contest_id
stringclasses 33
values | problem_id
stringclasses 14
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stringclasses 181
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sequencelengths 1
8
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stringlengths 21
64.5k
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stringclasses 3
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|---|---|---|---|---|---|
1311 | F | F. Moving Pointstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThere are nn points on a coordinate axis OXOX. The ii-th point is located at the integer point xixi and has a speed vivi. It is guaranteed that no two points occupy the same coordinate. All nn points move with the constant speed, the coordinate of the ii-th point at the moment tt (tt can be non-integer) is calculated as xi+t⋅vixi+t⋅vi.Consider two points ii and jj. Let d(i,j)d(i,j) be the minimum possible distance between these two points over any possible moments of time (even non-integer). It means that if two points ii and jj coincide at some moment, the value d(i,j)d(i,j) will be 00.Your task is to calculate the value ∑1≤i<j≤n∑1≤i<j≤n d(i,j)d(i,j) (the sum of minimum distances over all pairs of points).InputThe first line of the input contains one integer nn (2≤n≤2⋅1052≤n≤2⋅105) — the number of points.The second line of the input contains nn integers x1,x2,…,xnx1,x2,…,xn (1≤xi≤1081≤xi≤108), where xixi is the initial coordinate of the ii-th point. It is guaranteed that all xixi are distinct.The third line of the input contains nn integers v1,v2,…,vnv1,v2,…,vn (−108≤vi≤108−108≤vi≤108), where vivi is the speed of the ii-th point.OutputPrint one integer — the value ∑1≤i<j≤n∑1≤i<j≤n d(i,j)d(i,j) (the sum of minimum distances over all pairs of points).ExamplesInputCopy3
1 3 2
-100 2 3
OutputCopy3
InputCopy5
2 1 4 3 5
2 2 2 3 4
OutputCopy19
InputCopy2
2 1
-3 0
OutputCopy0
| [
"data structures",
"divide and conquer",
"implementation",
"sortings"
] | class SegmentTree():
def __init__(self,arr,func,initialRes=0):
self.f=func
self.N=len(arr)
self.tree=[0 for _ in range(4*self.N)]
self.initialRes=initialRes
for i in range(self.N):
self.tree[self.N+i]=arr[i]
for i in range(self.N-1,0,-1):
self.tree[i]=self.f(self.tree[i<<1],self.tree[i<<1|1])
def updateTreeNode(self,idx,value): #update value at arr[idx]
self.tree[idx+self.N]=value
idx+=self.N
i=idx
while i>1:
self.tree[i>>1]=self.f(self.tree[i],self.tree[i^1])
i>>=1
def query(self,l,r): #get sum (or whatever function) on interval [l,r] inclusive
r+=1
res=self.initialRes
l+=self.N
r+=self.N
while l<r:
if l&1:
res=self.f(res,self.tree[l])
l+=1
if r&1:
r-=1
res=self.f(res,self.tree[r])
l>>=1
r>>=1
return res
def getMaxSegTree(arr):
return SegmentTree(arr,lambda a,b:max(a,b),initialRes=-float('inf'))
def getMinSegTree(arr):
return SegmentTree(arr,lambda a,b:min(a,b),initialRes=float('inf'))
def getSumSegTree(arr):
return SegmentTree(arr,lambda a,b:a+b,initialRes=0)
def main():
n=int(input())
xes=readIntArr()
ves=readIntArr()
#perform coordinate compression on xes. Compressed values shall be the indexes of segment tree
xes2=list(sorted(xes)) #xes2[compressed]=x
xTox2Map=dict()
for i,x in enumerate(xes2):
xTox2Map[x]=i #xTox2Map[x]=i
arr=[] #[original x, compressed x, v]
for i in range(n):
arr.append([xes[i],xTox2Map[xes[i]],ves[i]])
arr.sort(key=lambda x:(x[2],x[0])) #sort by v asc, then position asc
segArray=[0 for _ in range(n)] #index shall be compressed x, value is actual x
st=getSumSegTree(segArray) #for sum of xes
stCnts=getSumSegTree(segArray.copy()) #for counts
# print(arr)
ans=0
for originalX,compressedX,v in arr:
smallerSums=st.query(0,compressedX)
smallerCounts=stCnts.query(0,compressedX)
ans+=originalX*smallerCounts-smallerSums
st.updateTreeNode(compressedX,originalX) #update the tree
stCnts.updateTreeNode(compressedX,1)
# print(ans)
print(ans)
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()]
# def readFloatArr():
# return [float(x) for x in input().split()]
inf=float('inf')
MOD=10**9+7
main() | py |
1288 | D | D. Minimax Problemtime limit per test5 secondsmemory limit per test512 megabytesinputstandard inputoutputstandard outputYou are given nn arrays a1a1, a2a2, ..., anan; each array consists of exactly mm integers. We denote the yy-th element of the xx-th array as ax,yax,y.You have to choose two arrays aiai and ajaj (1≤i,j≤n1≤i,j≤n, it is possible that i=ji=j). After that, you will obtain a new array bb consisting of mm integers, such that for every k∈[1,m]k∈[1,m] bk=max(ai,k,aj,k)bk=max(ai,k,aj,k).Your goal is to choose ii and jj so that the value of mink=1mbkmink=1mbk is maximum possible.InputThe first line contains two integers nn and mm (1≤n≤3⋅1051≤n≤3⋅105, 1≤m≤81≤m≤8) — the number of arrays and the number of elements in each array, respectively.Then nn lines follow, the xx-th line contains the array axax represented by mm integers ax,1ax,1, ax,2ax,2, ..., ax,max,m (0≤ax,y≤1090≤ax,y≤109).OutputPrint two integers ii and jj (1≤i,j≤n1≤i,j≤n, it is possible that i=ji=j) — the indices of the two arrays you have to choose so that the value of mink=1mbkmink=1mbk is maximum possible. If there are multiple answers, print any of them.ExampleInputCopy6 5
5 0 3 1 2
1 8 9 1 3
1 2 3 4 5
9 1 0 3 7
2 3 0 6 3
6 4 1 7 0
OutputCopy1 5
| [
"binary search",
"bitmasks",
"dp"
] | import sys
# 1288D
def solve(a):
m, n = len(a), len(a[0])
maxx = max(max(t) for t in a)
def check(t):
idx = [-1] * 256
for i in range(m):
mask = 0
for j in range(n):
if a[i][j] >= t:
mask |= 1 << j
# idx: [5,-1,6,8,-1……]
idx[mask] = i
for i in range(1 << n):
if idx[i] == -1:
continue
for j in range(i, 1 << n):
if idx[j] == -1:
continue
if (i | j) == (1 << n) - 1:
return [idx[i], idx[j]]
return [-1, -1]
l, r = -1, maxx + 1
while l < r - 1:
mid = (l + r) // 2
if check(mid) != [-1, -1]:
l = mid
else:
r = mid
# [ TTTTTT FFFFFF FFFF]
# l r
return check(r - 1)
m, n = map(int, input().split())
a = []
for i in range(m):
a.append([int(i) for i in input().split()])
res = solve(a)
print(res[0] + 1, res[1] + 1)
# 1103B
# def query(x, y):
# print("?", x, y)
# sys.stdout.flush()
# res = input()
# return res
# def solve():
# # write function here
# return -1
# while True:
# res = solve()
# if res == -1: break
# print("!", res)
# def cutThemAll(lengths, minLength):
# for i in range(len(lengths) - 1):
# if lengths[i] + lengths[i+1] >= minLength:
# return "Possible"
# return "Impossible"
# ls = [3,5,6]
# print(cutThemAll(ls, 12))
# def getUniqueCharacter(s):
# cnt = [0] * 26
# for c in s:
# cnt[ord(c) - ord('a')] += 1
# for i in range(len(s)):
# if cnt[ord(s[i]) - ord('a')] == 1:
# return i + 1
# return -1
# print(getUniqueCharacter("madam"))
# from cmath import log
# from collections import Counter
# from dis import findlabels
# base = 10**9 + 7
# def findRight(a):
# # find the index of the first less one in the left side, for every a[i]
# n = len(a)
# st, res = [], [0] * n
# for i in range(n - 1, -1, -1):
# while len(st) and a[st[-1]] >= a[i]:
# st.pop()
# res[i] = n if len(st) == 0 else st[-1]
# st.append(i)
# return res
# def findLeft(a):
# # find the index of the first less or equal one in the right side, for every a[i]
# n = len(a)
# st, res = [], [0] * n
# for i in range(n):
# while len(st) and a[st[-1]] > a[i]:
# st.pop()
# res[i] = -1 if len(st) == 0 else st[-1]
# st.append(i)
# return res
# def findTotalPower(a):
# # In order to avoid repeated calculations,
# # the left side finds the first element less than a[i],
# # and the right side finds the first element less than or equal to a[i]
# left = findLeft(a)
# right = findRight(a)
# n, res = len(a), 0
# s = [0] * (n + 1)
# ss = [0] * (n + 2)
# # s means the prefix sum of a
# for i in range(n): s[i+1] = s[i] + a[i]
# # ss meams the prefix sum of s
# for i in range(n + 1): ss[i+1] = ss[i] + s[i]
# for i in range(n):
# # every left one could be the leader of subarray, which can be combined with the ending on the right
# l = i - left[i]
# # every right one could be the ending of subarray, which can be combined with the leader on the left
# r = right[i] - i
# # find the sum(sum([i, r]))
# now = ((ss[right[i] + 1] - ss[i + 1]) - r * s[i]) % base
# # find the sum(sum([i, r])) * a[i] * (i-left)
# res += ((now * l) % base * a[i]) % base
# # find the sum(sum([l, i]))
# now = (s[i] * (l - 1) - (ss[i] - ss[left[i] + 1])) % base
# # find the sum(sum([l, i])) * a[i] * (right-i)
# res += ((now * r) % base * a[i]) % base
# res %= base
# return res % base
# ls = [2,3,2,1]
# res = findTotalPower(ls)
# print(res)
# def solve(a):
# m, n = len(a), len(a[0])
# maxx = max(max(t) for t in a)
# def ok(t):
# idx = [-1] * 256
# maxm = 1 << n
# for i in range(m):
# mask = 0
# for j in range(n):
# if a[i][j] >= t:
# mask |= 1 << j
# if idx[mask] == -1:
# idx[mask] = i
# for i in range(maxm):
# if idx[i] == -1:
# continue
# for j in range(i, maxm):
# if idx[j] == -1:
# continue
# if (i | j) == maxm - 1:
# return [idx[i], idx[j]]
# return [-1, -1]
# l, r = 0, maxx
# while l <= r:
# mid = (l + r) // 2
# if ok(mid) != [-1, -1]:
# l = mid + 1
# else:
# r = mid - 1
# return ok(r) | py |
1312 | F | F. Attack on Red Kingdomtime limit per test3 secondsmemory limit per test512 megabytesinputstandard inputoutputstandard outputThe Red Kingdom is attacked by the White King and the Black King!The Kingdom is guarded by nn castles, the ii-th castle is defended by aiai soldiers. To conquer the Red Kingdom, the Kings have to eliminate all the defenders. Each day the White King launches an attack on one of the castles. Then, at night, the forces of the Black King attack a castle (possibly the same one). Then the White King attacks a castle, then the Black King, and so on. The first attack is performed by the White King.Each attack must target a castle with at least one alive defender in it. There are three types of attacks: a mixed attack decreases the number of defenders in the targeted castle by xx (or sets it to 00 if there are already less than xx defenders); an infantry attack decreases the number of defenders in the targeted castle by yy (or sets it to 00 if there are already less than yy defenders); a cavalry attack decreases the number of defenders in the targeted castle by zz (or sets it to 00 if there are already less than zz defenders). The mixed attack can be launched at any valid target (at any castle with at least one soldier). However, the infantry attack cannot be launched if the previous attack on the targeted castle had the same type, no matter when and by whom it was launched. The same applies to the cavalry attack. A castle that was not attacked at all can be targeted by any type of attack.The King who launches the last attack will be glorified as the conqueror of the Red Kingdom, so both Kings want to launch the last attack (and they are wise enough to find a strategy that allows them to do it no matter what are the actions of their opponent, if such strategy exists). The White King is leading his first attack, and you are responsible for planning it. Can you calculate the number of possible options for the first attack that allow the White King to launch the last attack? Each option for the first attack is represented by the targeted castle and the type of attack, and two options are different if the targeted castles or the types of attack are different.InputThe first line contains one integer tt (1≤t≤10001≤t≤1000) — the number of test cases.Then, the test cases follow. Each test case is represented by two lines. The first line contains four integers nn, xx, yy and zz (1≤n≤3⋅1051≤n≤3⋅105, 1≤x,y,z≤51≤x,y,z≤5). The second line contains nn integers a1a1, a2a2, ..., anan (1≤ai≤10181≤ai≤1018).It is guaranteed that the sum of values of nn over all test cases in the input does not exceed 3⋅1053⋅105.OutputFor each test case, print the answer to it: the number of possible options for the first attack of the White King (or 00, if the Black King can launch the last attack no matter how the White King acts).ExamplesInputCopy3
2 1 3 4
7 6
1 1 2 3
1
1 1 2 2
3
OutputCopy2
3
0
InputCopy10
6 5 4 5
2 3 2 3 1 3
1 5 2 3
10
4 4 2 3
8 10 8 5
2 2 1 4
8 5
3 5 3 5
9 2 10
4 5 5 5
2 10 4 2
2 3 1 4
1 10
3 1 5 3
9 8 7
2 5 4 5
8 8
3 5 1 4
5 5 10
OutputCopy0
2
1
2
5
12
5
0
0
2
| [
"games",
"two pointers"
] | import sys
input = sys.stdin.readline
G0=[[[[0]*5500 for i in range(6)] for j in range(6)] for k in range(6)]
G1=[[[[0]*5500 for i in range(6)] for j in range(6)] for k in range(6)]
G2=[[[[0]*5500 for i in range(6)] for j in range(6)] for k in range(6)]
for x in range(1,6):
for y in range(1,6):
for z in range(1,6):
for i in range(1,5500):
s=G0[x][y][z][max(0,i-x)]
t=G1[x][y][z][max(0,i-y)]
u=G2[x][y][z][max(0,i-z)]
for j in range(5):
if j==s or j==t or j==u:
continue
else:
G0[x][y][z][i]=j
break
for j in range(5):
if j==s or j==u:
continue
else:
G1[x][y][z][i]=j
break
for j in range(5):
if j==s or j==t:
continue
else:
G2[x][y][z][i]=j
break
def lcm(x, y):
return (x * y) // math.gcd(x, y)
t=int(input())
for tests in range(t):
n,x,y,z=map(int,input().split())
B=list(map(int,input().split()))
A=[]
for a in B:
if a<=5400:
A.append(a)
else:
A.append(a%2520+2520)
XOR=0
for a in A:
XOR^=G0[x][y][z][a]
ANS=0
for a in A:
k=XOR^G0[x][y][z][a]
if G0[x][y][z][max(0,a-x)]==k:
ANS+=1
if G1[x][y][z][max(0,a-y)]==k:
ANS+=1
if G2[x][y][z][max(0,a-z)]==k:
ANS+=1
print(ANS)
| py |
1316 | B | B. String Modificationtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputVasya has a string ss of length nn. He decides to make the following modification to the string: Pick an integer kk, (1≤k≤n1≤k≤n). For ii from 11 to n−k+1n−k+1, reverse the substring s[i:i+k−1]s[i:i+k−1] of ss. For example, if string ss is qwer and k=2k=2, below is the series of transformations the string goes through: qwer (original string) wqer (after reversing the first substring of length 22) weqr (after reversing the second substring of length 22) werq (after reversing the last substring of length 22) Hence, the resulting string after modifying ss with k=2k=2 is werq. Vasya wants to choose a kk such that the string obtained after the above-mentioned modification is lexicographically smallest possible among all choices of kk. Among all such kk, he wants to choose the smallest one. Since he is busy attending Felicity 2020, he asks for your help.A string aa is lexicographically smaller than a string bb if and only if one of the following holds: aa is a prefix of bb, but a≠ba≠b; in the first position where aa and bb differ, the string aa has a letter that appears earlier in the alphabet than the corresponding letter in bb. InputEach test contains multiple test cases. The first line contains the number of test cases tt (1≤t≤50001≤t≤5000). The description of the test cases follows.The first line of each test case contains a single integer nn (1≤n≤50001≤n≤5000) — the length of the string ss.The second line of each test case contains the string ss of nn lowercase latin letters.It is guaranteed that the sum of nn over all test cases does not exceed 50005000.OutputFor each testcase output two lines:In the first line output the lexicographically smallest string s′s′ achievable after the above-mentioned modification. In the second line output the appropriate value of kk (1≤k≤n1≤k≤n) that you chose for performing the modification. If there are multiple values of kk that give the lexicographically smallest string, output the smallest value of kk among them.ExampleInputCopy6
4
abab
6
qwerty
5
aaaaa
6
alaska
9
lfpbavjsm
1
p
OutputCopyabab
1
ertyqw
3
aaaaa
1
aksala
6
avjsmbpfl
5
p
1
NoteIn the first testcase of the first sample; the string modification results for the sample abab are as follows : for k=1k=1 : abab for k=2k=2 : baba for k=3k=3 : abab for k=4k=4 : babaThe lexicographically smallest string achievable through modification is abab for k=1k=1 and 33. Smallest value of kk needed to achieve is hence 11. | [
"brute force",
"constructive algorithms",
"implementation",
"sortings",
"strings"
] | for _ in range(int(input())):
n=int(input())
s=input()
k=1
res=s
for j in range(1,n+1):
mov=n-j+1
t=s[:j-1]
final=s[j-1:]
if mov%2==0:
final+=t
else:
final=final+t[::-1]
if res>final:
k=j
res=final
print(res)
print(k) | py |
1285 | E | E. Delete a Segmenttime limit per test4 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThere are nn segments on a OxOx axis [l1,r1][l1,r1], [l2,r2][l2,r2], ..., [ln,rn][ln,rn]. Segment [l,r][l,r] covers all points from ll to rr inclusive, so all xx such that l≤x≤rl≤x≤r.Segments can be placed arbitrarily — be inside each other, coincide and so on. Segments can degenerate into points, that is li=rili=ri is possible.Union of the set of segments is such a set of segments which covers exactly the same set of points as the original set. For example: if n=3n=3 and there are segments [3,6][3,6], [100,100][100,100], [5,8][5,8] then their union is 22 segments: [3,8][3,8] and [100,100][100,100]; if n=5n=5 and there are segments [1,2][1,2], [2,3][2,3], [4,5][4,5], [4,6][4,6], [6,6][6,6] then their union is 22 segments: [1,3][1,3] and [4,6][4,6]. Obviously, a union is a set of pairwise non-intersecting segments.You are asked to erase exactly one segment of the given nn so that the number of segments in the union of the rest n−1n−1 segments is maximum possible.For example, if n=4n=4 and there are segments [1,4][1,4], [2,3][2,3], [3,6][3,6], [5,7][5,7], then: erasing the first segment will lead to [2,3][2,3], [3,6][3,6], [5,7][5,7] remaining, which have 11 segment in their union; erasing the second segment will lead to [1,4][1,4], [3,6][3,6], [5,7][5,7] remaining, which have 11 segment in their union; erasing the third segment will lead to [1,4][1,4], [2,3][2,3], [5,7][5,7] remaining, which have 22 segments in their union; erasing the fourth segment will lead to [1,4][1,4], [2,3][2,3], [3,6][3,6] remaining, which have 11 segment in their union. Thus, you are required to erase the third segment to get answer 22.Write a program that will find the maximum number of segments in the union of n−1n−1 segments if you erase any of the given nn segments.Note that if there are multiple equal segments in the given set, then you can erase only one of them anyway. So the set after erasing will have exactly n−1n−1 segments.InputThe first line contains one integer tt (1≤t≤1041≤t≤104) — the number of test cases in the test. Then the descriptions of tt test cases follow.The first of each test case contains a single integer nn (2≤n≤2⋅1052≤n≤2⋅105) — the number of segments in the given set. Then nn lines follow, each contains a description of a segment — a pair of integers lili, riri (−109≤li≤ri≤109−109≤li≤ri≤109), where lili and riri are the coordinates of the left and right borders of the ii-th segment, respectively.The segments are given in an arbitrary order.It is guaranteed that the sum of nn over all test cases does not exceed 2⋅1052⋅105.OutputPrint tt integers — the answers to the tt given test cases in the order of input. The answer is the maximum number of segments in the union of n−1n−1 segments if you erase any of the given nn segments.ExampleInputCopy3
4
1 4
2 3
3 6
5 7
3
5 5
5 5
5 5
6
3 3
1 1
5 5
1 5
2 2
4 4
OutputCopy2
1
5
| [
"brute force",
"constructive algorithms",
"data structures",
"dp",
"graphs",
"sortings",
"trees",
"two pointers"
] | 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 = int(input())
l, r = [], []
s = set()
for _ in range(n):
l0, r0 = map(int, input().split())
s.add(l0)
s.add(r0)
l.append(l0)
r.append(r0)
s = list(s)
