line of input consists of single integer nn — the number of news categories (1≤n≤2000001≤n≤200000).The second line of input consists of nn integers aiai — the number of publications of ii-th category selected by the batch algorithm (1≤ai≤1091≤ai≤109).The third line of input consists of nn integers titi — time it takes for targeted algorithm to find one new publication of category ii (1≤ti≤105)1≤ti≤105).OutputPrint one integer — the minimal required time for the targeted algorithm to get rid of categories with the same size.ExamplesInputCopy5
3 7 9 7 8
5 2 5 7 5
OutputCopy6
InputCopy5
1 2 3 4 5
1 1 1 1 1
OutputCopy0
NoteIn the first example; it is possible to find three publications of the second type; which will take 6 seconds.In the second example; all news categories contain a different number of publications. | [
"data structures",
"greedy",
"sortings"
] | import math
import heapq
import bisect
import functools
import itertools
import collections
import os
import sys
from io import BytesIO, IOBase
MOD0 = 10 ** 9 + 7
MOD1 = 998244353
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
read_str = lambda: input().strip()
read_int = lambda: int(input())
read_ints = lambda: map(int, input().split())
read_list = lambda: list(map(int, input().split()))
def solve():
n = read_int()
a = read_list()
b = read_list()
nums, heap, ans, cur_sum = sorted(zip(a, b), reverse=True), [], 0, 0
max_val, cur = nums[0][0], nums[-1][0]
while cur <= max_val:
while nums and nums[-1][0] == cur:
_, tmp = nums.pop()
heapq.heappush(heap, -tmp)
cur_sum += tmp
if cur == max_val and (not heap):
break
if heap:
cur_sum += heapq.heappop(heap)
ans += cur_sum
cur += 1
if cur > max_val:
max_val = cur
elif (not heap) and nums:
cur = nums[-1][0]
heap, cur_sum = [], 0
print(ans)
solve() | py |
1296 | E2 | E2. String Coloring (hard version)time limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputThis is a hard version of the problem. The actual problems are different; but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.You are given a string ss consisting of nn lowercase Latin letters.You have to color all its characters the minimum number of colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in ss).After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.The goal is to make the string sorted, i.e. all characters should be in alphabetical order.Your task is to find the minimum number of colors which you have to color the given string in so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.InputThe first line of the input contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the length of ss.The second line of the input contains the string ss consisting of exactly nn lowercase Latin letters.OutputIn the first line print one integer resres (1≤res≤n1≤res≤n) — the minimum number of colors in which you have to color the given string so that after coloring it can become sorted by some sequence of swaps.In the second line print any possible coloring that can be used to sort the string using some sequence of swaps described in the problem statement. The coloring is the array cc of length nn, where 1≤ci≤res1≤ci≤res and cici means the color of the ii-th character.ExamplesInputCopy9
abacbecfd
OutputCopy2
1 1 2 1 2 1 2 1 2
InputCopy8
aaabbcbb
OutputCopy2
1 2 1 2 1 2 1 1
InputCopy7
abcdedc
OutputCopy3
1 1 1 1 1 2 3
InputCopy5
abcde
OutputCopy1
1 1 1 1 1
| [
"data structures",
"dp"
] | import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from math import gcd, ceil
def prod(a, mod=10**9+7):
ans = 1
for each in a:
ans = (ans * each) % mod
return ans
def lcm(a, b): return a * b // gcd(a, b)
def binary(x, length=16):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
for _ in range(int(input()) if not True else 1):
n = int(input())
s = input()
ans = [0]*n
colors = ['a']*50
for i in range(n):
for j in range(50):
if ord(s[i]) >= ord(colors[j]):
colors[j] = s[i]
ans[i] = j+1
break
print(max(ans))
print(*ans) | py |
1296 | E1 | E1. String Coloring (easy version)time limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputThis is an easy version of the problem. The actual problems are different; but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.You are given a string ss consisting of nn lowercase Latin letters.You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in ss).After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.The goal is to make the string sorted, i.e. all characters should be in alphabetical order.Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.InputThe first line of the input contains one integer nn (1≤n≤2001≤n≤200) — the length of ss.The second line of the input contains the string ss consisting of exactly nn lowercase Latin letters.OutputIf it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of nn characters, the ii-th character should be '0' if the ii-th character is colored the first color and '1' otherwise).ExamplesInputCopy9
abacbecfd
OutputCopyYES
001010101
InputCopy8
aaabbcbb
OutputCopyYES
01011011
InputCopy7
abcdedc
OutputCopyNO
InputCopy5
abcde
OutputCopyYES
00000
| [
"constructive algorithms",
"dp",
"graphs",
"greedy",
"sortings"
] | from bisect import bisect_right
from collections import defaultdict
n=int(input())
s=input()
a=[0]*n
d=defaultdict(int)
ind=defaultdict(int)
for i,el in enumerate(s):
num=200*(ord(el)-ord("a"))+d[el]
d[el]+=1
a[i]=num
ind[num]=i
l=True
q=[False]*n
for i in range(n):
m=min(a[i:])
for j in range(n):
if a[j]==m:
break
if i==j:
if q[ind[m]]==False:
q[ind[m]]=0
elif q[ind[m]]==False:
s=set()
for el in a[i:j]:
if q[ind[el]]==False:
pass
else :
s.add(q[ind[el]])
if len(s)>1:
l=False
elif len(s)==0:
q[ind[m]]=0
for el in a[i:j]:
q[ind[el]]=1
else :
val=list(s)[0]
for el in a[i:j]:
q[ind[el]]=val
q[ind[m]]=(val+1)%2
a=a[:i]+[m]+a[i:j]+a[j+1:]
else :
s=set()
v=q[ind[m]]
for el in a[i:j]:
if q[ind[el]]==False:
pass
else :
s.add(q[ind[el]])
if len(s)>1:
l=False
elif len(s)==0:
for el in a[i:j]:
q[ind[el]]=(v+1)%2
else :
if v==list(s)[0]:
l=False
else:
for el in a[i:j]:
q[ind[el]]=(v+1)%2
a=a[:i]+[m]+a[i:j]+a[j+1:]
# print(ind[m],i,j,q,a)
if not l:
print("NO")
else :
print('YES')
res=""
for el in q:
res+=str(el)
print(res)
| py |
1320 | B | B. Navigation Systemtime limit per test2 secondsmemory limit per test512 megabytesinputstandard inputoutputstandard outputThe map of Bertown can be represented as a set of nn intersections, numbered from 11 to nn and connected by mm one-way roads. It is possible to move along the roads from any intersection to any other intersection. The length of some path from one intersection to another is the number of roads that one has to traverse along the path. The shortest path from one intersection vv to another intersection uu is the path that starts in vv, ends in uu and has the minimum length among all such paths.Polycarp lives near the intersection ss and works in a building near the intersection tt. Every day he gets from ss to tt by car. Today he has chosen the following path to his workplace: p1p1, p2p2, ..., pkpk, where p1=sp1=s, pk=tpk=t, and all other elements of this sequence are the intermediate intersections, listed in the order Polycarp arrived at them. Polycarp never arrived at the same intersection twice, so all elements of this sequence are pairwise distinct. Note that you know Polycarp's path beforehand (it is fixed), and it is not necessarily one of the shortest paths from ss to tt.Polycarp's car has a complex navigation system installed in it. Let's describe how it works. When Polycarp starts his journey at the intersection ss, the system chooses some shortest path from ss to tt and shows it to Polycarp. Let's denote the next intersection in the chosen path as vv. If Polycarp chooses to drive along the road from ss to vv, then the navigator shows him the same shortest path (obviously, starting from vv as soon as he arrives at this intersection). However, if Polycarp chooses to drive to another intersection ww instead, the navigator rebuilds the path: as soon as Polycarp arrives at ww, the navigation system chooses some shortest path from ww to tt and shows it to Polycarp. The same process continues until Polycarp arrives at tt: if Polycarp moves along the road recommended by the system, it maintains the shortest path it has already built; but if Polycarp chooses some other path, the system rebuilds the path by the same rules.Here is an example. Suppose the map of Bertown looks as follows, and Polycarp drives along the path [1,2,3,4][1,2,3,4] (s=1s=1, t=4t=4): Check the picture by the link http://tk.codeforces.com/a.png When Polycarp starts at 11, the system chooses some shortest path from 11 to 44. There is only one such path, it is [1,5,4][1,5,4]; Polycarp chooses to drive to 22, which is not along the path chosen by the system. When Polycarp arrives at 22, the navigator rebuilds the path by choosing some shortest path from 22 to 44, for example, [2,6,4][2,6,4] (note that it could choose [2,3,4][2,3,4]); Polycarp chooses to drive to 33, which is not along the path chosen by the system. When Polycarp arrives at 33, the navigator rebuilds the path by choosing the only shortest path from 33 to 44, which is [3,4][3,4]; Polycarp arrives at 44 along the road chosen by the navigator, so the system does not have to rebuild anything. Overall, we get 22 rebuilds in this scenario. Note that if the system chose [2,3,4][2,3,4] instead of [2,6,4][2,6,4] during the second step, there would be only 11 rebuild (since Polycarp goes along the path, so the system maintains the path [3,4][3,4] during the third step).The example shows us that the number of rebuilds can differ even if the map of Bertown and the path chosen by Polycarp stays the same. Given this information (the map and Polycarp's path), can you determine the minimum and the maximum number of rebuilds that could have happened during the journey?InputThe first line contains two integers nn and mm (2≤n≤m≤2⋅1052≤n≤m≤2⋅105) — the number of intersections and one-way roads in Bertown, respectively.Then mm lines follow, each describing a road. Each line contains two integers uu and vv (1≤u,v≤n1≤u,v≤n, u≠vu≠v) denoting a road from intersection uu to intersection vv. All roads in Bertown are pairwise distinct, which means that each ordered pair (u,v)(u,v) appears at most once in these mm lines (but if there is a road (u,v)(u,v), the road (v,u)(v,u) can also appear).The following line contains one integer kk (2≤k≤n2≤k≤n) — the number of intersections in Polycarp's path from home to his workplace.The last line contains kk integers p1p1, p2p2, ..., pkpk (1≤pi≤n1≤pi≤n, all these integers are pairwise distinct) — the intersections along Polycarp's path in the order he arrived at them. p1p1 is the intersection where Polycarp lives (s=p1s=p1), and pkpk is the intersection where Polycarp's workplace is situated (t=pkt=pk). It is guaranteed that for every i∈[1,k−1]i∈[1,k−1] the road from pipi to pi+1pi+1 exists, so the path goes along the roads of Bertown. OutputPrint two integers: the minimum and the maximum number of rebuilds that could have happened during the journey.ExamplesInputCopy6 9
1 5
5 4
1 2
2 3
3 4
4 1
2 6
6 4
4 2
4
1 2 3 4
OutputCopy1 2
InputCopy7 7
1 2
2 3
3 4
4 5
5 6
6 7
7 1
7
1 2 3 4 5 6 7
OutputCopy0 0
InputCopy8 13
8 7
8 6
7 5
7 4
6 5
6 4
5 3
5 2
4 3
4 2
3 1
2 1
1 8
5
8 7 5 2 1
OutputCopy0 3
| [
"dfs and similar",
"graphs",
"shortest paths"
] | import sys
input = sys.stdin.readline
from collections import deque
def bfs(root):
vis=[0]*n
queue=[root,-1]
queue=deque(queue)
vis[root]=1
d=0
while len(queue)!=1:
element = queue.popleft()
if element==-1:
queue.append(-1)
d+=1
continue
vis[element] = 1
dist[element] = d
for i in graph[element]:
if not vis[i]:
queue.append(i)
vis[i]=1
n,m=map(int,input().split())
graph=[[] for i in range(n)]
t=[[] for i in range(n)]
for i in range(m):
u,v=map(int,input().split())
graph[v-1].append(u-1)
t[u-1].append(v-1)
k=int(input())
p=list(map(int,input().split()))
dist=[0 for i in range(n)]
bfs(p[-1]-1)
# print (dist)
maxx=0
minn=0
for i in range(k-1):
if dist[p[i+1]-1]>dist[p[i]-1]-1:
maxx+=1
minn+=1
else:
count=0
# print (i,p[i],t[p[i]])
for j in t[p[i]-1]:
if dist[j]==dist[p[i]-1]-1:
count+=1
if count>=2:
maxx+=1
print (minn,maxx) | py |
1301 | D | D. Time to Runtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputBashar was practicing for the national programming contest. Because of sitting too much in front of the computer without doing physical movements and eating a lot Bashar became much fatter. Bashar is going to quit programming after the national contest and he is going to become an actor (just like his father); so he should lose weight.In order to lose weight, Bashar is going to run for kk kilometers. Bashar is going to run in a place that looks like a grid of nn rows and mm columns. In this grid there are two one-way roads of one-kilometer length between each pair of adjacent by side cells, one road is going from the first cell to the second one, and the other road is going from the second cell to the first one. So, there are exactly (4nm−2n−2m)(4nm−2n−2m) roads.Let's take, for example, n=3n=3 and m=4m=4. In this case, there are 3434 roads. It is the picture of this case (arrows describe roads):Bashar wants to run by these rules: He starts at the top-left cell in the grid; In one move Bashar may go up (the symbol 'U'), down (the symbol 'D'), left (the symbol 'L') or right (the symbol 'R'). More formally, if he stands in the cell in the row ii and in the column jj, i.e. in the cell (i,j)(i,j) he will move to: in the case 'U' to the cell (i−1,j)(i−1,j); in the case 'D' to the cell (i+1,j)(i+1,j); in the case 'L' to the cell (i,j−1)(i,j−1); in the case 'R' to the cell (i,j+1)(i,j+1); He wants to run exactly kk kilometers, so he wants to make exactly kk moves; Bashar can finish in any cell of the grid; He can't go out of the grid so at any moment of the time he should be on some cell; Bashar doesn't want to get bored while running so he must not visit the same road twice. But he can visit the same cell any number of times. Bashar asks you if it is possible to run by such rules. If it is possible, you should tell him how should he run.You should give him aa steps to do and since Bashar can't remember too many steps, aa should not exceed 30003000. In every step, you should give him an integer ff and a string of moves ss of length at most 44 which means that he should repeat the moves in the string ss for ff times. He will perform the steps in the order you print them.For example, if the steps are 22 RUD, 33 UUL then the moves he is going to move are RUD ++ RUD ++ UUL ++ UUL ++ UUL == RUDRUDUULUULUUL.Can you help him and give him a correct sequence of moves such that the total distance he will run is equal to kk kilometers or say, that it is impossible?InputThe only line contains three integers nn, mm and kk (1≤n,m≤5001≤n,m≤500, 1≤k≤1091≤k≤109), which are the number of rows and the number of columns in the grid and the total distance Bashar wants to run.OutputIf there is no possible way to run kk kilometers, print "NO" (without quotes), otherwise print "YES" (without quotes) in the first line.If the answer is "YES", on the second line print an integer aa (1≤a≤30001≤a≤3000) — the number of steps, then print aa lines describing the steps.To describe a step, print an integer ff (1≤f≤1091≤f≤109) and a string of moves ss of length at most 44. Every character in ss should be 'U', 'D', 'L' or 'R'.Bashar will start from the top-left cell. Make sure to move exactly kk moves without visiting the same road twice and without going outside the grid. He can finish at any cell.We can show that if it is possible to run exactly kk kilometers, then it is possible to describe the path under such output constraints.ExamplesInputCopy3 3 4
OutputCopyYES
2
2 R
2 L
InputCopy3 3 1000000000
OutputCopyNO
InputCopy3 3 8
OutputCopyYES
3
2 R
2 D
1 LLRR
InputCopy4 4 9
OutputCopyYES
1
3 RLD
InputCopy3 4 16
OutputCopyYES
8
3 R
3 L
1 D
3 R
1 D
1 U
3 L
1 D
NoteThe moves Bashar is going to move in the first example are: "RRLL".It is not possible to run 10000000001000000000 kilometers in the second example because the total length of the roads is smaller and Bashar can't run the same road twice.The moves Bashar is going to move in the third example are: "RRDDLLRR".The moves Bashar is going to move in the fifth example are: "RRRLLLDRRRDULLLD". It is the picture of his run (the roads on this way are marked with red and numbered in the order of his running): | [
"constructive algorithms",
"graphs",
"implementation"
] | import random, sys, os, math, gc
from collections import Counter, defaultdict, deque
from functools import lru_cache, reduce, cmp_to_key
from itertools import accumulate, combinations, permutations, product
from heapq import nsmallest, nlargest, heapify, heappop, heappush
from io import BytesIO, IOBase
from copy import deepcopy
from bisect import bisect_left, bisect_right
from math import factorial, gcd
from operator import mul, xor
from types import GeneratorType
# if "PyPy" in sys.version:
# import pypyjit; pypyjit.set_param('max_unroll_recursion=-1')
# sys.setrecursionlimit(2*10**5)
BUFSIZE = 8192
MOD = 10**9 + 7
MODD = 998244353
INF = float('inf')
D4 = [(1, 0), (0, 1), (-1, 0), (0, -1)]
D8 = [(1, 0), (1, 1), (0, 1), (-1, 1), (-1, 0), (-1, -1), (0, -1), (1, -1)]
def solve():
n, m, k = LII()
if k > 2 * (2 * n * m - n - m):
return print("NO")
ans = []
def run(s, c):
nonlocal k
if not s:
return
t = min(k, s)
if t:
k -= t
ans.append((t, c))
run(m - 1, "R")
for _ in range(m-1):
run(n - 1, "D")
run(n - 1, "U")
run(1, "L")
for _ in range(n-1):
run(1, "D")
run(m - 1, "R")
run(m - 1, "L")
run(n - 1, "U")
print("YES")
print(len(ans))
for i in ans:
print(*i)
def main():
t = 1
# t = II()
for _ in range(t):
solve()
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
def bitcnt(n):
c = (n & 0x5555555555555555) + ((n >> 1) & 0x5555555555555555)
c = (c & 0x3333333333333333) + ((c >> 2) & 0x3333333333333333)
c = (c & 0x0F0F0F0F0F0F0F0F) + ((c >> 4) & 0x0F0F0F0F0F0F0F0F)
c = (c & 0x00FF00FF00FF00FF) + ((c >> 8) & 0x00FF00FF00FF00FF)
c = (c & 0x0000FFFF0000FFFF) + ((c >> 16) & 0x0000FFFF0000FFFF)
c = (c & 0x00000000FFFFFFFF) + ((c >> 32) & 0x00000000FFFFFFFF)
return c
def lcm(x, y):
return x * y // gcd(x, y)
def lowbit(x):
return x & -x
def perm(n, r):
return factorial(n) // factorial(n - r) if n >= r else 0
def comb(n, r):
return factorial(n) // (factorial(r) * factorial(n - r)) if n >= r else 0
def probabilityMod(x, y, mod):
return x * pow(y, mod-2, mod) % mod
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index + 1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin = IOWrapper(sys.stdin)
# sys.stdout = IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def I():
return input()
def II():
return int(input())
def MI():
return map(int, input().split())
def LI():
return list(input().split())
def LII():
return list(map(int, input().split()))
def GMI():
return map(lambda x: int(x) - 1, input().split())
def LGMI():
return list(map(lambda x: int(x) - 1, input().split()))
def getGraph(n, m, directed=False):
d = [[] for _ in range(n)]
for _ in range(m):
u, v = LGMI()
d[u].append(v)
if not directed:
d[v].append(u)
return d
def getWeightedGraph(n, m, directed=False):
d = [[] for _ in range(n)]
for _ in range(m):
u, v, w = LII()
u -= 1; v -= 1
d[u].append((v, w))
if not directed:
d[v].append((u, w))
return d
if __name__ == "__main__":
main() | py |
1304 | D | D. Shortest and Longest LIStime limit per test3 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputGildong recently learned how to find the longest increasing subsequence (LIS) in O(nlogn)O(nlogn) time for a sequence of length nn. He wants to test himself if he can implement it correctly, but he couldn't find any online judges that would do it (even though there are actually many of them). So instead he's going to make a quiz for you about making permutations of nn distinct integers between 11 and nn, inclusive, to test his code with your output.The quiz is as follows.Gildong provides a string of length n−1n−1, consisting of characters '<' and '>' only. The ii-th (1-indexed) character is the comparison result between the ii-th element and the i+1i+1-st element of the sequence. If the ii-th character of the string is '<', then the ii-th element of the sequence is less than the i+1i+1-st element. If the ii-th character of the string is '>', then the ii-th element of the sequence is greater than the i+1i+1-st element.He wants you to find two possible sequences (not necessarily distinct) consisting of nn distinct integers between 11 and nn, inclusive, each satisfying the comparison results, where the length of the LIS of the first sequence is minimum possible, and the length of the LIS of the second sequence is maximum possible.InputEach test contains one or more test cases. The first line contains the number of test cases tt (1≤t≤1041≤t≤104).Each test case contains exactly one line, consisting of an integer and a string consisting of characters '<' and '>' only. The integer is nn (2≤n≤2⋅1052≤n≤2⋅105), the length of the permutation you need to find. The string is the comparison results explained in the description. The length of the string is n−1n−1.It is guaranteed that the sum of all nn in all test cases doesn't exceed 2⋅1052⋅105.OutputFor each test case, print two lines with nn integers each. The first line is the sequence with the minimum length of the LIS, and the second line is the sequence with the maximum length of the LIS. If there are multiple answers, print any one of them. Each sequence should contain all integers between 11 and nn, inclusive, and should satisfy the comparison results.It can be shown that at least one answer always exists.ExampleInputCopy3
