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stringlengths 6
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stringlengths 10
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float64 1k
8k
| memory_limit
float64 131k
1.05M
| wrong_status
stringclasses 2
values | wrong_cpu_time
float64 10
40k
| wrong_memory
float64 2.94k
3.37M
| wrong_code_size
int64 1
15.5k
| problem_description
stringlengths 1
4.75k
| wrong_code
stringlengths 1
6.92k
| acc_submission_id
stringlengths 10
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stringclasses 1
value | acc_cpu_time
float64 10
27.8k
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float64 2.94k
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stringlengths 19
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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s341233018 | p03545 | u378157957 | 2,000 | 262,144 | Wrong Answer | 28 | 9,332 | 634 | Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted. | import operator
vals = input()
nums = list(map(int, tuple(vals)))
def dump(a,b,c):
if a == operator.add:
a = '+'
else:
a = '-'
if b == operator.add:
b = '+'
else:
b = '-'
if c == operator.add:
c = '+'
else:
c = '-'
print(vals[0] + a + vals[1] + b + vals[2] + c + vals[3] + '=7')
for a in (operator.add, operator.sub):
for b in (operator.add, operator.sub):
for c in (operator.add, operator.sub):
if c(b(a(nums[0], nums[1]), nums[2]), nums[3]) == 7:
print(a, b, c)
dump(a, b, c)
break | s487971709 | Accepted | 27 | 9,032 | 234 | a, b, c, d = input()
for i in range(2**3):
op = bin(i)[2:].zfill(3).replace("0", "-").replace("1", "+")
equation = a + op[0] + b + op[1] + c + op[2] + d
if eval(equation) == 7:
print(equation + '=7')
break |
s698763617 | p03555 | u037221289 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 118 | You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise. | C1 = input()
C2 = input()
if C1[0] == C2[2] and C1[1] == C2[2] and C1[2] == C2[0]:
print('YES')
else:
print('NO') | s296938138 | Accepted | 17 | 2,940 | 119 | C1 = input()
C2 = input()
if C1[0] == C2[2] and C1[1] == C2[1] and C1[2] == C2[0]:
print('YES')
else:
print('NO')
|
s514023224 | p02612 | u153094838 | 2,000 | 1,048,576 | Wrong Answer | 34 | 9,144 | 123 | We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required. | # -*- coding: utf-8 -*-
"""
Created on Sun Jul 5 21:00:39 2020
@author: NEC-PCuser
"""
N = int(input())
print(N % 1000) | s870624073 | Accepted | 31 | 9,104 | 173 | # -*- coding: utf-8 -*-
"""
Created on Sun Jul 5 21:00:39 2020
@author: NEC-PCuser
"""
N = int(input())
if N % 1000 == 0:
print(0)
else:
print(1000 - (N % 1000)) |
s802792987 | p03816 | u620480037 | 2,000 | 262,144 | Wrong Answer | 46 | 14,564 | 61 | Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written. He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept. Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck. | N=input()
A=list(map(int,input().split()))
print(len(set(A))) | s534674228 | Accepted | 49 | 14,388 | 122 | N=input()
A=list(map(int,input().split()))
if len(set(A))%2==0:
ans=len(set(A))-1
else:
ans=len(set(A))
print(ans) |
s199613678 | p03455 | u400143507 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 384 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | enter = input().split()
def map_num(enter):
input_list = []
for num in enter:
if num != ' ':
input_list.append(int(num))
return input_list[0], input_list[1]
a, b = map_num(enter)
result = a * b
print(result)
if result % 2 == 0:
print('Even')
elif result % 2 != 0:
print('Odd')
| s958111268 | Accepted | 17 | 2,940 | 370 | enter = input().split()
def map_num(enter):
input_list = []
for num in enter:
if num != ' ':
input_list.append(int(num))
return input_list[0], input_list[1]
a, b = map_num(enter)
result = a * b
if result % 2 == 0:
print('Even')
elif result % 2 != 0:
print('Odd')
|
s476795605 | p03944 | u409542115 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 492 | There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white. Snuke plotted N points into the rectangle. The coordinate of the i-th (1 ≦ i ≦ N) point was (x_i, y_i). Then, he created an integer sequence a of length N, and for each 1 ≦ i ≦ N, he painted some region within the rectangle black, as follows: * If a_i = 1, he painted the region satisfying x < x_i within the rectangle. * If a_i = 2, he painted the region satisfying x > x_i within the rectangle. * If a_i = 3, he painted the region satisfying y < y_i within the rectangle. * If a_i = 4, he painted the region satisfying y > y_i within the rectangle. Find the area of the white region within the rectangle after he finished painting. | W, H, N = map(int, input().split())