s.sort()
d = dict()
for i in range(len(s)):
d[s[i]] = 2 * i
m = 2 * len(s)
x = [0] * m
lr = []
for l0, r0 in zip(l, r):
l1, r1 = d[l0], d[r0]
lr.append((l1, r1))
x[l1] += 1
x[r1 + 1] -= 1
c = 0
u = []
s = 0
for i in range(1, m):
x[i] += x[i - 1]
if not x[i]:
c += 1
u.append((s, i - 1))
s = i + 1
z = [0] * m
for l, r in u:
f = 0
for i in range(l, r + 1):
if x[i] == 1:
if f and not x[i + 1] == 1:
z[i + 1] = 1
if not x[i] == 1:
f = 1
z[r + 1] = 0
for i in range(1, m):
z[i] += z[i - 1]
ma = 0
for l, r in lr:
ma = max(ma, z[r + 1] - z[l])
ans0 = c + ma if n ^ c else n - 1
ans.append(ans0)
sys.stdout.write("\n".join(map(str, ans))) | py |
1311 | E | E. Construct the Binary Treetime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given two integers nn and dd. You need to construct a rooted binary tree consisting of nn vertices with a root at the vertex 11 and the sum of depths of all vertices equals to dd.A tree is a connected graph without cycles. A rooted tree has a special vertex called the root. A parent of a vertex vv is the last different from vv vertex on the path from the root to the vertex vv. The depth of the vertex vv is the length of the path from the root to the vertex vv. Children of vertex vv are all vertices for which vv is the parent. The binary tree is such a tree that no vertex has more than 22 children.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 only line of each test case contains two integers nn and dd (2≤n,d≤50002≤n,d≤5000) — the number of vertices in the tree and the required sum of depths of all vertices.It is guaranteed that the sum of nn and the sum of dd both does not exceed 50005000 (∑n≤5000,∑d≤5000∑n≤5000,∑d≤5000).OutputFor each test case, print the answer.If it is impossible to construct such a tree, print "NO" (without quotes) in the first line. Otherwise, print "{YES}" in the first line. Then print n−1n−1 integers p2,p3,…,pnp2,p3,…,pn in the second line, where pipi is the parent of the vertex ii. Note that the sequence of parents you print should describe some binary tree.ExampleInputCopy3
5 7
10 19
10 18
OutputCopyYES
1 2 1 3
YES
1 2 3 3 9 9 2 1 6
NO
NotePictures corresponding to the first and the second test cases of the example: | [
"brute force",
"constructive algorithms",
"trees"
] | import sys
from array import array # noqa: F401
from typing import List, Tuple, TypeVar, Generic, Sequence, Union # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
def main():
t = int(input())
for _ in range(t):
n, d = map(int, input().split())
depth = [0] + list(range(n))
child = [0] + [1] * (n - 1) + [0]
par = [-1, -1] + list(range(1, n))
cur_d = sum(depth)
if d > cur_d:
print('NO')
continue
for i in range(n, 0, -1):
if child[i]:
continue
child[par[i]] -= 1
p, delta = -1, 0
for j in range(1, n + 1):
if i != j and child[j] < 2 and cur_d - d >= depth[i] - depth[j] - 1 and depth[i] - depth[j] - 1 > delta:
delta = depth[i] - depth[j] - 1
p = j
if p != -1:
child[p] += 1
par[i] = p
depth[i] = depth[p] + 1
cur_d -= delta
else:
break
if cur_d == d:
print('YES')
print(*par[2:])
else:
print('NO')
if __name__ == '__main__':
main()
| py |
1305 | B | B. Kuroni and Simple Stringstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputNow that Kuroni has reached 10 years old; he is a big boy and doesn't like arrays of integers as presents anymore. This year he wants a Bracket sequence as a Birthday present. More specifically, he wants a bracket sequence so complex that no matter how hard he tries, he will not be able to remove a simple subsequence!We say that a string formed by nn characters '(' or ')' is simple if its length nn is even and positive, its first n2n2 characters are '(', and its last n2n2 characters are ')'. For example, the strings () and (()) are simple, while the strings )( and ()() are not simple.Kuroni will be given a string formed by characters '(' and ')' (the given string is not necessarily simple). An operation consists of choosing a subsequence of the characters of the string that forms a simple string and removing all the characters of this subsequence from the string. Note that this subsequence doesn't have to be continuous. For example, he can apply the operation to the string ')()(()))', to choose a subsequence of bold characters, as it forms a simple string '(())', delete these bold characters from the string and to get '))()'. Kuroni has to perform the minimum possible number of operations on the string, in such a way that no more operations can be performed on the remaining string. The resulting string does not have to be empty.Since the given string is too large, Kuroni is unable to figure out how to minimize the number of operations. Can you help him do it instead?A sequence of characters aa is a subsequence of a string bb if aa can be obtained from bb by deletion of several (possibly, zero or all) characters.InputThe only line of input contains a string ss (1≤|s|≤10001≤|s|≤1000) formed by characters '(' and ')', where |s||s| is the length of ss.OutputIn the first line, print an integer kk — the minimum number of operations you have to apply. Then, print 2k2k lines describing the operations in the following format:For each operation, print a line containing an integer mm — the number of characters in the subsequence you will remove.Then, print a line containing mm integers 1≤a1<a2<⋯<am1≤a1<a2<⋯<am — the indices of the characters you will remove. All integers must be less than or equal to the length of the current string, and the corresponding subsequence must form a simple string.If there are multiple valid sequences of operations with the smallest kk, you may print any of them.ExamplesInputCopy(()((
OutputCopy1
2
1 3
InputCopy)(
OutputCopy0
InputCopy(()())
OutputCopy1
4
1 2 5 6
NoteIn the first sample; the string is '(()(('. The operation described corresponds to deleting the bolded subsequence. The resulting string is '((('; and no more operations can be performed on it. Another valid answer is choosing indices 22 and 33, which results in the same final string.In the second sample, it is already impossible to perform any operations. | [
"constructive algorithms",
"greedy",
"strings",
"two pointers"
] | import sys
input = sys.stdin.readline
s = input()[:-1]
i, j = 0, len(s)-1
c = 0
a = [-1]
b = [-1]
while i < j:
if s[i] == '(':
if i+1 != a[-1]:
c += 1
a.append(i+1)
if s[j] == ')':
if j+1 != b[-1]:
c -= 1
b.append(j+1)
if c < 0:
i += 1
else:
j -= 1
c = min(len(a), len(b))
x = 2*c-2
if x == 0:
print(0)
else:
print(1)
print(x)
print(*a[1:c], *sorted(b[1:c])) | 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(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 |
1324 | D | D. Pair of Topicstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThe next lecture in a high school requires two topics to be discussed. The ii-th topic is interesting by aiai units for the teacher and by bibi units for the students.The pair of topics ii and jj (i<ji<j) is called good if ai+aj>bi+bjai+aj>bi+bj (i.e. it is more interesting for the teacher).Your task is to find the number of good pairs of topics.InputThe first line of the input contains one integer nn (2≤n≤2⋅1052≤n≤2⋅105) — the number of topics.The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109), where aiai is the interestingness of the ii-th topic for the teacher.The third line of the input contains nn integers b1,b2,…,bnb1,b2,…,bn (1≤bi≤1091≤bi≤109), where bibi is the interestingness of the ii-th topic for the students.OutputPrint one integer — the number of good pairs of topic.ExamplesInputCopy5
4 8 2 6 2
4 5 4 1 3
OutputCopy7
InputCopy4
1 3 2 4
1 3 2 4
OutputCopy0
| [
"binary search",
"data structures",
"sortings",
"two pointers"
] | n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = sorted([a[i] - b[i] for i in range(n)])
ans = 0
l = 0
r = n - 1
while l < r:
if c[l] + c[r] > 0:
ans += r - l
r -= 1
else:
l += 1
print(ans)
| py |
1325 | C | C. Ehab and Path-etic MEXstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given a tree consisting of nn nodes. You want to write some labels on the tree's edges such that the following conditions hold: Every label is an integer between 00 and n−2n−2 inclusive. All the written labels are distinct. The largest value among MEX(u,v)MEX(u,v) over all pairs of nodes (u,v)(u,v) is as small as possible. Here, MEX(u,v)MEX(u,v) denotes the smallest non-negative integer that isn't written on any edge on the unique simple path from node uu to node vv.InputThe first line contains the integer nn (2≤n≤1052≤n≤105) — the number of nodes in the tree.Each of the next n−1n−1 lines contains two space-separated integers uu and vv (1≤u,v≤n1≤u,v≤n) that mean there's an edge between nodes uu and vv. It's guaranteed that the given graph is a tree.OutputOutput n−1n−1 integers. The ithith of them will be the number written on the ithith edge (in the input order).ExamplesInputCopy3
1 2
1 3
OutputCopy0
1
InputCopy6
1 2
1 3
2 4
2 5
5 6
OutputCopy0
3
2
4
1NoteThe tree from the second sample: | [
"constructive algorithms",
"dfs and similar",
"greedy",
"trees"
] | import sys
input = lambda: sys.stdin.readline().rstrip()
N = int(input())
dat = [[] for _ in range(N)]
P = []
for _ in range(N-1):
a,b = map(int,input().split())
P.append((a,b))
dat[a-1].append(b-1)
dat[b-1].append(a-1)
k = []
for i in range(N):
k.append([len(dat[i]),i])
k.sort(reverse=True)
num = 0
i = []
for x in dat[k[0][1]]:
i.append([k[0][1],x])
vis = [0]*N
vis[k[0][1]] = 1
dic = {}
# if N==100000:
# exit()
while i:
k = []
for p,x in i:
if vis[x]==1:continue
dic[(p+1,x+1)] = num
dic[(x+1,p+1)] = num
num += 1
vis[x] = 1
for y in dat[x]:
if vis[y]==1:continue
k.append([x,y])
i = k
for i in P:
print(dic[i]) | py |
1305 | D | D. Kuroni and the Celebrationtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputThis is an interactive problem.After getting AC after 13 Time Limit Exceeded verdicts on a geometry problem; Kuroni went to an Italian restaurant to celebrate this holy achievement. Unfortunately, the excess sauce disoriented him, and he's now lost!The United States of America can be modeled as a tree (why though) with nn vertices. The tree is rooted at vertex rr, wherein lies Kuroni's hotel.Kuroni has a phone app designed to help him in such emergency cases. To use the app, he has to input two vertices uu and vv, and it'll return a vertex ww, which is the lowest common ancestor of those two vertices.However, since the phone's battery has been almost drained out from live-streaming Kuroni's celebration party, he could only use the app at most ⌊n2⌋⌊n2⌋ times. After that, the phone would die and there will be nothing left to help our dear friend! :(As the night is cold and dark, Kuroni needs to get back, so that he can reunite with his comfy bed and pillow(s). Can you help him figure out his hotel's location?InteractionThe interaction starts with reading a single integer nn (2≤n≤10002≤n≤1000), the number of vertices of the tree.Then you will read n−1n−1 lines, the ii-th of them has two integers xixi and yiyi (1≤xi,yi≤n1≤xi,yi≤n, xi≠yixi≠yi), denoting there is an edge connecting vertices xixi and yiyi. It is guaranteed that the edges will form a tree.Then you can make queries of type "? u v" (1≤u,v≤n1≤u,v≤n) to find the lowest common ancestor of vertex uu and vv.After the query, read the result ww as an integer.In case your query is invalid or you asked more than ⌊n2⌋⌊n2⌋ queries, the program will print −1−1 and will finish interaction. You will receive a Wrong answer verdict. Make sure to exit immediately to avoid getting other verdicts.When you find out the vertex rr, print "! rr" and quit after that. This query does not count towards the ⌊n2⌋⌊n2⌋ limit.Note that the tree 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.HacksTo hack, use the following format:The first line should contain two integers nn and rr (2≤n≤10002≤n≤1000, 1≤r≤n1≤r≤n), denoting the number of vertices and the vertex with Kuroni's hotel.The ii-th of the next n−1n−1 lines should contain two integers xixi and yiyi (1≤xi,yi≤n1≤xi,yi≤n) — denoting there is an edge connecting vertex xixi and yiyi.The edges presented should form a tree.ExampleInputCopy6
1 4
4 2
5 3
6 3
2 3
3
4
4
OutputCopy
? 5 6
? 3 1
? 1 2
! 4NoteNote 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.The image below demonstrates the tree in the sample test: | [
"constructive algorithms",
"dfs and similar",
"interactive",
"trees"
] | import sys
input = sys.stdin.buffer.readline
def dfs(n, a, g, p=None):
parent = [None for i in range(n+1)]
depth = [[a]]
parent[a] = a
if p is not None:
parent[p] = p
while True:
next_s = []
for x in depth[-1]:
for y in g[x]:
if parent[y] is None:
parent[y] = x
next_s.append(y)
if len(next_s)==0:
break
depth.append(next_s)
return depth, parent
def process(n, G):
g = [[] for i in range(n+1)]
checked = [0 for i in range(n+1)]
for a, b in G:
g[a].append(b)
g[b].append(a)
leaves = [i for i in range(1, n+1) if len(g[i])==1]
for a in leaves:
if checked[a]==0:
depth, parent = dfs(n, a, g)
b = None
while len(depth) > 0:
for b2 in depth[-1]:
if checked[b2]==0 and b is None:
b = b2
depth.pop()
if b is None:
b = leaves[0]
if True:
sys.stdout.write(f'? {a} {b}\n')
sys.stdout.flush()
r = int(input())
if (r==a or r==b) and len(g[r])==1:
sys.stdout.write(f'! {r}\n')
sys.stdout.flush()
return
depth1, parent1 = dfs(n, r, g)
a_list = [a]
while a_list[-1] != r:
a2 = a_list[-1]
a_list.append(parent1[a2])
b_list = [b]
while b_list[-1] != r:
b2 = b_list[-1]
b_list.append(parent1[b2])
a_list.pop()
if len(a_list) > 0:
a2 = a_list[-1]
depth2, parent2 = dfs(n, a2, g, p=r)
for row in depth2:
for x in row:
checked[x] = 1
b_list.pop()
if len(b_list) > 0:
b2 = b_list[-1]
depth3, parent3 = dfs(n, b2, g, p=r)
for row in depth3:
for x in row:
checked[x] = 1
answer = []
for i in range(1, n+1):
if checked[i]==0:
answer.append(i)
# print(answer)
while len(answer) > 1:
x = answer.pop()
y = answer.pop()
sys.stdout.write(f'? {x} {y}\n')
sys.stdout.flush()
r = int(input())
if r==x or r==y:
sys.stdout.write(f'! {r}\n')
sys.stdout.flush()
return
sys.stdout.write(f'! {answer[0]}\n')
sys.stdout.flush()
return
n = int(input())
G = []
for i in range(n-1):
a, b = [int(x) for x in input().split()]
G.append([a, b])
process(n, G)
| py |
1320 | C | C. World of Darkraft: Battle for Azathothtime limit per test2 secondsmemory limit per test512 megabytesinputstandard inputoutputstandard outputRoma is playing a new expansion for his favorite game World of Darkraft. He made a new character and is going for his first grind.Roma has a choice to buy exactly one of nn different weapons and exactly one of mm different armor sets. Weapon ii has attack modifier aiai and is worth caicai coins, and armor set jj has defense modifier bjbj and is worth cbjcbj coins.After choosing his equipment Roma can proceed to defeat some monsters. There are pp monsters he can try to defeat. Monster kk has defense xkxk, attack ykyk and possesses zkzk coins. Roma can defeat a monster if his weapon's attack modifier is larger than the monster's defense, and his armor set's defense modifier is larger than the monster's attack. That is, a monster kk can be defeated with a weapon ii and an armor set jj if ai>xkai>xk and bj>ykbj>yk. After defeating the monster, Roma takes all the coins from them. During the grind, Roma can defeat as many monsters as he likes. Monsters do not respawn, thus each monster can be defeated at most one.Thanks to Roma's excessive donations, we can assume that he has an infinite amount of in-game currency and can afford any of the weapons and armor sets. Still, he wants to maximize the profit of the grind. The profit is defined as the total coins obtained from all defeated monsters minus the cost of his equipment. Note that Roma must purchase a weapon and an armor set even if he can not cover their cost with obtained coins.Help Roma find the maximum profit of the grind.InputThe first line contains three integers nn, mm, and pp (1≤n,m,p≤2⋅1051≤n,m,p≤2⋅105) — the number of available weapons, armor sets and monsters respectively.The following nn lines describe available weapons. The ii-th of these lines contains two integers aiai and caicai (1≤ai≤1061≤ai≤106, 1≤cai≤1091≤cai≤109) — the attack modifier and the cost of the weapon ii.The following mm lines describe available armor sets. The jj-th of these lines contains two integers bjbj and cbjcbj (1≤bj≤1061≤bj≤106, 1≤cbj≤1091≤cbj≤109) — the defense modifier and the cost of the armor set jj.The following pp lines describe monsters. The kk-th of these lines contains three integers xk,yk,zkxk,yk,zk (1≤xk,yk≤1061≤xk,yk≤106, 1≤zk≤1031≤zk≤103) — defense, attack and the number of coins of the monster kk.OutputPrint a single integer — the maximum profit of the grind.ExampleInputCopy2 3 3
2 3
4 7
2 4
3 2
5 11
1 2 4
2 1 6
3 4 6
OutputCopy1
| [
"brute force",
"data structures",
"sortings"
] | import sys
from array import array
from bisect import *
input = lambda: sys.stdin.readline().rstrip("\r\n")
inp = lambda dtype: [dtype(x) for x in input().split()]
debug = lambda *x: print(*x, file=sys.stderr)
ceil1 = lambda a, b: (a + b - 1) // b
tests, Mint, Mlong = 1, 2 ** 31 - 1, 2 ** 63 - 1
out = []
max_ = 10 ** 6 + 1
class range_sum:
def __init__(self, n, default, func):
self.tree, self.n, self.lazy, self.h = [0] * (2 * n), n, [0] * (n), 20
self.default, self.func = default, func
def apply(self, p, val):
self.tree[p] += val
if p < self.n:
self.lazy[p] += val
# fix applied nodes and down nodes
def build(self, p):
while p > 1:
p >>= 1
self.tree[p] = self.func(self.tree[p << 1], self.tree[(p << 1) + 1]) + self.lazy[p]
# push lazy operations
def push(self, p):
for s in range(self.h, 0, -1):
i = p >> s
if self.lazy[i]:
self.apply(i * 2, self.lazy[i])
self.apply(i * 2 + 1, self.lazy[i])
self.lazy[i] = 0
def update(self, l, r, val):
'''[l,r)'''
l += self.n
r += self.n
l0, r0 = l, r
while l < r:
if l & 1:
self.apply(l, val)
l += 1
if r & 1:
r -= 1
self.apply(r, val)
l >>= 1
r >>= 1
self.build(l0)
self.build(r0 - 1)
def query(self, l, r):
'''[l,r)'''
l += self.n
r += self.n
# apply lazy nodes
self.push(l)
self.push(r - 1)
res = self.default
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
for _ in range(tests):
n, m, p = inp(int)
att = array('i', [Mint] * max_)
def_ = array('i', [Mint] * max_)
mdef = array('i', [-1] * p)
matt = array('i', [-1] * p)
mcoin = array('i', [-1] * p)
adj = [array('i') for _ in range(max_)]
defs = array('i')
tree = range_sum(m, -Mint, max)
for i in range(n):
a, ca = inp(int)
att[a] = min(att[a], ca)
for i in range(m):
b, cb = inp(int)
def_[b] = min(def_[b], cb)
for i in range(p):
mdef[i], matt[i], mcoin[i] = inp(int)
adj[mdef[i]].append(i)
for i in range(max_):
if def_[i] != Mint:
defs.append(i)
tree.update(len(defs) - 1, len(defs), -def_[i])
ans = -Mlong
for i in range(max_):
if att[i] != Mint:
ans = max(ans, tree.query(0, len(defs)) - att[i])
for j in adj[i]:
ix = bisect_right(defs, matt[j])
if ix < len(defs): tree.update(ix, len(defs), mcoin[j])
out.append(ans)
print('\n'.join(map(str, out)))
| py |
1295 | D | D. Same GCDstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given two integers aa and mm. Calculate the number of integers xx such that 0≤x<m0≤x<m and gcd(a,m)=gcd(a+x,m)gcd(a,m)=gcd(a+x,m).Note: gcd(a,b)gcd(a,b) is the greatest common divisor of aa and bb.InputThe first line contains the single integer TT (1≤T≤501≤T≤50) — the number of test cases.Next TT lines contain test cases — one per line. Each line contains two integers aa and mm (1≤a<m≤10101≤a<m≤1010).OutputPrint TT integers — one per test case. For each test case print the number of appropriate xx-s.ExampleInputCopy3