3 <<
7 >><>><
5 >>><
OutputCopy1 2 3
1 2 3
5 4 3 7 2 1 6
4 3 1 7 5 2 6
4 3 2 1 5
5 4 2 1 3
NoteIn the first case; 11 22 33 is the only possible answer.In the second case, the shortest length of the LIS is 22, and the longest length of the LIS is 33. In the example of the maximum LIS sequence, 44 '33' 11 77 '55' 22 '66' can be one of the possible LIS. | [
"constructive algorithms",
"graphs",
"greedy",
"two pointers"
] |
for _ in range(int(input())):
p=[el for el in input().split()]
n=int(p[0])
s=p[1]
ans=[0]*n
num=n
last=0
for i in range(n):
if i==n-1 or s[i]==">":
j=i
while j>=last :
ans[j]=num
num-=1
j-=1
last=i+1
print(*ans)
ans=[0]*(n)
num=1
last=0
for i in range(n):
if i==n-1 or s[i]=="<":
j=i
while j>=last :
ans[j]=num
num+=1
j-=1
last=i+1
print(*ans)
| py |
1141 | A | A. Game 23time limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputPolycarp plays "Game 23". Initially he has a number nn and his goal is to transform it to mm. In one move, he can multiply nn by 22 or multiply nn by 33. He can perform any number of moves.Print the number of moves needed to transform nn to mm. Print -1 if it is impossible to do so.It is easy to prove that any way to transform nn to mm contains the same number of moves (i.e. number of moves doesn't depend on the way of transformation).InputThe only line of the input contains two integers nn and mm (1≤n≤m≤5⋅1081≤n≤m≤5⋅108).OutputPrint the number of moves to transform nn to mm, or -1 if there is no solution.ExamplesInputCopy120 51840
OutputCopy7
InputCopy42 42
OutputCopy0
InputCopy48 72
OutputCopy-1
NoteIn the first example; the possible sequence of moves is: 120→240→720→1440→4320→12960→25920→51840.120→240→720→1440→4320→12960→25920→51840. The are 77 steps in total.In the second example, no moves are needed. Thus, the answer is 00.In the third example, it is impossible to transform 4848 to 7272. | [
"implementation",
"math"
] | import math
n,m=([int(x) for x in input().split()])
if n==m:
print(0)
else:
if (m/n)%2!=0 and (m/n)%3!=0:
print(-1)
else:
a=0
b=m/n
while b%2==0:
b=b//2
a=a+1
c = 0
d = m /n
while d % 3 == 0:
d = d // 3
c = c + 1
if math.gcd(int(b),int(d))==1:
print(a+c)
else:
print(-1) | py |
1284 | C | C. New Year and Permutationtime limit per test1 secondmemory limit per test1024 megabytesinputstandard inputoutputstandard outputRecall that the permutation is an array consisting of nn distinct integers from 11 to nn in arbitrary order. For example, [2,3,1,5,4][2,3,1,5,4] is a permutation, but [1,2,2][1,2,2] is not a permutation (22 appears twice in the array) and [1,3,4][1,3,4] is also not a permutation (n=3n=3 but there is 44 in the array).A sequence aa is a subsegment of a sequence bb if aa can be obtained from bb by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end. We will denote the subsegments as [l,r][l,r], where l,rl,r are two integers with 1≤l≤r≤n1≤l≤r≤n. This indicates the subsegment where l−1l−1 elements from the beginning and n−rn−r elements from the end are deleted from the sequence.For a permutation p1,p2,…,pnp1,p2,…,pn, we define a framed segment as a subsegment [l,r][l,r] where max{pl,pl+1,…,pr}−min{pl,pl+1,…,pr}=r−lmax{pl,pl+1,…,pr}−min{pl,pl+1,…,pr}=r−l. For example, for the permutation (6,7,1,8,5,3,2,4)(6,7,1,8,5,3,2,4) some of its framed segments are: [1,2],[5,8],[6,7],[3,3],[8,8][1,2],[5,8],[6,7],[3,3],[8,8]. In particular, a subsegment [i,i][i,i] is always a framed segments for any ii between 11 and nn, inclusive.We define the happiness of a permutation pp as the number of pairs (l,r)(l,r) such that 1≤l≤r≤n1≤l≤r≤n, and [l,r][l,r] is a framed segment. For example, the permutation [3,1,2][3,1,2] has happiness 55: all segments except [1,2][1,2] are framed segments.Given integers nn and mm, Jongwon wants to compute the sum of happiness for all permutations of length nn, modulo the prime number mm. Note that there exist n!n! (factorial of nn) different permutations of length nn.InputThe only line contains two integers nn and mm (1≤n≤2500001≤n≤250000, 108≤m≤109108≤m≤109, mm is prime).OutputPrint rr (0≤r<m0≤r<m), the sum of happiness for all permutations of length nn, modulo a prime number mm.ExamplesInputCopy1 993244853
OutputCopy1
InputCopy2 993244853
OutputCopy6
InputCopy3 993244853
OutputCopy32
InputCopy2019 993244853
OutputCopy923958830
InputCopy2020 437122297
OutputCopy265955509
NoteFor sample input n=3n=3; let's consider all permutations of length 33: [1,2,3][1,2,3], all subsegments are framed segment. Happiness is 66. [1,3,2][1,3,2], all subsegments except [1,2][1,2] are framed segment. Happiness is 55. [2,1,3][2,1,3], all subsegments except [2,3][2,3] are framed segment. Happiness is 55. [2,3,1][2,3,1], all subsegments except [2,3][2,3] are framed segment. Happiness is 55. [3,1,2][3,1,2], all subsegments except [1,2][1,2] are framed segment. Happiness is 55. [3,2,1][3,2,1], all subsegments are framed segment. Happiness is 66. Thus, the sum of happiness is 6+5+5+5+5+6=326+5+5+5+5+6=32. | [
"combinatorics",
"math"
] | # 10:47-
N,MOD = map(int, input().split())
fact = [1]*(N+1)
for i in range(1,N+1):
fact[i] = fact[i-1]*i%MOD
ans = 0
for i in range(1,N+1):
tmp = fact[i]*fact[N-i+1]*(N-i+1)
ans += tmp
ans %= MOD
print(ans)
| py |
1285 | A | A. Mezo Playing Zomatime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputToday; Mezo is playing a game. Zoma, a character in that game, is initially at position x=0x=0. Mezo starts sending nn commands to Zoma. There are two possible commands: 'L' (Left) sets the position x:=x−1x:=x−1; 'R' (Right) sets the position x:=x+1x:=x+1. Unfortunately, Mezo's controller malfunctions sometimes. Some commands are sent successfully and some are ignored. If the command is ignored then the position xx doesn't change and Mezo simply proceeds to the next command.For example, if Mezo sends commands "LRLR", then here are some possible outcomes (underlined commands are sent successfully): "LRLR" — Zoma moves to the left, to the right, to the left again and to the right for the final time, ending up at position 00; "LRLR" — Zoma recieves no commands, doesn't move at all and ends up at position 00 as well; "LRLR" — Zoma moves to the left, then to the left again and ends up in position −2−2. Mezo doesn't know which commands will be sent successfully beforehand. Thus, he wants to know how many different positions may Zoma end up at.InputThe first line contains nn (1≤n≤105)(1≤n≤105) — the number of commands Mezo sends.The second line contains a string ss of nn commands, each either 'L' (Left) or 'R' (Right).OutputPrint one integer — the number of different positions Zoma may end up at.ExampleInputCopy4
LRLR
OutputCopy5
NoteIn the example; Zoma may end up anywhere between −2−2 and 22. | [
"math"
] | n = int(input())
s = input()
print(n+1)
#234567812222222222222222222222
| py |
1296 | E2 | E2. String Coloring (hard version)time limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputThis is a hard version of the problem. The actual problems are different; but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.You are given a string ss consisting of nn lowercase Latin letters.You have to color all its characters the minimum number of colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in ss).After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.The goal is to make the string sorted, i.e. all characters should be in alphabetical order.Your task is to find the minimum number of colors which you have to color the given string in so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.InputThe first line of the input contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the length of ss.The second line of the input contains the string ss consisting of exactly nn lowercase Latin letters.OutputIn the first line print one integer resres (1≤res≤n1≤res≤n) — the minimum number of colors in which you have to color the given string so that after coloring it can become sorted by some sequence of swaps.In the second line print any possible coloring that can be used to sort the string using some sequence of swaps described in the problem statement. The coloring is the array cc of length nn, where 1≤ci≤res1≤ci≤res and cici means the color of the ii-th character.ExamplesInputCopy9
abacbecfd
OutputCopy2
1 1 2 1 2 1 2 1 2
InputCopy8
aaabbcbb
OutputCopy2
1 2 1 2 1 2 1 1
InputCopy7
abcdedc
OutputCopy3
1 1 1 1 1 2 3
InputCopy5
abcde
OutputCopy1
1 1 1 1 1
| [
"data structures",
"dp"
] | from math import *
from collections import *
import os
from io import BytesIO, IOBase
import sys
from bisect import *
from heapq import *
MOD = 1000000007
# Code by Big Dick Daddy Dick
def binpow(a, b, m):
a %= m
x = 1
while b > 0:
if b & 1:
x = x * a % m
a = a * a % m
b >>= 1
return x
def binser(arr, l, r, x):
while l < r:
mid = l + (r - l) // 2
# print(l, r, mid)
if arr[mid] == x:
return mid
elif arr[mid] < x:
l = mid + 1
else:
r = mid - 1
return mid
def lcm(a, b):
return (a * b) // gcd(a, b)
def sod(n):
l = list(str(n))
s = 0
for i in l:
s += int(i)
return s
def prime_factors(num):
l =[]
if num % 2:
l.append(2)
while num % 2 == 0:
num = num / 2
for i in range(3, int(sqrt(num)) + 1, 2):
if not num % i:
l.append(i)
while num % i == 0:
num = num / i
if num > 2:
l.append(num)
return l
def factmod(n, p):
f = defaultdict(int)
f[0] = 1
for i in range(1, n + 1):
f[i] = (f[i-1] * i) % MOD
"""
res = 1
while (n > 1):
if (n//p) % 2:
res = p - res
res = res * f[n%p] % p
n //= p
"""
return f
def largestPower(n, p):
# Initialize result
x = 0
# Calculate x = n/p + n/(p^2) + n/(p^3) + ....
while (n):
n //= p
x += n
return x
def modFact(n, p) :
if (n >= p) :
return 0
res = 1
isPrime = [1] * (n + 1)
i = 2
while(i * i <= n):
if (isPrime[i]):
for j in range(2 * i, n, i) :
isPrime[j] = 0
i += 1
# Consider all primes found by Sieve
for i in range(2, n):
if (isPrime[i]) :
k = largestPower(n, i)
# Multiply result with (i^k) % p
res = (res * binpow(i, k, p)) % p
return res
def drec(x, y):
if y == x + 1:
return 'R'
if y == x - 1:
return 'L'
if x < y:
return 'D'
return 'U'
def cellhash(x, y):
return (x - 1) * m + y
def bfs(src, dest):
q = deque([src])
vis[src] = True
while q:
i = q.popleft()
if i == dest:
return True
for j in ajl[i]:
if not vis[j]:
vis[j] = True
q.append(j)
return False
def bins(l, x, n):
i = bisect_left(l, x)
if i < n:
return i
if i:
return (i-1)
else:
return n
def cond(l):
for i in range(len(l) - 1):
if l[i] == str(int(l[i + 1]) - 1):
return False
return True
def isvowel(s):
if s in list("aeiou"):
return 1
return 0
def countOdd(L, R):
N = (R - L) // 2
# if either R or L is odd
if (R % 2 != 0 or L % 2 != 0):
N += 1
return N
def tst(A, B, C):
return ((A|B) & (B|C) & (C|A))
def palcheck(n, s):
i, j = 0, n - 1
while i <= j:
if s[i] == s[j]:
return False
i += 1
j -= 1
return True
def sakurajima(n):
if n < 9:
n = 10
l = [1] * (n + 1)
l[1] = 0
for i in range(2, int(n ** 0.5) + 1):
if l[i]:
for j in range(i * i, n + 1, i):
if not j % i:
l[j] = 0
return l
def prchck(n):
l = [1] * (n + 1)
l[1] = 0
for i in range(2, n + 1):
for j in range(2, int(sqrt(n)) + 1):
if j % i == 0:
l[j] = 1
return l
def ispal(s, n):
for i in range(n // 2):
if s[i] != s[n - i - 1]:
return False
return True
def dfs(i, tmp, vis):
vis[i] = True
if i == int(i):
for j in range(cnt[i]):
tmp.append(i)
for j in ajl[i]:
if not vis[j]:
tmp = dfs(j, tmp, vis)
return tmp
def panda(n, s):
x, y = '0', '0'
ans = []
ls, li = [], []
m = 0
cnt = 1
for i in range(n):
for j in range(m):
if s[i] >= ls[j]:
ans.append(li[j])
ls[j] = s[i]
break
else:
li.append(cnt)
ans.append(cnt)
ls.append(s[i])
cnt += 1
m += 1
# print(li, ls)
print(cnt - 1)
return ans
# Code by Big Dick Daddy Dick
# n = int(input())
# n, k = map(int, input().split())
# s = input()
# l = list(map(int, input().split()))
# memo = [[-1 for i in range(n + 1)] for j in range(2)]
input = sys.stdin.readline
t = 1
# t = int(input())
for _ in range(t):
n = int(input())
s = input()
print(*panda(n, s))
# print("Case #" + str(_ + 1) + ": " + str(ans)) | py |
1307 | C | C. Cow and Messagetime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputBessie the cow has just intercepted a text that Farmer John sent to Burger Queen! However; Bessie is sure that there is a secret message hidden inside.The text is a string ss of lowercase Latin letters. She considers a string tt as hidden in string ss if tt exists as a subsequence of ss whose indices form an arithmetic progression. For example, the string aab is hidden in string aaabb because it occurs at indices 11, 33, and 55, which form an arithmetic progression with a common difference of 22. Bessie thinks that any hidden string that occurs the most times is the secret message. Two occurrences of a subsequence of SS are distinct if the sets of indices are different. Help her find the number of occurrences of the secret message!For example, in the string aaabb, a is hidden 33 times, b is hidden 22 times, ab is hidden 66 times, aa is hidden 33 times, bb is hidden 11 time, aab is hidden 22 times, aaa is hidden 11 time, abb is hidden 11 time, aaab is hidden 11 time, aabb is hidden 11 time, and aaabb is hidden 11 time. The number of occurrences of the secret message is 66.InputThe first line contains a string ss of lowercase Latin letters (1≤|s|≤1051≤|s|≤105) — the text that Bessie intercepted.OutputOutput a single integer — the number of occurrences of the secret message.ExamplesInputCopyaaabb
OutputCopy6
InputCopyusaco
OutputCopy1
InputCopylol
OutputCopy2
NoteIn the first example; these are all the hidden strings and their indice sets: a occurs at (1)(1), (2)(2), (3)(3) b occurs at (4)(4), (5)(5) ab occurs at (1,4)(1,4), (1,5)(1,5), (2,4)(2,4), (2,5)(2,5), (3,4)(3,4), (3,5)(3,5) aa occurs at (1,2)(1,2), (1,3)(1,3), (2,3)(2,3) bb occurs at (4,5)(4,5) aab occurs at (1,3,5)(1,3,5), (2,3,4)(2,3,4) aaa occurs at (1,2,3)(1,2,3) abb occurs at (3,4,5)(3,4,5) aaab occurs at (1,2,3,4)(1,2,3,4) aabb occurs at (2,3,4,5)(2,3,4,5) aaabb occurs at (1,2,3,4,5)(1,2,3,4,5) Note that all the sets of indices are arithmetic progressions.In the second example, no hidden string occurs more than once.In the third example, the hidden string is the letter l. | [
"brute force",
"dp",
"math",
"strings"
] | import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
s = list(input().rstrip())
cnt = [0] * 26
x = [0] * (26 * 26)
for i in s:
for j in range(26):
x[(i - 97) + j * 26] += cnt[j]
cnt[i - 97] += 1
ans = max(max(x), max(cnt))
print(ans) | py |
1303 | B | B. National Projecttime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYour company was appointed to lay new asphalt on the highway of length nn. You know that every day you can either repair one unit of the highway (lay new asphalt over one unit of the highway) or skip repairing.Skipping the repair is necessary because of the climate. The climate in your region is periodical: there are gg days when the weather is good and if you lay new asphalt these days it becomes high-quality pavement; after that, the weather during the next bb days is bad, and if you lay new asphalt these days it becomes low-quality pavement; again gg good days, bb bad days and so on.You can be sure that you start repairing at the start of a good season, in other words, days 1,2,…,g1,2,…,g are good.You don't really care about the quality of the highway, you just want to make sure that at least half of the highway will have high-quality pavement. For example, if the n=5n=5 then at least 33 units of the highway should have high quality; if n=4n=4 then at least 22 units should have high quality.What is the minimum number of days is needed to finish the repair of the whole highway?InputThe first line contains a single integer TT (1≤T≤1041≤T≤104) — the number of test cases.Next TT lines contain test cases — one per line. Each line contains three integers nn, gg and bb (1≤n,g,b≤1091≤n,g,b≤109) — the length of the highway and the number of good and bad days respectively.OutputPrint TT integers — one per test case. For each test case, print the minimum number of days required to repair the whole highway if at least half of it should have high quality.ExampleInputCopy3
5 1 1
8 10 10
1000000 1 1000000
OutputCopy5
8
499999500000
NoteIn the first test case; you can just lay new asphalt each day; since days 1,3,51,3,5 are good.In the second test case, you can also lay new asphalt each day, since days 11-88 are good. | [
"math"
] | t = int(input())
for _ in range(t):
n, g, b = map(int, input().split())
if g >= n:
print(n)
continue
# nice way to get ceiling
half = (n+1) // 2
# multipler * chunk of g+b days + remainder good days
remainder = half % g
ans = (half//g) * (g+b) + remainder
if remainder == 0: # remove bad days excess
ans -= b
print(max(ans, n)) | py |
1285 | D | D. Dr. Evil Underscorestime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputToday; as a friendship gift, Bakry gave Badawy nn integers a1,a2,…,ana1,a2,…,an and challenged him to choose an integer XX such that the value max1≤i≤n(ai⊕X)max1≤i≤n(ai⊕X) is minimum possible, where ⊕⊕ denotes the bitwise XOR operation.As always, Badawy is too lazy, so you decided to help him and find the minimum possible value of max1≤i≤n(ai⊕X)max1≤i≤n(ai⊕X).InputThe first line contains integer nn (1≤n≤1051≤n≤105).The second line contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤230−10≤ai≤230−1).OutputPrint one integer — the minimum possible value of max1≤i≤n(ai⊕X)max1≤i≤n(ai⊕X).ExamplesInputCopy3
1 2 3
OutputCopy2
InputCopy2
1 5
OutputCopy4
NoteIn the first sample; we can choose X=3X=3.In the second sample, we can choose X=5X=5. | [
"bitmasks",
"brute force",
"dfs and similar",
"divide and conquer",
"dp",
"greedy",
"strings",
"trees"
] | # import time
# start = time.time()
def escolheBits(lista, bit):
if bit < 0 or len(lista) == 0:
return 0
bit1, bit0 = [], []
for i in lista:
if (i >> bit) & 1:
bit1.append(i)
else:
bit0.append(i)
if len(bit1) == 0:
return escolheBits(bit0, bit - 1)
if len(bit0) == 0:
return escolheBits(bit1, bit - 1)
return min(escolheBits(bit1, bit - 1), escolheBits(bit0, bit - 1)) + (1 << bit)
if __name__ == '__main__':
_ = input()
lista = list(map(int, input().split()))
print(escolheBits(lista, 30))
# print(f'end1 {time.time() - start}') | py |
1141 | D | D. Colored Bootstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThere are nn left boots and nn right boots. Each boot has a color which is denoted as a lowercase Latin letter or a question mark ('?'). Thus, you are given two strings ll and rr, both of length nn. The character lili stands for the color of the ii-th left boot and the character riri stands for the color of the ii-th right boot.A lowercase Latin letter denotes a specific color, but the question mark ('?') denotes an indefinite color. Two specific colors are compatible if they are exactly the same. An indefinite color is compatible with any (specific or indefinite) color.For example, the following pairs of colors are compatible: ('f', 'f'), ('?', 'z'), ('a', '?') and ('?', '?'). The following pairs of colors are not compatible: ('f', 'g') and ('a', 'z').Compute the maximum number of pairs of boots such that there is one left and one right boot in a pair and their colors are compatible.Print the maximum number of such pairs and the pairs themselves. A boot can be part of at most one pair.InputThe first line contains nn (1≤n≤1500001≤n≤150000), denoting the number of boots for each leg (i.e. the number of left boots and the number of right boots).The second line contains the string ll of length nn. It contains only lowercase Latin letters or question marks. The ii-th character stands for the color of the ii-th left boot.The third line contains the string rr of length nn. It contains only lowercase Latin letters or question marks. The ii-th character stands for the color of the ii-th right boot.OutputPrint kk — the maximum number of compatible left-right pairs of boots, i.e. pairs consisting of one left and one right boot which have compatible colors.The following kk lines should contain pairs aj,bjaj,bj (1≤aj,bj≤n1≤aj,bj≤n). The jj-th of these lines should contain the index ajaj of the left boot in the jj-th pair and index bjbj of the right boot in the jj-th pair. All the numbers ajaj should be distinct (unique), all the numbers bjbj should be distinct (unique).If there are many optimal answers, print any of them.ExamplesInputCopy10
codeforces
dodivthree
OutputCopy5
7 8
4 9
2 2
9 10
3 1
InputCopy7
abaca?b
zabbbcc
OutputCopy5
6 5
2 3
4 6
7 4
1 2
InputCopy9
bambarbia
hellocode
OutputCopy0
InputCopy10
code??????