x = [0] * N
y = [0] * N
a = [0] * N
for i in range(N):
x[i], y[i], a[i] = map(int, input().split())
X1 = 0
X2 = W
Y1 = 0
Y2 = H
entire = W * H
for i in range(N):
if a[i] == 1:
X1 = max(X1, x[i])
if a[i] == 2:
X2 = min(X2, x[i])
if a[i] == 3:
Y1 = max(Y1, y[i])
if a[i] == 4:
Y2 = min(Y2, y[i])
if X1 >= X2 or Y1 >= Y2:
print('0')
else:
entire = entire - (Y2-Y1) * (X2-X1)
print(entire)
| s873511853 | Accepted | 17 | 3,064 | 468 | W, H, N = map(int, input().split())
x = [0] * N
y = [0] * N
a = [0] * N
for i in range(N):
x[i], y[i], a[i] = map(int, input().split())
X1 = 0
X2 = W
Y1 = 0
Y2 = H
for i in range(N):
if a[i] == 1:
X1 = max(X1, x[i])
if a[i] == 2:
X2 = min(X2, x[i])
if a[i] == 3:
Y1 = max(Y1, y[i])
if a[i] == 4:
Y2 = min(Y2, y[i])
if X1 >= X2 or Y1 >= Y2:
print('0')
else:
entire = (Y2-Y1) * (X2-X1)
print(entire)
|
s013441745 | p02697 | u289162337 | 2,000 | 1,048,576 | Wrong Answer | 76 | 9,156 | 73 | You are going to hold a competition of one-to-one game called AtCoder Janken. _(Janken is the Japanese name for Rock-paper-scissors.)_ N players will participate in this competition, and they are given distinct integers from 1 through N. The arena has M playing fields for two players. You need to assign each playing field two distinct integers between 1 and N (inclusive). You cannot assign the same integer to multiple playing fields. The competition consists of N rounds, each of which proceeds as follows: * For each player, if there is a playing field that is assigned the player's integer, the player goes to that field and fight the other player who comes there. * Then, each player adds 1 to its integer. If it becomes N+1, change it to 1. You want to ensure that no player fights the same opponent more than once during the N rounds. Print an assignment of integers to the playing fields satisfying this condition. It can be proved that such an assignment always exists under the constraints given. | n, m = map(int, input().split())
for i in range(m):
print(i+1, 2*m+1-i) | s280167696 | Accepted | 80 | 9,288 | 187 | n, m = map(int, input().split())
if m%2 == 0:
x = m//2
y = m//2
else:
x = m//2
y = m//2+1
for i in range(x):
print(i+1, 2*x+1-i)
for i in range(y):
print(i+2*x+2, 2*y+2*x+1-i) |
s410159776 | p03150 | u638795007 | 2,000 | 1,048,576 | Wrong Answer | 40 | 4,804 | 1,096 | A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string. | def examA():
N = LI()
N.sort()
if N==[1,4,7,9]:
print("YES")
else:
print("NO")
return
def examB():
S = SI()
K = "keyence"
for i in range(8):
if S[:i]+S[-7+i:0]==K:
print("YES")
return
print("NO")
return
def examC():
ans = 0
print(ans)
return
def examD():
ans = 0
print(ans)
return
def examE():
ans = 0
print(ans)
return
def examF():
ans = 0
print(ans)
return
import sys,copy,bisect,itertools,heapq,math,random
from heapq import heappop,heappush,heapify
from collections import Counter,defaultdict,deque
def I(): return int(sys.stdin.readline())
def LI(): return list(map(int,sys.stdin.readline().split()))
def LSI(): return list(map(str,sys.stdin.readline().split()))
def LS(): return sys.stdin.readline().split()
def SI(): return sys.stdin.readline().strip()
global mod,mod2,inf,alphabet,_ep
mod = 10**9 + 7
mod2 = 998244353
inf = 10**18
_ep = 10**(-12)
alphabet = [chr(ord('a') + i) for i in range(26)]
if __name__ == '__main__':
examB()
"""
""" | s760841546 | Accepted | 41 | 4,804 | 1,148 | def examA():
N = LI()
N.sort()
if N==[1,4,7,9]:
print("YES")
else:
print("NO")
return
def examB():
S = SI()
K = "keyence"
for i in range(7):
if S[:i]+S[-7+i:]==K:
print("YES")
return
if S[:7]==K:
print("YES")
return
print("NO")
return
def examC():
ans = 0
print(ans)
return
def examD():
ans = 0
print(ans)
return
def examE():
ans = 0
print(ans)
return
def examF():
ans = 0
print(ans)
return
import sys,copy,bisect,itertools,heapq,math,random
from heapq import heappop,heappush,heapify
from collections import Counter,defaultdict,deque
def I(): return int(sys.stdin.readline())
def LI(): return list(map(int,sys.stdin.readline().split()))
def LSI(): return list(map(str,sys.stdin.readline().split()))
def LS(): return sys.stdin.readline().split()
def SI(): return sys.stdin.readline().strip()
global mod,mod2,inf,alphabet,_ep
mod = 10**9 + 7
mod2 = 998244353
inf = 10**18
_ep = 10**(-12)
alphabet = [chr(ord('a') + i) for i in range(26)]
if __name__ == '__main__':
examB()
"""
""" |
s392353301 | p03477 | u973108807 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 153 | A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R. Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan. Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right. | a,b,c,d = map(int, input().split())