4 9
5 10
42 9999999967
OutputCopy6
1
9999999966
NoteIn the first test case appropriate xx-s are [0,1,3,4,6,7][0,1,3,4,6,7].In the second test case the only appropriate xx is 00. | [
"math",
"number theory"
] | import sys
from collections import defaultdict, deque, Counter
from decimal import *
from heapq import heapify, heappop, heappush
import math
import random
import string
from copy import deepcopy
from itertools import combinations, permutations, product
from operator import mul, itemgetter
from functools import reduce
from bisect import bisect_left, bisect_right
def input():
return sys.stdin.readline().rstrip()
def getN():
return int(input())
def getNM():
return map(int, input().split())
def getList():
return list(map(int, input().split()))
def getListGraph():
return list(map(lambda x:int(x) - 1, input().split()))
def getArray(intn):
return [int(input()) for i in range(intn)]
mod = 10 ** 9 + 7
MOD = 998244353
# sys.setrecursionlimit(10000000)
# import pypyjit
# pypyjit.set_param('max_unroll_recursion=-1')
inf = float('inf')
eps = 10 ** (-12)
dy = [0, 1, 0, -1]
dx = [1, 0, -1, 0]
#############
# Main Code #
#############
# 素因数分解
def prime_factorize(n):
divisors = []
# 27(2 * 2 * 7)の7を出すためにtemp使う
temp = n
for i in range(2, int(math.sqrt(n)) + 1):
if temp % i == 0:
cnt = 0
while temp % i == 0:
cnt += 1
# 素因数を見つけるたびにtempを割っていく
temp //= i
divisors.append([i, cnt])
if temp != 1:
divisors.append([temp, 1])
if divisors == []:
divisors.append([n, 1])
return divisors
T = getN()
for _ in range(T):
A, M = getNM()
g = math.gcd(A, M)
A //= g
M //= g
c = M
# オイラーのトーシェント数の求め方
for k, _ in prime_factorize(M):
if k > 1:
c *= k - 1
c //= k
print(c) | 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"
] | t = int(input())
for _ in range(t):
n = int(input())
print(1,n-1) | py |
1286 | B | B. Numbers on Treetime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputEvlampiy was gifted a rooted tree. The vertices of the tree are numbered from 11 to nn. Each of its vertices also has an integer aiai written on it. For each vertex ii, Evlampiy calculated cici — the number of vertices jj in the subtree of vertex ii, such that aj<aiaj<ai. Illustration for the second example, the first integer is aiai and the integer in parentheses is ciciAfter the new year, Evlampiy could not remember what his gift was! He remembers the tree and the values of cici, but he completely forgot which integers aiai were written on the vertices.Help him to restore initial integers!InputThe first line contains an integer nn (1≤n≤2000)(1≤n≤2000) — the number of vertices in the tree.The next nn lines contain descriptions of vertices: the ii-th line contains two integers pipi and cici (0≤pi≤n0≤pi≤n; 0≤ci≤n−10≤ci≤n−1), where pipi is the parent of vertex ii or 00 if vertex ii is root, and cici is the number of vertices jj in the subtree of vertex ii, such that aj<aiaj<ai.It is guaranteed that the values of pipi describe a rooted tree with nn vertices.OutputIf a solution exists, in the first line print "YES", and in the second line output nn integers aiai (1≤ai≤109)(1≤ai≤109). If there are several solutions, output any of them. One can prove that if there is a solution, then there is also a solution in which all aiai are between 11 and 109109.If there are no solutions, print "NO".ExamplesInputCopy3
2 0
0 2
2 0
OutputCopyYES
1 2 1 InputCopy5
0 1
1 3
2 1
3 0
2 0
OutputCopyYES
2 3 2 1 2
| [
"constructive algorithms",
"data structures",
"dfs and similar",
"graphs",
"greedy",
"trees"
] | from sys import setrecursionlimit
setrecursionlimit(10 ** 6)
n = int(input())
c = [0 for i in range(n + 1)]
a = [0 for i in range(n + 1)]
g = [[] for i in range(n + 1)]
def dfs(x):
List = []
for i in g[x]:
List += dfs(i)
if c[x] > len(List):
raise ValueError
List.insert(c[x], x)
return List
for i in range(1,n + 1):
pi, ci = list(map(int, input().split(' ')))
c[i] = ci
g[pi].append(i)
try:
ans = dfs(0)
print("YES")
for i in range(0, n + 1):
a[ans[i]] = i
for i in range(1, n + 1):
print(a[i], end = ' ')
except ValueError:
print("NO") | 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())
ans = 0
mx = [0] * (10 ** 6 + 5)
mn = []
koltrue = 0
for _ in range(n):
l = list(map(int, input().split()))
mnn = l[1]
mxx = l[1]
t = False
for i in range(2, l[0] + 1):
if l[i-1] < l[i]:
t = True
mnn = min(mnn, l[i])
mxx = max(mxx, l[i])
if t:
koltrue += 1
mn.append('t')
else:
mn.append(mnn)
mx[mxx] += 1
#print(koltrue)
sufs = [0] * (10 ** 6 + 5)
for i in range(len(sufs) - 2, -1, -1):
sufs[i] = mx[i] + sufs[i+1]
#print(sufs[:10])
#print(mn)
for i in range(len(mn)):
if mn[i] == 't':
ans += n
else:
ans += koltrue
#print('#', mn[i], i)
ans += sufs[mn[i] + 1]
#print(ans)
print(ans)
# Thu Nov 24 2022 20:38:24 GMT+0300 (Moscow Standard Time)
| 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"
] | for _ in range(int(input())):
n,d=map(int,input().split())
l=list(map(int,input().split()))
m_=l[0]
w=1
i=1
while d>0:
if i<n:
if l[i]>0:
if d>=(l[i]*w):
m_=m_+l[i]
d=d-(l[i]*w)
l[i]=0
i+=1
w+=1
#print(d)
else:
q=d//w
m_=m_+q
d=0
else:
i+=1
w+=1
else:
break
print(m_)
| 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"
] | import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n, q = map(int, input().split())
s = [[0] * (n + 2) for _ in range(2)]
now = 0
ans = []
for _ in range(q):
r, c = map(int, input().split())
r -= 1
if not s[r][c]:
for i in range(c - 1, c + 2):
if s[r ^ 1][i]:
now += 1
else:
for i in range(c - 1, c + 2):
if s[r ^ 1][i]:
now -= 1
s[r][c] ^= 1
ans0 = "Yes" if not now else "No"
ans.append(ans0)
sys.stdout.write("\n".join(ans)) | 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"
] | from sys import stdin
input = stdin.readline
inp = lambda : list(map(int,input().split()))
"""
we can just solve it column independent
"""
def answer():
ans = 0
for j in range(m):
take = [0 for i in range(n)]
for i in range(n):
x = (a[i][j] - j - 1)
if(x >= 0 and x % m == 0):
v = (x // m)
if(v >= n):continue
if(v <= i):take[i - v] += 1
else:take[i + n - v] += 1
best = float('inf')
for i in range(n):
best = min(best , i + (n - take[i]))
ans += best
return ans
for T in range(1):
n , m = inp()
a = [inp() for i in range(n)]
print(answer())
| py |
1316 | A | A. Grade Allocationtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputnn students are taking an exam. The highest possible score at this exam is mm. Let aiai be the score of the ii-th student. You have access to the school database which stores the results of all students.You can change each student's score as long as the following conditions are satisfied: All scores are integers 0≤ai≤m0≤ai≤m The average score of the class doesn't change. You are student 11 and you would like to maximize your own score.Find the highest possible score you can assign to yourself such that all conditions are satisfied.InputEach test contains multiple test cases. The first line contains the number of test cases tt (1≤t≤2001≤t≤200). The description of the test cases follows.The first line of each test case contains two integers nn and mm (1≤n≤1031≤n≤103, 1≤m≤1051≤m≤105) — the number of students and the highest possible score respectively.The second line of each testcase contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤m0≤ai≤m) — scores of the students.OutputFor each testcase, output one integer — the highest possible score you can assign to yourself such that both conditions are satisfied._ExampleInputCopy2
4 10
1 2 3 4
4 5
1 2 3 4
OutputCopy10
5
NoteIn the first case; a=[1,2,3,4]a=[1,2,3,4], with average of 2.52.5. You can change array aa to [10,0,0,0][10,0,0,0]. Average remains 2.52.5, and all conditions are satisfied.In the second case, 0≤ai≤50≤ai≤5. You can change aa to [5,1,1,3][5,1,1,3]. You cannot increase a1a1 further as it will violate condition 0≤ai≤m0≤ai≤m. | [
"implementation"
] | t = int(input())
for i in range(t):
n, m = map(int, input().split())
a = list(map(int, input().split()))
if n == 1 or a[0] == m:
print(a[0])
continue
summary = sum(a[1:])
smth = summary / m
if smth >= 1:
a[0] = m
else:
a[0] += summary % m
if a[0] > m:
a[0] = m
print(a[0])
| py |
1324 | D | D. Pair of Topicstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThe next lecture in a high school requires two topics to be discussed. The ii-th topic is interesting by aiai units for the teacher and by bibi units for the students.The pair of topics ii and jj (i<ji<j) is called good if ai+aj>bi+bjai+aj>bi+bj (i.e. it is more interesting for the teacher).Your task is to find the number of good pairs of topics.InputThe first line of the input contains one integer nn (2≤n≤2⋅1052≤n≤2⋅105) — the number of topics.The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109), where aiai is the interestingness of the ii-th topic for the teacher.The third line of the input contains nn integers b1,b2,…,bnb1,b2,…,bn (1≤bi≤1091≤bi≤109), where bibi is the interestingness of the ii-th topic for the students.OutputPrint one integer — the number of good pairs of topic.ExamplesInputCopy5
4 8 2 6 2
4 5 4 1 3
OutputCopy7
InputCopy4
1 3 2 4
1 3 2 4
OutputCopy0
| [
"binary search",
"data structures",
"sortings",
"two pointers"
] | n = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
D = [0] * n
for i in range(n):
D[i] = A[i] - B[i]
D.sort()
# find pairs which gives Di + Dj > 0
total = 0
beg = 0
end = n - 1
while (end >= beg):
while (beg < end and D[beg] + D[end] <= 0):
beg += 1
total += max(end - beg, 0)
end -= 1
print(total)
| 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"
] | #Abir Kundu 31.01.2023 h
t=int (input ())
while t>0:
t-=1
n=int (input ())
if n%2==0:
print ("1"*(n//2)) # this 1s numbers will be maximum
else: #for odd
print ("7"+"1"*((n-3)//2))
# 7 needs 3 segments , we have n-3 segments
# odd -odd == even
# only for even segments 1s numbers will be maximum
# n//2 because 1 takes 2 segments to print | 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"
] | # 1286 A.Garland https://codeforces.com/problemset/problem/1286/A
#
def main():
n = int(input())
p = [int(j) for j in input().split()]
print(solve(n, p))
def solve(n, p):
dp = [[[float('inf')]*2 for _ in range(n+1)] for _ in range(n+1)]
dp[0][0][0] = 0
dp[0][0][1] = 0
for i in range(1, n+1):
for j in range(0, i+1):
if p[i-1]%2 == 0 and j>0:
dp[i][j][0] = min(dp[i-1][j-1][0], dp[i-1][j-1][1]+1)
if p[i-1]%2 == 1 or p[i-1] == 0:
dp[i][j][1] = min(dp[i-1][j][0]+1, dp[i-1][j][1])
return min(dp[n][n//2][0], dp[n][n//2][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"
] | ######################################################################################
#------------------------------------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('inputf.in', 'r')
sys.stdout = open('outputf.out', '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 MEX(nList, start):
nList = set(nList)
mex = start
while mex in nList:
mex += 1
return mex
def getClosest(val1, val2, target):
if (target - val1 >= val2 - target):
return val2
else:
return val1
def tobinary(a): return bin(a).replace('0b', '')
def nextPerfectSquare(N):
nextN = math.floor(math.sqrt(N)) + 1
return nextN * nextN
def modFact(n, p):
result = 1
for i in range(1, n + 1):
result *= i
result %= p
return result
def maxsubarrayproduct(arr):
n = len(arr)
max_ending_here = 1
min_ending_here = 1
max_so_far = 0
flag = 0
for i in range(0, n):
if arr[i] > 0:
max_ending_here = max_ending_here * arr[i]
min_ending_here = min (min_ending_here * arr[i], 1)
flag = 1
elif arr[i] == 0:
max_ending_here = 1
min_ending_here = 1
else:
temp = max_ending_here
max_ending_here = max (min_ending_here * arr[i], 1)
min_ending_here = temp * arr[i]
if (max_so_far < max_ending_here):
max_so_far = max_ending_here
if flag == 0 and max_so_far == 0:
return 0
return max_so_far
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
def sqrtl(x):
if (x == 0 or x == 1):
return x
start = 1
end = x//2
while (start <= end):
mid = (start + end) // 2
if (mid*mid == x):
return mid
if (mid * mid < x):
start = mid + 1
ans = mid
else:
end = mid-1
return ans
def ceill(a, b):
return (a // b) + int(a % b != 0)
def LcmOfArray(arr, idx):
# lcm(a,b) = (a*b/gcd(a,b))
if (idx == len(arr)-1):
return arr[idx]
a = arr[idx]
b = LcmOfArray(arr, idx+1)
return int(a*b/ math.gcd(a,b))
# #####################################################################################################
alphabets = list("abcdefghijklmnopqrstuvwxyz")
vowels = list("aeiou")
MOD1 = int(1e9 + 7)
MOD2 = 998244353
INF = int(1e17)
# ########################################################################
# #-----------------------------Code Here--------------------------------#
########################################################################
# vsInput()
for _ in range(inp()):
n = inp()
a = inlt()
i = 0
while i < n and a[i] >= i:
i += 1
buu = 1
while i < n:
if a[i - 1] <= a[i]:
a[i] = a[i - 1] - 1
if a[i] < 0:
print("NO")
buu = 0
break
i += 1
if buu:
print("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"
] | sequencias = int(input())
cont = 0
cresc = 0
ncresc = 0
maxi = []
mini = []
for x in range(sequencias):
sequencia = str(input()).split()
s = sequencia[1:]
sint = [int(p) for p in s]
for y in range(len(sint)):
if y > 0 and sint[y] > sint[y-1]:
cresc += 1
break
elif (y == len(sint) - 1):
maxi.append(max(sint))
mini.append(min(sint))
ncresc += 1
cont = cresc * cresc + (cresc * ncresc * 2)
mini.sort()
maxi.sort()
h = 0
for g in range(len(maxi)):
while h < ncresc and int(mini[h]) < int(maxi[g]):
h += 1
cont += h
print(cont)
| py |
1291 | A | A. Even But Not Eventime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputLet's define a number ebne (even but not even) if and only if its sum of digits is divisible by 22 but the number itself is not divisible by 22. For example, 1313, 12271227, 185217185217 are ebne numbers, while 1212, 22, 177013177013, 265918265918 are not. If you're still unsure what ebne numbers are, you can look at the sample notes for more clarification.You are given a non-negative integer ss, consisting of nn digits. You can delete some digits (they are not necessary consecutive/successive) to make the given number ebne. You cannot change the order of the digits, that is, after deleting the digits the remaining digits collapse. The resulting number shouldn't contain leading zeros. You can delete any number of digits between 00 (do not delete any digits at all) and n−1n−1.For example, if you are given s=s=222373204424185217171912 then one of possible ways to make it ebne is: 222373204424185217171912 →→ 2237344218521717191. The sum of digits of 2237344218521717191 is equal to 7070 and is divisible by 22, but number itself is not divisible by 22: it means that the resulting number is ebne.Find any resulting number that is ebne. If it's impossible to create an ebne number from the given number report about it.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 a single integer nn (1≤n≤30001≤n≤3000) — the number of digits in the original number.The second line of each test case contains a non-negative integer number ss, consisting of nn digits.It is guaranteed that ss does not contain leading zeros and the sum of nn over all test cases does not exceed 30003000.OutputFor each test case given in the input print the answer in the following format: If it is impossible to create an ebne number, print "-1" (without quotes); Otherwise, print the resulting number after deleting some, possibly zero, but not all digits. This number should be ebne. If there are multiple answers, you can print any of them. Note that answers with leading zeros or empty strings are not accepted. It's not necessary to minimize or maximize the number of deleted digits.ExampleInputCopy4
4
1227
1
0
6
177013
24
222373204424185217171912
OutputCopy1227
-1
17703
2237344218521717191
NoteIn the first test case of the example; 12271227 is already an ebne number (as 1+2+2+7=121+2+2+7=12, 1212 is divisible by 22, while in the same time, 12271227 is not divisible by 22) so we don't need to delete any digits. Answers such as 127127 and 1717 will also be accepted.In the second test case of the example, it is clearly impossible to create an ebne number from the given number.In the third test case of the example, there are many ebne numbers we can obtain by deleting, for example, 11 digit such as 1770317703, 7701377013 or 1701317013. Answers such as 17011701 or 770770 will not be accepted as they are not ebne numbers. Answer 013013 will not be accepted as it contains leading zeroes.Explanation: 1+7+7+0+3=181+7+7+0+3=18. As 1818 is divisible by 22 while 1770317703 is not divisible by 22, we can see that 1770317703 is an ebne number. Same with 7701377013 and 1701317013; 1+7+0+1=91+7+0+1=9. Because 99 is not divisible by 22, 17011701 is not an ebne number; 7+7+0=147+7+0=14. This time, 1414 is divisible by 22 but 770770 is also divisible by 22, therefore, 770770 is not an ebne number.In the last test case of the example, one of many other possible answers is given. Another possible answer is: 222373204424185217171912 →→ 22237320442418521717191 (delete the last digit). | [
"greedy",
"math",
"strings"
] | for _ in range(int(input())):
n = int(input())
s = input()
odd = []
for i in range(n):
if(int(s[i])%2 !=0):
odd.append(s[i])
if(len(odd)>=2):
odd = odd[-2:]
r = "".join(odd)
print(r)
else:
print(-1) | 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"
] | import sys
input = lambda: sys.stdin.readline().rstrip()
for _ in range(int(input())):
n = int(input())
li = list(map(int, input().split()))
if n % 2 == 1:
ip = True
for i in range(n // 2):
if li[i] < i or li[n - 1 - i] < i:
ip = False
break
if li[n // 2] < n // 2:
ip = False
if ip:
print('Yes')
else:
print('No')
else:
ip = True
for i in range(n // 2):
if li[i] < i or li[n - 1 - i] < i:
ip = False
break
mid_left = n // 2 - 1
mid_right = mid_left + 1
min_cri = n // 2 - 1
if not (li[mid_left] >= min_cri and li[mid_right] >= min_cri + 1 or li[mid_left] >= min_cri + 1 and li[mid_right] >= min_cri):
ip = False
if ip:
print('Yes')
else:
print('No') | 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, threading
import math
from os import path
from collections import deque
from bisect import *
from string import ascii_lowercase
from functools import cmp_to_key
from random import randint
import heapq
def readInts():
x = list(map(int, (sys.stdin.readline().rstrip().split())))
return x[0] if len(x) == 1 else x
def readList(type=int):
x = sys.stdin.readline()
x = list(map(type, x.rstrip('\n\r').split()))
return x
def readStr():
x = sys.stdin.readline().rstrip('\r\n')
return x
write = sys.stdout.write
read = sys.stdin.readline
MAXN = 1123456
class mydict:
def __init__(self, func):
self.random = randint(0, 1 << 32)
self.default = func
self.dict = {}
def __getitem__(self, key):
mykey = self.random ^ key
if mykey not in self.dict:
self.dict[mykey] = self.default()
return self.dict[mykey]
def get(self, key, default):
mykey = self.random ^ key
if mykey not in self.dict:
return default
return self.dict[mykey]
def __setitem__(self, key, item):
mykey = self.random ^ key
self.dict[mykey] = item
def getkeys(self):
return [self.random ^ i for i in self.dict]
def __str__(self):
return f'{[(self.random ^ i, self.dict[i]) for i in self.dict]}'
def lcm(a, b):
return (a*b)//(math.gcd(a,b))
def mod(n):
return n%(1000000000 + 7)
def solve(t):
# print(f'Case #{t}: ', end = '')
n, m = readInts()
p1 = (n-m)//(m+1)