??????test
OutputCopy10
6 2
1 6
7 3
3 5
4 8
9 7
5 1
2 4
10 9
8 10
| [
"greedy",
"implementation"
] | x=int(input())
a=list(input())
b=list(input())
a1=[[] for i in range(26)]
q=[]
b1=[[] for i in range(26)]
q1=[]
ans=[]
for i in range(len(a)):
if(a[i]=='?'):
q.append(i)
else:
a1[ord(a[i])-ord('a')].append(i)
for i in range(len(b)):
if(b[i]=='?'):
q1.append(i)
else:
b1[ord(b[i])-ord('a')].append(i)
for i in range(26):
if(a1[i]!=[]) and (b1[i]!=[]):
p=min([len(a1[i]),len(b1[i])])
for j in range(p):
ans.append([a1[i][len(a1[i])-1],b1[i][len(b1[i])-1]])
a1[i].pop(len(a1[i])-1)
b1[i].pop(len(b1[i])-1)
for i in range(26):
if(b1[i]!=[]):
p=min([len(q),len(b1[i])])
for j in range(p):
ans.append([q[len(q)-1],b1[i][len(b1[i])-1]])
q.pop(len(q)-1)
b1[i].pop(len(b1[i])-1)
if(a1[i]!=[]):
p=min([len(q1),len(a1[i])])
for j in range(p):
ans.append([a1[i][len(a1[i])-1],q1[len(q1)-1]])
q1.pop(len(q1)-1)
a1[i].pop(len(a1[i])-1)
p=min(len(q),len(q1))
for i in range(p):
ans.append([q[len(q)-1],q1[len(q1)-1]])
q.pop(len(q)-1)
q1.pop(len(q1)-1)
print(len(ans))
for i in range(len(ans)):
print(ans[i][0]+1,ans[i][1]+1) | py |
1296 | B | B. Food Buyingtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputMishka wants to buy some food in the nearby shop. Initially; he has ss burles on his card. Mishka can perform the following operation any number of times (possibly, zero): choose some positive integer number 1≤x≤s1≤x≤s, buy food that costs exactly xx burles and obtain ⌊x10⌋⌊x10⌋ burles as a cashback (in other words, Mishka spends xx burles and obtains ⌊x10⌋⌊x10⌋ back). The operation ⌊ab⌋⌊ab⌋ means aa divided by bb rounded down.It is guaranteed that you can always buy some food that costs xx for any possible value of xx.Your task is to say the maximum number of burles Mishka can spend if he buys food optimally.For example, if Mishka has s=19s=19 burles then the maximum number of burles he can spend is 2121. Firstly, he can spend x=10x=10 burles, obtain 11 burle as a cashback. Now he has s=10s=10 burles, so can spend x=10x=10 burles, obtain 11 burle as a cashback and spend it too.You have to answer tt independent test cases.InputThe first line of the input contains one integer tt (1≤t≤1041≤t≤104) — the number of test cases.The next tt lines describe test cases. Each test case is given on a separate line and consists of one integer ss (1≤s≤1091≤s≤109) — the number of burles Mishka initially has.OutputFor each test case print the answer on it — the maximum number of burles Mishka can spend if he buys food optimally.ExampleInputCopy6
1
10
19
9876
12345
1000000000
OutputCopy1
11
21
10973
13716
1111111111
| [
"math"
] | # https://codeforces.com/problemset/problem/1296/B
def check(n):
res = n
while n >= 10:
k = n % 10
n = n // 10
res += n
n = n + k
return res
for _ in range(int(input())):
print(check(int(input()))) | py |
1320 | A | A. Journey Planningtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputTanya wants to go on a journey across the cities of Berland. There are nn cities situated along the main railroad line of Berland, and these cities are numbered from 11 to nn. Tanya plans her journey as follows. First of all, she will choose some city c1c1 to start her journey. She will visit it, and after that go to some other city c2>c1c2>c1, then to some other city c3>c2c3>c2, and so on, until she chooses to end her journey in some city ck>ck−1ck>ck−1. So, the sequence of visited cities [c1,c2,…,ck][c1,c2,…,ck] should be strictly increasing.There are some additional constraints on the sequence of cities Tanya visits. Each city ii has a beauty value bibi associated with it. If there is only one city in Tanya's journey, these beauty values imply no additional constraints. But if there are multiple cities in the sequence, then for any pair of adjacent cities cici and ci+1ci+1, the condition ci+1−ci=bci+1−bcici+1−ci=bci+1−bci must hold.For example, if n=8n=8 and b=[3,4,4,6,6,7,8,9]b=[3,4,4,6,6,7,8,9], there are several three possible ways to plan a journey: c=[1,2,4]c=[1,2,4]; c=[3,5,6,8]c=[3,5,6,8]; c=[7]c=[7] (a journey consisting of one city is also valid). There are some additional ways to plan a journey that are not listed above.Tanya wants her journey to be as beautiful as possible. The beauty value of the whole journey is the sum of beauty values over all visited cities. Can you help her to choose the optimal plan, that is, to maximize the beauty value of the journey?InputThe first line contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of cities in Berland.The second line contains nn integers b1b1, b2b2, ..., bnbn (1≤bi≤4⋅1051≤bi≤4⋅105), where bibi is the beauty value of the ii-th city.OutputPrint one integer — the maximum beauty of a journey Tanya can choose.ExamplesInputCopy6
10 7 1 9 10 15
OutputCopy26
InputCopy1
400000
OutputCopy400000
InputCopy7
8 9 26 11 12 29 14
OutputCopy55
NoteThe optimal journey plan in the first example is c=[2,4,5]c=[2,4,5].The optimal journey plan in the second example is c=[1]c=[1].The optimal journey plan in the third example is c=[3,6]c=[3,6]. | [
"data structures",
"dp",
"greedy",
"math",
"sortings"
] | #! /bin/env python3
n = int(input())
b = list(map(int, input().split(' ')))
ans = -1000
ATshayu = {}
for i in range(n):
v = b[i] - i
try:
ATshayu[v] += b[i]
except:
ATshayu[v] = b[i]
finally:
ans = max(ans, ATshayu[v])
print(ans) | py |
1294 | D | D. MEX maximizingtime limit per test3 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputRecall that MEX of an array is a minimum non-negative integer that does not belong to the array. Examples: for the array [0,0,1,0,2][0,0,1,0,2] MEX equals to 33 because numbers 0,10,1 and 22 are presented in the array and 33 is the minimum non-negative integer not presented in the array; for the array [1,2,3,4][1,2,3,4] MEX equals to 00 because 00 is the minimum non-negative integer not presented in the array; for the array [0,1,4,3][0,1,4,3] MEX equals to 22 because 22 is the minimum non-negative integer not presented in the array. You are given an empty array a=[]a=[] (in other words, a zero-length array). You are also given a positive integer xx.You are also given qq queries. The jj-th query consists of one integer yjyj and means that you have to append one element yjyj to the array. The array length increases by 11 after a query.In one move, you can choose any index ii and set ai:=ai+xai:=ai+x or ai:=ai−xai:=ai−x (i.e. increase or decrease any element of the array by xx). The only restriction is that aiai cannot become negative. Since initially the array is empty, you can perform moves only after the first query.You have to maximize the MEX (minimum excluded) of the array if you can perform any number of such operations (you can even perform the operation multiple times with one element).You have to find the answer after each of qq queries (i.e. the jj-th answer corresponds to the array of length jj).Operations are discarded before each query. I.e. the array aa after the jj-th query equals to [y1,y2,…,yj][y1,y2,…,yj].InputThe first line of the input contains two integers q,xq,x (1≤q,x≤4⋅1051≤q,x≤4⋅105) — the number of queries and the value of xx.The next qq lines describe queries. The jj-th query consists of one integer yjyj (0≤yj≤1090≤yj≤109) and means that you have to append one element yjyj to the array.OutputPrint the answer to the initial problem after each query — for the query jj print the maximum value of MEX after first jj queries. Note that queries are dependent (the array changes after each query) but operations are independent between queries.ExamplesInputCopy7 3
0
1
2
2
0
0
10
OutputCopy1
2
3
3
4
4
7
InputCopy4 3
1
2
1
2
OutputCopy0
0
0
0
NoteIn the first example: After the first query; the array is a=[0]a=[0]: you don't need to perform any operations, maximum possible MEX is 11. After the second query, the array is a=[0,1]a=[0,1]: you don't need to perform any operations, maximum possible MEX is 22. After the third query, the array is a=[0,1,2]a=[0,1,2]: you don't need to perform any operations, maximum possible MEX is 33. After the fourth query, the array is a=[0,1,2,2]a=[0,1,2,2]: you don't need to perform any operations, maximum possible MEX is 33 (you can't make it greater with operations). After the fifth query, the array is a=[0,1,2,2,0]a=[0,1,2,2,0]: you can perform a[4]:=a[4]+3=3a[4]:=a[4]+3=3. The array changes to be a=[0,1,2,2,3]a=[0,1,2,2,3]. Now MEX is maximum possible and equals to 44. After the sixth query, the array is a=[0,1,2,2,0,0]a=[0,1,2,2,0,0]: you can perform a[4]:=a[4]+3=0+3=3a[4]:=a[4]+3=0+3=3. The array changes to be a=[0,1,2,2,3,0]a=[0,1,2,2,3,0]. Now MEX is maximum possible and equals to 44. After the seventh query, the array is a=[0,1,2,2,0,0,10]a=[0,1,2,2,0,0,10]. You can perform the following operations: a[3]:=a[3]+3=2+3=5a[3]:=a[3]+3=2+3=5, a[4]:=a[4]+3=0+3=3a[4]:=a[4]+3=0+3=3, a[5]:=a[5]+3=0+3=3a[5]:=a[5]+3=0+3=3, a[5]:=a[5]+3=3+3=6a[5]:=a[5]+3=3+3=6, a[6]:=a[6]−3=10−3=7a[6]:=a[6]−3=10−3=7, a[6]:=a[6]−3=7−3=4a[6]:=a[6]−3=7−3=4. The resulting array will be a=[0,1,2,5,3,6,4]a=[0,1,2,5,3,6,4]. Now MEX is maximum possible and equals to 77. | [
"data structures",
"greedy",
"implementation",
"math"
] | [n,x]=list(map(int,input().split(" ")))
arr=[0 for i in range(x)]
# mex=0
index=0
ans=[]
if 0:
none=0
else:
for i in range(n):
k=int(input())
arr[k%x]+=1
while(arr[index%x]!=0):
arr[index%x]-=1
index+=1
ans.append(index)
for i in ans:
print(i) | py |
1294 | B | B. Collecting Packagestime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputThere is a robot in a warehouse and nn packages he wants to collect. The warehouse can be represented as a coordinate grid. Initially, the robot stays at the point (0,0)(0,0). The ii-th package is at the point (xi,yi)(xi,yi). It is guaranteed that there are no two packages at the same point. It is also guaranteed that the point (0,0)(0,0) doesn't contain a package.The robot is semi-broken and only can move up ('U') and right ('R'). In other words, in one move the robot can go from the point (x,y)(x,y) to the point (x+1,yx+1,y) or to the point (x,y+1)(x,y+1).As we say above, the robot wants to collect all nn packages (in arbitrary order). He wants to do it with the minimum possible number of moves. If there are several possible traversals, the robot wants to choose the lexicographically smallest path.The string ss of length nn is lexicographically less than the string tt of length nn if there is some index 1≤j≤n1≤j≤n that for all ii from 11 to j−1j−1 si=tisi=ti and sj<tjsj<tj. It is the standard comparison of string, like in a dictionary. Most programming languages compare strings in this way.InputThe first line of the input contains an integer tt (1≤t≤1001≤t≤100) — the number of test cases. Then test cases follow.The first line of a test case contains one integer nn (1≤n≤10001≤n≤1000) — the number of packages.The next nn lines contain descriptions of packages. The ii-th package is given as two integers xixi and yiyi (0≤xi,yi≤10000≤xi,yi≤1000) — the xx-coordinate of the package and the yy-coordinate of the package.It is guaranteed that there are no two packages at the same point. It is also guaranteed that the point (0,0)(0,0) doesn't contain a package.The sum of all values nn over test cases in the test doesn't exceed 10001000.OutputPrint the answer for each test case.If it is impossible to collect all nn packages in some order starting from (0,00,0), print "NO" on the first line.Otherwise, print "YES" in the first line. Then print the shortest path — a string consisting of characters 'R' and 'U'. Among all such paths choose the lexicographically smallest path.Note that in this problem "YES" and "NO" can be only uppercase words, i.e. "Yes", "no" and "YeS" are not acceptable.ExampleInputCopy3
5
1 3
1 2
3 3
5 5
4 3
2
1 0
0 1
1
4 3
OutputCopyYES
RUUURRRRUU
NO
YES
RRRRUUU
NoteFor the first test case in the example the optimal path RUUURRRRUU is shown below: | [
"implementation",
"sortings"
] | def get(f): return f(input().strip())
def gets(f): return [*map(f, input().split())]
for _ in range(get(int)):
n = get(int)
xy = sorted(gets(int) for _ in range(n))
for i in range(1, n):
if xy[i][1] < xy[i - 1][1]:
print('NO')
break
else:
print('YES')
p = 0
q = 0
for x, y in xy:
print(end='R' * (x - p) + 'U' * (y - q))
p, q = x, y
print()
| py |
1311 | C | C. Perform the Combotime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou want to perform the combo on your opponent in one popular fighting game. The combo is the string ss consisting of nn lowercase Latin letters. To perform the combo, you have to press all buttons in the order they appear in ss. I.e. if s=s="abca" then you have to press 'a', then 'b', 'c' and 'a' again.You know that you will spend mm wrong tries to perform the combo and during the ii-th try you will make a mistake right after pipi-th button (1≤pi<n1≤pi<n) (i.e. you will press first pipi buttons right and start performing the combo from the beginning). It is guaranteed that during the m+1m+1-th try you press all buttons right and finally perform the combo.I.e. if s=s="abca", m=2m=2 and p=[1,3]p=[1,3] then the sequence of pressed buttons will be 'a' (here you're making a mistake and start performing the combo from the beginning), 'a', 'b', 'c', (here you're making a mistake and start performing the combo from the beginning), 'a' (note that at this point you will not perform the combo because of the mistake), 'b', 'c', 'a'.Your task is to calculate for each button (letter) the number of times you'll press it.You have to answer tt independent test cases.InputThe first line of the input contains one integer tt (1≤t≤1041≤t≤104) — the number of test cases.Then tt test cases follow.The first line of each test case contains two integers nn and mm (2≤n≤2⋅1052≤n≤2⋅105, 1≤m≤2⋅1051≤m≤2⋅105) — the length of ss and the number of tries correspondingly.The second line of each test case contains the string ss consisting of nn lowercase Latin letters.The third line of each test case contains mm integers p1,p2,…,pmp1,p2,…,pm (1≤pi<n1≤pi<n) — the number of characters pressed right during the ii-th try.It is guaranteed that the sum of nn and the sum of mm both does not exceed 2⋅1052⋅105 (∑n≤2⋅105∑n≤2⋅105, ∑m≤2⋅105∑m≤2⋅105).It is guaranteed that the answer for each letter does not exceed 2⋅1092⋅109.OutputFor each test case, print the answer — 2626 integers: the number of times you press the button 'a', the number of times you press the button 'b', ……, the number of times you press the button 'z'.ExampleInputCopy3
4 2
abca
1 3
10 5
codeforces
2 8 3 2 9
26 10
qwertyuioplkjhgfdsazxcvbnm
20 10 1 2 3 5 10 5 9 4
OutputCopy4 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 9 4 5 3 0 0 0 0 0 0 0 0 9 0 0 3 1 0 0 0 0 0 0 0
2 1 1 2 9 2 2 2 5 2 2 2 1 1 5 4 11 8 2 7 5 1 10 1 5 2
NoteThe first test case is described in the problem statement. Wrong tries are "a"; "abc" and the final try is "abca". The number of times you press 'a' is 44, 'b' is 22 and 'c' is 22.In the second test case, there are five wrong tries: "co", "codeforc", "cod", "co", "codeforce" and the final try is "codeforces". The number of times you press 'c' is 99, 'd' is 44, 'e' is 55, 'f' is 33, 'o' is 99, 'r' is 33 and 's' is 11. | [
"brute force"
] | from sys import stdin
input=lambda :stdin.readline()[:-1]
def solve():
n,m=map(int,input().split())
s=input()
p=list(map(lambda x:int(x)-1,input().split()))
cnt=[0]*n
for i in p:
cnt[i]+=1
tmp=1
ans=[0]*26
for i in range(n-1,-1,-1):
tmp+=cnt[i]
ans[ord(s[i])-97]+=tmp
print(*ans)
for _ in range(int(input())):
solve() | py |
1324 | B | B. Yet Another Palindrome Problemtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given an array aa consisting of nn integers.Your task is to determine if aa has some subsequence of length at least 33 that is a palindrome.Recall that an array bb is called a subsequence of the array aa if bb can be obtained by removing some (possibly, zero) elements from aa (not necessarily consecutive) without changing the order of remaining elements. For example, [2][2], [1,2,1,3][1,2,1,3] and [2,3][2,3] are subsequences of [1,2,1,3][1,2,1,3], but [1,1,2][1,1,2] and [4][4] are not.Also, recall that a palindrome is an array that reads the same backward as forward. In other words, the array aa of length nn is the palindrome if ai=an−i−1ai=an−i−1 for all ii from 11 to nn. For example, arrays [1234][1234], [1,2,1][1,2,1], [1,3,2,2,3,1][1,3,2,2,3,1] and [10,100,10][10,100,10] are palindromes, but arrays [1,2][1,2] and [1,2,3,1][1,2,3,1] are not.You have to answer tt independent test cases.InputThe first line of the input contains one integer tt (1≤t≤1001≤t≤100) — the number of test cases.Next 2t2t lines describe test cases. The first line of the test case contains one integer nn (3≤n≤50003≤n≤5000) — the length of aa. The second line of the test case contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤n1≤ai≤n), where aiai is the ii-th element of aa.It is guaranteed that the sum of nn over all test cases does not exceed 50005000 (∑n≤5000∑n≤5000).OutputFor each test case, print the answer — "YES" (without quotes) if aa has some subsequence of length at least 33 that is a palindrome and "NO" otherwise.ExampleInputCopy5
3
1 2 1
5
1 2 2 3 2
3
1 1 2
4
1 2 2 1
10
1 1 2 2 3 3 4 4 5 5
OutputCopyYES
YES
NO
YES
NO
NoteIn the first test case of the example; the array aa has a subsequence [1,2,1][1,2,1] which is a palindrome.In the second test case of the example, the array aa has two subsequences of length 33 which are palindromes: [2,3,2][2,3,2] and [2,2,2][2,2,2].In the third test case of the example, the array aa has no subsequences of length at least 33 which are palindromes.In the fourth test case of the example, the array aa has one subsequence of length 44 which is a palindrome: [1,2,2,1][1,2,2,1] (and has two subsequences of length 33 which are palindromes: both are [1,2,1][1,2,1]).In the fifth test case of the example, the array aa has no subsequences of length at least 33 which are palindromes. | [
"brute force",
"strings"
] | t = int(input())
while t>0:
n = int(input())
arr = list(map(int,input().split()))
f = 0
arr1 = []
for i in range(n):
arr1.append((arr[i],i))
arr1 = sorted(arr1)
for i in range(n-1):
if (arr1[i][0] == arr1[i+1][0]) and (arr1[i+1][1]-arr1[i][1])>1:
f = 1
break
if i<n-2:
if (arr1[i][0] == arr1[i+1][0]) and (arr1[i+1][0]==arr1[i+2][0]):
f = 1
break
if f>0:
print('YES')
else:
print('NO')
t = t-1 | py |
1141 | E | E. Superhero Battletime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputA superhero fights with a monster. The battle consists of rounds; each of which lasts exactly nn minutes. After a round ends, the next round starts immediately. This is repeated over and over again.Each round has the same scenario. It is described by a sequence of nn numbers: d1,d2,…,dnd1,d2,…,dn (−106≤di≤106−106≤di≤106). The ii-th element means that monster's hp (hit points) changes by the value didi during the ii-th minute of each round. Formally, if before the ii-th minute of a round the monster's hp is hh, then after the ii-th minute it changes to h:=h+dih:=h+di.The monster's initial hp is HH. It means that before the battle the monster has HH hit points. Print the first minute after which the monster dies. The monster dies if its hp is less than or equal to 00. Print -1 if the battle continues infinitely.InputThe first line contains two integers HH and nn (1≤H≤10121≤H≤1012, 1≤n≤2⋅1051≤n≤2⋅105). The second line contains the sequence of integers d1,d2,…,dnd1,d2,…,dn (−106≤di≤106−106≤di≤106), where didi is the value to change monster's hp in the ii-th minute of a round.OutputPrint -1 if the superhero can't kill the monster and the battle will last infinitely. Otherwise, print the positive integer kk such that kk is the first minute after which the monster is dead.ExamplesInputCopy1000 6
-100 -200 -300 125 77 -4
OutputCopy9
InputCopy1000000000000 5
-1 0 0 0 0
OutputCopy4999999999996
InputCopy10 4
-3 -6 5 4
OutputCopy-1
| [
"math"
] | k, n = map(int, input().split())
arr = [int(x) for x in input().split()]
sum = 0
for i in range(0, n):
sum = sum + arr[i]
if sum + k <= 0:
print(i + 1)
exit()
if sum >= 0 :
print(-1)
exit()
l = 1
h = 10000000000000
ans = 1000000000000000000
while l <= h :
mid = (l + h) >> 1
f = 0
pre = 0
for i in range(0, n):
if sum * (mid - 1) + arr[i] + k + pre <= 0:
f = 1
ans = min(ans, (mid - 1) * n + (i + 1))
break
pre = pre + arr[i]
if f == 1:
h = mid - 1
else:
l = mid + 1
print(ans) | py |
1324 | A | A. Yet Another Tetris Problemtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given some Tetris field consisting of nn columns. The initial height of the ii-th column of the field is aiai blocks. On top of these columns you can place only figures of size 2×12×1 (i.e. the height of this figure is 22 blocks and the width of this figure is 11 block). Note that you cannot rotate these figures.Your task is to say if you can clear the whole field by placing such figures.More formally, the problem can be described like this:The following process occurs while at least one aiai is greater than 00: You place one figure 2×12×1 (choose some ii from 11 to nn and replace aiai with ai+2ai+2); then, while all aiai are greater than zero, replace each aiai with ai−1ai−1. And your task is to determine if it is possible to clear the whole field (i.e. finish the described process), choosing the places for new figures properly.You have to answer tt independent test cases.InputThe first line of the input contains one integer tt (1≤t≤1001≤t≤100) — the number of test cases.The next 2t2t lines describe test cases. The first line of the test case contains one integer nn (1≤n≤1001≤n≤100) — the number of columns in the Tetris field. The second line of the test case contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1001≤ai≤100), where aiai is the initial height of the ii-th column of the Tetris field.OutputFor each test case, print the answer — "YES" (without quotes) if you can clear the whole Tetris field and "NO" otherwise.ExampleInputCopy4
3
1 1 3
4
1 1 2 1
2
11 11
1
100
OutputCopyYES
NO
YES
YES
NoteThe first test case of the example field is shown below:Gray lines are bounds of the Tetris field. Note that the field has no upper bound.One of the correct answers is to first place the figure in the first column. Then after the second step of the process; the field becomes [2,0,2][2,0,2]. Then place the figure in the second column and after the second step of the process, the field becomes [0,0,0][0,0,0].And the second test case of the example field is shown below:It can be shown that you cannot do anything to end the process.In the third test case of the example, you first place the figure in the second column after the second step of the process, the field becomes [0,2][0,2]. Then place the figure in the first column and after the second step of the process, the field becomes [0,0][0,0].In the fourth test case of the example, place the figure in the first column, then the field becomes [102][102] after the first step of the process, and then the field becomes [0][0] after the second step of the process. | [
"implementation",
"number theory"
] | for i in range(int(input())):
h = int(input())
a = sum(list(map(lambda x: int(x)%2, input().split())))
if(a==0 or a==h):
print("YES")
else:
print("NO") | py |