left = a+b
right = c+d
if left > right:
print('left')
elif right > left:
print('right')
else:
print('balanced') | s063698852 | Accepted | 17 | 3,060 | 132 | a,b,c,d = map(int, input().split())
L = a+b
R = c+d
if L > R:
print('Left')
elif R > L:
print('Right')
else:
print('Balanced') |
s980421415 | p03399 | u611047648 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 73 | You planned a trip using trains and buses. The train fare will be A yen (the currency of Japan) if you buy ordinary tickets along the way, and B yen if you buy an unlimited ticket. Similarly, the bus fare will be C yen if you buy ordinary tickets along the way, and D yen if you buy an unlimited ticket. Find the minimum total fare when the optimal choices are made for trains and buses. | a = input()
b = input()
c = input()
d = input()
print(min(a, b)+min(c,d)) | s866997460 | Accepted | 17 | 2,940 | 93 | a = int(input())
b = int(input())
c = int(input())
d = int(input())
print(min(a, b)+min(c,d)) |
s137537135 | p03957 | u940102677 | 1,000 | 262,144 | Time Limit Exceeded | 1,056 | 2,940 | 147 | This contest is `CODEFESTIVAL`, which can be shortened to the string `CF` by deleting some characters. Mr. Takahashi, full of curiosity, wondered if he could obtain `CF` from other strings in the same way. You are given a string s consisting of uppercase English letters. Determine whether the string `CF` can be obtained from the string s by deleting some characters. | s = input()
n = len(s)
i = 0
while i < n:
if s[i] == "C":
break
i += 1
while i < n:
if s[i] == "F":
print("Yes")
exit()
print("No") | s846389356 | Accepted | 17 | 2,940 | 158 | s = input()
n = len(s)
i = 0
while i < n:
if s[i] == "C":
break
i += 1
while i < n:
if s[i] == "F":
print("Yes")
exit()
i += 1
print("No") |
s744511556 | p02697 | u669173971 | 2,000 | 1,048,576 | Wrong Answer | 74 | 9,260 | 71 | You are going to hold a competition of one-to-one game called AtCoder Janken. _(Janken is the Japanese name for Rock-paper-scissors.)_ N players will participate in this competition, and they are given distinct integers from 1 through N. The arena has M playing fields for two players. You need to assign each playing field two distinct integers between 1 and N (inclusive). You cannot assign the same integer to multiple playing fields. The competition consists of N rounds, each of which proceeds as follows: * For each player, if there is a playing field that is assigned the player's integer, the player goes to that field and fight the other player who comes there. * Then, each player adds 1 to its integer. If it becomes N+1, change it to 1. You want to ensure that no player fights the same opponent more than once during the N rounds. Print an assignment of integers to the playing fields satisfying this condition. It can be proved that such an assignment always exists under the constraints given. | n,m=map(int, input().split())
for i in range(1,m+1):
print(i,n+1-i) | s219375029 | Accepted | 79 | 9,224 | 258 | n,m=map(int, input().split())
if m%2==1:
for i in range(m//2):
print(i+1,m-i)
for i in range(m//2+1):
print(m+i+1,2*m+1-i)
else:
for i in range(m//2):
print(i+1,m+1-i)
for i in range(m//2):
print(m+i+2,2*m+1-i) |
s946738458 | p02865 | u991269553 | 2,000 | 1,048,576 | Wrong Answer | 21 | 2,940 | 61 | How many ways are there to choose two distinct positive integers totaling N, disregarding the order? | n = int(input())
if n%2 != 0:
print(n/2)
else:
print(n/4)
| s057564070 | Accepted | 17 | 2,940 | 58 | import math
n = int(input())
a = n/2
print(math.ceil(a)-1) |
s418410384 | p03730 | u740284863 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 121 | We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`. | a,b,c=map(int,input().split())
for i in range(b):
if (i*a) % b == c:
print("Yes")
exit()
print("No")
| s233783229 | Accepted | 18 | 2,940 | 121 | a,b,c=map(int,input().split())
for i in range(b):
if (i*a) % b == c:
print("YES")
exit()
print("NO")
|
s744606092 | p03836 | u758411830 | 2,000 | 262,144 | Wrong Answer | 26 | 9,268 | 259 | Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him. | sx, sy, tx, ty = list(map(int, input().split()))
dx = tx - sx
dy = ty - sy
d1 = ['L'] + ['U']*(dy+1) + ['R']*(dx+1) + ['D']*1
d2 = ['L']*dx + ['D']*dy
d3 = ['D']*1 + ['R']*(dx+1) + ['U']*(dy+1) + ['L']*1
d4 = ['U']*dy + ['R']*dx
print(''.join(d1+d2+d3+d4))
| s158533043 | Accepted | 33 | 9,192 | 253 | sx, sy, tx, ty = list(map(int, input().split()))
dx = tx - sx
dy = ty - sy
d1 = ['U']*dy + ['R']*dx
d2 = ['D']*dy + ['L']*dx