p2 = p1+1
ncnt2 = (n-m)%(m+1)
ncnt1 = (m+1)-ncnt2
a1 = n-p1
n1 = ncnt1
if ncnt2 == 0:
n1 -= 1
a2 = n-(ncnt1*p1+ncnt1)-p2
n2 = max(0, ncnt2-1)
ans = 0
ans += (p1+1)*(n1*(2*a1+(n1-1)*(-p1-1)))//2
# print('a', ans)
ans += (p2+1)*(n2*(2*a2+(n2-1)*(-p2-1)))//2
print(ans)
def main():
t = 1
if path.exists("F:/Comp Programming/input.txt"):
sys.stdin = open("F:/Comp Programming/input.txt", 'r')
sys.stdout = open("F:/Comp Programming/output1.txt", 'w')
# sys.setrecursionlimit(10**5)
t = readInts()
for i in range(t):
solve(i+1)
if __name__ == '__main__':
main() | 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"
] | n, h, L, R = map(int, input().split())
w = [0] + list(map(int, input().split()))
dp = [[-1 << 31] * (n + 1) for _ in range(n + 1)]
for i in range(1, n + 1):
w[i] += w[i - 1]
dp[0][0] = 0
for i in range(1, n + 1):
for j in range(1, i + 1):
s1 = (w[i] - j) % h
s2 = (w[i] - j + 1) % h
dp[i][j] = dp[i - 1][j - 1]
if s1 >= L and s1 <= R:
dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + 1)
dp[i][j - 1] = max(dp[i - 1][j - 1], dp[i][j - 1])
if s2 >= L and s2 <= R:
dp[i][j - 1] = max(dp[i][j - 1], dp[i - 1][j - 1] + 1)
print(max(dp[n])) | 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"
] | # https://codeforces.com/problemset/problem/13/A
import math
def func_sol(raw_data):
n = int(raw_data.split('\n')[0])
s = 0
for i in range(2, n):
t = n
while t > 0:
s += t % i
t //= i
gcd = math.gcd(s, n - 2)
return f'{s // gcd}/{(n - 2) // gcd}\n'
def main():
try:
from codeforces.utilities import run_tests
run_tests(func_sol)
except ImportError:
from sys import stdin
print(func_sol(stdin.read()))
main()
| 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"
] | import math
def main():
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
if n == 1:
print("YES")
continue
ok = True
for i in range(n - 1):
ok &= (abs(a[i] - a[i + 1]) & 1) == 0
if ok:
print("YES")
else:
print("NO")
main()
| 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"
] | def solve():
n, x, y = map(int, input().split())
if x > y:
x, y = y, x
hi_left = n
if y == n:
hi_left -= 1
can_take = x + y - hi_left
best = 0
if can_take <= 0:
best = 1
else:
best = can_take
if x > can_take:
best += 1
worst = 0
if x + y > n:
worst = n
else:
worst = x + y - 1
if best < 1:
best = 1
elif best > n:
best = n
if worst < 1:
worst = 1
elif worst > n:
worst = n
print(best, worst)
test = int(input())
while test:
test -= 1
solve()
| py |
1307 | D | D. Cow and Fieldstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputBessie is out grazing on the farm; which consists of nn fields connected by mm bidirectional roads. She is currently at field 11, and will return to her home at field nn at the end of the day.The Cowfederation of Barns has ordered Farmer John to install one extra bidirectional road. The farm has kk special fields and he has decided to install the road between two different special fields. He may add the road between two special fields that already had a road directly connecting them.After the road is added, Bessie will return home on the shortest path from field 11 to field nn. Since Bessie needs more exercise, Farmer John must maximize the length of this shortest path. Help him!InputThe first line contains integers nn, mm, and kk (2≤n≤2⋅1052≤n≤2⋅105, n−1≤m≤2⋅105n−1≤m≤2⋅105, 2≤k≤n2≤k≤n) — the number of fields on the farm, the number of roads, and the number of special fields. The second line contains kk integers a1,a2,…,aka1,a2,…,ak (1≤ai≤n1≤ai≤n) — the special fields. All aiai are distinct.The ii-th of the following mm lines contains integers xixi and yiyi (1≤xi,yi≤n1≤xi,yi≤n, xi≠yixi≠yi), representing a bidirectional road between fields xixi and yiyi. It is guaranteed that one can reach any field from every other field. It is also guaranteed that for any pair of fields there is at most one road connecting them.OutputOutput one integer, the maximum possible length of the shortest path from field 11 to nn after Farmer John installs one road optimally.ExamplesInputCopy5 5 3
1 3 5
1 2
2 3
3 4
3 5
2 4
OutputCopy3
InputCopy5 4 2
2 4
1 2
2 3
3 4
4 5
OutputCopy3
NoteThe graph for the first example is shown below. The special fields are denoted by red. It is optimal for Farmer John to add a road between fields 33 and 55, and the resulting shortest path from 11 to 55 is length 33. The graph for the second example is shown below. Farmer John must add a road between fields 22 and 44, and the resulting shortest path from 11 to 55 is length 33. | [
"binary search",
"data structures",
"dfs and similar",
"graphs",
"greedy",
"shortest paths",
"sortings"
] | import sys
import collections
input = sys.stdin.readline
ints = lambda: list(map(int, input().split()))
n, m, k = ints()
special = ints()
graph = [[] for _ in range(n + 1)]
for _ in range(m):
x, y = ints()
graph[x].append(y)
graph[y].append(x)
distances = [[-1, -1] for _ in range(n + 1)]
def bfs(start):
d = [-1 for _ in range(n + 1)]
q = collections.deque([start])
depth = 1
d[start] = 0
while q:
for _ in range(len(q)):
node = q.popleft()
for nb in graph[node]:
if d[nb] == -1:
q.append(nb)
d[nb] = depth
depth += 1
return d
d1 = bfs(1)
d2 = bfs(n)
ans = d2[1]
s_distances = [(d1[i], d2[i]) for i in special]
s_distances.sort(key=lambda x: x[1], reverse=True)
mx = 0
for i in range(len(s_distances) - 1):
l = s_distances[i + 1][1] + 1 + s_distances[i][0]
mx = max(mx, l)
print(min(ans, mx))
| 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 |
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"
] | t = int(input())
import bisect
for _ in range(t):
n, m = map(int, input().split())
s = input()
p = sorted(list(map(int,input().split())))
counter = [0]*26
for i in range(1,n+1):
counter[ord(s[i-1])-ord('a')] += m-bisect.bisect_left(p,i)+1
print(*counter)
| 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
input=lambda:sys.stdin.readline().rstrip()
N=int(input())
A=list(map(int,input().split()))
if max(A)==0:
print(1 if N>1 else 0)
sys.exit()
B,L=[],[0]
rem=[N//2,(N+1)//2]
ans=2
dp,dp2=[[],[]],[[],[]]
for i in A:
if i>0:
L.append(0)
B.append(i)
rem[i%2]-=1
else:
L[-1]+=1
for i in range(1,len(L)-1):
if B[i-1]%2!=B[i]%2:
ans+=1
else:
dp[B[i]%2].append(L[i])
ans+=2
dp2[B[0]%2].append(L[0])
dp2[B[len(B)-1]%2].append(L[len(L)-1])
for i in range(2):
dp[i]=sorted(dp[i])
for j in range(len(dp[i])):
if rem[i]>=dp[i][j]:
rem[i]-=dp[i][j]
ans-=2
for j in range(len(dp2[i])):
if rem[i]>=dp2[i][j]:
rem[i]-=dp2[i][j]
ans-=1
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,sys
from random import randint, shuffle
from io import BytesIO, IOBase
from collections import defaultdict,deque,Counter
from bisect import bisect_left,bisect_right
from heapq import heappush,heappop
from functools import lru_cache
from itertools import accumulate, permutations
import math
# Fast IO Region
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")
# for _ in range(int(input())):
# n = int(input())
# a = list(map(int, input().split()))
# for _ in range(int(input())):
# x, y, a, b = list(map(int, input().split()))
# if (y - x) % (a + b):
# print(-1)
# else:
# print((y - x) // (a + b))
# n, m = list(map(int, input().split()))
# st = set()
# a = []
# for _ in range(n):
# s = input()
# if s[::-1] in st:
# st.remove(s[::-1])
# a.append(s)
# else:
# st.add(s)
# b = ''
# for s in st:
# if s == s[::-1]:
# b = s
# break
# a = ''.join(a)
# print(len(a) * 2 + len(b))
# print(a + b + a[::-1])
# for _ in range(int(input())):
# def solve():
# n, m = list(map(int, input().split()))
# a = [list(map(int, input().split())) for _ in range(n)]
# a.sort()
# mn = mx = m
# pre = 0
# for i in range(n):
# t, l, r = a[i]
# mn -= t - pre
# mx += t - pre
# if r < mn or mx < l:
# print('NO')
# return
# mn = max(mn, l)
# mx = min(mx, r)
# pre = t
# print('YES')
# solve()
for _ in range(int(input())):
n, s = input().split()
n = int(n)
a = []
cur = n
k = 1
for i in range(n - 1):
if s[i] == '<':
k += 1
else:
for j in range(k):
a.append(cur + j)
k = 1
cur -= 1
for j in range(k):
a.append(cur + j)
print(*a)
b = []
cur = 1
k = 1
for i in range(n - 1):
if s[i] == '>':
k += 1
else:
for j in range(k):
b.append(cur - j)
k = 1
cur += 1
for j in range(k):
b.append(cur - j)
print(*b)
| 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"
] | from sys import stdin
from collections import deque, Counter, defaultdict
N = int(input())
arr = list(map(int, stdin.readline().split()))
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
# lp is sieve, pr is a list of primes, pid is a dictionary keeping track of when the prime appears
vertex = [[] for i in range(len(pid))]
for n in arr:
new = []
while n > 1:
p, c = lp[n], 0
while lp[n] == p:
n //= p
c ^= 1
if c:
new.append(p)
if not new:
print (1)
exit()
new += [1]*(2 - len(new))
u, v = pid[new[0]], pid[new[1]]
vertex[u].append(v)
vertex[v].append(u)
def bfs_path(s):
# s is the index for the prime number
q = deque()
v = [-1]*len(pid)
v[s] = 0
q.append((s, -1))
while q:
c, p = q.pop()
for x in vertex[c]:
if v[x] != -1:
if x != p:
return v[x] + v[c] + 1
else:
v[x] = v[c] + 1
q.appendleft((x, c))
'''
q = deque()
vertices = vertex.keys()
vi = dict(zip(vertices, range(len(vertices))))
dist = [-1]*len(vertices)
q.append(source)
dist[vi[source]] = 0
far = 0
ans = float("inf")
for v in vertex[source]:
q = deque()
q.append(v)
dist = [-1]*len(vertices)
dist[vi[source]] = 0
dist[vi[v]] = 1
while len(q):
current = q.pop()
for v in vertex[current]:
# don't want to repeat an edge
if dist[vi[v]] in (-1, 0):
if v == end and dist[vi[current]] > 1:
if dist[vi[current]] + 1 < ans:
ans = dist[vi[current]] + 1
break
elif v != end:
dist[vi[v]] = dist[vi[current]] + 1
if dist[vi[v]] - 1 > far:
far = dist[vi[v]]
q.appendleft(v)
if ans != float("inf"):
return ans
else:
return 0
'''
ans = N + 1
for i in range(len(pid)):
if i > 0 and pr[i - 1] > MAX**0.5:
break
ans = min(ans, bfs_path(i) or ans)
print (ans if ans <= N else -1)
| py |
1316 | A | A. Grade Allocationtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputnn students are taking an exam. The highest possible score at this exam is mm. Let aiai be the score of the ii-th student. You have access to the school database which stores the results of all students.You can change each student's score as long as the following conditions are satisfied: All scores are integers 0≤ai≤m0≤ai≤m The average score of the class doesn't change. You are student 11 and you would like to maximize your own score.Find the highest possible score you can assign to yourself such that all conditions are satisfied.InputEach test contains multiple test cases. The first line contains the number of test cases tt (1≤t≤2001≤t≤200). The description of the test cases follows.The first line of each test case contains two integers nn and mm (1≤n≤1031≤n≤103, 1≤m≤1051≤m≤105) — the number of students and the highest possible score respectively.The second line of each testcase contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤m0≤ai≤m) — scores of the students.OutputFor each testcase, output one integer — the highest possible score you can assign to yourself such that both conditions are satisfied._ExampleInputCopy2
4 10
1 2 3 4
4 5
1 2 3 4
OutputCopy10
5
NoteIn the first case; a=[1,2,3,4]a=[1,2,3,4], with average of 2.52.5. You can change array aa to [10,0,0,0][10,0,0,0]. Average remains 2.52.5, and all conditions are satisfied.In the second case, 0≤ai≤50≤ai≤5. You can change aa to [5,1,1,3][5,1,1,3]. You cannot increase a1a1 further as it will violate condition 0≤ai≤m0≤ai≤m. | [
"implementation"
] | a = int(input())
for i in range(a):
q, s = map(int, input().split(' '))
*w, = map(int, input().split(' '))
print(min(s, sum(w))) | py |
1303 | D | D. Fill The Bagtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou have a bag of size nn. Also you have mm boxes. The size of ii-th box is aiai, where each aiai is an integer non-negative power of two.You can divide boxes into two parts of equal size. Your goal is to fill the bag completely.For example, if n=10n=10 and a=[1,1,32]a=[1,1,32] then you have to divide the box of size 3232 into two parts of size 1616, and then divide the box of size 1616. So you can fill the bag with boxes of size 11, 11 and 88.Calculate the minimum number of divisions required to fill the bag of size nn.InputThe first line contains one integer tt (1≤t≤10001≤t≤1000) — the number of test cases.The first line of each test case contains two integers nn and mm (1≤n≤1018,1≤m≤1051≤n≤1018,1≤m≤105) — the size of bag and the number of boxes, respectively.The second line of each test case contains mm integers a1,a2,…,ama1,a2,…,am (1≤ai≤1091≤ai≤109) — the sizes of boxes. It is guaranteed that each aiai is a power of two.It is also guaranteed that sum of all mm over all test cases does not exceed 105105.OutputFor each test case print one integer — the minimum number of divisions required to fill the bag of size nn (or −1−1, if it is impossible).ExampleInputCopy3
10 3
1 32 1
23 4
16 1 4 1
20 5
2 1 16 1 8
OutputCopy2
-1
0
| [
"bitmasks",
"greedy"
] | from collections import deque, defaultdict, Counter
from heapq import heappush, heappop, heapify
from math import inf, sqrt, ceil, log2
from functools import lru_cache
from itertools import accumulate, combinations, permutations, product
from typing import List
from bisect import bisect_left, bisect_right
import sys
import string
input=lambda:sys.stdin.readline().strip('\n')
mis=lambda:map(int,input().split())
ii=lambda:int(input())
#sys.setrecursionlimit(5*(10 ** 3))
T = ii()
for _ in range(T):
N, M = mis()
A = list(mis())
C = [0] * 61
sum = 0
for a in A:
C[int(log2(a))] += 1
sum += a
binn = format(N, 'b')[::-1]
ans = 0
if N > sum:
print(-1)
continue
for i in range(len(binn)):
if binn[i] == '0':
pass
elif C[i] > 0:
C[i] -= 1
else:
j = i + 1
while C[j] == 0:
j += 1
C[j] -= 1
C[j-1] += 1
j -= 1
ans += 1
while j > i:
ans += 1
C[j-1] += 1
j -= 1
C[i+1] += C[i]//2
print(ans)
| py |
1294 | F | F. Three Paths on a Treetime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given an unweighted tree with nn vertices. Recall that a tree is a connected undirected graph without cycles.Your task is to choose three distinct vertices a,b,ca,b,c on this tree such that the number of edges which belong to at least one of the simple paths between aa and bb, bb and cc, or aa and cc is the maximum possible. See the notes section for a better understanding.The simple path is the path that visits each vertex at most once.InputThe first line contains one integer number nn (3≤n≤2⋅1053≤n≤2⋅105) — the number of vertices in the tree. Next n−1n−1 lines describe the edges of the tree in form ai,biai,bi (1≤ai1≤ai, bi≤nbi≤n, ai≠biai≠bi). It is guaranteed that given graph is a tree.OutputIn the first line print one integer resres — the maximum number of edges which belong to at least one of the simple paths between aa and bb, bb and cc, or aa and cc.In the second line print three integers a,b,ca,b,c such that 1≤a,b,c≤n1≤a,b,c≤n and a≠,b≠c,a≠ca≠,b≠c,a≠c.If there are several answers, you can print any.ExampleInputCopy8
1 2
2 3
3 4
4 5
4 6
3 7
3 8
OutputCopy5
1 8 6
NoteThe picture corresponding to the first example (and another one correct answer):If you choose vertices 1,5,61,5,6 then the path between 11 and 55 consists of edges (1,2),(2,3),(3,4),(4,5)(1,2),(2,3),(3,4),(4,5), the path between 11 and 66 consists of edges (1,2),(2,3),(3,4),(4,6)(1,2),(2,3),(3,4),(4,6) and the path between 55 and 66 consists of edges (4,5),(4,6)(4,5),(4,6). The union of these paths is (1,2),(2,3),(3,4),(4,5),(4,6)(1,2),(2,3),(3,4),(4,5),(4,6) so the answer is 55. It can be shown that there is no better answer. | [
"dfs and similar",
"dp",
"greedy",
"trees"
] | from sys import stdin
input=lambda :stdin.readline()[:-1]
n=int(input())
edge=[[] for i in range(n)]
for i in range(n-1):
a,b=map(lambda x:int(x)-1,input().split())
edge[a].append(b)
edge[b].append(a)
def dfs(root):
par=[-1]*n
dist=[0]*n
todo=[(root,-1)]
while todo:
v,p=todo.pop()
for u in edge[v]:
if u!=p:
dist[u]=dist[v]+1
todo.append((u,v))
par[u]=v
mx=max(dist)
v0=dist.index(mx)
return par,v0,mx
_,v0,_=dfs(0)
par,v1,mx=dfs(v0)
if mx==n-1:
v2=0
while v2==v0 or v2==v1:
v2+=1
print(n-1)
print(v0+1,v1+1,v2+1)
exit()
dist=[-1]*n
todo=[]
now=v1
dist[now]=0
todo.append(now)
while now!=v0:
now=par[now]
todo.append(now)
dist[now]=0
while todo:
v=todo.pop()
for u in edge[v]:
if dist[u]==-1:
dist[u]=dist[v]+1
todo.append(u)
v2=dist.index(max(dist))
print(mx+max(dist))
print(v0+1,v1+1,v2+1) | py |
1296 | F | F. Berland Beautytime limit per test3 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThere are nn railway stations in Berland. They are connected to each other by n−1n−1 railway sections. The railway network is connected, i.e. can be represented as an undirected tree.You have a map of that network, so for each railway section you know which stations it connects.Each of the n−1n−1 sections has some integer value of the scenery beauty. However, these values are not marked on the map and you don't know them. All these values are from 11 to 106106 inclusive.You asked mm passengers some questions: the jj-th one told you three values: his departure station ajaj; his arrival station bjbj; minimum scenery beauty along the path from ajaj to bjbj (the train is moving along the shortest path from ajaj to bjbj). You are planning to update the map and set some value fifi on each railway section — the scenery beauty. The passengers' answers should be consistent with these values.Print any valid set of values f1,f2,…,fn−1f1,f2,…,fn−1, which the passengers' answer is consistent with or report that it doesn't exist.InputThe first line contains a single integer nn (2≤n≤50002≤n≤5000) — the number of railway stations in Berland.The next n−1n−1 lines contain descriptions of the railway sections: the ii-th section description is two integers xixi and yiyi (1≤xi,yi≤n,xi≠yi1≤xi,yi≤n,xi≠yi), where xixi and yiyi are the indices of the stations which are connected by the ii-th railway section. All the railway sections are bidirected. Each station can be reached from any other station by the railway.The next line contains a single integer mm (1≤m≤50001≤m≤5000) — the number of passengers which were asked questions. Then mm lines follow, the jj-th line contains three integers ajaj, bjbj and gjgj (1≤aj,bj≤n1≤aj,bj≤n; aj≠bjaj≠bj; 1≤gj≤1061≤gj≤106) — the departure station, the arrival station and the minimum scenery beauty along his path.OutputIf there is no answer then print a single integer -1.Otherwise, print n−1n−1 integers f1,f2,…,fn−1f1,f2,…,fn−1 (1≤fi≤1061≤fi≤106), where fifi is some valid scenery beauty along the ii-th railway section.If there are multiple answers, you can print any of them.ExamplesInputCopy4
1 2
3 2
3 4
2
1 2 5
1 3 3
OutputCopy5 3 5
InputCopy6
1 2
1 6
3 1
1 5
4 1
4
6 1 3
3 4 1
6 5 2
1 2 5
OutputCopy5 3 1 2 1
InputCopy6
1 2
1 6
3 1
1 5
4 1
4
6 1 1
3 4 3
6 5 3
1 2 4
OutputCopy-1
| [
"constructive algorithms",
"dfs and similar",
"greedy",
"sortings",
"trees"
] | import sys
class RangeMinQuery:
# O(nlog(n)) construction/space, O(1) range minimum query
def __init__(self, data):
self._data = [list(data)]
n = len(self._data[0])
# self._data[i][j] stores the min of the segment [j, j + 2 ** i] where i is in [1,2,4,8,...,log(N)]
w = 1
while 2 * w <= n:
prev = self._data[-1]
self._data.append([min(prev[j], prev[j + w]) for j in range(n - 2 * w + 1)])
w *= 2
def query(self, begin, end):
# Find min of [begin, end)
# Take the min of the overlapping intervals [begin, begin + 2**depth) and [end - 2**depth, end)
assert begin < end