1287 | B | B. Hypersettime limit per test3 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputBees Alice and Alesya gave beekeeper Polina famous card game "Set" as a Christmas present. The deck consists of cards that vary in four features across three options for each kind of feature: number of shapes; shape, shading, and color. In this game, some combinations of three cards are said to make up a set. For every feature — color, number, shape, and shading — the three cards must display that feature as either all the same, or pairwise different. The picture below shows how sets look.Polina came up with a new game called "Hyperset". In her game, there are nn cards with kk features, each feature has three possible values: "S", "E", or "T". The original "Set" game can be viewed as "Hyperset" with k=4k=4.Similarly to the original game, three cards form a set, if all features are the same for all cards or are pairwise different. The goal of the game is to compute the number of ways to choose three cards that form a set.Unfortunately, winter holidays have come to an end, and it's time for Polina to go to school. Help Polina find the number of sets among the cards lying on the table.InputThe first line of each test contains two integers nn and kk (1≤n≤15001≤n≤1500, 1≤k≤301≤k≤30) — number of cards and number of features.Each of the following nn lines contains a card description: a string consisting of kk letters "S", "E", "T". The ii-th character of this string decribes the ii-th feature of that card. All cards are distinct.OutputOutput a single integer — the number of ways to choose three cards that form a set.ExamplesInputCopy3 3
SET
ETS
TSE
OutputCopy1InputCopy3 4
SETE
ETSE
TSES
OutputCopy0InputCopy5 4
SETT
TEST
EEET
ESTE
STES
OutputCopy2NoteIn the third example test; these two triples of cards are sets: "SETT"; "TEST"; "EEET" "TEST"; "ESTE", "STES" | [
"brute force",
"data structures",
"implementation"
] | n, k=[int(v) for v in input().split()]
w=[]
q=set()
for j in range(n):
w.append(input())
q=set(w)
c=set(["S", "E", "T"])
res=0
for j in range(n):
for z in range(j+1, n):
eta=[]
for l in range(k):
if w[j][l]==w[z][l]:
eta.append(w[j][l])
else:
y=[w[j][l], w[z][l]]
if "S" not in y:
eta.append("S")
elif "E" not in y:
eta.append("E")
else:
eta.append("T")
eta="".join([str(v) for v in eta])
#print(eta)
if eta in q:
res+=1
print(res//3) | py |
1313 | A | A. Fast Food Restauranttime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputTired of boring office work; Denis decided to open a fast food restaurant.On the first day he made aa portions of dumplings, bb portions of cranberry juice and cc pancakes with condensed milk.The peculiarity of Denis's restaurant is the procedure of ordering food. For each visitor Denis himself chooses a set of dishes that this visitor will receive. When doing so, Denis is guided by the following rules: every visitor should receive at least one dish (dumplings, cranberry juice, pancakes with condensed milk are all considered to be dishes); each visitor should receive no more than one portion of dumplings, no more than one portion of cranberry juice and no more than one pancake with condensed milk; all visitors should receive different sets of dishes. What is the maximum number of visitors Denis can feed?InputThe first line contains an integer tt (1≤t≤5001≤t≤500) — the number of test cases to solve.Each of the remaining tt lines contains integers aa, bb and cc (0≤a,b,c≤100≤a,b,c≤10) — the number of portions of dumplings, the number of portions of cranberry juice and the number of condensed milk pancakes Denis made.OutputFor each test case print a single integer — the maximum number of visitors Denis can feed.ExampleInputCopy71 2 10 0 09 1 72 2 32 3 23 2 24 4 4OutputCopy3045557NoteIn the first test case of the example, Denis can feed the first visitor with dumplings, give the second a portion of cranberry juice, and give the third visitor a portion of cranberry juice and a pancake with a condensed milk.In the second test case of the example, the restaurant Denis is not very promising: he can serve no customers.In the third test case of the example, Denise can serve four visitors. The first guest will receive a full lunch of dumplings, a portion of cranberry juice and a pancake with condensed milk. The second visitor will get only dumplings. The third guest will receive a pancake with condensed milk, and the fourth guest will receive a pancake and a portion of dumplings. Please note that Denis hasn't used all of the prepared products, but is unable to serve more visitors. | [
"brute force",
"greedy",
"implementation"
] | r = ""
for t in range(int(input())):
a, b, c = sorted(list(map(int, input().split())), reverse=True)
suma = 0
if a > 0:
a -= 1
suma += 1
if b > 0:
b -= 1
suma += 1
if c > 0:
c -= 1
suma += 1
if a > 0 and b > 0:
a -= 1
b -= 1
suma += 1
if a > 0 and c > 0:
a -= 1
c -= 1
suma += 1
if b > 0 and c > 0:
b -= 1
c -= 1
suma += 1
if a > 0 and b > 0 and c > 0:
suma += 1
r += f"{suma}\n"
print(r)
| py |
1288 | A | A. Deadlinetime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputAdilbek was assigned to a special project. For Adilbek it means that he has nn days to run a special program and provide its results. But there is a problem: the program needs to run for dd days to calculate the results.Fortunately, Adilbek can optimize the program. If he spends xx (xx is a non-negative integer) days optimizing the program, he will make the program run in ⌈dx+1⌉⌈dx+1⌉ days (⌈a⌉⌈a⌉ is the ceiling function: ⌈2.4⌉=3⌈2.4⌉=3, ⌈2⌉=2⌈2⌉=2). The program cannot be run and optimized simultaneously, so the total number of days he will spend is equal to x+⌈dx+1⌉x+⌈dx+1⌉.Will Adilbek be able to provide the generated results in no more than nn days?InputThe first line contains a single integer TT (1≤T≤501≤T≤50) — the number of test cases.The next TT lines contain test cases – one per line. Each line contains two integers nn and dd (1≤n≤1091≤n≤109, 1≤d≤1091≤d≤109) — the number of days before the deadline and the number of days the program runs.OutputPrint TT answers — one per test case. For each test case print YES (case insensitive) if Adilbek can fit in nn days or NO (case insensitive) otherwise.ExampleInputCopy3
1 1
4 5
5 11
OutputCopyYES
YES
NO
NoteIn the first test case; Adilbek decides not to optimize the program at all; since d≤nd≤n.In the second test case, Adilbek can spend 11 day optimizing the program and it will run ⌈52⌉=3⌈52⌉=3 days. In total, he will spend 44 days and will fit in the limit.In the third test case, it's impossible to fit in the limit. For example, if Adilbek will optimize the program 22 days, it'll still work ⌈112+1⌉=4⌈112+1⌉=4 days. | [
"binary search",
"brute force",
"math",
"ternary search"
] | def time(x, d):
return(x+d/(x+1))
t = int(input())
for xx in range(t):
n, d = map(int,input().split())
l = 0
r = d - 1
ans = "NO"
if d <= n:
ans = "YES"
else:
while l < r:
m1 = (l+((r-l)//3))
m2 = (l+((2*(r-l))//3))
tm1 = time(m1, d)
tm2 = time(m2, d)
# print(m1, tm1, m2, tm2, " ", l, r)
if tm1 <= n or tm2 <= n:
ans = "YES"
break
if tm1 == tm2:
l = m1
r = m2
elif tm1 > tm2:
l = m1 + 1
else:
r = m2 - 1
print(ans) | py |
1320 | B | B. Navigation Systemtime limit per test2 secondsmemory limit per test512 megabytesinputstandard inputoutputstandard outputThe map of Bertown can be represented as a set of nn intersections, numbered from 11 to nn and connected by mm one-way roads. It is possible to move along the roads from any intersection to any other intersection. The length of some path from one intersection to another is the number of roads that one has to traverse along the path. The shortest path from one intersection vv to another intersection uu is the path that starts in vv, ends in uu and has the minimum length among all such paths.Polycarp lives near the intersection ss and works in a building near the intersection tt. Every day he gets from ss to tt by car. Today he has chosen the following path to his workplace: p1p1, p2p2, ..., pkpk, where p1=sp1=s, pk=tpk=t, and all other elements of this sequence are the intermediate intersections, listed in the order Polycarp arrived at them. Polycarp never arrived at the same intersection twice, so all elements of this sequence are pairwise distinct. Note that you know Polycarp's path beforehand (it is fixed), and it is not necessarily one of the shortest paths from ss to tt.Polycarp's car has a complex navigation system installed in it. Let's describe how it works. When Polycarp starts his journey at the intersection ss, the system chooses some shortest path from ss to tt and shows it to Polycarp. Let's denote the next intersection in the chosen path as vv. If Polycarp chooses to drive along the road from ss to vv, then the navigator shows him the same shortest path (obviously, starting from vv as soon as he arrives at this intersection). However, if Polycarp chooses to drive to another intersection ww instead, the navigator rebuilds the path: as soon as Polycarp arrives at ww, the navigation system chooses some shortest path from ww to tt and shows it to Polycarp. The same process continues until Polycarp arrives at tt: if Polycarp moves along the road recommended by the system, it maintains the shortest path it has already built; but if Polycarp chooses some other path, the system rebuilds the path by the same rules.Here is an example. Suppose the map of Bertown looks as follows, and Polycarp drives along the path [1,2,3,4][1,2,3,4] (s=1s=1, t=4t=4): Check the picture by the link http://tk.codeforces.com/a.png When Polycarp starts at 11, the system chooses some shortest path from 11 to 44. There is only one such path, it is [1,5,4][1,5,4]; Polycarp chooses to drive to 22, which is not along the path chosen by the system. When Polycarp arrives at 22, the navigator rebuilds the path by choosing some shortest path from 22 to 44, for example, [2,6,4][2,6,4] (note that it could choose [2,3,4][2,3,4]); Polycarp chooses to drive to 33, which is not along the path chosen by the system. When Polycarp arrives at 33, the navigator rebuilds the path by choosing the only shortest path from 33 to 44, which is [3,4][3,4]; Polycarp arrives at 44 along the road chosen by the navigator, so the system does not have to rebuild anything. Overall, we get 22 rebuilds in this scenario. Note that if the system chose [2,3,4][2,3,4] instead of [2,6,4][2,6,4] during the second step, there would be only 11 rebuild (since Polycarp goes along the path, so the system maintains the path [3,4][3,4] during the third step).The example shows us that the number of rebuilds can differ even if the map of Bertown and the path chosen by Polycarp stays the same. Given this information (the map and Polycarp's path), can you determine the minimum and the maximum number of rebuilds that could have happened during the journey?InputThe first line contains two integers nn and mm (2≤n≤m≤2⋅1052≤n≤m≤2⋅105) — the number of intersections and one-way roads in Bertown, respectively.Then mm lines follow, each describing a road. Each line contains two integers uu and vv (1≤u,v≤n1≤u,v≤n, u≠vu≠v) denoting a road from intersection uu to intersection vv. All roads in Bertown are pairwise distinct, which means that each ordered pair (u,v)(u,v) appears at most once in these mm lines (but if there is a road (u,v)(u,v), the road (v,u)(v,u) can also appear).The following line contains one integer kk (2≤k≤n2≤k≤n) — the number of intersections in Polycarp's path from home to his workplace.The last line contains kk integers p1p1, p2p2, ..., pkpk (1≤pi≤n1≤pi≤n, all these integers are pairwise distinct) — the intersections along Polycarp's path in the order he arrived at them. p1p1 is the intersection where Polycarp lives (s=p1s=p1), and pkpk is the intersection where Polycarp's workplace is situated (t=pkt=pk). It is guaranteed that for every i∈[1,k−1]i∈[1,k−1] the road from pipi to pi+1pi+1 exists, so the path goes along the roads of Bertown. OutputPrint two integers: the minimum and the maximum number of rebuilds that could have happened during the journey.ExamplesInputCopy6 9
1 5
5 4
1 2
2 3
3 4
4 1
2 6
6 4
4 2
4
1 2 3 4
OutputCopy1 2
InputCopy7 7
1 2
2 3
3 4
4 5
5 6
6 7
7 1
7
1 2 3 4 5 6 7
OutputCopy0 0
InputCopy8 13
8 7
8 6
7 5
7 4
6 5
6 4
5 3
5 2
4 3
4 2
3 1
2 1
1 8
5
8 7 5 2 1
OutputCopy0 3
| [
"dfs and similar",
"graphs",
"shortest paths"
] | '''
t3=
8->6|7(1)->5|4(2)->2|3(3)
'''
from sys import stdin
input=stdin.readline
from collections import defaultdict,deque
n,m=map(int,input().strip().split())
g=defaultdict(list)
for _ in range(m):
x,y=map(int,input().strip().split())
g[y-1].append(x-1)
a=int(input())
l=list(map(lambda s:int(s)-1,input().strip().split()))
# print(l)
d=[float("inf")]*n
d[l[-1]]=0
par=defaultdict(list)
q=deque([l[-1]])
par2=defaultdict(set)
while q:
t=q.popleft()
# print(q)
for i in g[t]:
if d[t]+1<d[i]:
d[i]=d[t]+1
par[i]=[t]
q.append(i)
elif d[t]+1==d[i]:
par[i].append(t)
mn=0
mx=len(l)-1
for i in range(len(l)-1):
if d[l[i+1]]!=d[l[i]]-1:
mn+=1
if len(par[l[i]])==1 and par[l[i]][0]==l[i+1]:
mx-=1
print(mn,mx)
| py |
1311 | B | B. WeirdSorttime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given an array aa of length nn.You are also given a set of distinct positions p1,p2,…,pmp1,p2,…,pm, where 1≤pi<n1≤pi<n. The position pipi means that you can swap elements a[pi]a[pi] and a[pi+1]a[pi+1]. You can apply this operation any number of times for each of the given positions.Your task is to determine if it is possible to sort the initial array in non-decreasing order (a1≤a2≤⋯≤ana1≤a2≤⋯≤an) using only allowed swaps.For example, if a=[3,2,1]a=[3,2,1] and p=[1,2]p=[1,2], then we can first swap elements a[2]a[2] and a[3]a[3] (because position 22 is contained in the given set pp). We get the array a=[3,1,2]a=[3,1,2]. Then we swap a[1]a[1] and a[2]a[2] (position 11 is also contained in pp). We get the array a=[1,3,2]a=[1,3,2]. Finally, we swap a[2]a[2] and a[3]a[3] again and get the array a=[1,2,3]a=[1,2,3], sorted in non-decreasing order.You can see that if a=[4,1,2,3]a=[4,1,2,3] and p=[3,2]p=[3,2] then you cannot sort the array.You have to answer tt independent test cases.InputThe first line of the input contains one integer tt (1≤t≤1001≤t≤100) — the number of test cases.Then tt test cases follow. The first line of each test case contains two integers nn and mm (1≤m<n≤1001≤m<n≤100) — the number of elements in aa and the number of elements in pp. The second line of the test case contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1001≤ai≤100). The third line of the test case contains mm integers p1,p2,…,pmp1,p2,…,pm (1≤pi<n1≤pi<n, all pipi are distinct) — the set of positions described in the problem statement.OutputFor each test case, print the answer — "YES" (without quotes) if you can sort the initial array in non-decreasing order (a1≤a2≤⋯≤ana1≤a2≤⋯≤an) using only allowed swaps. Otherwise, print "NO".ExampleInputCopy6
3 2
3 2 1
1 2
4 2
4 1 2 3
3 2
5 1
1 2 3 4 5
1
4 2
2 1 4 3
1 3
4 2
4 3 2 1
1 3
5 2
2 1 2 3 3
1 4
OutputCopyYES
NO
YES
YES
NO
YES
| [
"dfs and similar",
"sortings"
] | T=int(input(''))
while(T>0):
[n,m]=list(map(int,input('').split()))
arr=list([int(x) for x in input('').split()])
arr2=[0]*n
vis=[0]*n
for i in range(0,n):
arr2[i]=arr[i]
vis=[0]*n
arr2.sort()
temp=list(int(x) for x in input('').split())
for i in temp:
i=i-1
vis[i]=1
i,f=n-1,0
while(i>-1):
if(arr[i] != arr2[i]):
j=i-1
while(j>-1 and arr[j]!=arr2[i]):
j=j-1
while(j<i):
if(vis[j]==0):
f=1
break
arr[j],arr[j+1]=arr[j+1],arr[i]
j=j+1
if(f):
break
i=i-1
if(f):
print("No")
else:
print("Yes")
T-=1 | py |
1299 | C | C. Water Balancetime limit per test3 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThere are nn water tanks in a row, ii-th of them contains aiai liters of water. The tanks are numbered from 11 to nn from left to right.You can perform the following operation: choose some subsegment [l,r][l,r] (1≤l≤r≤n1≤l≤r≤n), and redistribute water in tanks l,l+1,…,rl,l+1,…,r evenly. In other words, replace each of al,al+1,…,aral,al+1,…,ar by al+al+1+⋯+arr−l+1al+al+1+⋯+arr−l+1. For example, if for volumes [1,3,6,7][1,3,6,7] you choose l=2,r=3l=2,r=3, new volumes of water will be [1,4.5,4.5,7][1,4.5,4.5,7]. You can perform this operation any number of times.What is the lexicographically smallest sequence of volumes of water that you can achieve?As a reminder:A sequence aa is lexicographically smaller than a sequence bb of the same length if and only if the following holds: in the first (leftmost) position where aa and bb differ, the sequence aa has a smaller element than the corresponding element in bb.InputThe first line contains an integer nn (1≤n≤1061≤n≤106) — the number of water tanks.The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1061≤ai≤106) — initial volumes of water in the water tanks, in liters.Because of large input, reading input as doubles is not recommended.OutputPrint the lexicographically smallest sequence you can get. In the ii-th line print the final volume of water in the ii-th tank.Your answer is considered correct if the absolute or relative error of each aiai does not exceed 10−910−9.Formally, let your answer be a1,a2,…,ana1,a2,…,an, and the jury's answer be b1,b2,…,bnb1,b2,…,bn. Your answer is accepted if and only if |ai−bi|max(1,|bi|)≤10−9|ai−bi|max(1,|bi|)≤10−9 for each ii.ExamplesInputCopy4
7 5 5 7
OutputCopy5.666666667
5.666666667
5.666666667
7.000000000
InputCopy5
7 8 8 10 12
OutputCopy7.000000000
8.000000000
8.000000000
10.000000000
12.000000000
InputCopy10
3 9 5 5 1 7 5 3 8 7
OutputCopy3.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
7.500000000
7.500000000
NoteIn the first sample; you can get the sequence by applying the operation for subsegment [1,3][1,3].In the second sample, you can't get any lexicographically smaller sequence. | [
"data structures",
"geometry",
"greedy"
] | import io,os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n=int(input())
a=list(map(int,input().split()))
stack=[]
for aa in a:
stack.append([aa,1])
while len(stack)>1 and stack[-1][0]/stack[-1][1]<stack[-2][0]/stack[-2][1]:
stack[-2]=[stack[-2][0]+stack[-1][0],stack[-2][1]+stack[-1][1]]
stack.pop()
b=[]
for block,l in stack:
b.extend([str(block/l)]*l)
print("\n".join(b)) | py |
1325 | A | A. EhAb AnD gCdtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given a positive integer xx. Find any such 22 positive integers aa and bb such that GCD(a,b)+LCM(a,b)=xGCD(a,b)+LCM(a,b)=x.As a reminder, GCD(a,b)GCD(a,b) is the greatest integer that divides both aa and bb. Similarly, LCM(a,b)LCM(a,b) is the smallest integer such that both aa and bb divide it.It's guaranteed that the solution always exists. If there are several such pairs (a,b)(a,b), you can output any of them.InputThe first line contains a single integer tt (1≤t≤100)(1≤t≤100) — the number of testcases.Each testcase consists of one line containing a single integer, xx (2≤x≤109)(2≤x≤109).OutputFor each testcase, output a pair of positive integers aa and bb (1≤a,b≤109)1≤a,b≤109) such that GCD(a,b)+LCM(a,b)=xGCD(a,b)+LCM(a,b)=x. It's guaranteed that the solution always exists. If there are several such pairs (a,b)(a,b), you can output any of them.ExampleInputCopy2
2
14
OutputCopy1 1
6 4
NoteIn the first testcase of the sample; GCD(1,1)+LCM(1,1)=1+1=2GCD(1,1)+LCM(1,1)=1+1=2.In the second testcase of the sample, GCD(6,4)+LCM(6,4)=2+12=14GCD(6,4)+LCM(6,4)=2+12=14. | [
"constructive algorithms",
"greedy",
"number theory"
] | for _ in range(int(input())):
print(1,int(input())-1) | py |
1290 | B | B. Irreducible Anagramstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputLet's call two strings ss and tt anagrams of each other if it is possible to rearrange symbols in the string ss to get a string, equal to tt.Let's consider two strings ss and tt which are anagrams of each other. We say that tt is a reducible anagram of ss if there exists an integer k≥2k≥2 and 2k2k non-empty strings s1,t1,s2,t2,…,sk,tks1,t1,s2,t2,…,sk,tk that satisfy the following conditions: If we write the strings s1,s2,…,sks1,s2,…,sk in order, the resulting string will be equal to ss; If we write the strings t1,t2,…,tkt1,t2,…,tk in order, the resulting string will be equal to tt; For all integers ii between 11 and kk inclusive, sisi and titi are anagrams of each other. If such strings don't exist, then tt is said to be an irreducible anagram of ss. Note that these notions are only defined when ss and tt are anagrams of each other.For example, consider the string s=s= "gamegame". Then the string t=t= "megamage" is a reducible anagram of ss, we may choose for example s1=s1= "game", s2=s2= "gam", s3=s3= "e" and t1=t1= "mega", t2=t2= "mag", t3=t3= "e": On the other hand, we can prove that t=t= "memegaga" is an irreducible anagram of ss.You will be given a string ss and qq queries, represented by two integers 1≤l≤r≤|s|1≤l≤r≤|s| (where |s||s| is equal to the length of the string ss). For each query, you should find if the substring of ss formed by characters from the ll-th to the rr-th has at least one irreducible anagram.InputThe first line contains a string ss, consisting of lowercase English characters (1≤|s|≤2⋅1051≤|s|≤2⋅105).The second line contains a single integer qq (1≤q≤1051≤q≤105) — the number of queries.Each of the following qq lines contain two integers ll and rr (1≤l≤r≤|s|1≤l≤r≤|s|), representing a query for the substring of ss formed by characters from the ll-th to the rr-th.OutputFor each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.ExamplesInputCopyaaaaa
3
1 1
2 4
5 5
OutputCopyYes
No
Yes
InputCopyaabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
OutputCopyNo
Yes
Yes
Yes
No
No
NoteIn the first sample; in the first and third queries; the substring is "a"; which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand; in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s1=s1= "a", s2=s2= "aa", t1=t1= "a", t2=t2= "aa" to show that it is a reducible anagram.In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram. | [
"binary search",
"constructive algorithms",
"data structures",
"strings",
"two pointers"
] | import sys
input=sys.stdin.readline
s=list(input().rstrip())
n=len(s)
cnt=[[0]*(n+1) for i in range(26)]
for i in range(1,n+1):
cnt[ord(s[i-1])-ord("a")][i]+=1
for j in range(26):
cnt[j][i]+=cnt[j][i-1]
q=int(input())
for _ in range(q):
l,r=map(int,input().split())
if s[r-1]!=s[l-1]:
print("Yes")
continue
if r-l+1==1:
print("Yes")
continue
flag=0
for i in range(26):
flag+=(cnt[i][r]-cnt[i][l-1]>0)
if flag>=3:
print("Yes")
else:
print("No") | py |
1288 | E | E. Messenger Simulatortime limit per test3 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputPolycarp is a frequent user of the very popular messenger. He's chatting with his friends all the time. He has nn friends, numbered from 11 to nn.Recall that a permutation of size nn is an array of size nn such that each integer from 11 to nn occurs exactly once in this array.So his recent chat list can be represented with a permutation pp of size nn. p1p1 is the most recent friend Polycarp talked to, p2p2 is the second most recent and so on.Initially, Polycarp's recent chat list pp looks like 1,2,…,n1,2,…,n (in other words, it is an identity permutation).After that he receives mm messages, the jj-th message comes from the friend ajaj. And that causes friend ajaj to move to the first position in a permutation, shifting everyone between the first position and the current position of ajaj by 11. Note that if the friend ajaj is in the first position already then nothing happens.For example, let the recent chat list be p=[4,1,5,3,2]p=[4,1,5,3,2]: if he gets messaged by friend 33, then pp becomes [3,4,1,5,2][3,4,1,5,2]; if he gets messaged by friend 44, then pp doesn't change [4,1,5,3,2][4,1,5,3,2]; if he gets messaged by friend 22, then pp becomes [2,4,1,5,3][2,4,1,5,3]. For each friend consider all position he has been at in the beginning and after receiving each message. Polycarp wants to know what were the minimum and the maximum positions.InputThe first line contains two integers nn and mm (1≤n,m≤3⋅1051≤n,m≤3⋅105) — the number of Polycarp's friends and the number of received messages, respectively.The second line contains mm integers a1,a2,…,ama1,a2,…,am (1≤ai≤n1≤ai≤n) — the descriptions of the received messages.OutputPrint nn pairs of integers. For each friend output the minimum and the maximum positions he has been in the beginning and after receiving each message.ExamplesInputCopy5 4
3 5 1 4