d3 = ['L'] + ['U']*(dy+1) + ['R']*(dx+1) + ['D']
d4 = ['R'] + ['D']*(dy+1) + ['L']*(dx+1) + ['U']
print(''.join(d1+d2+d3+d4))
|
s423421089 | p03605 | u131405882 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 72 | It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N? | N = input()
if N[0] == 9 or N[1] == 9:
print('Yes')
else:
print('No') | s694114163 | Accepted | 17 | 2,940 | 77 | N = input()
if N[0] == '9' or N[1] == '9':
print('Yes')
else:
print('No') |
s509997606 | p03485 | u717501752 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 89 | You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer. | a,b=map(float,input().split())
if (a+b)%2==1:
print((a+b)/2+0.5)
else:
print((a+b)/2) | s975260752 | Accepted | 17 | 2,940 | 99 | a,b=map(float,input().split())
if (a+b)%2==1:
print(int((a+b)/2+0.5))
else:
print(int((a+b)/2)) |
s679200722 | p03369 | u284854859 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 50 | In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen. | li = input().split()
print(700+100*li.count('o')) | s919944948 | Accepted | 17 | 2,940 | 47 | s=list(input())
print(700+100*(s.count('o'))) |
s420072978 | p03606 | u413165887 | 2,000 | 262,144 | Wrong Answer | 20 | 2,940 | 102 | Joisino is working as a receptionist at a theater. The theater has 100000 seats, numbered from 1 to 100000. According to her memo, N groups of audiences have come so far, and the i-th group occupies the consecutive seats from Seat l_i to Seat r_i (inclusive). How many people are sitting at the theater now? | n = int(input())
r = 0
for _ in range(n):
l, r = map(int, input().split())
r += l-r+1
print(r) | s166339497 | Accepted | 20 | 2,940 | 117 | n = int(input())
result = 0
for _ in range(n):
l, r = map(int, input().split())
result += r-l+1
print(result) |
s254833215 | p03377 | u408375121 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 111 | There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals. | A, B, X = map(int, input().split())
if A > X:
print('No')
elif X <= A + B:
print('Yes')
else:
print('No') | s359288946 | Accepted | 17 | 2,940 | 94 | A, B, X = map(int, input().split())
if A > X or X > A + B:
print('NO')
else:
print('YES')
|
s542811361 | p03543 | u716660050 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 100 | We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**? | N=list(input())
ans=False
for i in N:
if N.count(i)>2:
ans=True
break
print(ans) | s005018003 | Accepted | 17 | 3,060 | 171 | N=input()
l=['000','111','222','333','444','555','666','777','888','999']
for i in l:
if i == N[0:3] or i==N[1:len(N)]:
print('Yes')
exit()
print('No') |
s886883449 | p03456 | u963009166 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 115 | AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number. | import math
a, b = map(int, input().split())
if math.sqrt(a * b) % 1 == 0:
print('Yes')
else:
print('No')
| s191001763 | Accepted | 17 | 2,940 | 110 | import math
a, b = input().split()
if math.sqrt(int(a + b)) % 1 == 0:
print('Yes')
else:
print('No')
|
s975763343 | p03433 | u272557899 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 97 | E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins. | n = int(input())
a = int(input())
t = n % 500 - a
if a <= 0:
print("Yes")
else:
print("No") | s869411699 | Accepted | 18 | 2,940 | 98 | n = int(input())
a = int(input())
t = n % 500 - a
if t <= 0:
print("Yes")
else:
print("No")
|
s324389268 | p02612 | u826138795 | 2,000 | 1,048,576 | Wrong Answer | 31 | 9,008 | 83 | We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required. | N=input()
i=(int(N[0])+1)*1000
j=i-int(N)
if j==1000:
print(0)
else:
print(j-i) | s861665483 | Accepted | 28 | 9,112 | 117 | N=input()
if len(N)>3:
i=(int(N[:-3])+1)*1000
else:
i=1000
j=i-int(N)
if j>999:
print(0)
else:
print(j) |
s085913643 | p03495 | u547167033 | 2,000 | 262,144 | Wrong Answer | 151 | 39,344 | 197 | Takahashi has N balls. Initially, an integer A_i is written on the i-th ball. He would like to rewrite the integer on some balls so that there are at most K different integers written on the N balls. Find the minimum number of balls that Takahashi needs to rewrite the integers on them. | from collections import Counter
n,k=map(int,input().split())
a=list(map(int,input().split()))
c=Counter(a).most_common()
ans=0
idx=1
while k>0:
k-=c[-idx][1]
ans+=c[-idx][1]
idx+=1
print(ans) | s299737884 | Accepted | 237 | 39,344 | 197 | from collections import Counter
n,k=map(int,input().split())
a=list(map(int,input().split()))
c=Counter(a).most_common()
ans=0
idx=1
m=len(c)
while m>k:
m-=1
ans+=c[-idx][1]
idx+=1
print(ans) |