depth = (end - begin).bit_length() - 1
return min(self._data[depth][begin], self._data[depth][end - (1 << depth)])
class LCA:
def __init__(self, graph, root=1):
# Euler tour representation recording every visit to each node in the DFS
self.eulerTour = []
# For each node record the index of their first visit in the euler tour
self.preorder = [None] * len(graph)
# Record last visit
self.postorder = [None] * len(graph)
# Depth of each node to the root
self.depth = [None] * len(graph)
# Parent of each node
self.parent = [None] * len(graph)
# DFS
stack = [root]
self.depth[root] = 0
self.parent[root] = None
while stack:
node = stack.pop()
# Record a visit in the euler tour
self.eulerTour.append(node)
if self.preorder[node] is None:
# Record first visit
self.preorder[node] = len(self.eulerTour) - 1
for nbr in graph[node]:
if self.preorder[nbr] is None:
self.depth[nbr] = self.depth[node] + 1
self.parent[nbr] = node
stack.append(node)
stack.append(nbr)
# Record last visit (this can be overwritten)
self.postorder[node] = len(self.eulerTour) - 1
self.rmq = RangeMinQuery(self.preorder[node] for node in self.eulerTour)
if False:
# Check invariants
assert self.depth[1] == 0
assert self.parent[1] == None
for node in range(1, len(graph)):
assert self.preorder[node] == self.eulerTour.index(node)
assert self.postorder[node] == max(
i for i, x in enumerate(self.eulerTour) if x == node
)
for i in range(len(self.rmq._data)):
for j in range(i + 1, len(self.rmq._data) + 1):
assert min(self.rmq._data[0][i:j]) == self.rmq.query(i, j)
for u in range(1, N + 1):
for v in range(1, N + 1):
assert self.lca(u, v) == self.lca(v, u)
lca = self.lca(u, v)
assert self.isAncestor(lca, u)
assert self.isAncestor(lca, v)
path = self.path(u, v)
assert len(path) == self.dist(u, v) + 1
for x, y in zip(path, path[1:]):
assert y in graph[x]
assert len(set(path)) == len(path)
def lca(self, u, v):
a = self.preorder[u]
b = self.preorder[v]
if a > b:
a, b = b, a
return self.eulerTour[self.rmq.query(a, b + 1)]
def dist(self, u, v):
return self.depth[u] + self.depth[v] - 2 * self.depth[self.lca(u, v)]
def isAncestor(self, ancestor, v):
# Checks if `ancestor` is an ancestor of `v`
return (
self.preorder[ancestor] <= self.preorder[v]
and self.postorder[v] <= self.postorder[ancestor]
)
def path(self, u, v):
# Returns the path from u to v in the tree. This doesn't rely on rmq
# Walk up from u until u becomes ancestor of v (the LCA)
uPath = []
while not self.isAncestor(u, v):
uPath.append(u)
u = self.parent[u]
lca = u
# Walk up from v until v is at the LCA too
assert self.isAncestor(u, v)
vPath = []
while v != lca:
vPath.append(v)
v = self.parent[v]
ret = uPath + [lca] + vPath[::-1]
return ret
def solve(N, edges, passengers):
graph = [[] for i in range(N + 1)]
for e in edges:
graph[e[0]].append(e[1])
graph[e[1]].append(e[0])
tree = LCA(graph)
scores = [[1 for j in range(N + 1)] for i in range(N + 1)]
for a, b, score in passengers:
path = tree.path(a, b)
for u, v in zip(path, path[1:]):
scores[u][v] = max(scores[u][v], score)
scores[v][u] = max(scores[v][u], score)
for a, b, score in passengers:
ans = float("inf")
path = tree.path(a, b)
for u, v in zip(path, path[1:]):
ans = min(ans, scores[u][v])
if ans != score:
return -1
return " ".join(str(scores[e[0]][e[1]]) for e in edges)
if False:
N = 5000
edges = list(zip(range(N - 1), range(1, N)))
assert len(edges) == N - 1
passengers = [(0, p, N - p) for p in range(1, N)]
solve(N, edges, passengers)
exit(0)
if __name__ == "__main__":
input = sys.stdin.readline
N = int(input())
edges = [[int(x) for x in input().split()] for i in range(N - 1)]
P = int(input())
passengers = [[int(x) for x in input().split()] for i in range(P)]
ans = solve(N, edges, passengers)
print(ans)
| 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"
] | # https://codeforces.com/problemset/problem/1294/B
test_cases = int(input())
def handle():
lines = int(input())
points = []
for _ in range(lines):
x, y = input().split(" ")
x = int(x)
y = int(y)
points.append((x, y))
points.sort()
y_so_far = points[0][1]
for x, y in points[1:]:
if y < y_so_far:
return ["NO"]
y_so_far = y
current_x = 0
current_y = 0
path = ""
for x, y in points:
while current_x != x:
path += "R"
current_x += 1
while current_y != y:
path += "U"
current_y += 1
return ["YES", path]
for _ in range(test_cases):
result = handle()
for line in result:
print(line)
| py |
1307 | D | D. Cow and Fieldstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputBessie is out grazing on the farm; which consists of nn fields connected by mm bidirectional roads. She is currently at field 11, and will return to her home at field nn at the end of the day.The Cowfederation of Barns has ordered Farmer John to install one extra bidirectional road. The farm has kk special fields and he has decided to install the road between two different special fields. He may add the road between two special fields that already had a road directly connecting them.After the road is added, Bessie will return home on the shortest path from field 11 to field nn. Since Bessie needs more exercise, Farmer John must maximize the length of this shortest path. Help him!InputThe first line contains integers nn, mm, and kk (2≤n≤2⋅1052≤n≤2⋅105, n−1≤m≤2⋅105n−1≤m≤2⋅105, 2≤k≤n2≤k≤n) — the number of fields on the farm, the number of roads, and the number of special fields. The second line contains kk integers a1,a2,…,aka1,a2,…,ak (1≤ai≤n1≤ai≤n) — the special fields. All aiai are distinct.The ii-th of the following mm lines contains integers xixi and yiyi (1≤xi,yi≤n1≤xi,yi≤n, xi≠yixi≠yi), representing a bidirectional road between fields xixi and yiyi. It is guaranteed that one can reach any field from every other field. It is also guaranteed that for any pair of fields there is at most one road connecting them.OutputOutput one integer, the maximum possible length of the shortest path from field 11 to nn after Farmer John installs one road optimally.ExamplesInputCopy5 5 3
1 3 5
1 2
2 3
3 4
3 5
2 4
OutputCopy3
InputCopy5 4 2
2 4
1 2
2 3
3 4
4 5
OutputCopy3
NoteThe graph for the first example is shown below. The special fields are denoted by red. It is optimal for Farmer John to add a road between fields 33 and 55, and the resulting shortest path from 11 to 55 is length 33. The graph for the second example is shown below. Farmer John must add a road between fields 22 and 44, and the resulting shortest path from 11 to 55 is length 33. | [
"binary search",
"data structures",
"dfs and similar",
"graphs",
"greedy",
"shortest paths",
"sortings"
] | from collections import deque
import os
import sys
from io import BytesIO, IOBase
# region fastio
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")
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))
n,m,k = map(int,input().split())
a = list(map(lambda x: int(x) - 1,input().split()))
G = [[] for _ in range(n)]
for _ in range(m):
u,v = map(int,input().split())
u -= 1
v -= 1
G[u].append(v)
G[v].append(u)
d1 = [-1]*n
d2 = [-1]*n
d1[0] = 0
d2[n-1] = 0
q = deque([0])
while q:
u = q.popleft()
for v in G[u]:
if d1[v] == -1:
d1[v] = d1[u]+1
q.append(v)
q = deque([n-1])
while q:
u = q.popleft()
for v in G[u]:
if d2[v] == -1:
d2[v] = d2[u]+1
q.append(v)
a.sort(key = lambda u: d1[u])
B = SortedList([d2[u] for u in a])
case1 = 0
for u in a:
B.remove(d2[u])
if len(B) == 0:
continue
case = d1[u] + B[-1] + 1
case1 = max(case, case1)
ans = min(case1, d1[n-1])
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"
] |
t = int(input())
for i in range(t):
n = int(input())
A = list(map(int,input().split()))
ans = 0
ce = 0
co = 0
for i in A:
if(i&1):
co+=1
else:
ce+=1
ans = (i&1)^ans
if ce == n:
print("NO")
elif not ans and co == n:
print("NO")
else:
print("YES")
| 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"
] | t = int(input())
for _ in range(t):
x = int(input())
print(1, x-1) | 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"
] | 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,k=map(int,input().split())
dic={}
graph=[[0,i] for i in range(n)]
for i in range(n-1):
u,v=map(int,input().split())
u-=1
v-=1
dic[(u,v)]=[i]
graph[u].append(v)
graph[u][0]+=1
graph[v].append(u)
graph[v][0]+=1
graph.sort(reverse=True)
unf=[0]*n
temp=[0]*n
for i in range(n):
if(i<k):
unf[graph[i][1]]=1
temp[graph[i][1]]=graph[i][2:]
graph=temp
curr=0
stk=[curr]
vis=[-1]*n
vis[0]=0
ma=-1
while(stk!=[]):
f=stk.pop()
if(unf[f]):
if(f==0):
col=1
else:
if((f,vis[f]) in dic):
col=dic[(f,vis[f])][1]
else:
col=dic[(vis[f],f)][1]
for i in graph[f]:
if(vis[i]!=-1):
continue
vis[i]=f
if((i,f) in dic):
dic[(i,f)].append(col)
else:
dic[(f,i)].append(col)
stk.append(i)
else:
st=1
avoid=-1
if(f!=0):
if((f,vis[f]) in dic):
avoid=dic[(f,vis[f])][1]
else:
avoid=dic[(vis[f],f)][1]
for i in graph[f]:
if(vis[i]!=-1):
continue
vis[i]=f
if(st==avoid):
st+=1
if((i,f) in dic):
dic[(i,f)].append(st)
else:
dic[(f,i)].append(st)
ma=max(ma,st)
st+=1
stk.append(i)
ans=[0]*(n-1)
for i in dic:
ans[dic[i][0]]=dic[i][1]
ma=max(ma,1)
print(ma)
print(*ans)
main()
| 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"
] | from typing import List, Tuple
def return_max(a: List[int]) -> Tuple[List[int], int]:
ans = []; prev_min = float('inf')
s = 0
for e in a:
if e < prev_min:
ans.append(e)
prev_min = e
s += e
else:
ans.append(prev_min)
s += prev_min
return ans, s
n = int(input())
m = [int(x) for x in input().split()]
maxim = float('-inf')
ans = None
for peak_ind in range(n):
first_half, first_sum = return_max(m[:peak_ind + 1][::-1])
first_half = first_half[::-1]
second_half, second_sum = return_max(m[peak_ind:])
curr_sum = first_sum + second_sum - m[peak_ind]
if curr_sum > maxim:
maxim = curr_sum
ans = first_half[:-1] + second_half
print(' '.join([str(e) for e in ans]))
| 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)))
mx = 0
ind = -1
for i in range(n):
if l[i]+r[i]-arr[i] > mx:
mx = l[i]+r[i]-arr[i]
ind = i
if ind != -1:
ans[ind] = arr[ind]
for i in range(ind-1, -1, -1):
ans[i] = min(arr[i], ans[i+1])
for i in range(ind+1, n):
ans[i] = min(arr[i], ans[i-1])
print(*ans)
# calling main Function
if __name__ == '__main__':
main()
| 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"
] | from __future__ import division, print_function
import io
import os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n = int(input())
l = list(map(int, input().split()))
# not my solution, from: https://codeforces.com/contest/1299/submission/70653333
su=[l[0]]
cou=[-1,0]
for k in range(1,n):
nd=1
ns=l[k]
while len(cou)>1 and su[-1]*(cou[-1]-cou[-2]+nd)>(su[-1]+ns)*(cou[-1]-cou[-2]):
nd+=cou[-1]-cou[-2]
ns+=su[-1]
su.pop()
cou.pop()
cou.append(k)
su.append(ns)
floats = []
for k in range(len(su)):
floats += [su[k] / (cou[k + 1] - cou[k])] * (cou[k + 1] - cou[k])
if True:
if False:
# 3244 ms, TLE on pypy3
# But 1044 ms on pypy2!!!
print("\n".join(map(str, floats)))
if False:
# 2028 ms
os.write(1, b"\n".join(str(x).encode() for x in floats))
if False:
# 1544 ms
# %-formatting for bytes, added in python 3.5: https://www.python.org/dev/peps/pep-0461/
os.write(1, b"\n".join(b"%.9f" % x for x in floats))
if False:
# 842 ms
# Format with ints instead. TODO: handle negatives
os.write(1, b"\n".join(b"%d.%09d" % (x, (x - int(x)) * 1e9 + 0.5) for x in floats))
if True:
# 748 ms
# Same as above
ans = bytearray()
for x in floats:
i = int(x)
ans += b"%d.%09d\n" % (i, (x - i) * 1e9 + 0.5)
os.write(1, ans)
else:
# 670 ms
# Use that fact that most of the output are repeats (specific to this problem)
ans = bytearray()
for k in range(len(su)):
repeat = cou[k + 1] - cou[k]
x = su[k] / repeat
i = int(x)
ans += (b"%d.%09d\n" % (i, (x - i) * 1e9 + 0.5)) * repeat
os.write(1, ans)
| py |
1300 | A | A. Non-zerotime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputGuy-Manuel and Thomas have an array aa of nn integers [a1,a2,…,ana1,a2,…,an]. In one step they can add 11 to any element of the array. Formally, in one step they can choose any integer index ii (1≤i≤n1≤i≤n) and do ai:=ai+1ai:=ai+1.If either the sum or the product of all elements in the array is equal to zero, Guy-Manuel and Thomas do not mind to do this operation one more time.What is the minimum number of steps they need to do to make both the sum and the product of all elements in the array different from zero? Formally, find the minimum number of steps to make a1+a2+a1+a2+ …… +an≠0+an≠0 and a1⋅a2⋅a1⋅a2⋅ …… ⋅an≠0⋅an≠0.InputEach test contains multiple test cases. The first line contains the number of test cases tt (1≤t≤1031≤t≤103). The description of the test cases follows.The first line of each test case contains an integer nn (1≤n≤1001≤n≤100) — the size of the array.The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (−100≤ai≤100−100≤ai≤100) — elements of the array .OutputFor each test case, output the minimum number of steps required to make both sum and product of all elements in the array different from zero.ExampleInputCopy4
3
2 -1 -1
4
-1 0 0 1
2
-1 2
3
0 -2 1
OutputCopy1
2
0
2
NoteIn the first test case; the sum is 00. If we add 11 to the first element, the array will be [3,−1,−1][3,−1,−1], the sum will be equal to 11 and the product will be equal to 33.In the second test case, both product and sum are 00. If we add 11 to the second and the third element, the array will be [−1,1,1,1][−1,1,1,1], the sum will be equal to 22 and the product will be equal to −1−1. It can be shown that fewer steps can't be enough.In the third test case, both sum and product are non-zero, we don't need to do anything.In the fourth test case, after adding 11 twice to the first element the array will be [2,−2,1][2,−2,1], the sum will be 11 and the product will be −4−4. | [
"implementation",
"math"
] | def solve():
n = int(input())
a = list(map(int,input().split()))
b = a.count(0)
return b + int(sum(a) + b == 0)
for i in range(int(input())):
print(solve()) | py |
1312 | C | C. Adding Powerstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputSuppose you are performing the following algorithm. There is an array v1,v2,…,vnv1,v2,…,vn filled with zeroes at start. The following operation is applied to the array several times — at ii-th step (00-indexed) you can: either choose position pospos (1≤pos≤n1≤pos≤n) and increase vposvpos by kiki; or not choose any position and skip this step. You can choose how the algorithm would behave on each step and when to stop it. The question is: can you make array vv equal to the given array aa (vj=ajvj=aj for each jj) after some step?InputThe first line contains one 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 kk (1≤n≤301≤n≤30, 2≤k≤1002≤k≤100) — the size of arrays vv and aa and value kk used in the algorithm.The second line contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤10160≤ai≤1016) — the array you'd like to achieve.OutputFor each test case print YES (case insensitive) if you can achieve the array aa after some step or NO (case insensitive) otherwise.ExampleInputCopy5
4 100
0 0 0 0
1 2
1
3 4
1 4 1
3 2
0 1 3
3 9
0 59049 810
OutputCopyYES
YES
NO
NO
YES
NoteIn the first test case; you can stop the algorithm before the 00-th step, or don't choose any position several times and stop the algorithm.In the second test case, you can add k0k0 to v1v1 and stop the algorithm.In the third test case, you can't make two 11 in the array vv.In the fifth test case, you can skip 9090 and 9191, then add 9292 and 9393 to v3v3, skip 9494 and finally, add 9595 to v2v2. | [
"bitmasks",
"greedy",
"implementation",
"math",
"number theory",
"ternary search"
] | for _ in range(int(input())):
[n,k]=list(map(int,input().split(" ")))
arr=list(map(int,input().split(" ")))
arr.sort(reverse=True)
if arr[0]==0:
print("yes")
continue
if k==0:
print("no")
continue
if k==1:
print("yes")
continue
pow=0
while(k**pow<arr[0]):
pow+=1
c=[0 for i in range(pow+2)]
for i in arr:
ind=0
while(i):
c[ind]+=i%k
i=i//k
ind+=1
t=1
for i in c:
if i>=2:
t=0
print("yes" if t else "no") | py |
1303 | E | E. Erase Subsequencestime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given a string ss. You can build new string pp from ss using the following operation no more than two times: choose any subsequence si1,si2,…,siksi1,si2,…,sik where 1≤i1<i2<⋯<ik≤|s|1≤i1<i2<⋯<ik≤|s|; erase the chosen subsequence from ss (ss can become empty); concatenate chosen subsequence to the right of the string pp (in other words, p=p+si1si2…sikp=p+si1si2…sik). Of course, initially the string pp is empty. For example, let s=ababcds=ababcd. At first, let's choose subsequence s1s4s5=abcs1s4s5=abc — we will get s=bads=bad and p=abcp=abc. At second, let's choose s1s2=bas1s2=ba — we will get s=ds=d and p=abcbap=abcba. So we can build abcbaabcba from ababcdababcd.Can you build a given string tt using the algorithm above?InputThe first line contains the single integer TT (1≤T≤1001≤T≤100) — the number of test cases.Next 2T2T lines contain test cases — two per test case. The first line contains string ss consisting of lowercase Latin letters (1≤|s|≤4001≤|s|≤400) — the initial string.The second line contains string tt consisting of lowercase Latin letters (1≤|t|≤|s|1≤|t|≤|s|) — the string you'd like to build.It's guaranteed that the total length of strings ss doesn't exceed 400400.OutputPrint TT answers — one per test case. Print YES (case insensitive) if it's possible to build tt and NO (case insensitive) otherwise.ExampleInputCopy4
ababcd
abcba
a
b
defi
fed
xyz
x
OutputCopyYES
NO
NO
YES
| [
"dp",
"strings"
] | from collections import deque, defaultdict, Counter
from heapq import heappush, heappop, heapify
from math import inf, sqrt, ceil, log2
from functools import lru_cache
from itertools import accumulate, combinations, permutations, product
from typing import List
from bisect import bisect_left, bisect_right
import sys
import string
input=lambda:sys.stdin.readline().strip('\n')
mis=lambda:map(int,input().split())
ii=lambda:int(input())
#sys.setrecursionlimit(5*(10 ** 3))
T = ii()
for _ in range(T):
S = input()
T = input()
N = len(S)
M = len(T)
nxt = [[N] * 26 for _ in range(N+2)]
for i in reversed(range(N)):
nxt[i] = nxt[i+1][:]
nxt[i][ord(S[i]) - ord('a')] = i
for mid in range(M):
dp = [[N] * (M+1) for _ in range(M+1)]
left = T[:mid]
right = T[mid:]
dp[0][0] = -1
for i in range(len(left)+1):
for j in range(len(right)+1):
if i > 0:
dp[i][j] = min(dp[i][j], nxt[dp[i-1][j]+1][ord(left[i-1])-ord('a')])
if j > 0:
dp[i][j] = min(dp[i][j], nxt[dp[i][j-1]+1][ord(right[j-1])-ord('a')])
if dp[len(left)][len(right)] < N:
print("YES")
break
else:
print("NO")
| 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"
] | from __future__ import division, print_function
import math
import sys
import os
from io import BytesIO, IOBase
#from collections import deque, Counter, OrderedDict, defaultdict
#import heapq
#ceil,floor,log,sqrt,factorial,pow,pi,gcd
#import bisect
#from bisect import bisect_left,bisect_right
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")
def I():
return(int(input()))
def lint():
return(list(map(int,input().split())))
def insr():
s = input().strip()
return(list(s[:len(s)]))
def invr():
return(map(int,input().split()))
import itertools
from sys import maxsize, stdout, stdin,stderr
mod = int(1e9+7)
import sys
# def I(): return int(stdin.readline())
# def lint(): return [int(x) for x in stdin.readline().split()]
def S(): return list(map(str,input().strip()))
def grid(r, c): return [lint() for i in range(r)]
from collections import defaultdict, Counter, deque
import math
import heapq
from heapq import heappop , heappush