OutputCopy1 3
2 5
1 4
1 5
1 5
InputCopy4 3
1 2 4
OutputCopy1 3
1 2
3 4
1 4
NoteIn the first example; Polycarp's recent chat list looks like this: [1,2,3,4,5][1,2,3,4,5] [3,1,2,4,5][3,1,2,4,5] [5,3,1,2,4][5,3,1,2,4] [1,5,3,2,4][1,5,3,2,4] [4,1,5,3,2][4,1,5,3,2] So, for example, the positions of the friend 22 are 2,3,4,4,52,3,4,4,5, respectively. Out of these 22 is the minimum one and 55 is the maximum one. Thus, the answer for the friend 22 is a pair (2,5)(2,5).In the second example, Polycarp's recent chat list looks like this: [1,2,3,4][1,2,3,4] [1,2,3,4][1,2,3,4] [2,1,3,4][2,1,3,4] [4,2,1,3][4,2,1,3] | [
"data structures"
] | import sys
class segmenttree:
def __init__(self, n, default=0, func=lambda a, b: a + b):
self.tree, self.n, self.func, self.default = [0] * (2 * n), n, func, default
def fill(self, arr):
self.tree[self.n:] = arr
for i in range(self.n - 1, 0, -1):
self.tree[i] = self.func(self.tree[i << 1], self.tree[(i << 1) + 1])
# get interval[l,r)
def query(self, l, r):
res = self.default
l += self.n
r += self.n
while l < r:
if l & 1:
res = self.func(res, self.tree[l])
l += 1
if r & 1:
r -= 1
res = self.func(res, self.tree[r])
l >>= 1
r >>= 1
return res
def __setitem__(self, ix, val):
ix += self.n
self.tree[ix] = val
while ix > 1:
self.tree[ix >> 1] = self.func(self.tree[ix], self.tree[ix ^ 1])
ix >>= 1
def __getitem__(self, item):
return self.tree[item + self.n]
input = lambda: sys.stdin.buffer.readline().decode().strip()
n, m = map(int, input().split())
a, mi, ma, lst = [int(x) - 1 for x in input().split()], list(range(1, n + 1)), list(range(1, n + 1)), [-1] * n
tree = segmenttree(m + n)
for i in range(m):
mi[a[i]] = 1
if lst[a[i]] != -1:
ma[a[i]] = max(ma[a[i]], tree.query(lst[a[i]], i))
tree[lst[a[i]]] = 0
else:
ma[a[i]] += tree.query(m + a[i] + 1, n + m)
tree[a[i] + m] = tree[i] = 1
lst[a[i]] = i
for i in range(n):
if lst[i] == -1:
ma[i] += tree.query(m + i + 1, n + m)
else:
ma[i] = max(ma[i], tree.query(lst[i], m))
print(*[f'{mi[i]} {ma[i]}' for i in range(n)], sep='\n')
| py |
1292 | C | C. Xenon's Attack on the Gangstime limit per test3 secondsmemory limit per test512 megabytesinputstandard inputoutputstandard outputINSPION FullBand Master - INSPION INSPION - IOLITE-SUNSTONEOn another floor of the A.R.C. Markland-N; the young man Simon "Xenon" Jackson, takes a break after finishing his project early (as always). Having a lot of free time, he decides to put on his legendary hacker "X" instinct and fight against the gangs of the cyber world.His target is a network of nn small gangs. This network contains exactly n−1n−1 direct links, each of them connecting two gangs together. The links are placed in such a way that every pair of gangs is connected through a sequence of direct links.By mining data, Xenon figured out that the gangs used a form of cross-encryption to avoid being busted: every link was assigned an integer from 00 to n−2n−2 such that all assigned integers are distinct and every integer was assigned to some link. If an intruder tries to access the encrypted data, they will have to surpass SS password layers, with SS being defined by the following formula:S=∑1≤u<v≤nmex(u,v)S=∑1≤u<v≤nmex(u,v)Here, mex(u,v)mex(u,v) denotes the smallest non-negative integer that does not appear on any link on the unique simple path from gang uu to gang vv.Xenon doesn't know the way the integers are assigned, but it's not a problem. He decides to let his AI's instances try all the passwords on his behalf, but before that, he needs to know the maximum possible value of SS, so that the AIs can be deployed efficiently.Now, Xenon is out to write the AI scripts, and he is expected to finish them in two hours. Can you find the maximum possible SS before he returns?InputThe first line contains an integer nn (2≤n≤30002≤n≤3000), the number of gangs in the network.Each of the next n−1n−1 lines contains integers uiui and vivi (1≤ui,vi≤n1≤ui,vi≤n; ui≠viui≠vi), indicating there's a direct link between gangs uiui and vivi.It's guaranteed that links are placed in such a way that each pair of gangs will be connected by exactly one simple path.OutputPrint the maximum possible value of SS — the number of password layers in the gangs' network.ExamplesInputCopy3
1 2
2 3
OutputCopy3
InputCopy5
1 2
1 3
1 4
3 5
OutputCopy10
NoteIn the first example; one can achieve the maximum SS with the following assignment: With this assignment, mex(1,2)=0mex(1,2)=0, mex(1,3)=2mex(1,3)=2 and mex(2,3)=1mex(2,3)=1. Therefore, S=0+2+1=3S=0+2+1=3.In the second example, one can achieve the maximum SS with the following assignment: With this assignment, all non-zero mex value are listed below: mex(1,3)=1mex(1,3)=1 mex(1,5)=2mex(1,5)=2 mex(2,3)=1mex(2,3)=1 mex(2,5)=2mex(2,5)=2 mex(3,4)=1mex(3,4)=1 mex(4,5)=3mex(4,5)=3 Therefore, S=1+2+1+2+1+3=10S=1+2+1+2+1+3=10. | [
"combinatorics",
"dfs and similar",
"dp",
"greedy",
"trees"
] | ''' Testing Python performance @Pajenegod's solution '''
INF = 10 ** 10
def main():
#print = out.append
''' Cook your dish here! '''
# Read input and build the graph
n = get_int()
coupl = [[] for _ in range(n)]
for _ in range(n - 1):
u, v = get_list()
coupl[u-1].append(v-1)
coupl[v-1].append(u-1)
# Relabel to speed up n^2 operations later on
bfs = [0]
found = [0] * n
found[0] = 1
for node in bfs:
for nei in coupl[node]:
if not found[nei]:
found[nei] = 1
bfs.append(nei)
new_label = [0] * n
for i in range(n):
new_label[bfs[i]] = i
coupl = [coupl[i] for i in bfs]
for c in coupl:
c[:] = [new_label[x] for x in c]
##### DP using multisource bfs
DP = [0] * (n * n)
size = [1] * (n * n)
P = [-1] * (n * n)
# Create the bfs ordering
bfs = [root * n + root for root in range(n)]
for ind in bfs:
P[ind] = ind
for ind in bfs:
node, root = divmod(ind, n)
for nei in coupl[node]:
ind2 = nei * n + root
if P[ind2] == -1:
bfs.append(ind2)
P[ind2] = ind
del bfs[:n]
# Do the DP
for ind in reversed(bfs):
node, root = divmod(ind, n)
ind2 = root * n + node
pind = P[ind]
parent = pind // n
# Update size of (root, parent)
size[pind] += size[ind]
# Update DP value of (root, parent)
DP[pind] = max(DP[pind], max(DP[ind], DP[ind2]) + size[ind] * size[ind2])
print(max(DP[root * n + root] for root in range(n)))
''' Pythonista fLite 1.1 '''
import sys
#from collections import defaultdict, Counter, deque
# from bisect import bisect_left, bisect_right
# from functools import reduce
# import math
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
out = []
get_int = lambda: int(input())
get_list = lambda: list(map(int, input().split()))
main()
#[main() for _ in range(int(input()))]
print(*out, sep='\n')
| py |
1292 | E | E. Rin and The Unknown Flowertime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputMisoilePunch♪ - 彩This is an interactive problem!On a normal day at the hidden office in A.R.C. Markland-N; Rin received an artifact, given to her by the exploration captain Sagar.After much analysis, she now realizes that this artifact contains data about a strange flower, which has existed way before the New Age. However, the information about its chemical structure has been encrypted heavily.The chemical structure of this flower can be represented as a string pp. From the unencrypted papers included, Rin already knows the length nn of that string, and she can also conclude that the string contains at most three distinct letters: "C" (as in Carbon), "H" (as in Hydrogen), and "O" (as in Oxygen).At each moment, Rin can input a string ss of an arbitrary length into the artifact's terminal, and it will return every starting position of ss as a substring of pp.However, the artifact has limited energy and cannot be recharged in any way, since the technology is way too ancient and is incompatible with any current A.R.C.'s devices. To be specific: The artifact only contains 7575 units of energy. For each time Rin inputs a string ss of length tt, the artifact consumes 1t21t2 units of energy. If the amount of energy reaches below zero, the task will be considered failed immediately, as the artifact will go black forever. Since the artifact is so precious yet fragile, Rin is very nervous to attempt to crack the final data. Can you give her a helping hand?InteractionThe interaction starts with a single integer tt (1≤t≤5001≤t≤500), the number of test cases. The interaction for each testcase is described below:First, read an integer nn (4≤n≤504≤n≤50), the length of the string pp.Then you can make queries of type "? s" (1≤|s|≤n1≤|s|≤n) to find the occurrences of ss as a substring of pp.After the query, you need to read its result as a series of integers in a line: The first integer kk denotes the number of occurrences of ss as a substring of pp (−1≤k≤n−1≤k≤n). If k=−1k=−1, it means you have exceeded the energy limit or printed an invalid query, and you need to terminate immediately, to guarantee a "Wrong answer" verdict, otherwise you might get an arbitrary verdict because your solution will continue to read from a closed stream. The following kk integers a1,a2,…,aka1,a2,…,ak (1≤a1<a2<…<ak≤n1≤a1<a2<…<ak≤n) denote the starting positions of the substrings that match the string ss.When you find out the string pp, print "! pp" to finish a test case. This query doesn't consume any energy. The interactor will return an integer 11 or 00. If the interactor returns 11, you can proceed to the next test case, or terminate the program if it was the last testcase.If the interactor returns 00, it means that your guess is incorrect, and you should to terminate to guarantee a "Wrong answer" verdict.Note that in every test case the string pp is fixed beforehand and will not change during the queries, i.e. the interactor is not adaptive.After printing any query do not forget to print end of line and flush the output. Otherwise, you might get Idleness limit exceeded. To do this, use: fflush(stdout) or cout.flush() in C++; System.out.flush() in Java; flush(output) in Pascal; stdout.flush() in Python; see the documentation for other languages.HacksFor hack, use the following format. Note that you can only hack with one test case:The first line should contain a single integer tt (t=1t=1).The second line should contain an integer nn (4≤n≤504≤n≤50) — the string's size.The third line should contain a string of size nn, consisting of characters "C", "H" and "O" only. This is the string contestants will have to find out.ExamplesInputCopy1
4
2 1 2
1 2
0
1OutputCopy
? C
? CH
? CCHO
! CCHH
InputCopy2
5
0
2 2 3
1
8
1 5
1 5
1 3
2 1 2
1OutputCopy
? O
? HHH
! CHHHH
? COO
? COOH
? HCCOO
? HH
! HHHCCOOH
NoteNote that the example interaction contains extra empty lines so that it's easier to read. The real interaction doesn't contain any empty lines and you shouldn't print any extra empty lines as well. | [
"constructive algorithms",
"greedy",
"interactive",
"math"
] | import sys
t = int(input())
def get():
ans = ""
for i in range(n): ans = ans + S[i]
return ans
def query(pat):
print("Querying ", pat, file = sys.stderr)
print("? " + pat, flush = True)
A = list(map(int, input().split()))
for k in A[1:]:
for i in range(len(pat)):
S[k-1+i] = pat[i]
return A[0]
def answer():
ans = "! "
for i in range(n): ans = ans + S[i]
print("Answering ", ans, file = sys.stderr)
print(ans, flush = True)
input()
for q in range(t):
n = int(input())
S = ['?'] * n
Ptrns = [ "CO", "CH", "HO", "HC" ]
for pat in Ptrns: query(pat)
print("After initial queries ", get(), file = sys.stderr)
known = None
for i in range(n):
if S[i]!='?':
known = i
break
if known is None:
if query("OOO")>0:
if query("OOOC")>0:
for i in range(n):
if S[i]=='?': S[i] = 'C'
else:
for i in range(n):
if S[i]=='?': S[i] = 'H'
else:
if query("CCC")>0:
for i in range(n):
if S[i]=='?': S[i] = 'O'
elif query("HHH")>0:
for i in range(n):
if S[i]=='?': S[i] = 'O'
else:
S[0] = S[1] = 'O'
if query("OOHH")>0:
S[2] = S[3] = 'H'
else:
S[2] = S[3] = 'C'
answer()
continue
suf = S[known] + S[known+1]
while known > 0:
query(S[known] + suf)
if S[known-1]=='?':
for i in range(known):
S[i] = 'O'
break
known-=1
suf = S[known] + suf
r = known
while r < n and S[r]!='?': r+=1
pref = ""
for i in range(r): pref = pref + S[i]
while r < n:
print("I have now ", pref, file = sys.stderr)
if S[r-1]=='O':
query(pref + "O")
if S[r]=='?':
if r==n-1: # ostatni znak
if query(pref + "C")==0:
S[r] = 'H'
else: # jeszcze nie
query(pref+ "CC")
if S[r]=='?': S[r] = S[r+1] = 'H'
else:
S[r]=S[r-1]
while r < n and S[r]!='?':
pref = pref + S[r]
r+=1
answer()
| py |
1305 | F | F. Kuroni and the Punishmenttime limit per test2.5 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputKuroni is very angry at the other setters for using him as a theme! As a punishment; he forced them to solve the following problem:You have an array aa consisting of nn positive integers. An operation consists of choosing an element and either adding 11 to it or subtracting 11 from it, such that the element remains positive. We say the array is good if the greatest common divisor of all its elements is not 11. Find the minimum number of operations needed to make the array good.Unable to match Kuroni's intellect, the setters failed to solve the problem. Help them escape from Kuroni's punishment!InputThe first line contains an integer nn (2≤n≤2⋅1052≤n≤2⋅105) — the number of elements in the array.The second line contains nn integers a1,a2,…,ana1,a2,…,an. (1≤ai≤10121≤ai≤1012) — the elements of the array.OutputPrint a single integer — the minimum number of operations required to make the array good.ExamplesInputCopy3
6 2 4
OutputCopy0
InputCopy5
9 8 7 3 1
OutputCopy4
NoteIn the first example; the first array is already good; since the greatest common divisor of all the elements is 22.In the second example, we may apply the following operations: Add 11 to the second element, making it equal to 99. Subtract 11 from the third element, making it equal to 66. Add 11 to the fifth element, making it equal to 22. Add 11 to the fifth element again, making it equal to 33. The greatest common divisor of all elements will then be equal to 33, so the array will be good. It can be shown that no sequence of three or less operations can make the array good. | [
"math",
"number theory",
"probabilities"
] | import random
n = int(input())
a = list(map(int, input().split()))
limit = min(8, n)
iterations = [x for x in range(n)]
random.shuffle(iterations)
random.shuffle(iterations)
iterations = iterations[:limit]
def factorization(x):
primes = []
i = 2
while i * i <= x:
if x % i == 0:
primes.append(i)
while x % i == 0: x //= i
i = i + 1
if x > 1: primes.append(x)
return primes
def solve_with_fixed_gcd(arr, gcd):
result = 0
for x in arr:
if x < gcd: result += (gcd - x)
else:
remainder = x % gcd
result += min(remainder, gcd - remainder)
return result
answer = float("inf")
prime_list = set()
for index in iterations:
for x in range(-1, 2):
tmp = factorization(a[index]-x)
for z in tmp: prime_list.add(z)
for prime in prime_list:
answer = min(answer, solve_with_fixed_gcd(a, prime))
if answer == 0: break
print(answer) | py |
1141 | G | G. Privatization of Roads in Treelandtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputTreeland consists of nn cities and n−1n−1 roads. Each road is bidirectional and connects two distinct cities. From any city you can get to any other city by roads. Yes, you are right — the country's topology is an undirected tree.There are some private road companies in Treeland. The government decided to sell roads to the companies. Each road will belong to one company and a company can own multiple roads.The government is afraid to look unfair. They think that people in a city can consider them unfair if there is one company which owns two or more roads entering the city. The government wants to make such privatization that the number of such cities doesn't exceed kk and the number of companies taking part in the privatization is minimal.Choose the number of companies rr such that it is possible to assign each road to one company in such a way that the number of cities that have two or more roads of one company is at most kk. In other words, if for a city all the roads belong to the different companies then the city is good. Your task is to find the minimal rr that there is such assignment to companies from 11 to rr that the number of cities which are not good doesn't exceed kk. The picture illustrates the first example (n=6,k=2n=6,k=2). The answer contains r=2r=2 companies. Numbers on the edges denote edge indices. Edge colors mean companies: red corresponds to the first company, blue corresponds to the second company. The gray vertex (number 33) is not good. The number of such vertices (just one) doesn't exceed k=2k=2. It is impossible to have at most k=2k=2 not good cities in case of one company. InputThe first line contains two integers nn and kk (2≤n≤200000,0≤k≤n−12≤n≤200000,0≤k≤n−1) — the number of cities and the maximal number of cities which can have two or more roads belonging to one company.The following n−1n−1 lines contain roads, one road per line. Each line contains a pair of integers xixi, yiyi (1≤xi,yi≤n1≤xi,yi≤n), where xixi, yiyi are cities connected with the ii-th road.OutputIn the first line print the required rr (1≤r≤n−11≤r≤n−1). In the second line print n−1n−1 numbers c1,c2,…,cn−1c1,c2,…,cn−1 (1≤ci≤r1≤ci≤r), where cici is the company to own the ii-th road. If there are multiple answers, print any of them.ExamplesInputCopy6 2
1 4
4 3
3 5
3 6
5 2
OutputCopy2
1 2 1 1 2 InputCopy4 2
3 1
1 4
1 2
OutputCopy1
1 1 1 InputCopy10 2
10 3
1 2
1 3
1 4
2 5
2 6
2 7
3 8
3 9
OutputCopy3
1 1 2 3 2 3 1 3 1 | [
"binary search",
"constructive algorithms",
"dfs and similar",
"graphs",
"greedy",
"trees"
] | import sys, collections, math, bisect, heapq, random, functools
input = sys.stdin.readline
out = sys.stdout.flush
def solve():
n,k = map(int,input().split())
edges = [[] for i in range(n + 1)]
degree = [0] * (n + 1)
e = []
for i in range(n - 1):
u,v = map(int,input().split())
e.append([u,v])
edges[u].append(v)
edges[v].append(u)
degree[u] += 1
degree[v] += 1
a = list(degree[i] for i in range(1,n + 1))
a.sort(reverse = True)
tar = a[k]
color = collections.defaultdict(lambda :-1)
queue = collections.deque()
queue.append([1,1])
vis = {1}
while queue:
cur,c = queue.popleft()
for next_ in edges[cur]:
if next_ in vis:
continue
vis.add(next_)
if color[(cur,next_)] == -1:
color[(cur,next_)] = c
color[(next_,cur)] = c
c += 1
if c > tar:
c -= tar
queue.append([next_,c])
print(tar)
ans = []
for u,v in e:
ans.append(color[(u,v)])
print(*ans)
if __name__ == '__main__':
solve() | py |
1324 | E | E. Sleeping Scheduletime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputVova had a pretty weird sleeping schedule. There are hh hours in a day. Vova will sleep exactly nn times. The ii-th time he will sleep exactly after aiai hours from the time he woke up. You can assume that Vova woke up exactly at the beginning of this story (the initial time is 00). Each time Vova sleeps exactly one day (in other words, hh hours).Vova thinks that the ii-th sleeping time is good if he starts to sleep between hours ll and rr inclusive.Vova can control himself and before the ii-th time can choose between two options: go to sleep after aiai hours or after ai−1ai−1 hours.Your task is to say the maximum number of good sleeping times Vova can obtain if he acts optimally.InputThe first line of the input contains four integers n,h,ln,h,l and rr (1≤n≤2000,3≤h≤2000,0≤l≤r<h1≤n≤2000,3≤h≤2000,0≤l≤r<h) — the number of times Vova goes to sleep, the number of hours in a day and the segment of the good sleeping time.The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai<h1≤ai<h), where aiai is the number of hours after which Vova goes to sleep the ii-th time.OutputPrint one integer — the maximum number of good sleeping times Vova can obtain if he acts optimally.ExampleInputCopy7 24 21 23
16 17 14 20 20 11 22
OutputCopy3
NoteThe maximum number of good times in the example is 33.The story starts from t=0t=0. Then Vova goes to sleep after a1−1a1−1 hours, now the time is 1515. This time is not good. Then Vova goes to sleep after a2−1a2−1 hours, now the time is 15+16=715+16=7. This time is also not good. Then Vova goes to sleep after a3a3 hours, now the time is 7+14=217+14=21. This time is good. Then Vova goes to sleep after a4−1a4−1 hours, now the time is 21+19=1621+19=16. This time is not good. Then Vova goes to sleep after a5a5 hours, now the time is 16+20=1216+20=12. This time is not good. Then Vova goes to sleep after a6a6 hours, now the time is 12+11=2312+11=23. This time is good. Then Vova goes to sleep after a7a7 hours, now the time is 23+22=2123+22=21. This time is also good. | [
"dp",
"implementation"
] | from bisect import bisect_left, bisect_right
from collections import Counter, deque
from functools import lru_cache
from math import factorial, comb, sqrt, gcd, lcm
from copy import deepcopy
import heapq
from sys import stdin, stdout
# 加快读入速度, 但是注意后面的换行符(\n)
# 如果是使用 input().split() 或者 int(input()) 之类的, 换行符就去掉了
input = stdin.readline
# 由记忆化搜索改 dp
def f(cur_index, s):
if cur_index == len(L):
return 0
s += L[cur_index]
return max(f(cur_index + 1, s), f(cur_index + 1, s - 1)) + (1 if left <= s % h <= right else 0)
def main():
n, h, left, right = map(int, input().split())
L = list(map(int, input().split()))
dp = [[0] * h for _ in range(n)]
walked = [[False] * h for _ in range(n)]
walked[0][L[0] % h] = True
if left <= L[0] % h <= right:
dp[0][L[0] % h] = 1
walked[0][(L[0] - 1) % h] = True
if left <= (L[0] - 1) % h <= right:
dp[0][(L[0] - 1) % h] = 1
for i in range(len(L) - 1):
for j in range(h):
if walked[i][j] is True:
walked[i + 1][(j + L[i + 1]) % h] = True
dp[i + 1][(j + L[i + 1]) % h] = max(dp[i + 1][(j + L[i + 1]) % h],
dp[i][j] + (1 if left <= (j + L[i + 1]) % h <= right else 0))
walked[i + 1][(j + L[i + 1] - 1) % h] = True
dp[i + 1][(j + L[i + 1] - 1) % h] = max(dp[i + 1][(j + L[i + 1] - 1) % h],
dp[i][j] + (1 if left <= (j + L[i + 1] - 1) % h <= right else 0))
ans = 0
for j in range(h):
ans = max(ans, dp[-1][j])
print(ans)
if __name__ == "__main__":
main()
| py |
13 | A | A. Numberstime limit per test1 secondmemory limit per test64 megabytesinputstdinoutputstdoutLittle Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.Now he wonders what is an average value of sum of digits of the number A written in all bases from 2 to A - 1.Note that all computations should be done in base 10. You should find the result as an irreducible fraction; written in base 10.InputInput contains one integer number A (3 ≤ A ≤ 1000).OutputOutput should contain required average value in format «X/Y», where X is the numerator and Y is the denominator.ExamplesInputCopy5OutputCopy7/3InputCopy3OutputCopy2/1NoteIn the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively. | [
"implementation",
"math"
] | from math import log,ceil,gcd
n = int(input())
a = 0
for i in range(2,n):
m = n
while m:
m,r = divmod(m,i)
a += r
g = gcd(a,n-2)
print(f"{a//g}/{(n-2)//g}") | py |