s324363307 | p03543 | u013756322 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 115 | We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**? | n = input()
s = set(n)
print(s)
if (len(s) <= 2 and n.count(s.pop()) % 2 == 1):
print('Yes')
else:
print('No')
| s106697684 | Accepted | 17 | 2,940 | 85 | n = input()
if (n[0] * 3 in n or n[-1] * 3 in n):
print('Yes')
else:
print('No')
|
s898771741 | p03400 | u102960641 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 105 | Some number of chocolate pieces were prepared for a training camp. The camp had N participants and lasted for D days. The i-th participant (1 \leq i \leq N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on. As a result, there were X chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces. Find the number of chocolate pieces prepared at the beginning of the camp. | n = int(input())
d,x = map(int, input().split())
for i in range(n):
x += (d-1) // int(input())
print(x) | s615780924 | Accepted | 18 | 2,940 | 117 | n = int(input())
d,x = map(int, input().split())
for i in range(n):
a = (d-1) // int(input()) +1
x += a
print(x)
|
s741447110 | p03997 | u772649753 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 67 | You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid. | a = int(input())
b = int(input())
h = int(input())
print((a+b)*h/2) | s312110251 | Accepted | 17 | 2,940 | 72 | a = int(input())
b = int(input())
h = int(input())
print(int((a+b)*h/2)) |
s191247881 | p03971 | u735335967 | 2,000 | 262,144 | Wrong Answer | 112 | 3,988 | 337 | There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these. Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests. * A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B. * An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students. A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these. Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass. | n, a, b = map(int, input().split())
s = input()
print(len(s))
b_cnt = 0
a_cnt = 0
for i in range(n):
if s[i] == "a" and (a_cnt+b_cnt) < a+b:
a_cnt += 1
print("Yes")
elif s[i] == "b" and (a_cnt+b_cnt) < a+b and b_cnt < b:
#print(b_cnt)
b_cnt += 1
print("Yes")
else:
print("No")
| s918082265 | Accepted | 142 | 4,784 | 404 | n, a, b = map(int, input().split())
s = input()
total_cnt = 0
b_cnt = 0
ans = ["No"]*n
for i in range(n):
if total_cnt >= a+b:
break
if s[i] == "c":
continue
if s[i] == "a":
total_cnt += 1
ans[i] = "Yes"
if s[i] == "b":
b_cnt += 1
if b_cnt <= b:
total_cnt += 1
ans[i] = "Yes"
for i in range(n):
print(ans[i])
|
s538471344 | p03400 | u075303794 | 2,000 | 262,144 | Wrong Answer | 28 | 9,160 | 191 | Some number of chocolate pieces were prepared for a training camp. The camp had N participants and lasted for D days. The i-th participant (1 \leq i \leq N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on. As a result, there were X chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces. Find the number of chocolate pieces prepared at the beginning of the camp. | N=int(input())
D,X=map(int,input().split())
A=[int(input()) for _ in range(N)]
ans=0
for i in range(N):
for j in range(N):
if 30>=A[i]*j:
ans+=1
else:
break
print(ans+X) | s371302432 | Accepted | 27 | 9,112 | 189 | N=int(input())
D,X=map(int,input().split())
A=[int(input()) for _ in range(N)]
ans=0
for i in range(N):
for j in range(D):
if D>A[i]*j:
ans+=1
else:
break
print(ans+X) |
s204755160 | p02663 | u897633035 | 2,000 | 1,048,576 | Wrong Answer | 26 | 9,168 | 123 | In this problem, we use the 24-hour clock. Takahashi gets up exactly at the time H_1 : M_1 and goes to bed exactly at the time H_2 : M_2. (See Sample Inputs below for clarity.) He has decided to study for K consecutive minutes while he is up. What is the length of the period in which he can start studying? | h1, m1, h2, m2, k = map(int, input().split())
a = (h2*60+m2)-(h1*60+m1)
if(a>=k):
print((h2*60+m2)-k)
else :
print(0) | s658084693 | Accepted | 23 | 9,164 | 113 | h1, m1, h2, m2, k = map(int, input().split())
a = (h2*60+m2)-(h1*60+m1)
if(a>k):
print(a-k)
else :
print(0) |
s600563969 | p03645 | u001024152 | 2,000 | 262,144 | Wrong Answer | 654 | 21,192 | 347 | In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands. For convenience, we will call them Island 1, Island 2, ..., Island N. There are M kinds of regular boat services between these islands. Each service connects two islands. The i-th service connects Island a_i and Island b_i. Cat Snuke is on Island 1 now, and wants to go to Island N. However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services. Help him. | N, M = map(int, input().split())
starts = set([])