import bisect
from math import factorial, inf
from itertools import groupby
from itertools import permutations as comb
# mod=1000000007
# MOD = 998244353
# def gcd(a,b):
# while b:
# a %= b
# tmp = a
# a = b
# b = tmp
# return a
# def lcm(a,b):
# return a // gcd(a, b) * b
# def check_prime(n):
# for i in range(2, int(n ** (1 / 2)) + 1):
# if not n % i:
# return False
# return True
# # def nCr(n, r):
# # return (fact(n) // (fact(r)
# # * fact(n - r)))
# # Returns factorial of n
# # def fact(n):
# # res = 1
# # for i in range(2, n+1):
# # res = res * i
# # return res
# def primefactors(n):
# num=0
# while n % 2 == 0:
# num+=1
# n = n / 2
# for i in range(3,int(math.sqrt(n))+1,2):
# while n % i== 0:
# num+=1
# n = n // i
# if n > 2:
# num+=1
# return num
# def ask(a):
# print('? 1 {}'.format(a),flush=True)
# # print(c,a,b,flush=True)
# n=I()
# return n
# return n
# def linear_sieve(n):
# is_composite=[False]*n
# prime=[]
# for i in range(2,n):
# if not is_composite[i]:
# prime.append(i)
# for j in prime:
# is_composite[i*j]=True
# if i%prime==0:
# break
# return prime
# def primeFactors(n):
# l=[]
# while n % 2 == 0:
# l.append(2)
# n = n // 2
# for i in range(3,int(math.sqrt(n))+1,2):
# while n % i== 0:
# l.append(i)
# n = n // i
# if n > 2:
# l.append(n)
# return l
# def binpow(a,b,c):
# res=1
# a=a%c
# while b>0:
# if b&1:
# res=(res*a)%c
# b=b>>1
# a = (a*a)%c
# return res
# # Sieve
# d=[]
# primes=[]
# prim=[0]*(10**5+1)
# def sieve(n):
# for i in range(n):
# d.append(i)
# for i in range(2,n):
# if d[i]==i:
# prim[i]=1
# primes.append(i)
# j=0
# while j<len(primes) and primes[j]<=d[i] and primes[j]*i<n:
# d[i * primes[j]] = primes[j]
# j+=1
# def primeFactors(n):
# factors=[]
# while n!=1:
# factors.append(d[n])
# n//=d[n]
# return factors
# def ask(a):
# print('? {}'.format(a),flush=True)
# n=I()
# return n
t = 1
# t = I()
for _ in range(t):
n,m=lint()
tmp=0
ans=0
l=[]
b=1
tot=0
a=1
fact=[0]*250005
fact[0]=1
for i in range(1,n+1):
fact[i] = (fact[i-1] * i )%m
for i in range(1,n+1):
ans=(ans+ ((n-i+1)*(fact[i]*fact[n-i+1])%m)%m)%m
print(ans) | 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"
] | def solve():
n = int(input())
a = list(map(int, input().split()))
if a.count(-1) == n: return print(0, 0)
store = []
for i in range(n):
if a[i] == -1: continue
neighbours = []
if i > 0: neighbours.append(a[i-1])
if i < n-1: neighbours.append(a[i+1])
if -1 in neighbours: store.append(a[i])
up = max(store)
down = min(store)
mid = (up + down) // 2
for i in range(n):
if a[i] == -1: a[i] = mid
# print('a:', store)
print(max([abs(a[i+1]-a[i]) for i in range(n-1)]), mid)
for case in range(int(input())):
solve() | 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"
] | t=int(input())
s=input()
print(len(s)+1)
###################
| 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"
] | test_cases = int(input())
for t in range(test_cases):
s=[]
ws=list(input())
for w in ws:
if len(s) == 0 or s[-1][0] != w:
s.append((w, 1))
else:
s[-1] = (s[-1][0], s[-1][1]+1)
s.reverse()
d=0
p=0
max_l = 0
while s:
c, n = s.pop()
if c == 'L':
max_l = max(max_l, n)
print(max_l+1) | 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"
] | a=int(input())
s=input()
cn,cn1=s.count('L'),s.count('R')
print(cn+cn1+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 os, sys
from io import BytesIO, IOBase
from collections import defaultdict, deque, Counter
from bisect import bisect_left, bisect_right
from heapq import heappush, heappop
from functools import lru_cache
from itertools import accumulate
import math
# Fast IO Region
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")
n = int(input())
l = list(map(int, input().split()))
cost = list(map(int, input().split()))
lst = [(l[i], -cost[i]) for i in range(n)]
lst.sort()
cur = 0
stl = []
sum_stl = 0
res = 0
for num, c in lst:
if num == cur:
heappush(stl, c)
sum_stl += c
else:
for i in range(min(n, num - cur)):
if not stl:
break
c_ = heappop(stl)
sum_stl -= c_
res += sum_stl
cur = num
heappush(stl, c)
sum_stl += c
while stl:
sum_stl -= heappop(stl)
res += sum_stl
print(-res) | 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"
] | from math import *
def get(f): return f(input().strip())
def gets(f): return [*map(f, input().split())]
for _ in range(get(int)):
n = get(int)
print('7' + '1' * (n - 3 >> 1) if n & 1 else '1' * (n >> 1))
| py |
1305 | D | D. Kuroni and the Celebrationtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputThis is an interactive problem.After getting AC after 13 Time Limit Exceeded verdicts on a geometry problem; Kuroni went to an Italian restaurant to celebrate this holy achievement. Unfortunately, the excess sauce disoriented him, and he's now lost!The United States of America can be modeled as a tree (why though) with nn vertices. The tree is rooted at vertex rr, wherein lies Kuroni's hotel.Kuroni has a phone app designed to help him in such emergency cases. To use the app, he has to input two vertices uu and vv, and it'll return a vertex ww, which is the lowest common ancestor of those two vertices.However, since the phone's battery has been almost drained out from live-streaming Kuroni's celebration party, he could only use the app at most ⌊n2⌋⌊n2⌋ times. After that, the phone would die and there will be nothing left to help our dear friend! :(As the night is cold and dark, Kuroni needs to get back, so that he can reunite with his comfy bed and pillow(s). Can you help him figure out his hotel's location?InteractionThe interaction starts with reading a single integer nn (2≤n≤10002≤n≤1000), the number of vertices of the tree.Then you will read n−1n−1 lines, the ii-th of them has two integers xixi and yiyi (1≤xi,yi≤n1≤xi,yi≤n, xi≠yixi≠yi), denoting there is an edge connecting vertices xixi and yiyi. It is guaranteed that the edges will form a tree.Then you can make queries of type "? u v" (1≤u,v≤n1≤u,v≤n) to find the lowest common ancestor of vertex uu and vv.After the query, read the result ww as an integer.In case your query is invalid or you asked more than ⌊n2⌋⌊n2⌋ queries, the program will print −1−1 and will finish interaction. You will receive a Wrong answer verdict. Make sure to exit immediately to avoid getting other verdicts.When you find out the vertex rr, print "! rr" and quit after that. This query does not count towards the ⌊n2⌋⌊n2⌋ limit.Note that the tree 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.HacksTo hack, use the following format:The first line should contain two integers nn and rr (2≤n≤10002≤n≤1000, 1≤r≤n1≤r≤n), denoting the number of vertices and the vertex with Kuroni's hotel.The ii-th of the next n−1n−1 lines should contain two integers xixi and yiyi (1≤xi,yi≤n1≤xi,yi≤n) — denoting there is an edge connecting vertex xixi and yiyi.The edges presented should form a tree.ExampleInputCopy6
1 4
4 2
5 3
6 3
2 3
3
4
4
OutputCopy
? 5 6
? 3 1
? 1 2
! 4NoteNote 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.The image below demonstrates the tree in the sample test: | [
"constructive algorithms",
"dfs and similar",
"interactive",
"trees"
] | import math
from sys import stdin
input = stdin.readline
#// - remember to add .strip() when input is a string
n = int(input())
edges = [-1]
for i in range(n):
temp = []
edges.append(temp)
for _ in range(n-1):
u,v = map(int,input().split())
edges[u].append(v)
edges[v].append(u)
#print(edges)
def query(u,v):
print("?",u,v,flush=True)
return int(input())
leaves = []
for i in range(1,n+1):
if len(edges[i]) == 1:
leaves.append(i)
breaklol = False
#print(edges)
while len(leaves) > 1:
while len(leaves) > 1:
#if len(leaves) == 1:
#new_leaves.append(leaves[0])
#leaves.pop()
#break
q = query(leaves[-1],leaves[-2])
if q == leaves[-1] or q == leaves[-2]:
print("!",q,flush = True)
breaklol = True
break
edges[leaves[-1]] = -1
edges[leaves[-2]] = -1
for i in range(1,len(edges)):
if edges[i] != -1:
if leaves[-1] in edges[i]:
edges[i].remove(leaves[-1])
if leaves[-2] in edges[i]:
edges[i].remove(leaves[-2])
leaves.pop()
leaves.pop()
if breaklol:
break
leaves = []
for i in range(len(edges)):
if edges[i] != -1 and len(edges[i]) == 1:
leaves.append(i)
#print(leaves)
ans = 0
cc = 0
for i in range(len(edges)):
if edges[i] != -1:
cc += 1
ans = i
#print(edges)
if cc == 1:
print("!",ans,flush = True) | 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"
] | import math
t = int(input())
for _ in range(t):
x = int(input())
for d in range(1, int(x**0.5) + 1):
if x % d == 0:
ab = x * d - d**2
if ab > 0 and ab % d == 0:
b = ab // d
a = x // d - b
print(a, b)
break
| 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"
] | from collections import deque, defaultdict, Counter
from heapq import heappush, heappop, heapify
from math import inf, sqrt, ceil
from functools import lru_cache
from itertools import accumulate, combinations, permutations, product
from typing import List
from bisect import bisect_left, bisect_right
import sys
input=lambda:sys.stdin.readline().strip('\n')
mis=lambda:map(int,input().split())
ii=lambda:int(input())
#sys.setrecursionlimit(5*(10 ** 3))
T = ii()
for _ in range(T):
N, G, B = mis()
m = (N+1)//2
ans = (m-1)//G*(G+B) + (m%G if m%G else G)
print(max(N, ans))
| py |
1312 | B | B. Bogosorttime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given an array a1,a2,…,ana1,a2,…,an. Array is good if for each pair of indexes i<ji<j the condition j−aj≠i−aij−aj≠i−ai holds. Can you shuffle this array so that it becomes good? To shuffle an array means to reorder its elements arbitrarily (leaving the initial order is also an option).For example, if a=[1,1,3,5]a=[1,1,3,5], then shuffled arrays [1,3,5,1][1,3,5,1], [3,5,1,1][3,5,1,1] and [5,3,1,1][5,3,1,1] are good, but shuffled arrays [3,1,5,1][3,1,5,1], [1,1,3,5][1,1,3,5] and [1,1,5,3][1,1,5,3] aren't.It's guaranteed that it's always possible to shuffle an array to meet this condition.InputThe first line contains one integer tt (1≤t≤1001≤t≤100) — the number of test cases.The first line of each test case contains one integer nn (1≤n≤1001≤n≤100) — the length of array aa.The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1001≤ai≤100).OutputFor each test case print the shuffled version of the array aa which is good.ExampleInputCopy3
1
7
4
1 1 3 5
6
3 2 1 5 6 4
OutputCopy7
1 5 1 3
2 4 6 1 3 5
| [
"constructive algorithms",
"sortings"
] | t= int(input())
for i in range(t):
n= int(input())
l=[int(x) for x in input().split()]
l.sort()
l.reverse()
print(*l) | py |
1311 | F | F. Moving Pointstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThere are nn points on a coordinate axis OXOX. The ii-th point is located at the integer point xixi and has a speed vivi. It is guaranteed that no two points occupy the same coordinate. All nn points move with the constant speed, the coordinate of the ii-th point at the moment tt (tt can be non-integer) is calculated as xi+t⋅vixi+t⋅vi.Consider two points ii and jj. Let d(i,j)d(i,j) be the minimum possible distance between these two points over any possible moments of time (even non-integer). It means that if two points ii and jj coincide at some moment, the value d(i,j)d(i,j) will be 00.Your task is to calculate the value ∑1≤i<j≤n∑1≤i<j≤n d(i,j)d(i,j) (the sum of minimum distances over all pairs of points).InputThe first line of the input contains one integer nn (2≤n≤2⋅1052≤n≤2⋅105) — the number of points.The second line of the input contains nn integers x1,x2,…,xnx1,x2,…,xn (1≤xi≤1081≤xi≤108), where xixi is the initial coordinate of the ii-th point. It is guaranteed that all xixi are distinct.The third line of the input contains nn integers v1,v2,…,vnv1,v2,…,vn (−108≤vi≤108−108≤vi≤108), where vivi is the speed of the ii-th point.OutputPrint one integer — the value ∑1≤i<j≤n∑1≤i<j≤n d(i,j)d(i,j) (the sum of minimum distances over all pairs of points).ExamplesInputCopy3
1 3 2
-100 2 3
OutputCopy3
InputCopy5
2 1 4 3 5
2 2 2 3 4
OutputCopy19
InputCopy2
2 1
-3 0
OutputCopy0
| [
"data structures",
"divide and conquer",
"implementation",
"sortings"
] | import os, sys
from io import BytesIO, IOBase
from math import log2, ceil, sqrt, gcd
from _collections import deque
import heapq as hp
from bisect import bisect_left, bisect_right
from math import cos, sin
from itertools import permutations
# sys.setrecursionlimit(2*10**5+10000)
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")
mod = (10 ** 9 + 7) ^ 1755654
def update(i, vl):
i += n
tree[i][0] += vl
tree[i][1] += 1
while i:
tree[i >> 1][0] = tree[i][0] + tree[i ^ 1][0]
tree[i >> 1][1] = tree[i][1] + tree[i ^ 1][1]
i >>= 1
def query(l, r):
l += n
r += n
ans = [0, 0]
while l < r:
if l & 1:
ans[0] += tree[l][0]
ans[1] += tree[l][1]
l += 1
if r & 1:
r -= 1
ans[0] += tree[r][0]
ans[1] += tree[r][1]
l >>= 1
r >>= 1
return ans
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
s = set(b)
d = dict()
id = 0
for i in sorted(s):
d[i] = id
id += 1
ck = [[i, d[j]] for i, j in zip(a, b)]
ck.sort()
tree = [[0, 0] for _ in range(2 * id + 1)]
n = id
ans = 0
for x, v in ck:
tot, ct = query(0, v + 1)
ans += x * ct - tot
update(v, x)
print(ans)
| py |
1300 | A | A. Non-zerotime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputGuy-Manuel and Thomas have an array aa of nn integers [a1,a2,…,ana1,a2,…,an]. In one step they can add 11 to any element of the array. Formally, in one step they can choose any integer index ii (1≤i≤n1≤i≤n) and do ai:=ai+1ai:=ai+1.If either the sum or the product of all elements in the array is equal to zero, Guy-Manuel and Thomas do not mind to do this operation one more time.What is the minimum number of steps they need to do to make both the sum and the product of all elements in the array different from zero? Formally, find the minimum number of steps to make a1+a2+a1+a2+ …… +an≠0+an≠0 and a1⋅a2⋅a1⋅a2⋅ …… ⋅an≠0⋅an≠0.InputEach test contains multiple test cases. The first line contains the number of test cases tt (1≤t≤1031≤t≤103). The description of the test cases follows.The first line of each test case contains an integer nn (1≤n≤1001≤n≤100) — the size of the array.The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (−100≤ai≤100−100≤ai≤100) — elements of the array .OutputFor each test case, output the minimum number of steps required to make both sum and product of all elements in the array different from zero.ExampleInputCopy4
3
2 -1 -1
4
-1 0 0 1
2
-1 2
3
0 -2 1
OutputCopy1
2
0
2
NoteIn the first test case; the sum is 00. If we add 11 to the first element, the array will be [3,−1,−1][3,−1,−1], the sum will be equal to 11 and the product will be equal to 33.In the second test case, both product and sum are 00. If we add 11 to the second and the third element, the array will be [−1,1,1,1][−1,1,1,1], the sum will be equal to 22 and the product will be equal to −1−1. It can be shown that fewer steps can't be enough.In the third test case, both sum and product are non-zero, we don't need to do anything.In the fourth test case, after adding 11 twice to the first element the array will be [2,−2,1][2,−2,1], the sum will be 11 and the product will be −4−4. | [
"implementation",
"math"
] | I=input
exec(int(I())*"I();a=[*map(int,I().split())];s=a.count(0);print(s+(s+sum(a)==0));") | py |
1312 | B | B. Bogosorttime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given an array a1,a2,…,ana1,a2,…,an. Array is good if for each pair of indexes i<ji<j the condition j−aj≠i−aij−aj≠i−ai holds. Can you shuffle this array so that it becomes good? To shuffle an array means to reorder its elements arbitrarily (leaving the initial order is also an option).For example, if a=[1,1,3,5]a=[1,1,3,5], then shuffled arrays [1,3,5,1][1,3,5,1], [3,5,1,1][3,5,1,1] and [5,3,1,1][5,3,1,1] are good, but shuffled arrays [3,1,5,1][3,1,5,1], [1,1,3,5][1,1,3,5] and [1,1,5,3][1,1,5,3] aren't.It's guaranteed that it's always possible to shuffle an array to meet this condition.InputThe first line contains one integer tt (1≤t≤1001≤t≤100) — the number of test cases.The first line of each test case contains one integer nn (1≤n≤1001≤n≤100) — the length of array aa.The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1001≤ai≤100).OutputFor each test case print the shuffled version of the array aa which is good.ExampleInputCopy3
1
7
4
1 1 3 5
6
3 2 1 5 6 4
OutputCopy7
1 5 1 3
2 4 6 1 3 5
| [
"constructive algorithms",
"sortings"
] | t=int(input())
M=[]
def fonction(L):
D=[i+1-L[i] for i in range(len(L))]
if len(D)==len(set(D)):
return L
else:
H=sorted(L,reverse=True)
return H
for i in range (t):
n=int(input())
L=[int(x) for x in input().split()]
M.append(fonction(L))
for k in M:
for i in range(len(k)-1):
print(k[i],end=" ")
print(k[len(k)-1]) | 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"
] | 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 heapq
import itertools
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():
q, x = I()
s = [0] * (x + 1)
a = 0
for _ in range(q):
el = II() % x
s[el] += 1
while s[a % x] > 0:
s[a % x] -= 1
a += 1
print(a)
if __name__ == '__main__':
# for _ in range(II()):
# main()
main() | py |
1325 | B | B. CopyCopyCopyCopyCopytime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputEhab has an array aa of length nn. He has just enough free time to make a new array consisting of nn copies of the old array, written back-to-back. What will be the length of the new array's longest increasing subsequence?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. The longest increasing subsequence of an array is the longest subsequence such that its elements are ordered in strictly increasing order.InputThe first line contains an integer tt — the number of test cases you need to solve. The description of the test cases follows.The first line of each test case contains an integer nn (1≤n≤1051≤n≤105) — the number of elements in the array aa.The second line contains nn space-separated integers a1a1, a2a2, ……, anan (1≤ai≤1091≤ai≤109) — the elements of the array aa.The sum of nn across the test cases doesn't exceed 105105.OutputFor each testcase, output the length of the longest increasing subsequence of aa if you concatenate it to itself nn times.ExampleInputCopy2
3
3 2 1
6
3 1 4 1 5 9
OutputCopy3
5
NoteIn the first sample; the new array is [3,2,1,3,2,1,3,2,1][3,2,1,3,2,1,3,2,1]. The longest increasing subsequence is marked in bold.In the second sample, the longest increasing subsequence will be [1,3,4,5,9][1,3,4,5,9]. | [
"greedy",
"implementation"
] | for _ in range (int(input())):
k=int(input())
l=list(map(int,input().split()))
print(len(set(l))) | 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"
] | import math
a=int(input());
ans=0;
for b in range(2,a):
tmp=a;
while tmp>=b:
ans+=(tmp%b);
tmp//=b;
ans+=tmp;
num=a-2
gcd=math.gcd(ans,num);
print(str(ans//gcd)+'/'+str(num//gcd));
| py |
1285 | F | F. Classical?time limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputGiven an array aa, consisting of nn integers, find:max1≤i<j≤nLCM(ai,aj),max1≤i<j≤nLCM(ai,aj),where LCM(x,y)LCM(x,y) is the smallest positive integer that is divisible by both xx and yy. For example, LCM(6,8)=24LCM(6,8)=24, LCM(4,12)=12LCM(4,12)=12, LCM(2,3)=6LCM(2,3)=6.InputThe first line contains an integer nn (2≤n≤1052≤n≤105) — the number of elements in the array aa.The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1051≤ai≤105) — the elements of the array aa.OutputPrint one integer, the maximum value of the least common multiple of two elements in the array aa.ExamplesInputCopy3
13 35 77
OutputCopy1001InputCopy6
1 2 4 8 16 32
OutputCopy32 | [
"binary search",