1316 | D | D. Nash Matrixtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputNash designed an interesting yet simple board game where a player is simply required to follow instructions written on the cell where the player currently stands. This board game is played on the n×nn×n board. Rows and columns of this board are numbered from 11 to nn. The cell on the intersection of the rr-th row and cc-th column is denoted by (r,c)(r,c).Some cells on the board are called blocked zones. On each cell of the board, there is written one of the following 55 characters — UU, DD, LL, RR or XX — instructions for the player. Suppose that the current cell is (r,c)(r,c). If the character is RR, the player should move to the right cell (r,c+1)(r,c+1), for LL the player should move to the left cell (r,c−1)(r,c−1), for UU the player should move to the top cell (r−1,c)(r−1,c), for DD the player should move to the bottom cell (r+1,c)(r+1,c). Finally, if the character in the cell is XX, then this cell is the blocked zone. The player should remain in this cell (the game for him isn't very interesting from now on).It is guaranteed that the characters are written in a way that the player will never have to step outside of the board, no matter at which cell he starts.As a player starts from a cell, he moves according to the character in the current cell. The player keeps moving until he lands in a blocked zone. It is also possible that the player will keep moving infinitely long.For every of the n2n2 cells of the board Alice, your friend, wants to know, how will the game go, if the player starts in this cell. For each starting cell of the board, she writes down the cell that the player stops at, or that the player never stops at all. She gives you the information she has written: for each cell (r,c)(r,c) she wrote: a pair (xx,yy), meaning if a player had started at (r,c)(r,c), he would end up at cell (xx,yy). or a pair (−1−1,−1−1), meaning if a player had started at (r,c)(r,c), he would keep moving infinitely long and would never enter the blocked zone. It might be possible that Alice is trying to fool you and there's no possible grid that satisfies all the constraints Alice gave you. For the given information Alice provided you, you are required to decipher a possible board, or to determine that such a board doesn't exist. If there exist several different boards that satisfy the provided information, you can find any of them.InputThe first line of the input contains a single integer nn (1≤n≤1031≤n≤103) — the side of the board.The ii-th of the next nn lines of the input contains 2n2n integers x1,y1,x2,y2,…,xn,ynx1,y1,x2,y2,…,xn,yn, where (xj,yj)(xj,yj) (1≤xj≤n,1≤yj≤n1≤xj≤n,1≤yj≤n, or (xj,yj)=(−1,−1)(xj,yj)=(−1,−1)) is the pair written by Alice for the cell (i,j)(i,j). OutputIf there doesn't exist a board satisfying the information that Alice gave you, print a single line containing INVALID. Otherwise, in the first line print VALID. In the ii-th of the next nn lines, print the string of nn characters, corresponding to the characters in the ii-th row of the suitable board you found. Each character of a string can either be UU, DD, LL, RR or XX. If there exist several different boards that satisfy the provided information, you can find any of them.ExamplesInputCopy2
1 1 1 1
2 2 2 2
OutputCopyVALID
XL
RX
InputCopy3
-1 -1 -1 -1 -1 -1
-1 -1 2 2 -1 -1
-1 -1 -1 -1 -1 -1
OutputCopyVALID
RRD
UXD
ULLNoteFor the sample test 1 :The given grid in output is a valid one. If the player starts at (1,1)(1,1), he doesn't move any further following XX and stops there. If the player starts at (1,2)(1,2), he moves to left following LL and stops at (1,1)(1,1). If the player starts at (2,1)(2,1), he moves to right following RR and stops at (2,2)(2,2). If the player starts at (2,2)(2,2), he doesn't move any further following XX and stops there. The simulation can be seen below : For the sample test 2 : The given grid in output is a valid one, as a player starting at any cell other than the one at center (2,2)(2,2), keeps moving in an infinitely long cycle and never stops. Had he started at (2,2)(2,2), he wouldn't have moved further following instruction XX .The simulation can be seen below : | [
"constructive algorithms",
"dfs and similar",
"graphs",
"implementation"
] |
d4i=[0,-1,0,1]
d4j=[-1,0,1,0]
direction=['L','U','R','D']
back=['R','D','L','U']
def main():
# Observations:
# if ri,cj==xi,yj, i,j is x
# if ri,cj==-1,-1, i,j is cycle:
# Steps:
# Mark cycles and Xs
# DFS from all X only to nodes pointing to x
# Fill cycles by pointing to an adjacent cycle square
# Check that all cells are marked
n=int(input())
finalState=[[] for _ in range(n)] #[xi,yj]
ans=[[None for _ in range(n)] for __ in range(n)]
for i in range(n):
temp=readIntArr()
for j in range(n):
finalState[i].append([temp[2*j]-1,temp[2*j+1]-1]) #0-indexing
endPoints=[]
cyclePoints=set()
for i in range(n):
for j in range(n):
x,y=finalState[i][j]
if x==i and y==j:
ans[i][j]='X'
endPoints.append([x,y])
elif x==-2: #due to 0-indexing
ans[i][j]='c' #cycle
cyclePoints.add((i,j))
for endX,endY in endPoints:
stack=[[endX,endY]]
while stack:
i,j=stack.pop()
for z in range(4):
ii=i+d4i[z]
if not 0<=ii<n:
continue
jj=j+d4j[z]
if not 0<=jj<n:
continue
if ans[ii][jj]==None and finalState[ii][jj]==[endX,endY]:
ans[ii][jj]=back[z]
stack.append([ii,jj])
# Fill Cycles (point each cycle point to another cycle point)
for ci,cj in cyclePoints:
for z in range(4):
ii=ci+d4i[z]
jj=cj+d4j[z]
if (ii,jj) in cyclePoints:
ans[ci][cj]=direction[z]
break
ok=True
outcomes={'U','D','R','L','X'}
for i in range(n):
for j in range(n):
if ans[i][j] not in outcomes:
ok=False
if ok:
print('VALID')
multiLineArrayOfArraysPrint(ans)
else:
print('INVALID')
return
import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
# import sys
# input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
def oneLineArrayPrint(arr):
print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
print('\n'.join([''.join([str(x) for x in y]) for y in arr]))
def readIntArr():
return [int(x) for x in input().split()]
inf=float('inf')
MOD=10**9+7
for _ in range(1):
main() | py |
1291 | B | B. Array Sharpeningtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou're given an array a1,…,ana1,…,an of nn non-negative integers.Let's call it sharpened if and only if there exists an integer 1≤k≤n1≤k≤n such that a1<a2<…<aka1<a2<…<ak and ak>ak+1>…>anak>ak+1>…>an. In particular, any strictly increasing or strictly decreasing array is sharpened. For example: The arrays [4][4], [0,1][0,1], [12,10,8][12,10,8] and [3,11,15,9,7,4][3,11,15,9,7,4] are sharpened; The arrays [2,8,2,8,6,5][2,8,2,8,6,5], [0,1,1,0][0,1,1,0] and [2,5,6,9,8,8][2,5,6,9,8,8] are not sharpened. You can do the following operation as many times as you want: choose any strictly positive element of the array, and decrease it by one. Formally, you can choose any ii (1≤i≤n1≤i≤n) such that ai>0ai>0 and assign ai:=ai−1ai:=ai−1.Tell if it's possible to make the given array sharpened using some number (possibly zero) of these operations.InputThe input consists of multiple test cases. The first line contains a single integer tt (1≤t≤15 0001≤t≤15 000) — the number of test cases. The description of the test cases follows.The first line of each test case contains a single integer nn (1≤n≤3⋅1051≤n≤3⋅105).The second line of each test case contains a sequence of nn non-negative integers a1,…,ana1,…,an (0≤ai≤1090≤ai≤109).It is guaranteed that the sum of nn over all test cases does not exceed 3⋅1053⋅105.OutputFor each test case, output a single line containing "Yes" (without quotes) if it's possible to make the given array sharpened using the described operations, or "No" (without quotes) otherwise.ExampleInputCopy10
1
248618
3
12 10 8
6
100 11 15 9 7 8
4
0 1 1 0
2
0 0
2
0 1
2
1 0
2
1 1
3
0 1 0
3
1 0 1
OutputCopyYes
Yes
Yes
No
No
Yes
Yes
Yes
Yes
No
NoteIn the first and the second test case of the first test; the given array is already sharpened.In the third test case of the first test; we can transform the array into [3,11,15,9,7,4][3,11,15,9,7,4] (decrease the first element 9797 times and decrease the last element 44 times). It is sharpened because 3<11<153<11<15 and 15>9>7>415>9>7>4.In the fourth test case of the first test, it's impossible to make the given array sharpened. | [
"greedy",
"implementation"
] | def solve():
n = int(input())
aseq = read_ints()
i = peak = 0
while i < n and aseq[i] >= i:
peak = i
i += 1
prev = aseq[peak]
for i in range(peak+1, n):
if aseq[i] >= prev:
if prev - 1 < 0:
return False
prev -= 1
else:
prev = aseq[i]
return True
def main():
t = int(input())
output = []
for _ in range(t):
ans = solve()
output.append('Yes' if ans else 'No')
print_lines(output)
def read_ints(): return [int(c) for c in input().split()]
def print_lines(lst): print('\n'.join(map(str, lst)))
if __name__ == "__main__":
from os import environ as env
if 'COMPUTERNAME' in env and 'L2A6HRI' in env['COMPUTERNAME']:
import sys
sys.stdout = open('out.txt', 'w')
sys.stdin = open('in.txt', 'r')
main()
| py |
1296 | E1 | E1. String Coloring (easy version)time limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputThis is an easy version of the problem. The actual problems are different; but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.You are given a string ss consisting of nn lowercase Latin letters.You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in ss).After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.The goal is to make the string sorted, i.e. all characters should be in alphabetical order.Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.InputThe first line of the input contains one integer nn (1≤n≤2001≤n≤200) — the length of ss.The second line of the input contains the string ss consisting of exactly nn lowercase Latin letters.OutputIf it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of nn characters, the ii-th character should be '0' if the ii-th character is colored the first color and '1' otherwise).ExamplesInputCopy9
abacbecfd
OutputCopyYES
001010101
InputCopy8
aaabbcbb
OutputCopyYES
01011011
InputCopy7
abcdedc
OutputCopyNO
InputCopy5
abcde
OutputCopyYES
00000
| [
"constructive algorithms",
"dp",
"graphs",
"greedy",
"sortings"
] | from __future__ import division, print_function
import os,sys
import re
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
from math import ceil, floor, factorial
# from math import log,sqrt,cos,tan,sin,radians
from bisect import bisect_left, bisect_right
from collections import defaultdict
# from collections import deque, Counter, OrderedDict
# from heapq import nsmallest, nlargest, heapify, heappop, heappush, heapreplace
# from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
# from decimal import *
# from itertools import permutations
# ======================== Functions declaration Starts ========================
abc='abcdefghijklmnopqrstuvwxyz'
abd={'a':0,'b':1,'c':2,'d':3,'e':4,'f':5,'g':6,'h':7,'i':8,'j':9,'k':10,'l':11,'m':12,'n':13,'o':14,'p':15,'q':16,'r':17,'s':18,'t':19,'u':20,'v':21,'w':22,'x':23,'y':24,'z':25}
M=1000000007
# M=998244353
INF = float("inf")
PI = 3.141592653589793
def copy2d(lst): return [x[:] for x in lst] #Copy 2D list... Avoid Using Deepcopy
def isPowerOfTwo(x): return (x and (not(x & (x - 1))) )
LB = bisect_left # Lower bound
UB = bisect_right # Upper bound
def BS(a, x): # Binary Search
i = bisect_left(a, x)
if i != len(a) and a[i] == x:
return i
else:
return -1
def gcd(x, y):
while y:
x, y = y, x % y
return x
def lcm(x, y):
return (x*y)//gcd(x,y)
# import threading
# def dmain():
# sys.setrecursionlimit(1000000)
# threading.stack_size(1024000)
# thread = threading.Thread(target=main)
# thread.start()
# ========================= Functions declaration Ends =========================
def dfs(node, col, visited, graph, cond):
visited[node] = col
# if not cond[0]:
# return
for i in graph[node]:
if visited[i] == -1:
dfs(i, 1-col, visited, graph, cond)
elif visited[i] == col:
cond[0] = False
def main():
TestCases = 1
for _ in range(TestCases):
n = int(input())
s = input()
arr = []
for i in range(n):
arr.append((s[i],i))
graph = [[] for i in range(n)]
is_sorted = False
for i in range(n):
if is_sorted:
break
is_sorted = True
for j in range(n-i-1):
if arr[j][0] > arr[j+1][0]:
arr[j], arr[j+1] = arr[j+1], arr[j]
u, v = arr[j][1], arr[j+1][1]
graph[u].append(v)
graph[v].append(u)
is_sorted = False
# print(graph)
visited = [-1]*n
cond = [True]
for i in range(n):
if visited[i] == -1:
dfs(i, 0, visited, graph, cond)
if not cond[0]:
break
if cond[0]:
print('YES')
print(*visited, sep='')
else:
print('NO')
# =============================== Region Fastio ===============================
if not os.path.isdir('C:/users/acer'):
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# =============================== Endregion ===============================
if __name__ == "__main__":
#read()
main()
#dmain()
| py |
1295 | C | C. Obtain The Stringtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given two strings ss and tt consisting of lowercase Latin letters. Also you have a string zz which is initially empty. You want string zz to be equal to string tt. You can perform the following operation to achieve this: append any subsequence of ss at the end of string zz. A subsequence is a sequence that can be derived from the given sequence by deleting zero or more elements without changing the order of the remaining elements. For example, if z=acz=ac, s=abcdes=abcde, you may turn zz into following strings in one operation: z=acacez=acace (if we choose subsequence aceace); z=acbcdz=acbcd (if we choose subsequence bcdbcd); z=acbcez=acbce (if we choose subsequence bcebce). Note that after this operation string ss doesn't change.Calculate the minimum number of such operations to turn string zz into string tt. InputThe first line contains the integer TT (1≤T≤1001≤T≤100) — the number of test cases.The first line of each testcase contains one string ss (1≤|s|≤1051≤|s|≤105) consisting of lowercase Latin letters.The second line of each testcase contains one string tt (1≤|t|≤1051≤|t|≤105) consisting of lowercase Latin letters.It is guaranteed that the total length of all strings ss and tt in the input does not exceed 2⋅1052⋅105.OutputFor each testcase, print one integer — the minimum number of operations to turn string zz into string tt. If it's impossible print −1−1.ExampleInputCopy3
aabce
ace
abacaba
aax
ty
yyt
OutputCopy1
-1
3
| [
"dp",
"greedy",
"strings"
] | import math
from collections import Counter, deque
from sys import stdout
import time
from math import factorial, log, gcd
import sys
from decimal import Decimal
import heapq
def S():
return sys.stdin.readline().split()
def I():
return [int(i) for i in sys.stdin.readline().split()]
def II():
return int(sys.stdin.readline())
def IS():
return sys.stdin.readline().replace('\n', '')
def main():
s = IS()
n = len(s)
t = IS()
alphabet = set(s)
if alphabet | set(t) != alphabet:
print(-1)
return
suf_m = dict([(i, [-1] * n) for i in alphabet])
suf_m[s[-1]][-1] = n - 1
for i in range(n - 2, -1, -1):
el = s[i]
for a in alphabet:
if a != el:
suf_m[a][i] = suf_m[a][i + 1]
else:
suf_m[el][i] = i
idx = -1
op = 0
for i in range(len(t)):
el = t[i]
idx = suf_m[el][idx + 1]
if idx == n - 1:
op += 1
idx = -1
elif idx == -1:
op += 1
idx = suf_m[el][0]
if idx != -1:
op += 1
print(op)
if __name__ == '__main__':
for _ in range(II()):
main()
# main() | py |
1294 | C | C. Product of Three Numberstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given one integer number nn. Find three distinct integers a,b,ca,b,c such that 2≤a,b,c2≤a,b,c and a⋅b⋅c=na⋅b⋅c=n or say that it is impossible to do it.If there are several answers, you can print any.You have to answer tt independent test cases.InputThe first line of the input contains one integer tt (1≤t≤1001≤t≤100) — the number of test cases.The next nn lines describe test cases. The ii-th test case is given on a new line as one integer nn (2≤n≤1092≤n≤109).OutputFor each test case, print the answer on it. Print "NO" if it is impossible to represent nn as a⋅b⋅ca⋅b⋅c for some distinct integers a,b,ca,b,c such that 2≤a,b,c2≤a,b,c.Otherwise, print "YES" and any possible such representation.ExampleInputCopy5
64
32
97
2
12345
OutputCopyYES
2 4 8
NO
NO
NO
YES
3 5 823
| [
"greedy",
"math",
"number theory"
] | def factor(n: int) -> bool:
count = 0
mul = []
i = 2
while(i*i <= n):
if (n%i==0):
j = findOhterFactor(i, n)
if(j):
mul.append(i)
mul.append(j)
mul.append(n//i//j)
return mul
i+=1
return []
def findOhterFactor(i: int, n: int) -> int:
j = 2
f = n//i
while (j*j<=f):
if(f%j==0):
o = f//j
if (j != i and o!=i and o!=j):
return j
j+=1
return 0
def solve() -> None:
t = int(input())
for __ in range(t):
n = int(input())
arr = factor(n)
if (len(arr)):
print("Yes")
print(' '.join(map(str, arr)))
else:
print("No")
if __name__ == '__main__':
solve()
| py |
1301 | C | C. Ayoub's functiontime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputAyoub thinks that he is a very smart person; so he created a function f(s)f(s), where ss is a binary string (a string which contains only symbols "0" and "1"). The function f(s)f(s) is equal to the number of substrings in the string ss that contains at least one symbol, that is equal to "1".More formally, f(s)f(s) is equal to the number of pairs of integers (l,r)(l,r), such that 1≤l≤r≤|s|1≤l≤r≤|s| (where |s||s| is equal to the length of string ss), such that at least one of the symbols sl,sl+1,…,srsl,sl+1,…,sr is equal to "1". For example, if s=s="01010" then f(s)=12f(s)=12, because there are 1212 such pairs (l,r)(l,r): (1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4),(2,5),(3,4),(3,5),(4,4),(4,5)(1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4),(2,5),(3,4),(3,5),(4,4),(4,5).Ayoub also thinks that he is smarter than Mahmoud so he gave him two integers nn and mm and asked him this problem. For all binary strings ss of length nn which contains exactly mm symbols equal to "1", find the maximum value of f(s)f(s).Mahmoud couldn't solve the problem so he asked you for help. Can you help him? InputThe input consists of multiple test cases. The first line contains a single integer tt (1≤t≤1051≤t≤105) — the number of test cases. The description of the test cases follows.The only line for each test case contains two integers nn, mm (1≤n≤1091≤n≤109, 0≤m≤n0≤m≤n) — the length of the string and the number of symbols equal to "1" in it.OutputFor every test case print one integer number — the maximum value of f(s)f(s) over all strings ss of length nn, which has exactly mm symbols, equal to "1".ExampleInputCopy5
3 1
3 2
3 3
4 0
5 2
OutputCopy4
5
6
0
12
NoteIn the first test case; there exists only 33 strings of length 33, which has exactly 11 symbol, equal to "1". These strings are: s1=s1="100", s2=s2="010", s3=s3="001". The values of ff for them are: f(s1)=3,f(s2)=4,f(s3)=3f(s1)=3,f(s2)=4,f(s3)=3, so the maximum value is 44 and the answer is 44.In the second test case, the string ss with the maximum value is "101".In the third test case, the string ss with the maximum value is "111".In the fourth test case, the only string ss of length 44, which has exactly 00 symbols, equal to "1" is "0000" and the value of ff for that string is 00, so the answer is 00.In the fifth test case, the string ss with the maximum value is "01010" and it is described as an example in the problem statement. | [
"binary search",
"combinatorics",
"greedy",
"math",
"strings"
] | import sys
from math import *
from collections import *
inp = lambda: sys.stdin.buffer.readline().decode().strip()
out=sys.stdout.write
# n=int(inp())
# arr=list(map(int,inp().split()))
for _ in range(int(inp())):
n,m=map(int,inp().split())
ones=m
zeros=n-m
buckets=ones+1
total=(n*(n+1))//2
h=ceil(zeros/buckets)
l=max(0,h-1)
#x.h+y.l=zeros
#x+y=buckets
y=buckets*h-zeros
x=buckets-y
remove=(x*h*(h+1))//2+(y*l*(l+1))//2
print(total-remove) | py |
1285 | D | D. Dr. Evil Underscorestime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputToday; as a friendship gift, Bakry gave Badawy nn integers a1,a2,…,ana1,a2,…,an and challenged him to choose an integer XX such that the value max1≤i≤n(ai⊕X)max1≤i≤n(ai⊕X) is minimum possible, where ⊕⊕ denotes the bitwise XOR operation.As always, Badawy is too lazy, so you decided to help him and find the minimum possible value of max1≤i≤n(ai⊕X)max1≤i≤n(ai⊕X).InputThe first line contains integer nn (1≤n≤1051≤n≤105).The second line contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤230−10≤ai≤230−1).OutputPrint one integer — the minimum possible value of max1≤i≤n(ai⊕X)max1≤i≤n(ai⊕X).ExamplesInputCopy3
1 2 3
OutputCopy2
InputCopy2
1 5
OutputCopy4
NoteIn the first sample; we can choose X=3X=3.In the second sample, we can choose X=5X=5. | [
"bitmasks",
"brute force",
"dfs and similar",
"divide and conquer",
"dp",
"greedy",
"strings",
"trees"
] | if __name__ == '__main__':
n = input()
a = input()
ll = a.split(' ')
l = [int(x) for x in ll]
# max_val = max(l)
# max_bit = 0
# while max_val:
# max_val >>= 1
# max_bit += 1
def solve(arr: list, bit: int) -> int:
if not arr or bit < 0:
return 0
zero = []
one = []
for x in arr:
if (x >> bit) & 1:
one.append(x)
else:
zero.append(x)
if not one:
return solve(zero, bit - 1)
if not zero:
return solve(one, bit - 1)
return min(solve(zero, bit - 1), solve(one, bit - 1)) + (1 << bit)
ans = solve(l, 30)
print(ans)
| py |
1286 | C2 | C2. Madhouse (Hard version)time limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputThis problem is different with easy version only by constraints on total answers lengthIt is an interactive problemVenya joined a tour to the madhouse; in which orderlies play with patients the following game. Orderlies pick a string ss of length nn, consisting only of lowercase English letters. The player can ask two types of queries: ? l r – ask to list all substrings of s[l..r]s[l..r]. Substrings will be returned in random order, and in every substring, all characters will be randomly shuffled. ! s – guess the string picked by the orderlies. This query can be asked exactly once, after that the game will finish. If the string is guessed correctly, the player wins, otherwise he loses. The player can ask no more than 33 queries of the first type.To make it easier for the orderlies, there is an additional limitation: the total number of returned substrings in all queries of the first type must not exceed ⌈0.777(n+1)2⌉⌈0.777(n+1)2⌉ (⌈x⌉⌈x⌉ is xx rounded up).Venya asked you to write a program, which will guess the string by interacting with the orderlies' program and acting by the game's rules.Your program should immediately terminate after guessing the string using a query of the second type. In case your program guessed the string incorrectly, or it violated the game rules, it will receive verdict Wrong answer.Note that in every test case the string is fixed beforehand and will not change during the game, which means that the interactor is not adaptive.InputFirst line contains number nn (1≤n≤1001≤n≤100) — the length of the picked string.InteractionYou start the interaction by reading the number nn.To ask a query about a substring from ll to rr inclusively (1≤l≤r≤n1≤l≤r≤n), you should output? l ron a separate line. After this, all substrings of s[l..r]s[l..r] will be returned in random order, each substring exactly once. In every returned substring all characters will be randomly shuffled.In the case, if you ask an incorrect query, ask more than 33 queries of the first type or there will be more than ⌈0.777(n+1)2⌉⌈0.777(n+1)2⌉ substrings returned in total, you will receive verdict Wrong answer.To guess the string ss, you should output! son a separate line.After printing each query, do not forget to flush the output. Otherwise, you will get Idleness limit exceeded. To flush the output, you can use: fflush(stdout) or cout.flush() in C++; System.out.flush() in Java; flush(output) in Pascal; stdout.flush() in Python; see documentation for other languages. If you received - (dash) as an answer to any query, you need to terminate your program with exit code 0 (for example, by calling exit(0)). This means that there was an error in the interaction protocol. If you don't terminate with exit code 0, you can receive any unsuccessful verdict. Hack formatTo hack a solution, use the following format:The first line should contain one integer nn (1≤n≤1001≤n≤100) — the length of the string, and the following line should contain the string ss.ExampleInputCopy4
a
aa
a
cb
b
c
cOutputCopy? 1 2
? 3 4
? 4 4
! aabc | [
"brute force",
"constructive algorithms",
"hashing",
"interactive",
"math"
] | from sys import stdout
class String2:
def __init__(self, s, b=None):
if b is not None:
self.b = b
return
self.b = [0] * 26
for i in s:
self.b[ord(i) - ord('a')] += 1
def __add__(self, other):
b = self.b.copy()
for i in range(26):
b[i] += other.b[i]