ends = set([])
for _ in range(M):
a, b = map(int, input().split())
if a == 1 or b == 1:
starts.add(a if b == 1 else b)
if a == N or b == N:
ends.add(a if b == N else b)
print(starts, ends)
ans = "POSSIBLE" if len(starts&ends)>=1 else "IMPOSSIBLE"
print(ans) | s833699167 | Accepted | 607 | 19,024 | 318 | N, M = map(int, input().split())
starts = set([])
ends = set([])
for _ in range(M):
a, b = map(int, input().split())
if a == 1 or b == 1:
starts.add(a if b == 1 else b)
if a == N or b == N:
ends.add(a if b == N else b)
ans = "POSSIBLE" if len(starts&ends)>=1 else "IMPOSSIBLE"
print(ans) |
s900761872 | p03921 | u163320134 | 2,000 | 262,144 | Wrong Answer | 464 | 9,532 | 1,083 | On a planet far, far away, M languages are spoken. They are conveniently numbered 1 through M. For _CODE FESTIVAL 20XX_ held on this planet, N participants gathered from all over the planet. The i-th (1≦i≦N) participant can speak K_i languages numbered L_{i,1}, L_{i,2}, ..., L_{i,{}K_i}. Two participants A and B can _communicate_ with each other if and only if one of the following conditions is satisfied: * There exists a language that both A and B can speak. * There exists a participant X that both A and B can communicate with. Determine whether all N participants can communicate with all other participants. | class UnionFind():
def __init__(self,n):
self.n=n
self.root=[-1]*(n+1)
self.rank=[0]*(n+1)
def FindRoot(self,x):
if self.root[x]<0:
return x
else:
self.root[x]=self.FindRoot(self.root[x])
return self.root[x]
def Unite(self,x,y):
x=self.FindRoot(x)
y=self.FindRoot(y)
if x==y:
return
else:
if self.rank[x]>self.rank[y]:
self.root[x]+=self.root[y]
self.root[y]=x
elif self.rank[x]<=self.rank[y]:
self.root[y]+=self.root[x]
self.root[x]=y
if self.rank[x]==self.rank[y]:
self.rank[y]+=1
def isSameGroup(self,x,y):
return self.FindRoot(x)==self.FindRoot(y)
def Count(self,x):
return -self.root[self.FindRoot(x)]
n,m=map(int,input().split())
t=UnionFind(m)
for _ in range(n):
l=list(map(int,input().split()))
if l[0]==1:
t.Unite(0,l[1])
else:
for i in range(l[0]-1):
t.Unite(l[1],l[2+i])
s=set()
for i in range(m):
r=t.FindRoot(i)
if r==i:
continue
else:
s.add(r)
if len(s)==1:
print('Yes')
else:
print('No') | s688641045 | Accepted | 684 | 11,452 | 1,006 | class UnionFind():
def __init__(self,n):
self.n=n
self.root=[-1]*(n+1)
self.rank=[0]*(n+1)
def FindRoot(self,x):
if self.root[x]<0:
return x
else:
self.root[x]=self.FindRoot(self.root[x])
return self.root[x]
def Unite(self,x,y):
x=self.FindRoot(x)
y=self.FindRoot(y)
if x==y:
return
else:
if self.rank[x]>self.rank[y]:
self.root[x]+=self.root[y]
self.root[y]=x
elif self.rank[x]<=self.rank[y]:
self.root[y]+=self.root[x]
self.root[x]=y
if self.rank[x]==self.rank[y]:
self.rank[y]+=1
def isSameGroup(self,x,y):
return self.FindRoot(x)==self.FindRoot(y)
def Count(self,x):
return -self.root[self.FindRoot(x)]
n,m=map(int,input().split())
t=UnionFind(n+m)
for i in range(n):
l=list(map(int,input().split()))
for j in range(l[0]):
t.Unite(i+1,n+l[1+j])
s=set()
for i in range(n):
r=t.FindRoot(i+1)
s.add(r)
if len(s)==1:
print('YES')
else:
print('NO') |
s652258421 | p03643 | u319612498 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 28 | This contest, _AtCoder Beginner Contest_ , is abbreviated as _ABC_. When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example, ABC680 is the 680th round of ABC. What is the abbreviation for the N-th round of ABC? Write a program to output the answer. | n=input()
print("ABC"+n[1:]) | s516085600 | Accepted | 17 | 2,940 | 29 | n=input()
print("ABC"+n[0:])
|
s770405599 | p03943 | u045939752 | 2,000 | 262,144 | Wrong Answer | 23 | 3,064 | 90 | Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students. | x = map(int, input().split())
if 2 * max(x) == sum(x):
print("Yes")
else:
print("No")
| s757926583 | Accepted | 24 | 3,064 | 98 | x = [ int(i) for i in input().split()]
if 2 * max(x) == sum(x):
print("Yes")
else:
print("No") |
s110126828 | p02741 | u896451538 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 97 | Print the K-th element of the following sequence of length 32: 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51 | 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51 | s826655587 | Accepted | 17 | 3,060 | 135 | s = [1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51]
k = int(input())
print(s[k-1]) |
s739370397 | p03080 | u688055251 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,064 | 150 | There are N people numbered 1 to N. Each person wears a red hat or a blue hat. You are given a string s representing the colors of the people. Person i wears a red hat if s_i is `R`, and a blue hat if s_i is `B`. Determine if there are more people wearing a red hat than people wearing a blue hat. | N=int(input())