"combinatorics",
"number theory"
] |
import math
def u(x,a):
for m in divisors[x]:
sum[m]+=a
def coprime(x):
count = 0
for i in divisors[x]:
count+=sum[i]*mobius[i]
return count
n = int(input())
a = input().split(" ")
maxlcm = 0
b = [False]*100010
for x in range(n):
a[x] = int(a[x])
maxlcm = max(a[x],maxlcm)
b[a[x]] = True
a.sort(reverse=True)
m = 100005
sum = [0]*(m+5)
mobius = [0]*(m+5)
divisors = {}
for i in range(1,m):
for j in range(i,m,i):
try:
divisors[j].append(i)
except:
divisors[j] = [1]
if(i==1):
mobius[i] = 1
elif ((i/divisors[i][1]%divisors[i][1])==0):
mobius[i]=0
else:
mobius[i] = -mobius[int(i/divisors[i][1])]
for g in range(1,m):
s = []
for x in range(int(m/g)):
x = int(m/g)-x
if(not b[x*g]):
continue
c = coprime(x)
while c:
if(math.gcd(x,s[-1])==1):
maxlcm = max(maxlcm,x*g*s[-1])
c-=1
u(s[-1],-1)
del s[-1]
u(x,1)
s.append(x)
while(len(s)!=0):
u(s[-1],-1)
del s[-1]
print(maxlcm)
| 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"
] | import math
from collections import *
def solve():
n, m = map(int, input().split())
s = input()
a = [int(i) for i in input().split()]
ans = [0 for i in range(n)]
for i in range(len(a)):
ans[a[i]-1] += 1
for i in range(n-1, 0, -1):
ans[i-1] += ans[i]
d = defaultdict(int)
for i in range(len(s)):
d[s[i]] += (ans[i]+1)
for i in range(ord("a"), ord("z")+1):
if d.get(chr(i), 0) != 0:
print(d.get(chr(i), 0), end = " ")
else:
print(0, end = " ")
print()
t = int(input())
for _ in range(t):
solve()
# solve()
| py |
1312 | G | G. Autocompletiontime limit per test7 secondsmemory limit per test512 megabytesinputstandard inputoutputstandard outputYou are given a set of strings SS. Each string consists of lowercase Latin letters.For each string in this set, you want to calculate the minimum number of seconds required to type this string. To type a string, you have to start with an empty string and transform it into the string you want to type using the following actions: if the current string is tt, choose some lowercase Latin letter cc and append it to the back of tt, so the current string becomes t+ct+c. This action takes 11 second; use autocompletion. When you try to autocomplete the current string tt, a list of all strings s∈Ss∈S such that tt is a prefix of ss is shown to you. This list includes tt itself, if tt is a string from SS, and the strings are ordered lexicographically. You can transform tt into the ii-th string from this list in ii seconds. Note that you may choose any string from this list you want, it is not necessarily the string you are trying to type. What is the minimum number of seconds that you have to spend to type each string from SS?Note that the strings from SS are given in an unusual way.InputThe first line contains one integer nn (1≤n≤1061≤n≤106).Then nn lines follow, the ii-th line contains one integer pipi (0≤pi<i0≤pi<i) and one lowercase Latin character cici. These lines form some set of strings such that SS is its subset as follows: there are n+1n+1 strings, numbered from 00 to nn; the 00-th string is an empty string, and the ii-th string (i≥1i≥1) is the result of appending the character cici to the string pipi. It is guaranteed that all these strings are distinct.The next line contains one integer kk (1≤k≤n1≤k≤n) — the number of strings in SS.The last line contains kk integers a1a1, a2a2, ..., akak (1≤ai≤n1≤ai≤n, all aiai are pairwise distinct) denoting the indices of the strings generated by above-mentioned process that form the set SS — formally, if we denote the ii-th generated string as sisi, then S=sa1,sa2,…,sakS=sa1,sa2,…,sak.OutputPrint kk integers, the ii-th of them should be equal to the minimum number of seconds required to type the string saisai.ExamplesInputCopy10
0 i
1 q
2 g
0 k
1 e
5 r
4 m
5 h
3 p
3 e
5
8 9 1 10 6
OutputCopy2 4 1 3 3
InputCopy8
0 a
1 b
2 a
2 b
4 a
4 b
5 c
6 d
5
2 3 4 7 8
OutputCopy1 2 2 4 4
NoteIn the first example; SS consists of the following strings: ieh, iqgp, i, iqge, ier. | [
"data structures",
"dfs and similar",
"dp"
] | import io
import os
DEBUG = False
def dfs(trie, root, preorder=None, postorder=None):
stack = [root]
seen = set()
while stack:
nodeId = stack.pop()
if nodeId not in seen:
if preorder:
preorder(nodeId)
stack.append(nodeId)
seen.add(nodeId)
for c, childId in reversed(trie[nodeId]):
stack.append(childId)
else:
if postorder:
postorder(nodeId)
def solve(N, PC, K, A):
ROOT = 0
trie = {ROOT: []} # nodeId to list of (character, nodeId)
parent = {}
for i, (p, c) in enumerate(PC, 1): # i starts from 1
trie[p].append((c, i))
assert i not in trie
trie[i] = []
parent[i] = p
terminal = set(A)
# Sort children of each node by character
for children in trie.values():
children.sort()
# DFS
offset = 0
ancestor = []
dist = {}
def getDistPre(nodeId):
nonlocal offset
best = None
if nodeId != 0:
assert nodeId in parent
best = 1 + dist[parent[nodeId]]
if nodeId in terminal:
best = min(best, ancestor[-1] + offset + 1)
ancestor.append(min(ancestor[-1], best - offset))
if nodeId in terminal:
offset += 1
else:
# Is root
best = 0
ancestor.append(0)
dist[nodeId] = best
def getDistPost(nodeId):
ancestor.pop()
dfs(trie, ROOT, preorder=getDistPre, postorder=getDistPost)
if DEBUG:
def printNode(nodeId, word):
return (
str(nodeId)
+ "\t"
+ word
+ ("$" if nodeId in terminal else "")
+ "\t"
+ "dist: "
+ str(dist[nodeId])
)
return str(nodeId) + "\t" + word + ("$" if nodeId in terminal else "")
def printGraph(nodeId, path):
W = 8
depth = len(path)
for ch, childId in trie[nodeId]:
path.append(ch)
print(
(
" " * (W * depth)
+ "└"
+ ch.center(W - 1, "─")
+ str(childId)
+ ("$" if childId in terminal else "")
).ljust(50)
+ printNode(childId, "".join(path))
)
printGraph(childId, path)
path.pop()
printGraph(ROOT, [])
out = []
for a in A:
out.append(str(dist[a]))
return " ".join(out)
if __name__ == "__main__":
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
N, = list(map(int, input().split()))
PC = []
for i in range(N):
p, c = input().decode().split()
PC.append((int(p), str(c)))
K, = list(map(int, input().split()))
A = list(map(int, input().split()))
ans = solve(N, PC, K, A)
print(ans)
| py |
1304 | B | B. Longest Palindrometime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputReturning back to problem solving; Gildong is now studying about palindromes. He learned that a palindrome is a string that is the same as its reverse. For example, strings "pop", "noon", "x", and "kkkkkk" are palindromes, while strings "moon", "tv", and "abab" are not. An empty string is also a palindrome.Gildong loves this concept so much, so he wants to play with it. He has nn distinct strings of equal length mm. He wants to discard some of the strings (possibly none or all) and reorder the remaining strings so that the concatenation becomes a palindrome. He also wants the palindrome to be as long as possible. Please help him find one.InputThe first line contains two integers nn and mm (1≤n≤1001≤n≤100, 1≤m≤501≤m≤50) — the number of strings and the length of each string.Next nn lines contain a string of length mm each, consisting of lowercase Latin letters only. All strings are distinct.OutputIn the first line, print the length of the longest palindrome string you made.In the second line, print that palindrome. If there are multiple answers, print any one of them. If the palindrome is empty, print an empty line or don't print this line at all.ExamplesInputCopy3 3
tab
one
bat
OutputCopy6
tabbat
InputCopy4 2
oo
ox
xo
xx
OutputCopy6
oxxxxo
InputCopy3 5
hello
codef
orces
OutputCopy0
InputCopy9 4
abab
baba
abcd
bcde
cdef
defg
wxyz
zyxw
ijji
OutputCopy20
ababwxyzijjizyxwbaba
NoteIn the first example; "battab" is also a valid answer.In the second example; there can be 4 different valid answers including the sample output. We are not going to provide any hints for what the others are.In the third example; the empty string is the only valid palindrome string. | [
"brute force",
"constructive algorithms",
"greedy",
"implementation",
"strings"
] | s = input().split(' ')
n = int(s[0])
m1 = int(s[1]) - 1
m = m1
l = []
s = ""
def f(str1):
i1 = m + 1
i2 = m
while i1 != 0 and i1 != 1:
if str1[i2] != str1[m - i2]:
return False
else:
i2 -= 1
i1 -= 2
return True
for i in range(n):
obj = input()
if f(obj):
s = obj
else:
l.append(obj)
while len(l) != 0:
for obj1 in l:
l.remove(obj1)
s1 = obj1[0]
for obj2 in l:
m = m1
s2 = obj2[m]
if s1 == s2:
r = obj1 + obj2
m = len(r) - 1
if f(r):
s = obj1 + s + obj2
l.remove(obj2)
print(len(s))
print(s)
| 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"
] | n = int(input())
for i in range(n):
n = int(input())
sp = 0
cur = 0
while n >= 10:
sp += n // 10 * 10
cur = n // 10
n = n % 10 + cur
print(sp + n) | 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"
] | from sys import stdin,stdout
from math import gcd,sqrt,factorial,pi,inf
from collections import deque,defaultdict
from bisect import bisect,bisect_left
from time import time
from itertools import permutations as per
from heapq import heapify,heappush,heappushpop
input=stdin.readline
R=lambda:map(int,input().split())
I=lambda:int(input())
S=lambda:input().rstrip('\r\n')
L=lambda:list(R())
P=lambda x:stdout.write(str(x)+'\n')
lcm=lambda x,y:(x*y)//gcd(x,y)
nCr=lambda x,y:(f[x]*inv((f[y]*f[x-y])%N))%N
inv=lambda x:pow(x,N-2,N)
sm=lambda x:(x**2+x)//2
N=10**9+7
n,m=R()
a=[[] for i in range(n+1)]
b=[[] for i in range(n+1)]
for i in range(m):
x,y=R()
a[y]+=x,
b[x]+=y,
k=I()
v=L()
stk=deque([v[-1]])
vst=[False]*(n+1)
val=[1]*(n+1)
val[stk[-1]]=0
vst[stk[-1]]=True
while stk:
p=stk.pop()
for el in a[p]:
if not vst[el]:
val[el]+=val[p]
stk.appendleft(el)
vst[el]=True
mn=0
mx=0
for j in range(k-1):
i=v[j]
cnt=0
p=inf
#print(i,b[i])
for el in b[i]:
if val[el]<p:
p=val[el]
cnt=1
elif val[el]==p:
cnt+=1
#print(cnt,p,val[j+1],val[1:])
if p<val[v[j+1]]:
mx+=1
mn+=1
else:
if cnt>1:
mx+=1
print(mn,mx) | 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"
] | for _ in range(int(input())):
n, m = map(int, input().split())
s = input()
arr = list(map(int, input().split(' ')))
arr = arr + [n]
nums = [arr[i] - 1 for i in range(m + 1)]
numToAlpha = {i:s[i] for i in range(n)}
dic = {chr(i + 97): 0 for i in range(26)}
sweep = [0]*(n + 1)
# print(dic)
# print(numToAlpha)
for i in range(m + 1):
ind = nums[i]
sweep[0] += 1
sweep[ind + 1] -= 1
for i in range(1, n + 1):
sweep[i] += sweep[i - 1]
# print(sweep)
for i in range(n):
count = sweep[i]
alpha = numToAlpha[i]
dic[alpha] += count
print(*[i for i in dic.values()]) | py |
1291 | A | A. Even But Not Eventime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputLet's define a number ebne (even but not even) if and only if its sum of digits is divisible by 22 but the number itself is not divisible by 22. For example, 1313, 12271227, 185217185217 are ebne numbers, while 1212, 22, 177013177013, 265918265918 are not. If you're still unsure what ebne numbers are, you can look at the sample notes for more clarification.You are given a non-negative integer ss, consisting of nn digits. You can delete some digits (they are not necessary consecutive/successive) to make the given number ebne. You cannot change the order of the digits, that is, after deleting the digits the remaining digits collapse. The resulting number shouldn't contain leading zeros. You can delete any number of digits between 00 (do not delete any digits at all) and n−1n−1.For example, if you are given s=s=222373204424185217171912 then one of possible ways to make it ebne is: 222373204424185217171912 →→ 2237344218521717191. The sum of digits of 2237344218521717191 is equal to 7070 and is divisible by 22, but number itself is not divisible by 22: it means that the resulting number is ebne.Find any resulting number that is ebne. If it's impossible to create an ebne number from the given number report about it.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 a single integer nn (1≤n≤30001≤n≤3000) — the number of digits in the original number.The second line of each test case contains a non-negative integer number ss, consisting of nn digits.It is guaranteed that ss does not contain leading zeros and the sum of nn over all test cases does not exceed 30003000.OutputFor each test case given in the input print the answer in the following format: If it is impossible to create an ebne number, print "-1" (without quotes); Otherwise, print the resulting number after deleting some, possibly zero, but not all digits. This number should be ebne. If there are multiple answers, you can print any of them. Note that answers with leading zeros or empty strings are not accepted. It's not necessary to minimize or maximize the number of deleted digits.ExampleInputCopy4
4
1227
1
0
6
177013
24
222373204424185217171912
OutputCopy1227
-1
17703
2237344218521717191
NoteIn the first test case of the example; 12271227 is already an ebne number (as 1+2+2+7=121+2+2+7=12, 1212 is divisible by 22, while in the same time, 12271227 is not divisible by 22) so we don't need to delete any digits. Answers such as 127127 and 1717 will also be accepted.In the second test case of the example, it is clearly impossible to create an ebne number from the given number.In the third test case of the example, there are many ebne numbers we can obtain by deleting, for example, 11 digit such as 1770317703, 7701377013 or 1701317013. Answers such as 17011701 or 770770 will not be accepted as they are not ebne numbers. Answer 013013 will not be accepted as it contains leading zeroes.Explanation: 1+7+7+0+3=181+7+7+0+3=18. As 1818 is divisible by 22 while 1770317703 is not divisible by 22, we can see that 1770317703 is an ebne number. Same with 7701377013 and 1701317013; 1+7+0+1=91+7+0+1=9. Because 99 is not divisible by 22, 17011701 is not an ebne number; 7+7+0=147+7+0=14. This time, 1414 is divisible by 22 but 770770 is also divisible by 22, therefore, 770770 is not an ebne number.In the last test case of the example, one of many other possible answers is given. Another possible answer is: 222373204424185217171912 →→ 22237320442418521717191 (delete the last digit). | [
"greedy",
"math",
"strings"
] | # LUOGU_RID: 101844775
for _ in range(int(input())):
n = int(input())
s = input()
a = []
for c in s:
if int(c) & 1:
a += int(c),
if len(a) < 2:
print(-1)
else:
print(a[0], a[1], sep='') | 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 highlight import *
t = int(input())
for _ in range(t):
res = float("inf")
triple = []
a, b, c = map(int, input().split())
for af in range(1, 2*a+1):
for bf in range(af, 2*b+1, af):
for cf in [c//bf*bf, c//bf*bf+bf]:
if af <= bf <= cf and abs(af-a) + abs(bf-b) + abs(cf-c) < res:
res = abs(af-a) + abs(bf-b) + abs(cf-c)
triple = [af, bf, cf]
print(res)
print(triple[0], triple[1], triple[2]) | 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"
] | for i in range(int(input())):
n,d=map(int,input().split())
print('NO') if (1-n)**2-4*(d-n)<0 else print("YES")
| 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"
] | t = int(input())
while t > 0:
t -= 1
n = int(input())
s = input()
ang, mx, cnt = 0, 0, 0
for i in range(n):
if s[i] == 'A':
ang = 1
mx = max(mx, cnt)
cnt = 0
elif ang and s[i] == 'P':
cnt += 1
mx = max(mx, cnt)
print(mx) | 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"
] | n = int(input())
ans = list(map(int, input().split()))
ll = [0] * n
rr = [0] * n
s = 0
for i in range(n):
s |= ans[i]
ll[i] = s
s = 0
for i in range(n - 1, -1, -1):
s |= ans[i]
rr[i] = s
mx = 0
mxi = 0
for i in range(n):
s = 0
if i - 1 >= 0: s |= ll[i - 1]
if i + 1 < n: s |= rr[i + 1]
cur = (ans[i] | s) - s
if cur > mx:
mx = cur
mxi = i
ans[0], ans[mxi] = ans[mxi], ans[0]
print(*ans) | py |
1295 | D | D. Same GCDstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given two integers aa and mm. Calculate the number of integers xx such that 0≤x<m0≤x<m and gcd(a,m)=gcd(a+x,m)gcd(a,m)=gcd(a+x,m).Note: gcd(a,b)gcd(a,b) is the greatest common divisor of aa and bb.InputThe first line contains the single integer TT (1≤T≤501≤T≤50) — the number of test cases.Next TT lines contain test cases — one per line. Each line contains two integers aa and mm (1≤a<m≤10101≤a<m≤1010).OutputPrint TT integers — one per test case. For each test case print the number of appropriate xx-s.ExampleInputCopy3
4 9
5 10
42 9999999967
OutputCopy6
1
9999999966
NoteIn the first test case appropriate xx-s are [0,1,3,4,6,7][0,1,3,4,6,7].In the second test case the only appropriate xx is 00. | [
"math",
"number theory"
] | import sys
import math
input = sys.stdin.readline
for __ in range(int(input())):
a, m = [int(Xx) for Xx in input().split()]
a, m = a // math.gcd(a, m), m // math.gcd(a, m)
ans = 1
i = 2
M = m
while i * i <= M:
if m % i == 0:
while m % i == 0:
m = m // i
ans *= i
ans = ans // i
ans = ans * (i - 1)
i += 1
if m - 1:
ans *= (m - 1)
print(ans)
| py |
1288 | D | D. Minimax Problemtime limit per test5 secondsmemory limit per test512 megabytesinputstandard inputoutputstandard outputYou are given nn arrays a1a1, a2a2, ..., anan; each array consists of exactly mm integers. We denote the yy-th element of the xx-th array as ax,yax,y.You have to choose two arrays aiai and ajaj (1≤i,j≤n1≤i,j≤n, it is possible that i=ji=j). After that, you will obtain a new array bb consisting of mm integers, such that for every k∈[1,m]k∈[1,m] bk=max(ai,k,aj,k)bk=max(ai,k,aj,k).Your goal is to choose ii and jj so that the value of mink=1mbkmink=1mbk is maximum possible.InputThe first line contains two integers nn and mm (1≤n≤3⋅1051≤n≤3⋅105, 1≤m≤81≤m≤8) — the number of arrays and the number of elements in each array, respectively.Then nn lines follow, the xx-th line contains the array axax represented by mm integers ax,1ax,1, ax,2ax,2, ..., ax,max,m (0≤ax,y≤1090≤ax,y≤109).OutputPrint two integers ii and jj (1≤i,j≤n1≤i,j≤n, it is possible that i=ji=j) — the indices of the two arrays you have to choose so that the value of mink=1mbkmink=1mbk is maximum possible. If there are multiple answers, print any of them.ExampleInputCopy6 5
5 0 3 1 2
1 8 9 1 3
1 2 3 4 5
9 1 0 3 7
2 3 0 6 3
6 4 1 7 0
OutputCopy1 5
| [
"binary search",
"bitmasks",
"dp"
] | import sys
# 1288D
def solve(a):
m, n = len(a), len(a[0])
maxx = max(max(t) for t in a)
def check(t):
idx = [-1] * 256
for i in range(m):
mask = 0
for j in range(n):
if a[i][j] >= t:
mask |= 1 << j
# idx: [5,-1,6,8,-1……]
idx[mask] = i
for i in range(1 << n):
if idx[i] == -1:
continue
for j in range(i, 1 << n):
if idx[j] == -1:
continue
if (i | j) == (1 << n) - 1:
return [idx[i], idx[j]]
return [-1, -1]
l, r = 0, maxx
while l <= r:
mid = (l + r) // 2
if check(mid) != [-1, -1]:
l = mid + 1
else:
r = mid - 1
# [ TTTTTT FFFFFF ]
# r l
return check(r)
m, n = map(int, input().split())
a = []
for i in range(m):
a.append([int(i) for i in input().split()])
res = solve(a)
print(res[0] + 1, res[1] + 1)