return String2('', b)
def __sub__(self, other):
b = self.b.copy()
for i in range(26):
b[i] -= other.b[i]
return b
def __mul__(self, other):
ans = String2('', self.b)
for i in range(26):
ans.b[i] *= other
return ans
def req(l, r, k=0):
print('?', l, r)
v = [''.join(sorted(input())) for i in range((r - l + 1) * (r - l + 2) // 2)]
stdout.flush()
return v
def compute(v):
bukvi = [[0] * (n + 2) for _ in range(26)]
for el in v:
cur = len(el)
for e in el:
bukvi[ord(e) - ord('a')][cur] += 1
return bukvi
def compute2(bukvi):
bukvis = [set() for i in range(n + 2)]
for i in range(26):
prev = bukvi[i][1]
for j in range(1, n // 2 + n % 2 + 1):
while bukvi[i][j] != prev:
bukvis[j].add(chr(ord('a') + i))
prev += 1
return bukvis
def solve(va, va2):
for i in va2:
va.remove(i)
va.sort(key=len)
s = va[0]
for i in range(1, len(va)):
for j in range(26):
if va[i].count(chr(ord('a') + j)) != va[i - 1].count(chr(ord('a') + j)):
s += chr(ord('a') + j)
return s
def check(v, s, s2, f):
s3 = String2(v[0])
for i in range(1, len(v)):
s3 = s3 + String2(v[i])
le = len(v[0])
cur = String2('', String2('', f - String2(s)) - String2(s2)) * le
for i in range(le - 2):
cur = cur + (String2(s[i]) * (i + 1)) + (String2(s2[-i-1]) * (i + 1))
cur = cur + (String2(s[le - 2]) * (le - 1)) + (String2(s[le - 1:]) * le)
e = cur - s3
for i in range(26):
if e[i]:
return chr(ord('a') + i)
def main():
if n == 1:
va = req(1, 1)
print('!', va[0])
return
elif n == 2:
va2 = req(1, 1)
va3 = req(2, 2)
print('!', va2[0] + va3[0])
return
elif n == 3:
va = req(1, 1)
va2 = req(2, 2)
va3 = req(3, 3)
print('!', va[0] + va2[0] + va3[0])
return
va = req(1, n)
va2 = req(1, max(n // 2 + n % 2, 2))
va3 = req(2, max(n // 2 + n % 2, 2))
# bukvi2 = compute(va2)
# bukvi3 = compute(va3)
ma = [[] for i in range(n * 2)]
for i in va:
ma[len(i)].append(i)
a = String2(''.join(ma[1]))
s = solve(va2, va3)
s2 = ''
for i in range(2, n // 2 + 1):
s2 = check(ma[i], s, s2, a) + s2
se = String2('', a - String2(s)) - String2(s2)
for i in range(len(se)):
if se[i]:
s += chr(ord('a') + i)
break
print('!', s + s2)
n = int(input())
main()
| py |
1325 | D | D. Ehab the Xorcisttime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputGiven 2 integers uu and vv, find the shortest array such that bitwise-xor of its elements is uu, and the sum of its elements is vv.InputThe only line contains 2 integers uu and vv (0≤u,v≤1018)(0≤u,v≤1018).OutputIf there's no array that satisfies the condition, print "-1". Otherwise:The first line should contain one integer, nn, representing the length of the desired array. The next line should contain nn positive integers, the array itself. If there are multiple possible answers, print any.ExamplesInputCopy2 4
OutputCopy2
3 1InputCopy1 3
OutputCopy3
1 1 1InputCopy8 5
OutputCopy-1InputCopy0 0
OutputCopy0NoteIn the first sample; 3⊕1=23⊕1=2 and 3+1=43+1=4. There is no valid array of smaller length.Notice that in the fourth sample the array is empty. | [
"bitmasks",
"constructive algorithms",
"greedy",
"number theory"
] | [u,v]=list(map(int,input().split(" ")))
if u==0 and v==0:
print(0)
elif u>v or u%2!=v%2:
print(-1)
elif u==v:
print(1)
print(v)
else:
t=(v-u)//2
if t&u:
print(3)
print(str(t)+" "+str(t)+" "+str(u))
else:
print(2)
print(str(t)+" "+str(t^u)) | py |
1315 | A | A. Dead Pixeltime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputScreen resolution of Polycarp's monitor is a×ba×b pixels. Unfortunately, there is one dead pixel at his screen. It has coordinates (x,y)(x,y) (0≤x<a,0≤y<b0≤x<a,0≤y<b). You can consider columns of pixels to be numbered from 00 to a−1a−1, and rows — from 00 to b−1b−1.Polycarp wants to open a rectangular window of maximal size, which doesn't contain the dead pixel. The boundaries of the window should be parallel to the sides of the screen.Print the maximal area (in pixels) of a window that doesn't contain the dead pixel inside itself.InputIn the first line you are given an integer tt (1≤t≤1041≤t≤104) — the number of test cases in the test. In the next lines you are given descriptions of tt test cases.Each test case contains a single line which consists of 44 integers a,b,xa,b,x and yy (1≤a,b≤1041≤a,b≤104; 0≤x<a0≤x<a; 0≤y<b0≤y<b) — the resolution of the screen and the coordinates of a dead pixel. It is guaranteed that a+b>2a+b>2 (e.g. a=b=1a=b=1 is impossible).OutputPrint tt integers — the answers for each test case. Each answer should contain an integer equal to the maximal possible area (in pixels) of a rectangular window, that doesn't contain the dead pixel.ExampleInputCopy6
8 8 0 0
1 10 0 3
17 31 10 4
2 1 0 0
5 10 3 9
10 10 4 8
OutputCopy56
6
442
1
45
80
NoteIn the first test case; the screen resolution is 8×88×8, and the upper left pixel is a dead pixel. Here you can see one of two possible layouts of the maximal window. | [
"implementation"
] | for _ in range(int(input(''))):
k=list(map(int,input().split()))
a,b,x,y=k[0],k[1],k[2],k[3]
l=[(a-x-1)*b,a*(b-y-1),a*y,b*x]
print(max(l)) | py |
1303 | B | B. National Projecttime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYour company was appointed to lay new asphalt on the highway of length nn. You know that every day you can either repair one unit of the highway (lay new asphalt over one unit of the highway) or skip repairing.Skipping the repair is necessary because of the climate. The climate in your region is periodical: there are gg days when the weather is good and if you lay new asphalt these days it becomes high-quality pavement; after that, the weather during the next bb days is bad, and if you lay new asphalt these days it becomes low-quality pavement; again gg good days, bb bad days and so on.You can be sure that you start repairing at the start of a good season, in other words, days 1,2,…,g1,2,…,g are good.You don't really care about the quality of the highway, you just want to make sure that at least half of the highway will have high-quality pavement. For example, if the n=5n=5 then at least 33 units of the highway should have high quality; if n=4n=4 then at least 22 units should have high quality.What is the minimum number of days is needed to finish the repair of the whole highway?InputThe first line contains a single integer TT (1≤T≤1041≤T≤104) — the number of test cases.Next TT lines contain test cases — one per line. Each line contains three integers nn, gg and bb (1≤n,g,b≤1091≤n,g,b≤109) — the length of the highway and the number of good and bad days respectively.OutputPrint TT integers — one per test case. For each test case, print the minimum number of days required to repair the whole highway if at least half of it should have high quality.ExampleInputCopy3
5 1 1
8 10 10
1000000 1 1000000
OutputCopy5
8
499999500000
NoteIn the first test case; you can just lay new asphalt each day; since days 1,3,51,3,5 are good.In the second test case, you can also lay new asphalt each day, since days 11-88 are good. | [
"math"
] | t= int(input())
for counter in range(t):
array = [int(x) for x in input().split()]
allowed = array[0] // 2
finished = 0
g = array[1]
b = array[2]
cycle = g + b
progress = cycle * (allowed // b)
ans = progress
leftover = allowed % b
remaining = array[0] - progress
if remaining <= g + leftover:
ans += remaining
else:
ans += cycle
progress += g + leftover
remaining = array[0] - progress
if remaining % g == 0:
ans += (remaining // g) * cycle - b
else:
ans += (remaining // g) * cycle + (remaining % g)
print(ans)
| py |
1304 | D | D. Shortest and Longest LIStime limit per test3 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputGildong recently learned how to find the longest increasing subsequence (LIS) in O(nlogn)O(nlogn) time for a sequence of length nn. He wants to test himself if he can implement it correctly, but he couldn't find any online judges that would do it (even though there are actually many of them). So instead he's going to make a quiz for you about making permutations of nn distinct integers between 11 and nn, inclusive, to test his code with your output.The quiz is as follows.Gildong provides a string of length n−1n−1, consisting of characters '<' and '>' only. The ii-th (1-indexed) character is the comparison result between the ii-th element and the i+1i+1-st element of the sequence. If the ii-th character of the string is '<', then the ii-th element of the sequence is less than the i+1i+1-st element. If the ii-th character of the string is '>', then the ii-th element of the sequence is greater than the i+1i+1-st element.He wants you to find two possible sequences (not necessarily distinct) consisting of nn distinct integers between 11 and nn, inclusive, each satisfying the comparison results, where the length of the LIS of the first sequence is minimum possible, and the length of the LIS of the second sequence is maximum possible.InputEach test contains one or more test cases. The first line contains the number of test cases tt (1≤t≤1041≤t≤104).Each test case contains exactly one line, consisting of an integer and a string consisting of characters '<' and '>' only. The integer is nn (2≤n≤2⋅1052≤n≤2⋅105), the length of the permutation you need to find. The string is the comparison results explained in the description. The length of the string is n−1n−1.It is guaranteed that the sum of all nn in all test cases doesn't exceed 2⋅1052⋅105.OutputFor each test case, print two lines with nn integers each. The first line is the sequence with the minimum length of the LIS, and the second line is the sequence with the maximum length of the LIS. If there are multiple answers, print any one of them. Each sequence should contain all integers between 11 and nn, inclusive, and should satisfy the comparison results.It can be shown that at least one answer always exists.ExampleInputCopy3
3 <<
7 >><>><
5 >>><
OutputCopy1 2 3
1 2 3
5 4 3 7 2 1 6
4 3 1 7 5 2 6
4 3 2 1 5
5 4 2 1 3
NoteIn the first case; 11 22 33 is the only possible answer.In the second case, the shortest length of the LIS is 22, and the longest length of the LIS is 33. In the example of the maximum LIS sequence, 44 '33' 11 77 '55' 22 '66' can be one of the possible LIS. | [
"constructive algorithms",
"graphs",
"greedy",
"two pointers"
] | import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#######################################
import sys,threading,math,functools
# sys.setrecursionlimit(600000)
# threading.stack_size(10**8)
def primes(n):
ans=[]
for i in range(2,int(n**0.5)+1):
if n%i==0:
ans.append(i)
while(n%i==0):
n//=i
if n==1:
break
if n!=1:
ans.append(n)
return ans
def countSetBits(n):
ans=0
while(n):
ans+=n%2
n//=2
return ans
def factors(n):
ans=[]
rev=[]
for i in range(1,int(n**0.5)+1):
if n%i==0:
if i*i==n:
ans.append(i)
continue
ans.append(i)
rev.append(n//i)
return ans+rev[::-1]
#list(map(int,input().split()))
for _ in range(int(input())):
[n,s]=list(input().split())
n=int(n)
k=s.count(">")
#sequence starts from k+1
a=[k+1]
i,j=k-1,k+1
x=0
temp=[]
for t in s:
if t==">":
a.append(i+1)
i-=1
if x:
temp.append(x)
x=0
else:
a.append(j+1)
j+=1
x+=1
# print(a)
# print(x)
if x:
temp.append(x)
b=[0 for i in range(n)]
completed=n
i=0
# print(a)
# print(temp)
while(i<n):
# print(i,b)
if a[i]<=k+1:
b[i]=a[i]
i+=1
else:
for j in range(temp[0]):
b[i+j]=completed-temp[0]+j+1
i+=temp[0]
completed-=temp[0]
temp.pop(0)
print(*b)
print(*a)
| py |
1301 | B | B. Motarack's Birthdaytime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputDark is going to attend Motarack's birthday. Dark decided that the gift he is going to give to Motarack is an array aa of nn non-negative integers.Dark created that array 10001000 years ago, so some elements in that array disappeared. Dark knows that Motarack hates to see an array that has two adjacent elements with a high absolute difference between them. He doesn't have much time so he wants to choose an integer kk (0≤k≤1090≤k≤109) and replaces all missing elements in the array aa with kk.Let mm be the maximum absolute difference between all adjacent elements (i.e. the maximum value of |ai−ai+1||ai−ai+1| for all 1≤i≤n−11≤i≤n−1) in the array aa after Dark replaces all missing elements with kk.Dark should choose an integer kk so that mm is minimized. Can you help him?InputThe input consists of multiple test cases. The first line contains a single integer tt (1≤t≤1041≤t≤104) — the number of test cases. The description of the test cases follows.The first line of each test case contains one integer nn (2≤n≤1052≤n≤105) — the size of the array aa.The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (−1≤ai≤109−1≤ai≤109). If ai=−1ai=−1, then the ii-th integer is missing. It is guaranteed that at least one integer is missing in every test case.It is guaranteed, that the sum of nn for all test cases does not exceed 4⋅1054⋅105.OutputPrint the answers for each test case in the following format:You should print two integers, the minimum possible value of mm and an integer kk (0≤k≤1090≤k≤109) that makes the maximum absolute difference between adjacent elements in the array aa equal to mm.Make sure that after replacing all the missing elements with kk, the maximum absolute difference between adjacent elements becomes mm.If there is more than one possible kk, you can print any of them.ExampleInputCopy7
5
-1 10 -1 12 -1
5
-1 40 35 -1 35
6
-1 -1 9 -1 3 -1
2
-1 -1
2
0 -1
4
1 -1 3 -1
7
1 -1 7 5 2 -1 5
OutputCopy1 11
5 35
3 6
0 42
0 0
1 2
3 4
NoteIn the first test case after replacing all missing elements with 1111 the array becomes [11,10,11,12,11][11,10,11,12,11]. The absolute difference between any adjacent elements is 11. It is impossible to choose a value of kk, such that the absolute difference between any adjacent element will be ≤0≤0. So, the answer is 11.In the third test case after replacing all missing elements with 66 the array becomes [6,6,9,6,3,6][6,6,9,6,3,6]. |a1−a2|=|6−6|=0|a1−a2|=|6−6|=0; |a2−a3|=|6−9|=3|a2−a3|=|6−9|=3; |a3−a4|=|9−6|=3|a3−a4|=|9−6|=3; |a4−a5|=|6−3|=3|a4−a5|=|6−3|=3; |a5−a6|=|3−6|=3|a5−a6|=|3−6|=3. So, the maximum difference between any adjacent elements is 33. | [
"binary search",
"greedy",
"ternary search"
] | import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
t = int(input())
ans = []
inf = pow(10, 9) + 1
for _ in range(t):
n = int(input())
a = [0] + list(map(int, input().split())) + [0]
ma, mi = -inf, inf
for i in range(1, n + 1):
if not a[i] == -1 and min(a[i - 1], a[i + 1]) == -1:
ma = max(ma, a[i])
mi = min(mi, a[i])
m0 = 0
for i in range(1, n):
if min(a[i], a[i + 1]) > -1:
m0 = max(m0, abs(a[i] - a[i + 1]))
m, k = 0, 0
if mi <= ma:
m = max((ma - mi + 1) // 2, m0)
k = mi + (ma - mi + 1) // 2
ans.append(" ".join(map(str, (m, k))))
sys.stdout.write("\n".join(ans)) | py |
1305 | E | E. Kuroni and the Score Distributiontime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputKuroni is the coordinator of the next Mathforces round written by the "Proof by AC" team. All the preparation has been done; and he is discussing with the team about the score distribution for the round.The round consists of nn problems, numbered from 11 to nn. The problems are ordered in increasing order of difficulty, no two problems have the same difficulty. A score distribution for the round can be denoted by an array a1,a2,…,ana1,a2,…,an, where aiai is the score of ii-th problem. Kuroni thinks that the score distribution should satisfy the following requirements: The score of each problem should be a positive integer not exceeding 109109. A harder problem should grant a strictly higher score than an easier problem. In other words, 1≤a1<a2<⋯<an≤1091≤a1<a2<⋯<an≤109. The balance of the score distribution, defined as the number of triples (i,j,k)(i,j,k) such that 1≤i<j<k≤n1≤i<j<k≤n and ai+aj=akai+aj=ak, should be exactly mm. Help the team find a score distribution that satisfies Kuroni's requirement. In case such a score distribution does not exist, output −1−1.InputThe first and single line contains two integers nn and mm (1≤n≤50001≤n≤5000, 0≤m≤1090≤m≤109) — the number of problems and the required balance.OutputIf there is no solution, print a single integer −1−1.Otherwise, print a line containing nn integers a1,a2,…,ana1,a2,…,an, representing a score distribution that satisfies all the requirements. If there are multiple answers, print any of them.ExamplesInputCopy5 3
OutputCopy4 5 9 13 18InputCopy8 0
OutputCopy10 11 12 13 14 15 16 17
InputCopy4 10
OutputCopy-1
NoteIn the first example; there are 33 triples (i,j,k)(i,j,k) that contribute to the balance of the score distribution. (1,2,3)(1,2,3) (1,3,4)(1,3,4) (2,4,5)(2,4,5) | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | n,m=map(int,input().split())
ans=[1]
if n==1 and m==0:
print(*ans)
exit()
elif n==2 and m==0:
ans.append(2)
print(*ans)
exit()
suma=0
ans=[1,2]
for i in range(3,n+1):
if suma+(i-1)//2<m:
ans.append(i)
suma+=(i-1)//2
else:
ans.append(i-1+(i-2*(m-suma)))
suma=m
break
if suma<m:
print(-1)
exit()
ceil=10**9
maxa=max(ans)
while len(ans)<n:
ans.append(ceil)
ceil-=maxa+1
ans=sorted(ans)
print(*ans) | py |
1296 | A | A. Array with Odd Sumtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given an array aa consisting of nn integers.In one move, you can choose two indices 1≤i,j≤n1≤i,j≤n such that i≠ji≠j and set ai:=ajai:=aj. You can perform such moves any number of times (possibly, zero). You can choose different indices in different operations. The operation := is the operation of assignment (i.e. you choose ii and jj and replace aiai with ajaj).Your task is to say if it is possible to obtain an array with an odd (not divisible by 22) sum of elements.You have to answer tt independent test cases.InputThe first line of the input contains one integer tt (1≤t≤20001≤t≤2000) — the number of test cases.The next 2t2t lines describe test cases. The first line of the test case contains one integer nn (1≤n≤20001≤n≤2000) — the number of elements in aa. The second line of the test case contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤20001≤ai≤2000), where aiai is the ii-th element of aa.It is guaranteed that the sum of nn over all test cases does not exceed 20002000 (∑n≤2000∑n≤2000).OutputFor each test case, print the answer on it — "YES" (without quotes) if it is possible to obtain the array with an odd sum of elements, and "NO" otherwise.ExampleInputCopy5
2
2 3
4
2 2 8 8
3
3 3 3
4
5 5 5 5
4
1 1 1 1
OutputCopyYES
NO
YES
NO
NO
| [
"math"
] | for i in range(int(input())):
Number=int(input())
listNum=list(map(int,input().split()))
count=sum(listNum)
odd=0
even=0
for j in range(Number):
if(listNum[j]%2==1):
odd+=listNum[j]
else:
even+=listNum[j]
if(count%2==1):
print('YES')
else:
if(odd!=0 and even!=0):
print('YES')
else:
print('NO')
| py |
1325 | E | E. Ehab's REAL Number Theory Problemtime limit per test3 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given an array aa of length nn that has a special condition: every element in this array has at most 7 divisors. Find the length of the shortest non-empty subsequence of this array product of whose elements is a perfect square.A sequence aa is a subsequence of an array bb if aa can be obtained from bb by deletion of several (possibly, zero or all) elements.InputThe first line contains an integer nn (1≤n≤1051≤n≤105) — the length of aa.The second line contains nn integers a1a1, a2a2, ……, anan (1≤ai≤1061≤ai≤106) — the elements of the array aa.OutputOutput the length of the shortest non-empty subsequence of aa product of whose elements is a perfect square. If there are several shortest subsequences, you can find any of them. If there's no such subsequence, print "-1".ExamplesInputCopy3
1 4 6
OutputCopy1InputCopy4
2 3 6 6
OutputCopy2InputCopy3
6 15 10
OutputCopy3InputCopy4
2 3 5 7
OutputCopy-1NoteIn the first sample; you can choose a subsequence [1][1].In the second sample, you can choose a subsequence [6,6][6,6].In the third sample, you can choose a subsequence [6,15,10][6,15,10].In the fourth sample, there is no such subsequence. | [
"brute force",
"dfs and similar",
"graphs",
"number theory",
"shortest paths"
] | import sys
MAX = 1_000_005
lp = [0] * MAX
pr = []
pid = {1: 0}
for i in range(2, MAX):
if not lp[i]:
lp[i] = i
pr.append(i)
pid[i] = len(pr)
for p in pr:
if p > lp[i] or i * p >= MAX:
break
lp[i * p] = p
n = int(input())
a = list(map(int, input().split()))
g = [[] for _ in range(len(pid))]
e = set()
for x in a:
f = []
while x > 1:
p, c = lp[x], 0
while lp[x] == p:
x //= p
c ^= 1
if c:
f.append(p)
if not f:
print(1)
sys.exit(0)
f += [1] * (2 - len(f))
u, v = pid[f[0]], pid[f[1]]
if (u, v) in e or (v, u) in e:
print ('hahahah')
print(2)
sys.exit(0)
g[u].append(v)
g[v].append(u)
def bfs(s):
d = [-1] * len(pid)
d[s] = 0
q = [(s, -1)]
while q:
q2 = []
for u, p in q:
for v in g[u]:
if d[v] != -1:
if v != p:
return d[u] + d[v] + 1
else:
d[v] = d[u] + 1
q2.append((v, u))
q = q2
ans = n + 1
for i in range(len(pid)):
if i > 0 and pr[i - 1] ** 2 >= MAX:
break
ans = min(ans, bfs(i) or ans)
print(ans if ans <= n else -1)
| py |
1296 | C | C. Yet Another Walking Robottime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputThere is a robot on a coordinate plane. Initially; the robot is located at the point (0,0)(0,0). Its path is described as a string ss of length nn consisting of characters 'L', 'R', 'U', 'D'.Each of these characters corresponds to some move: 'L' (left): means that the robot moves from the point (x,y)(x,y) to the point (x−1,y)(x−1,y); 'R' (right): means that the robot moves from the point (x,y)(x,y) to the point (x+1,y)(x+1,y); 'U' (up): means that the robot moves from the point (x,y)(x,y) to the point (x,y+1)(x,y+1); 'D' (down): means that the robot moves from the point (x,y)(x,y) to the point (x,y−1)(x,y−1). The company that created this robot asked you to optimize the path of the robot somehow. To do this, you can remove any non-empty substring of the path. But this company doesn't want their customers to notice the change in the robot behavior. It means that if before the optimization the robot ended its path at the point (xe,ye)(xe,ye), then after optimization (i.e. removing some single substring from ss) the robot also ends its path at the point (xe,ye)(xe,ye).This optimization is a low-budget project so you need to remove the shortest possible non-empty substring to optimize the robot's path such that the endpoint of his path doesn't change. It is possible that you can't optimize the path. Also, it is possible that after the optimization the target path is an empty string (i.e. deleted substring is the whole string ss).Recall that the substring of ss is such string that can be obtained from ss by removing some amount of characters (possibly, zero) from the prefix and some amount of characters (possibly, zero) from the suffix. For example, the substrings of "LURLLR" are "LU", "LR", "LURLLR", "URL", but not "RR" and "UL".You have to answer tt independent test cases.InputThe first line of the input contains one integer tt (1≤t≤10001≤t≤1000) — the number of test cases.The next 2t2t lines describe test cases. Each test case is given on two lines. The first line of the test case contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the length of the robot's path. The second line of the test case contains one string ss consisting of nn characters 'L', 'R', 'U', 'D' — the robot's path.It is guaranteed that the sum of nn over all test cases does not exceed 2⋅1052⋅105 (∑n≤2⋅105∑n≤2⋅105).OutputFor each test case, print the answer on it. If you cannot remove such non-empty substring that the endpoint of the robot's path doesn't change, print -1. Otherwise, print two integers ll and rr such that 1≤l≤r≤n1≤l≤r≤n — endpoints of the substring you remove. The value r−l+1r−l+1 should be minimum possible. If there are several answers, print any of them.ExampleInputCopy4
4
LRUD
4
LURD
5
RRUDU
5
LLDDR
OutputCopy1 2
1 4
3 4
-1
| [
"data structures",
"implementation"
] | import sys
from math import sqrt, gcd, factorial, ceil, floor, pi, inf