s=input().split()
y=0
v=0
for i in s:
if i=='R':
y+=1
else:
v+=1
if y>v:
print('Yes')
else:
print('No') | s803347098 | Accepted | 18 | 2,940 | 154 | N=int(input())
s=input()
y=0
v=0
for i in s:
if i=='R':
y+=1
elif i=='B':
v+=1
if y>v:
print('Yes')
elif v>=y:
print('No') |
s149007095 | p04029 | u036492525 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 31 | There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total? | a=int(input())
print((1+a)*a/2) | s383842157 | Accepted | 17 | 2,940 | 36 | a=int(input())
print(int((1+a)*a/2)) |
s675786380 | p03853 | u757030836 | 2,000 | 262,144 | Wrong Answer | 19 | 3,060 | 179 | There is an image with a height of H pixels and a width of W pixels. Each of the pixels is represented by either `.` or `*`. The character representing the pixel at the i-th row from the top and the j-th column from the left, is denoted by C_{i,j}. Extend this image vertically so that its height is doubled. That is, print a image with a height of 2H pixels and a width of W pixels where the pixel at the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division is rounded down). | h,w =map(int,input().split())
c =[[str(i) for i in input()] for _ in range(h)]
for i in range(2*h):
if i <h:
print("".join(c[i]))
elif i >= h:
print("".join(c[i-h]))
| s196511590 | Accepted | 18 | 3,060 | 93 | h,w = map(int,input().split())
for i in range(h):
c = input()
print(c)
print(c)
|
s194965297 | p03448 | u897328029 | 2,000 | 262,144 | Wrong Answer | 55 | 3,064 | 355 | You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different. | a_500 = int(input().split()[0])
b_100 = int(input().split()[0])
c_50 = int(input().split()[0])
x = int(input().split()[0])
all_coins = []
count = 0
for a_n in range(a_500+1):
for b_n in range(b_100+1):
for c_n in range(c_50+1):
hoge = 500*a_n + 100*b_n + 50*c_50
if hoge == x:
count += 1
print(count) | s395927815 | Accepted | 54 | 3,064 | 356 | a_500 = int(input().split()[0])
b_100 = int(input().split()[0])
c_50 = int(input().split()[0])
x = int(input().split()[0])
all_coins = []
count = 0
for a_n in range(a_500+1):
for b_n in range(b_100+1):
for c_n in range(c_50+1):
total = 500*a_n + 100*b_n + 50*c_n
if total == x:
count += 1
print(count) |
s742678445 | p02665 | u515740713 | 2,000 | 1,048,576 | Wrong Answer | 2,232 | 710,472 | 569 | Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1. | import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
N, *A=map(int,read().split())
bound = [0]*(N+1)
bound[0] = 1
if N == 0:
if A[0] == 1:
print(1)
else:
print(-1)
sys.exit()
for i in range(1,N+1):
bound[i] = (bound[i-1]-A[i-1]) *2
if bound[i] <= 0:
print(-1)
sys.exit()
print(bound)
cum_sum = 0
for i in range(N,-1,-1):
cum_sum += A[i]
bound[i] = min(bound[i],cum_sum)
print(sum(bound)) | s460042104 | Accepted | 803 | 674,396 | 567 | import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
N, *A=map(int,read().split())
bound = [0]*(N+1)
bound[0] = 1
if N == 0:
if A[0] == 1:
print(1)
else:
print(-1)
sys.exit()
for i in range(1,N+1):
bound[i] = (bound[i-1]-A[i-1]) *2
if bound[i] < A[i]:
print(-1)
sys.exit()
cum_sum = 0
for i in range(N,-1,-1):
cum_sum += A[i]
bound[i] = min(bound[i],cum_sum)
print(sum(bound))
|
s338370570 | p03448 | u670180528 | 2,000 | 262,144 | Wrong Answer | 49 | 2,940 | 108 | You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different. | a,b,c,x=map(int,open(0))
print(sum(500*i+100*j+50*k==x for i in range(a)for j in range(b)for k in range(c))) | s886095012 | Accepted | 50 | 2,940 | 114 | a,b,c,x=map(int,open(0))
print(sum(500*i+100*j+50*k==x for i in range(a+1)for j in range(b+1)for k in range(c+1))) |
s238381245 | p03495 | u059210959 | 2,000 | 262,144 | Wrong Answer | 601 | 51,596 | 654 | Takahashi has N balls. Initially, an integer A_i is written on the i-th ball. He would like to rewrite the integer on some balls so that there are at most K different integers written on the N balls. Find the minimum number of balls that Takahashi needs to rewrite the integers on them. | # encoding:utf-8
import copy
import numpy as np
import random
n,k = map(int,input().split())
a = [int(i) for i in input().split()]
count_dic = {}
for i in range(len(a)):
if a[i] in count_dic.keys():
count_dic[a[i]] += 1
else:
count_dic[a[i]] = 1
# print(type(count_dic[1]))
print(count_dic)
count_list = sorted(count_dic.items(),key=lambda x: x[0])
ans = 0
# print(len(count_list))
while len(count_list)>k:
# print(count_list)
delte_tuple = count_list[-1]
add_tuple = count_list[0]