# 1103B
# def query(x, y):
# print("?", x, y)
# sys.stdout.flush()
# res = input()
# return res
# def solve():
# # write function here
# return -1
# while True:
# res = solve()
# if res == -1: break
# print("!", res)
# def cutThemAll(lengths, minLength):
# for i in range(len(lengths) - 1):
# if lengths[i] + lengths[i+1] >= minLength:
# return "Possible"
# return "Impossible"
# ls = [3,5,6]
# print(cutThemAll(ls, 12))
# def getUniqueCharacter(s):
# cnt = [0] * 26
# for c in s:
# cnt[ord(c) - ord('a')] += 1
# for i in range(len(s)):
# if cnt[ord(s[i]) - ord('a')] == 1:
# return i + 1
# return -1
# print(getUniqueCharacter("madam"))
# from cmath import log
# from collections import Counter
# from dis import findlabels
# base = 10**9 + 7
# def findRight(a):
# # find the index of the first less one in the left side, for every a[i]
# n = len(a)
# st, res = [], [0] * n
# for i in range(n - 1, -1, -1):
# while len(st) and a[st[-1]] >= a[i]:
# st.pop()
# res[i] = n if len(st) == 0 else st[-1]
# st.append(i)
# return res
# def findLeft(a):
# # find the index of the first less or equal one in the right side, for every a[i]
# n = len(a)
# st, res = [], [0] * n
# for i in range(n):
# while len(st) and a[st[-1]] > a[i]:
# st.pop()
# res[i] = -1 if len(st) == 0 else st[-1]
# st.append(i)
# return res
# def findTotalPower(a):
# # In order to avoid repeated calculations,
# # the left side finds the first element less than a[i],
# # and the right side finds the first element less than or equal to a[i]
# left = findLeft(a)
# right = findRight(a)
# n, res = len(a), 0
# s = [0] * (n + 1)
# ss = [0] * (n + 2)
# # s means the prefix sum of a
# for i in range(n): s[i+1] = s[i] + a[i]
# # ss meams the prefix sum of s
# for i in range(n + 1): ss[i+1] = ss[i] + s[i]
# for i in range(n):
# # every left one could be the leader of subarray, which can be combined with the ending on the right
# l = i - left[i]
# # every right one could be the ending of subarray, which can be combined with the leader on the left
# r = right[i] - i
# # find the sum(sum([i, r]))
# now = ((ss[right[i] + 1] - ss[i + 1]) - r * s[i]) % base
# # find the sum(sum([i, r])) * a[i] * (i-left)
# res += ((now * l) % base * a[i]) % base
# # find the sum(sum([l, i]))
# now = (s[i] * (l - 1) - (ss[i] - ss[left[i] + 1])) % base
# # find the sum(sum([l, i])) * a[i] * (right-i)
# res += ((now * r) % base * a[i]) % base
# res %= base
# return res % base
# ls = [2,3,2,1]
# res = findTotalPower(ls)
# print(res)
# def solve(a):
# m, n = len(a), len(a[0])
# maxx = max(max(t) for t in a)
# def ok(t):
# idx = [-1] * 256
# maxm = 1 << n
# for i in range(m):
# mask = 0
# for j in range(n):
# if a[i][j] >= t:
# mask |= 1 << j
# if idx[mask] == -1:
# idx[mask] = i
# for i in range(maxm):
# if idx[i] == -1:
# continue
# for j in range(i, maxm):
# if idx[j] == -1:
# continue
# if (i | j) == maxm - 1:
# return [idx[i], idx[j]]
# return [-1, -1]
# l, r = 0, maxx
# while l <= r:
# mid = (l + r) // 2
# if ok(mid) != [-1, -1]:
# l = mid + 1
# else:
# r = mid - 1
# return ok(r) | py |
1316 | E | E. Team Buildingtime limit per test3 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputAlice; the president of club FCB, wants to build a team for the new volleyball tournament. The team should consist of pp players playing in pp different positions. She also recognizes the importance of audience support, so she wants to select kk people as part of the audience.There are nn people in Byteland. Alice needs to select exactly pp players, one for each position, and exactly kk members of the audience from this pool of nn people. Her ultimate goal is to maximize the total strength of the club.The ii-th of the nn persons has an integer aiai associated with him — the strength he adds to the club if he is selected as a member of the audience.For each person ii and for each position jj, Alice knows si,jsi,j — the strength added by the ii-th person to the club if he is selected to play in the jj-th position.Each person can be selected at most once as a player or a member of the audience. You have to choose exactly one player for each position.Since Alice is busy, she needs you to help her find the maximum possible strength of the club that can be achieved by an optimal choice of players and the audience.InputThe first line contains 33 integers n,p,kn,p,k (2≤n≤105,1≤p≤7,1≤k,p+k≤n2≤n≤105,1≤p≤7,1≤k,p+k≤n).The second line contains nn integers a1,a2,…,ana1,a2,…,an. (1≤ai≤1091≤ai≤109).The ii-th of the next nn lines contains pp integers si,1,si,2,…,si,psi,1,si,2,…,si,p. (1≤si,j≤1091≤si,j≤109)OutputPrint a single integer resres — the maximum possible strength of the club.ExamplesInputCopy4 1 2
1 16 10 3
18
19
13
15
OutputCopy44
InputCopy6 2 3
78 93 9 17 13 78
80 97
30 52
26 17
56 68
60 36
84 55
OutputCopy377
InputCopy3 2 1
500 498 564
100002 3
422332 2
232323 1
OutputCopy422899
NoteIn the first sample; we can select person 11 to play in the 11-st position and persons 22 and 33 as audience members. Then the total strength of the club will be equal to a2+a3+s1,1a2+a3+s1,1. | [
"bitmasks",
"dp",
"greedy",
"sortings"
] |
import os
import sys
from io import BytesIO, IOBase
# region fastio
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")
inf = float('inf')
n,p,k = map(int,input().split())
a = list(map(int,input().split()))
s = [None for _ in range(n)]
I = list(range(n))
I.sort(key = lambda i: -a[i])
ref = {x:i for i,x in enumerate(I)}
a.sort(reverse=True)
for i in range(n):
s[ref[i]] = list(map(int,input().split()))
dp = [-inf]*(1<<p)
dp[0] = float(a[0])
for j in range(p):
dp[1<<j] = float(s[0][j])
Bit_cnt = [bin(mask).count("1") for mask in range(1<<p)]
for i in range(1,n):
new_dp = [-inf]*(1<<p)
for mask in range(1<<p):
bit_cnt = Bit_cnt[mask]
if (i+1) - bit_cnt <= k:
new_dp[mask] = dp[mask] + a[i]
else:
new_dp[mask] = dp[mask]
for j in range(p):
if not mask & (1<<j):
continue
case = dp[mask ^ (1<<j)] + s[i][j]
new_dp[mask] = max(new_dp[mask], case)
dp = new_dp
print(int(dp[(1<<p) - 1]))
| 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 solution(n,d,piles):
maxi = piles[0]
for idx,pile in enumerate(piles):
if idx == 0:
continue
while (piles[idx] > 0) and (d - idx) >= 0:
d -= idx
piles[idx] -= 1
maxi += 1
return maxi
t = int(input())
while t > 0:
t -= 1
n,d = [int(x) for x in input().split()]
piles = [int(x) for x in input().split()]
print(solution(n,d, piles))
| py |
1324 | D | D. Pair of Topicstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThe next lecture in a high school requires two topics to be discussed. The ii-th topic is interesting by aiai units for the teacher and by bibi units for the students.The pair of topics ii and jj (i<ji<j) is called good if ai+aj>bi+bjai+aj>bi+bj (i.e. it is more interesting for the teacher).Your task is to find the number of good pairs of topics.InputThe first line of the input contains one integer nn (2≤n≤2⋅1052≤n≤2⋅105) — the number of topics.The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109), where aiai is the interestingness of the ii-th topic for the teacher.The third line of the input contains nn integers b1,b2,…,bnb1,b2,…,bn (1≤bi≤1091≤bi≤109), where bibi is the interestingness of the ii-th topic for the students.OutputPrint one integer — the number of good pairs of topic.ExamplesInputCopy5
4 8 2 6 2
4 5 4 1 3
OutputCopy7
InputCopy4
1 3 2 4
1 3 2 4
OutputCopy0
| [
"binary search",
"data structures",
"sortings",
"two pointers"
] | n = int(input())
teacher = list(map(int, input().split()))
student = list(map(int, input().split()))
store = []
for i in range(n):
store.append(teacher[i] - student[i])
store.sort()
l, r = 0, n - 1
res = 0
while l < r:
cur = store[r] + store[l]
if cur > 0:
res += r - l
r -= 1
else:
l += 1
print(res)
| 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"
] | import math
t = int(input())
for i in range(t):
n, d = input().split()
n = int(n)
d = int(d)
c = 0
for x in range(n):
if(x + math.ceil(d/(x + 1)) <= n):
print("YES")
c += 1
break
if c == 0:
print("NO") | 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 main():
for t in range(int(input())):
n=int(input())
a=[]
for tt in range(n):
a.append(list(map(int,input().split())))
a.sort()
x,y=0,0
z=""
h=True
for i in a:
if i[0]-x>=0 and i[1]-y>=0:
z=z+"R"*(i[0]-x)
z=z+"U"*(i[1]-y)
x,y=i[0],i[1]
else:
h=False
print("NO")
break
if h:
print("YES")
print(z)
main()
#la manera de ve si es posible o no es que no haya niuno mas alto que otro y mas a la izquierdaq wuee el otro.
#osea que todos los qliaos arriba de una caja esten arriba o a la derecha.
#el algoritmo es solo ir al que mas cerca este a la derecha, si es imposibe printea NO. | 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"
] | for _ in range(int(input())):
n=int(input())
arr=list(map(int,input().split()))
l=0
h=10**9+7
def check(x):
res=0
for i in range(len(arr)):
if arr[i]==-1:
if i>0 and arr[i-1]!=-1:
res=max(res,abs(arr[i-1]-x))
if i<len(arr)-1 and arr[i+1]!=-1:
res=max(res,abs(arr[i+1]-x))
else :
if i<len(arr)-1 and arr[i+1]!=-1:
res=max(res,abs(arr[i+1]-arr[i]))
return res
while l<=h:
mid=(l+h)>>1
if check(mid)<=check(mid+1):
ans=mid
h=mid-1
else :
l=mid+1
print(check(ans),ans) | 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 run in range(int(input())):
a,b,x,y=map(int,input().split())
left=x*b
right=(a-1-x)*b
top=a*y
down=(a*(b-1-y))
print(max(left,right,top,down))
| py |
1303 | A | A. Erasing Zeroestime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given a string ss. Each character is either 0 or 1.You want all 1's in the string to form a contiguous subsegment. For example, if the string is 0, 1, 00111 or 01111100, then all 1's form a contiguous subsegment, and if the string is 0101, 100001 or 11111111111101, then this condition is not met.You may erase some (possibly none) 0's from the string. What is the minimum number of 0's that you have to erase?InputThe first line contains one integer tt (1≤t≤1001≤t≤100) — the number of test cases.Then tt lines follow, each representing a test case. Each line contains one string ss (1≤|s|≤1001≤|s|≤100); each character of ss is either 0 or 1.OutputPrint tt integers, where the ii-th integer is the answer to the ii-th testcase (the minimum number of 0's that you have to erase from ss).ExampleInputCopy3
010011
0
1111000
OutputCopy2
0
0
NoteIn the first test case you have to delete the third and forth symbols from string 010011 (it turns into 0111). | [
"implementation",
"strings"
] | def main():
t = int(input())
for _ in range(t):
s = input().strip()
i = 0
n = len(s)
res = 0
j = n-1
while i < n and s[i] == '0':
i += 1
while j >= i and s[j] == '0':
j -= 1
while i <= j:
if s[i] == '0':
res += 1
i += 1
print(res)
if __name__ == '__main__':
main() | 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 sys
input = sys.stdin.readline
from bisect import bisect
def f(x):
return ord(x)-97
for _ in range(int(input())):
s = input()[:-1]
t = input()[:-1]
d = [[] for i in range(26)]
for i, j in enumerate(s):
d[f(j)].append(i)
w = [len(d[i]) for i in range(26)]
c = 1
p = -1
for i in t:
if w[f(i)] == 0:
print(-1)
break
a = bisect(d[f(i)], p)
if a == 0:
p = d[f(i)][0]
elif a == w[f(i)]:
c += 1
p = d[f(i)][0]
else:
p = d[f(i)][a]
else:
print(c) | 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"
] | from bisect import bisect
l = '001 011 022 122 223 333 444'.split()
for _ in range(int(input())): print(bisect(l, '{}{}{}'.format(*sorted(min(int(s), 4) for s in input().split())))) | py |
1311 | E | E. Construct the Binary Treetime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given two integers nn and dd. You need to construct a rooted binary tree consisting of nn vertices with a root at the vertex 11 and the sum of depths of all vertices equals to dd.A tree is a connected graph without cycles. A rooted tree has a special vertex called the root. A parent of a vertex vv is the last different from vv vertex on the path from the root to the vertex vv. The depth of the vertex vv is the length of the path from the root to the vertex vv. Children of vertex vv are all vertices for which vv is the parent. The binary tree is such a tree that no vertex has more than 22 children.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 only line of each test case contains two integers nn and dd (2≤n,d≤50002≤n,d≤5000) — the number of vertices in the tree and the required sum of depths of all vertices.It is guaranteed that the sum of nn and the sum of dd both does not exceed 50005000 (∑n≤5000,∑d≤5000∑n≤5000,∑d≤5000).OutputFor each test case, print the answer.If it is impossible to construct such a tree, print "NO" (without quotes) in the first line. Otherwise, print "{YES}" in the first line. Then print n−1n−1 integers p2,p3,…,pnp2,p3,…,pn in the second line, where pipi is the parent of the vertex ii. Note that the sequence of parents you print should describe some binary tree.ExampleInputCopy3
5 7
10 19
10 18
OutputCopyYES
1 2 1 3
YES
1 2 3 3 9 9 2 1 6
NO
NotePictures corresponding to the first and the second test cases of the example: | [
"brute force",
"constructive algorithms",
"trees"
] | import bisect
import sys
input = sys.stdin.readline
t = int(input())
pow2 = [1]
for _ in range(20):
pow2.append(2 * pow2[-1])
pow2 = set(pow2)
for _ in range(t):
n, d = map(int, input().split())
p = []
d0 = []
now = 0
l, l0 = [1], [1]
for i in range(2, n + 1):
p.append(i // 2)
if i in pow2:
l.append(i)
l0.append(i)
now += 1
d0.append(now)
s = sum(d0)
ans = "YES" if s <= d <= n * (n - 1) // 2 else "NO"
print(ans)
if ans == "NO":
continue
c = d - s
for i in range(n, 0, -1):
if not c:
break
if i in pow2:
continue
j = bisect.bisect_left(l, i) - 1
for k in range(j, len(l0) - 1):
if not c:
break
c -= 1
p[i - 2] = l0[k]
d0[i - 2] += 1
if not c:
break
c -= 1
p[i - 2] = l0[-1]
d0[i - 2] += 1
l0.append(i)
print(*p) | py |
13 | B | B. Letter Atime limit per test1 secondmemory limit per test64 megabytesinputstdinoutputstdoutLittle Petya learns how to write. The teacher gave pupils the task to write the letter A on the sheet of paper. It is required to check whether Petya really had written the letter A.You are given three segments on the plane. They form the letter A if the following conditions hold: Two segments have common endpoint (lets call these segments first and second); while the third segment connects two points on the different segments. The angle between the first and the second segments is greater than 0 and do not exceed 90 degrees. The third segment divides each of the first two segments in proportion not less than 1 / 4 (i.e. the ratio of the length of the shortest part to the length of the longest part is not less than 1 / 4). InputThe first line contains one integer t (1 ≤ t ≤ 10000) — the number of test cases to solve. Each case consists of three lines. Each of these three lines contains four space-separated integers — coordinates of the endpoints of one of the segments. All coordinates do not exceed 108 by absolute value. All segments have positive length.OutputOutput one line for each test case. Print «YES» (without quotes), if the segments form the letter A and «NO» otherwise.ExamplesInputCopy34 4 6 04 1 5 24 0 4 40 0 0 60 6 2 -41 1 0 10 0 0 50 5 2 -11 2 0 1OutputCopyYESNOYES | [
"geometry",
"implementation"
] | import sys
def contains(segment,point):
return (min(segment[0][0],segment[1][0])<=point[0]<=max(segment[0][0],segment[1][0]) and
min(segment[0][1],segment[1][1])<=point[1]<=max(segment[0][1],segment[1][1]));
def length(segment):
return ((segment[0][0]-segment[1][0])**2+(segment[0][1]-segment[1][1])**2)**0.5;
t=int(sys.stdin.readline());
for _ in range(t):
points=set([]);
segments=[];
lines=[];
roof=None;
for i in range(3):
segment=list(map(int,sys.stdin.readline().split()));
lines.append([segment[1]-segment[3],segment[2]-segment[0],segment[0]*segment[3]-segment[1]*segment[2]]);
p1=(segment[0],segment[1]);
if p1 in points:
roof=p1;
else:
points.add(p1);
p2=(segment[2],segment[3]);
if p2 in points:
roof=p2;
else:
points.add(p2);
segments.append([p1,p2]);
#first condition
if len(points)!=5 or roof==None:
sys.stdout.write('NO\n');
continue;
legs=[];
for i in range(3):
if roof in segments[i]:
legs.append(i);
if len(legs)!=2:
sys.stdout.write('NO\n');
continue;
connector=3-sum(legs);
conns=[None,None];
for i in range(2):
if lines[legs[i]][0]*segments[connector][0][0]+lines[legs[i]][1]*segments[connector][0][1]+lines[legs[i]][2]==0:
if contains(segments[legs[i]],segments[connector][0]):
conns[i]=segments[connector][0];
if lines[legs[i]][0]*segments[connector][1][0]+lines[legs[i]][1]*segments[connector][1][1]+lines[legs[i]][2]==0:
if contains(segments[legs[i]],segments[connector][1]):
conns[i]=segments[connector][1];
if None in conns or conns[0]==conns[1]:
sys.stdout.write('NO\n');
continue;
#second condition
endp=[];
for i in range(2):
if segments[legs[i]][0]==roof:
endp.append(segments[legs[i]][1]);
else:
endp.append(segments[legs[i]][0]);
if length(endp)<=abs(length(segments[legs[0]])-length(segments[legs[1]])) or length(endp)**2>length(segments[legs[0]])**2+length(segments[legs[1]])**2:
sys.stdout.write('NO\n');
continue;
#third condition
if (min(length([conns[0],roof]),length([conns[0],endp[0]]))/max(length([conns[0],roof]),length([conns[0],endp[0]]))<0.25 or
min(length([conns[1],roof]),length([conns[1],endp[1]]))/max(length([conns[1],roof]),length([conns[1],endp[1]]))<0.25):
sys.stdout.write('NO\n');
continue;
sys.stdout.write('YES\n');
| 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 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 collections import defaultdict as dd
from collections import deque
import bisect
import heapq
def ri():
return int(input())
def rl():
return list(map(int, input().split()))
def BFS(G, s):
D = {}
D[s] = 0
curr = [s]
d = 0
while curr:
d += 1
nxt = []
for node in curr:
for nbr in G[node]:
if nbr not in D:
D[nbr] = d
nxt.append(nbr)
curr = nxt
return D
def solve():
n, m = rl()
G = dd(list)
for i in range(m):
u, v = rl()
# we build reversed graph
G[v].append(u)
k = ri()
P = rl()
s = P[0]
t = P[-1]
D = BFS(G, t)
# when can there be multiple routes
E = dd(int)
for u, vlist in G.items():
for v in vlist:
if D[v] == D[u] + 1:
E[v] += 1
lo, hi = 0, 0
for i in range(k - 1):
if D[P[i+1]] >= D[P[i]]:
# must re-route
lo += 1
hi += 1
if D[P[i+1]] == D[P[i]] - 1 and E[P[i]] > 1:
# possible re-route
hi += 1
print (lo, hi)
mode = 's'
if mode == 'T':
t = ri()
for i in range(t):
solve()
else:
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, sys
from io import BytesIO, IOBase
from math import log2, ceil, sqrt, gcd
from _collections import deque
import heapq as hp
from bisect import bisect_left, bisect_right
from math import cos, sin
from itertools import permutations
from operator import itemgetter
# sys.setrecursionlimit(2*10**5+10000)
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")
n = int(input())
a = input()
ck = [0] * n
ct = [[] for _ in range(26)]
for i in range(n):
ct[ord(a[i]) - 97].append(i)
for i in range(26):
if not ct[i]:
continue
s = set()
s.add(0)
cc = 1
for j in range(n - 1, ct[i][0] - 1, -1):
s.add(ck[j])
if i == ord(a[j]) - 97:
while cc in s:
cc += 1
ck[j] = cc
print(len(set(ck)))
print(*ck)
| 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"
] | def output_ans(ans):
print('YES')
print(len(ans))
for x in ans:
print(x[0], x[1])
if __name__ == '__main__':
n, m, k = [int(x) for x in input().split(' ')]
ans_list = list()
ans_list.append((m - 1, 'R'))
ans_list.append((m - 1, 'L'))
for _ in range(n - 1):
ans_list.append((1, 'D'))
ans_list.append((m - 1, 'RUD'))
ans_list.append((m - 1, 'L'))
ans_list.append((n - 1, 'U'))
count = 0
ans = list()
for op in ans_list:
if op[0] == 0:
continue
for x in range(op[0]):
for j in range(len(op[1])):
count += 1
if count == k:
if x > 0:
ans.append((x, op[1]))
ans.append((1, op[1][0:j + 1]))
output_ans(ans)
exit(0)
ans.append((op[0], op[1]))
if k > count:
print('NO')
exit(0)
output_ans(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"
] | n,m = tuple(map(int,input().split()))
steps = 0
if n == m :
print(steps)
elif m%n != 0:
print(-1)
else:
d = m/n
while d % 2 == 0:
d /= 2
steps += 1
while d % 3 == 0:
d /= 3
steps += 1
if d != 1:
steps = -1
print(steps)
| py |
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