from collections import deque, Counter, OrderedDict, defaultdict
from heapq import heapify, heappush, heappop
#sys.setrecursionlimit(10**5)
from functools import lru_cache
#@lru_cache(None)
#======================================================#
input = lambda: sys.stdin.readline()
I = lambda: int(input().strip())
S = lambda: input().strip()
M = lambda: map(int,input().strip().split())
L = lambda: list(map(int,input().strip().split()))
#======================================================#
#======================================================#
def primelist():
L = [False for i in range((10**10)+1)]
primes = [False for i in range((10**10)+1)]
for i in range(2,(10**10)+1):
if not L[i]:
primes[i]=True
for j in range(i,(10**10)+1,i):
L[j]=True
return primes
def isPrime(n):
p = primelist()
return p[n]
#======================================================#
def bst(arr,x):
low,high = 0,len(arr)-1
ans = -1
while low<=high:
mid = (low+high)//2
#if arr[mid]==x:
# return mid
if arr[mid]<=x:
low = mid+1
else:
high = mid-1
ans = mid
return ans
def factors(x):
l1 = []
l2 = []
for i in range(1,int(sqrt(x))+1):
if x%i==0:
if (i*i)==x:
l1.append(i)
else:
l1.append(i)
l2.append(x//i)
for i in range(len(l2)//2):
l2[i],l2[len(l2)-1-i]=l2[len(l2)-1-i],l2[i]
return l1+l2
#======================================================#
def power(n,x):
if x==0:
return 1
k = (10**9)+7
if n==1:
return 1
if x==1:
return n
ans = power(n,x//2)
if x%2==0:
return (ans*ans)%k
return (ans*ans*n)%k
#======================================================#
for _ in range(I()):
n = I()
s = S()
d = {(0,0):-1}
ans = -1
res = inf
l,r,u,b = 0,0,0,0
for i in range(n):
if s[i]=='L':
l+=1
elif s[i]=='R':
r+=1
elif s[i]=='U':
u+=1
else:
b+=1
if (l-r,u-b) in d:
if (i-d[(l-r,u-b)])<res:
ans = i
res = i-d[(l-r,u-b)]
d[(l-r,u-b)]=i
if ans==-1:
print(-1)
else:
print(ans-res+2,ans+1)
| py |
1313 | C2 | C2. Skyscrapers (hard version)time limit per test3 secondsmemory limit per test512 megabytesinputstandard inputoutputstandard outputThis is a harder version of the problem. In this version n≤500000n≤500000The outskirts of the capital are being actively built up in Berland. The company "Kernel Panic" manages the construction of a residential complex of skyscrapers in New Berlskva. All skyscrapers are built along the highway. It is known that the company has already bought nn plots along the highway and is preparing to build nn skyscrapers, one skyscraper per plot.Architects must consider several requirements when planning a skyscraper. Firstly, since the land on each plot has different properties, each skyscraper has a limit on the largest number of floors it can have. Secondly, according to the design code of the city, it is unacceptable for a skyscraper to simultaneously have higher skyscrapers both to the left and to the right of it.Formally, let's number the plots from 11 to nn. Then if the skyscraper on the ii-th plot has aiai floors, it must hold that aiai is at most mimi (1≤ai≤mi1≤ai≤mi). Also there mustn't be integers jj and kk such that j<i<kj<i<k and aj>ai<akaj>ai<ak. Plots jj and kk are not required to be adjacent to ii.The company wants the total number of floors in the built skyscrapers to be as large as possible. Help it to choose the number of floors for each skyscraper in an optimal way, i.e. in such a way that all requirements are fulfilled, and among all such construction plans choose any plan with the maximum possible total number of floors.InputThe first line contains a single integer nn (1≤n≤5000001≤n≤500000) — the number of plots.The second line contains the integers m1,m2,…,mnm1,m2,…,mn (1≤mi≤1091≤mi≤109) — the limit on the number of floors for every possible number of floors for a skyscraper on each plot.OutputPrint nn integers aiai — the number of floors in the plan for each skyscraper, such that all requirements are met, and the total number of floors in all skyscrapers is the maximum possible.If there are multiple answers possible, print any of them.ExamplesInputCopy51 2 3 2 1OutputCopy1 2 3 2 1 InputCopy310 6 8OutputCopy10 6 6 NoteIn the first example, you can build all skyscrapers with the highest possible height.In the second test example, you cannot give the maximum height to all skyscrapers as this violates the design code restriction. The answer [10,6,6][10,6,6] is optimal. Note that the answer of [6,6,8][6,6,8] also satisfies all restrictions, but is not optimal. | [
"data structures",
"dp",
"greedy"
] | # Author Name: Ajay Meena
# Codeforce : https://codeforces.com/profile/majay1638
import sys
import math
import bisect
import heapq
from bisect import bisect_right
from sys import stdin, stdout
from collections import deque
# -------------- INPUT FUNCTIONS ------------------
def get_ints_in_variables(): return map(
int, sys.stdin.readline().strip().split())
def get_int(): return int(sys.stdin.readline())
def get_ints_in_list(): return list(
map(int, sys.stdin.readline().strip().split()))
def get_list_of_list(n): return [list(
map(int, sys.stdin.readline().strip().split())) for _ in range(n)]
def get_string(): return sys.stdin.readline().strip()
# -------- SOME CUSTOMIZED FUNCTIONS-----------
def myceil(x, y): return (x + y - 1) // y
# -------------- SOLUTION FUNCTION ------------------
def Solution():
# Write Your Code Here
pass
def helper(arr, n):
l = [0 for _ in range(n)]
stk = deque()
for i in range(n):
while len(stk) and arr[stk[-1]] >= arr[i]:
stk.pop()
if len(stk) == 0:
l[i] = (i+1)*arr[i]
else:
j = stk[-1]
l[i] = (((i-j))*arr[i])+l[j]
stk.append(i)
return l
def main():
# Take input Here and Call solution function
n = get_int()
arr = get_ints_in_list()
if n == 1:
print(arr[0])
return
# ans = [0 for _ in range(n)]
l = helper(arr, n)
r = list(reversed(helper(list(reversed(arr)), n)))
ind = max([_ for _ in range(n)], key=lambda i: l[i]+r[i]-arr[i])
for i in range(ind-1, -1, -1):
arr[i] = min(arr[i], arr[i+1])
for i in range(ind+1, n):
arr[i] = min(arr[i], arr[i-1])
print(*arr)
# calling main Function
if __name__ == '__main__':
main()
| py |
1307 | B | B. Cow and Friendtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputBessie has way too many friends because she is everyone's favorite cow! Her new friend Rabbit is trying to hop over so they can play! More specifically; he wants to get from (0,0)(0,0) to (x,0)(x,0) by making multiple hops. He is only willing to hop from one point to another point on the 2D plane if the Euclidean distance between the endpoints of a hop is one of its nn favorite numbers: a1,a2,…,ana1,a2,…,an. What is the minimum number of hops Rabbit needs to get from (0,0)(0,0) to (x,0)(x,0)? Rabbit may land on points with non-integer coordinates. It can be proved that Rabbit can always reach his destination.Recall that the Euclidean distance between points (xi,yi)(xi,yi) and (xj,yj)(xj,yj) is (xi−xj)2+(yi−yj)2−−−−−−−−−−−−−−−−−−√(xi−xj)2+(yi−yj)2.For example, if Rabbit has favorite numbers 11 and 33 he could hop from (0,0)(0,0) to (4,0)(4,0) in two hops as shown below. Note that there also exists other valid ways to hop to (4,0)(4,0) in 22 hops (e.g. (0,0)(0,0) →→ (2,−5–√)(2,−5) →→ (4,0)(4,0)). Here is a graphic for the first example. Both hops have distance 33, one of Rabbit's favorite numbers. In other words, each time Rabbit chooses some number aiai and hops with distance equal to aiai in any direction he wants. The same number can be used multiple times.InputThe input consists of multiple test cases. The first line contains an integer tt (1≤t≤10001≤t≤1000) — the number of test cases. Next 2t2t lines contain test cases — two lines per test case.The first line of each test case contains two integers nn and xx (1≤n≤1051≤n≤105, 1≤x≤1091≤x≤109) — the number of favorite numbers and the distance Rabbit wants to travel, respectively.The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) — Rabbit's favorite numbers. It is guaranteed that the favorite numbers are distinct.It is guaranteed that the sum of nn over all the test cases will not exceed 105105.OutputFor each test case, print a single integer — the minimum number of hops needed.ExampleInputCopy42 41 33 123 4 51 552 1015 4OutputCopy2
3
1
2
NoteThe first test case of the sample is shown in the picture above. Rabbit can hop to (2,5–√)(2,5), then to (4,0)(4,0) for a total of two hops. Each hop has a distance of 33, which is one of his favorite numbers.In the second test case of the sample, one way for Rabbit to hop 33 times is: (0,0)(0,0) →→ (4,0)(4,0) →→ (8,0)(8,0) →→ (12,0)(12,0).In the third test case of the sample, Rabbit can hop from (0,0)(0,0) to (5,0)(5,0).In the fourth test case of the sample, Rabbit can hop: (0,0)(0,0) →→ (5,102–√)(5,102) →→ (10,0)(10,0). | [
"geometry",
"greedy",
"math"
] | import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
t = int(input())
ans = []
for _ in range(t):
n, x = map(int, input().split())
a = list(map(int, input().split()))
a.sort(reverse = True)
s = set(a)
if x in s:
ans0 = 1
elif a[0] > x:
ans0 = 2
else:
ans0 = x // a[0] + min(x % a[0], 1)
ans.append(ans0)
sys.stdout.write("\n".join(map(str, ans))) | py |
1295 | A | A. Display The Numbertime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou have a large electronic screen which can display up to 998244353998244353 decimal digits. The digits are displayed in the same way as on different electronic alarm clocks: each place for a digit consists of 77 segments which can be turned on and off to compose different digits. The following picture describes how you can display all 1010 decimal digits:As you can see, different digits may require different number of segments to be turned on. For example, if you want to display 11, you have to turn on 22 segments of the screen, and if you want to display 88, all 77 segments of some place to display a digit should be turned on.You want to display a really large integer on the screen. Unfortunately, the screen is bugged: no more than nn segments can be turned on simultaneously. So now you wonder what is the greatest integer that can be displayed by turning on no more than nn segments.Your program should be able to process tt different test cases.InputThe first line contains one integer tt (1≤t≤1001≤t≤100) — the number of test cases in the input.Then the test cases follow, each of them is represented by a separate line containing one integer nn (2≤n≤1052≤n≤105) — the maximum number of segments that can be turned on in the corresponding testcase.It is guaranteed that the sum of nn over all test cases in the input does not exceed 105105.OutputFor each test case, print the greatest integer that can be displayed by turning on no more than nn segments of the screen. Note that the answer may not fit in the standard 3232-bit or 6464-bit integral data type.ExampleInputCopy2
3
4
OutputCopy7
11
| [
"greedy"
] | for _ in range(int(input())):
n = int(input())
if n <= 3:
if n == 3:
print(7)
else:
print(1)
else:
if n % 2 == 0:
print('1' * (n // 2))
else:
print('7' + ('1' * ((n - 3) // 2))) | py |
1141 | F1 | F1. Same Sum Blocks (Easy)time limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThis problem is given in two editions; which differ exclusively in the constraints on the number nn.You are given an array of integers a[1],a[2],…,a[n].a[1],a[2],…,a[n]. A block is a sequence of contiguous (consecutive) elements a[l],a[l+1],…,a[r]a[l],a[l+1],…,a[r] (1≤l≤r≤n1≤l≤r≤n). Thus, a block is defined by a pair of indices (l,r)(l,r).Find a set of blocks (l1,r1),(l2,r2),…,(lk,rk)(l1,r1),(l2,r2),…,(lk,rk) such that: They do not intersect (i.e. they are disjoint). Formally, for each pair of blocks (li,ri)(li,ri) and (lj,rj(lj,rj) where i≠ji≠j either ri<ljri<lj or rj<lirj<li. For each block the sum of its elements is the same. Formally, a[l1]+a[l1+1]+⋯+a[r1]=a[l2]+a[l2+1]+⋯+a[r2]=a[l1]+a[l1+1]+⋯+a[r1]=a[l2]+a[l2+1]+⋯+a[r2]= ⋯=⋯= a[lk]+a[lk+1]+⋯+a[rk].a[lk]+a[lk+1]+⋯+a[rk]. The number of the blocks in the set is maximum. Formally, there does not exist a set of blocks (l′1,r′1),(l′2,r′2),…,(l′k′,r′k′)(l1′,r1′),(l2′,r2′),…,(lk′′,rk′′) satisfying the above two requirements with k′>kk′>k. The picture corresponds to the first example. Blue boxes illustrate blocks. Write a program to find such a set of blocks.InputThe first line contains integer nn (1≤n≤501≤n≤50) — the length of the given array. The second line contains the sequence of elements a[1],a[2],…,a[n]a[1],a[2],…,a[n] (−105≤ai≤105−105≤ai≤105).OutputIn the first line print the integer kk (1≤k≤n1≤k≤n). The following kk lines should contain blocks, one per line. In each line print a pair of indices li,rili,ri (1≤li≤ri≤n1≤li≤ri≤n) — the bounds of the ii-th block. You can print blocks in any order. If there are multiple answers, print any of them.ExamplesInputCopy7
4 1 2 2 1 5 3
OutputCopy3
7 7
2 3
4 5
InputCopy11
-5 -4 -3 -2 -1 0 1 2 3 4 5
OutputCopy2
3 4
1 1
InputCopy4
1 1 1 1
OutputCopy4
4 4
1 1
2 2
3 3
| [
"greedy"
] | from bisect import *
from collections import *
import sys
import io, os
import math
import random
from heapq import *
gcd = math.gcd
sqrt = math.sqrt
maxint=10**21
def ceil(a, b):
a = -a
k = a // b
k = -k
return k
# arr=list(map(int, input().split()))
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def strinp(testcases):
k = 5
if (testcases == -1 or testcases == 1):
k = 1
f = str(input())
f = f[2:len(f) - k]
return f
def main():
n=int(input())
arr=list(map(int, input().split()))
pref=[0]*n
prev=0
for i in range(n):
pref[i]=prev+arr[i]
prev=pref[i]
dic={}
for i in range(n):
for j in range(i,n):
a=0
if(i!=0):
a=pref[i-1]
if((pref[j]-a) in dic):
dic[pref[j]-a].append([j+1,i+1])
else:
dic[pref[j]-a]=[[j+1,i+1]]
ans=0
curr=-1
for i in dic:
dic[i].sort()
lis=dic[i]
temp=[lis[0]]
k=len(lis)
ed=lis[0][0]
for j in range(1,k):
if(lis[j][1]>ed):
ed=lis[j][0]
temp.append(lis[j])
if(len(temp)>ans):
ans=len(temp)
curr=temp
print(ans)
for i in range(ans):
curr[i].reverse()
print(*curr[i])
main()
| py |
1296 | C | C. Yet Another Walking Robottime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputThere is a robot on a coordinate plane. Initially; the robot is located at the point (0,0)(0,0). Its path is described as a string ss of length nn consisting of characters 'L', 'R', 'U', 'D'.Each of these characters corresponds to some move: 'L' (left): means that the robot moves from the point (x,y)(x,y) to the point (x−1,y)(x−1,y); 'R' (right): means that the robot moves from the point (x,y)(x,y) to the point (x+1,y)(x+1,y); 'U' (up): means that the robot moves from the point (x,y)(x,y) to the point (x,y+1)(x,y+1); 'D' (down): means that the robot moves from the point (x,y)(x,y) to the point (x,y−1)(x,y−1). The company that created this robot asked you to optimize the path of the robot somehow. To do this, you can remove any non-empty substring of the path. But this company doesn't want their customers to notice the change in the robot behavior. It means that if before the optimization the robot ended its path at the point (xe,ye)(xe,ye), then after optimization (i.e. removing some single substring from ss) the robot also ends its path at the point (xe,ye)(xe,ye).This optimization is a low-budget project so you need to remove the shortest possible non-empty substring to optimize the robot's path such that the endpoint of his path doesn't change. It is possible that you can't optimize the path. Also, it is possible that after the optimization the target path is an empty string (i.e. deleted substring is the whole string ss).Recall that the substring of ss is such string that can be obtained from ss by removing some amount of characters (possibly, zero) from the prefix and some amount of characters (possibly, zero) from the suffix. For example, the substrings of "LURLLR" are "LU", "LR", "LURLLR", "URL", but not "RR" and "UL".You have to answer tt independent test cases.InputThe first line of the input contains one integer tt (1≤t≤10001≤t≤1000) — the number of test cases.The next 2t2t lines describe test cases. Each test case is given on two lines. The first line of the test case contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the length of the robot's path. The second line of the test case contains one string ss consisting of nn characters 'L', 'R', 'U', 'D' — the robot's path.It is guaranteed that the sum of nn over all test cases does not exceed 2⋅1052⋅105 (∑n≤2⋅105∑n≤2⋅105).OutputFor each test case, print the answer on it. If you cannot remove such non-empty substring that the endpoint of the robot's path doesn't change, print -1. Otherwise, print two integers ll and rr such that 1≤l≤r≤n1≤l≤r≤n — endpoints of the substring you remove. The value r−l+1r−l+1 should be minimum possible. If there are several answers, print any of them.ExampleInputCopy4
4
LRUD
4
LURD
5
RRUDU
5
LLDDR
OutputCopy1 2
1 4
3 4
-1
| [
"data structures",
"implementation"
] | import sys;sc = sys.stdin.readline;out=sys.stdout.write
for _ in range(int(sc())):
n=int(sc());s=str(input());p=[0,0];d={(0,0):1};r,l,ind=n,0,1
for e in range(n):
if s[e]=='L':p[0]-=1
elif s[e]=='R':p[0]+=1
elif s[e]=='U':p[1]+=1
else:p[1]-=1
p=tuple(p)
if p in d and r-l>ind-d[p]:l,r=d[p],ind
ind+=1;d[p]=ind;p=list(p)
if l: out(str(l)+str(" ")+str(r)+"\n")
else: out("-1\n") | py |
1299 | C | C. Water Balancetime limit per test3 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThere are nn water tanks in a row, ii-th of them contains aiai liters of water. The tanks are numbered from 11 to nn from left to right.You can perform the following operation: choose some subsegment [l,r][l,r] (1≤l≤r≤n1≤l≤r≤n), and redistribute water in tanks l,l+1,…,rl,l+1,…,r evenly. In other words, replace each of al,al+1,…,aral,al+1,…,ar by al+al+1+⋯+arr−l+1al+al+1+⋯+arr−l+1. For example, if for volumes [1,3,6,7][1,3,6,7] you choose l=2,r=3l=2,r=3, new volumes of water will be [1,4.5,4.5,7][1,4.5,4.5,7]. You can perform this operation any number of times.What is the lexicographically smallest sequence of volumes of water that you can achieve?As a reminder:A sequence aa is lexicographically smaller than a sequence bb of the same length if and only if the following holds: in the first (leftmost) position where aa and bb differ, the sequence aa has a smaller element than the corresponding element in bb.InputThe first line contains an integer nn (1≤n≤1061≤n≤106) — the number of water tanks.The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1061≤ai≤106) — initial volumes of water in the water tanks, in liters.Because of large input, reading input as doubles is not recommended.OutputPrint the lexicographically smallest sequence you can get. In the ii-th line print the final volume of water in the ii-th tank.Your answer is considered correct if the absolute or relative error of each aiai does not exceed 10−910−9.Formally, let your answer be a1,a2,…,ana1,a2,…,an, and the jury's answer be b1,b2,…,bnb1,b2,…,bn. Your answer is accepted if and only if |ai−bi|max(1,|bi|)≤10−9|ai−bi|max(1,|bi|)≤10−9 for each ii.ExamplesInputCopy4
7 5 5 7
OutputCopy5.666666667
5.666666667
5.666666667
7.000000000
InputCopy5
7 8 8 10 12
OutputCopy7.000000000
8.000000000
8.000000000
10.000000000
12.000000000
InputCopy10
3 9 5 5 1 7 5 3 8 7
OutputCopy3.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
7.500000000
7.500000000
NoteIn the first sample; you can get the sequence by applying the operation for subsegment [1,3][1,3].In the second sample, you can't get any lexicographically smaller sequence. | [
"data structures",
"geometry",
"greedy"
] | import sys
input = sys.stdin.buffer.readline
n = int(input())
a = list(map(int, input().split()))
stk = []
for x in a:
s = x
c = 1
while stk and stk[-1][0] * c >= s * stk[-1][1]:
s += stk[-1][0]
c += stk[-1][1]
stk.pop()
stk.append((s, c))
sys.stdout.write(''.join('{}\n'.format(s / c) * c for s, c in stk))
| py |
1313 | C1 | C1. Skyscrapers (easy version)time limit per test1 secondmemory limit per test512 megabytesinputstandard inputoutputstandard outputThis is an easier version of the problem. In this version n≤1000n≤1000The outskirts of the capital are being actively built up in Berland. The company "Kernel Panic" manages the construction of a residential complex of skyscrapers in New Berlskva. All skyscrapers are built along the highway. It is known that the company has already bought nn plots along the highway and is preparing to build nn skyscrapers, one skyscraper per plot.Architects must consider several requirements when planning a skyscraper. Firstly, since the land on each plot has different properties, each skyscraper has a limit on the largest number of floors it can have. Secondly, according to the design code of the city, it is unacceptable for a skyscraper to simultaneously have higher skyscrapers both to the left and to the right of it.Formally, let's number the plots from 11 to nn. Then if the skyscraper on the ii-th plot has aiai floors, it must hold that aiai is at most mimi (1≤ai≤mi1≤ai≤mi). Also there mustn't be integers jj and kk such that j<i<kj<i<k and aj>ai<akaj>ai<ak. Plots jj and kk are not required to be adjacent to ii.The company wants the total number of floors in the built skyscrapers to be as large as possible. Help it to choose the number of floors for each skyscraper in an optimal way, i.e. in such a way that all requirements are fulfilled, and among all such construction plans choose any plan with the maximum possible total number of floors.InputThe first line contains a single integer nn (1≤n≤10001≤n≤1000) — the number of plots.The second line contains the integers m1,m2,…,mnm1,m2,…,mn (1≤mi≤1091≤mi≤109) — the limit on the number of floors for every possible number of floors for a skyscraper on each plot.OutputPrint nn integers aiai — the number of floors in the plan for each skyscraper, such that all requirements are met, and the total number of floors in all skyscrapers is the maximum possible.If there are multiple answers possible, print any of them.ExamplesInputCopy51 2 3 2 1OutputCopy1 2 3 2 1 InputCopy310 6 8OutputCopy10 6 6 NoteIn the first example, you can build all skyscrapers with the highest possible height.In the second test example, you cannot give the maximum height to all skyscrapers as this violates the design code restriction. The answer [10,6,6][10,6,6] is optimal. Note that the answer of [6,6,8][6,6,8] also satisfies all restrictions, but is not optimal. | [
"brute force",
"data structures",
"dp",
"greedy"
] | import re
import functools
import random
import sys
import os
import math
from collections import Counter, defaultdict, deque
from functools import lru_cache, reduce
from itertools import accumulate, combinations, permutations
from heapq import nsmallest, nlargest, heappushpop, heapify, heappop, heappush
from io import BytesIO, IOBase
from copy import deepcopy
import threading
from typing import *
from operator import add, xor, mul, ior, iand, itemgetter
import bisect
BUFSIZE = 4096
inf = float('inf')
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin = IOWrapper(sys.stdin)
sys.stdout = IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def I():
return input()
def II():
return int(input())
def MII():
return map(int, input().split())
def LI():
return list(input().split())
def LII():
return list(map(int, input().split()))
def GMI():
return map(lambda x: int(x) - 1, input().split())
def LGMI():
return list(map(lambda x: int(x) - 1, input().split()))
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
n = II()
nums = LII()
idx = -1
res = 0
for i in range(n):
ans = nums[i]
tmp = nums[i]
for j in range(i-1, -1, -1):
tmp = min(tmp, nums[j])
ans += tmp
tmp = nums[i]
for j in range(i+1, n):
tmp = min(tmp, nums[j])
ans += tmp
if ans > res: res = ans; idx = i
final = [0] * n
final[idx] = nums[idx]
tmp = nums[idx]
for j in range(idx-1, -1, -1):
tmp = min(tmp, nums[j])
final[j] = tmp
tmp = nums[idx]
for j in range(idx+1, n):
tmp = min(tmp, nums[j])
final[j] = tmp
print(*final) | py |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.