count_list[0] = tuple([add_tuple[0],add_tuple[1]+delte_tuple[1]])
ans += delte_tuple[1]
del count_list[-1]
print(ans) | s674108123 | Accepted | 554 | 48,776 | 670 | # encoding:utf-8
import copy
import numpy as np
import random
n,k = map(int,input().split())
a = [int(i) for i in input().split()]
count_dic = {}
for i in range(len(a)):
if a[i] in count_dic.keys():
count_dic[a[i]] += 1
else:
count_dic[a[i]] = 1
# print(type(count_dic[1]))
# print(count_dic)
count_list = sorted(count_dic.items(),key=lambda x: x[1],reverse=True)
ans = 0
# print(len(count_list))
while len(count_list)>k:
# print(count_list)
delte_tuple = count_list[-1]
add_tuple = count_list[0]
count_list[0] = tuple([add_tuple[0],add_tuple[1]+delte_tuple[1]])
ans += delte_tuple[1]
del count_list[-1]
print(ans)
|
s267678081 | p02613 | u127157764 | 2,000 | 1,048,576 | Wrong Answer | 146 | 16,188 | 210 | Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format. | s = []
n = int(input())
for i in range(n):
s.append(input())
print("AC x " + str(s.count("AC")))
print("WA x " + str(s.count("WA")))
print("TlE x " + str(s.count("TLE")))
print("RE x " + str(s.count("RE"))) | s112245346 | Accepted | 147 | 16,272 | 210 | s = []
n = int(input())
for i in range(n):
s.append(input())
print("AC x " + str(s.count("AC")))
print("WA x " + str(s.count("WA")))
print("TLE x " + str(s.count("TLE")))
print("RE x " + str(s.count("RE"))) |
s771696222 | p02261 | u840247626 | 1,000 | 131,072 | Wrong Answer | 20 | 5,592 | 355 | Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance). | n = int(input())
c = input().split()
cb = c[:]
for i in range(n):
for j in range(n-1,i,-1):
if cb[j][1] < cb[j-1][1]:
cb[j-1],cb[j] = cb[j],cb[j-1]
print(*cb)
print('Stable')
cs = c[:]
for i in range(n):
m = min(range(i,n), key=lambda j: cs[j][1])
if i != m:
cs[i],cs[m] = cs[m],cs[i]
print(*cs)
print(('' if cb == cs else 'Not ') + 'Stable')
| s768696650 | Accepted | 20 | 5,608 | 354 | n = int(input())
c = input().split()
cb = c[:]
for i in range(n):
for j in range(n-1,i,-1):
if cb[j][1] < cb[j-1][1]:
cb[j-1],cb[j] = cb[j],cb[j-1]
print(*cb)
print('Stable')
cs = c[:]
for i in range(n):
m = min(range(i,n), key=lambda j: cs[j][1])
if i != m:
cs[i],cs[m] = cs[m],cs[i]
print(*cs)
print('Stable' if cb == cs else 'Not stable')
|
s677365425 | p04043 | u714300041 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 89 | Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order. | A, B, C = map(int, input().split())
if A*B*C == 5*5*7:
print("Yes")
else:
print("No") | s976201489 | Accepted | 17 | 2,940 | 122 | A, B, C = map(int, input().split())
if A*B*C == 5*5*7 and A >= 5 and B >= 5 and C >= 5:
print("YES")
else:
print("NO") |
s893577182 | p02603 | u503111914 | 2,000 | 1,048,576 | Wrong Answer | 29 | 9,184 | 178 | To become a millionaire, M-kun has decided to make money by trading in the next N days. Currently, he has 1000 yen and no stocks - only one kind of stock is issued in the country where he lives. He is famous across the country for his ability to foresee the future. He already knows that the price of one stock in the next N days will be as follows: * A_1 yen on the 1-st day, A_2 yen on the 2-nd day, ..., A_N yen on the N-th day. In the i-th day, M-kun can make the following trade **any number of times** (possibly zero), **within the amount of money and stocks that he has at the time**. * Buy stock: Pay A_i yen and receive one stock. * Sell stock: Sell one stock for A_i yen. What is the maximum possible amount of money that M-kun can have in the end by trading optimally? | N = int(input())
A = list(map(int,input().split()))
ans = 1000
a = A[0]
for i in range(N):
if A[i] - a > 0:
ans += (ans // a) * (A[i] - a)
a = A[i]
print(ans) | s557199397 | Accepted | 27 | 9,156 | 208 | N = int(input())
A = list(map(int,input().split()))
ans = 1000
m = A[0]
M = A[0]
for i in range(N):
if A[i] < m:
m = A[i]
else:
ans += (ans//m) * (A[i] - m)
m = A[i]
print(ans) |
s584689073 | p03711 | u530383736 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 219 | Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group. | # -*- coding: utf-8 -*-
x,y=[int(i) for i in input().rstrip().split()]
g1={1,3,5,7,8,10,12}
g2={4,6,9,11}
g3={2}
for G in [g1,g2,g3] :
if all( n in G for n in [x,y] ):
print("YES")
break
else:
print("NO") | s371388822 | Accepted | 17 | 3,064 | 219 | # -*- coding: utf-8 -*-
x,y=[int(i) for i in input().rstrip().split()]
g1={1,3,5,7,8,10,12}
g2={4,6,9,11}
g3={2}
for G in [g1,g2,g3] :
if all( n in G for n in [x,y] ):
print("Yes")
break
else:
print("No") |
s672966951 | p03044 | u098012509 | 2,000 | 1,048,576 | Wrong Answer | 234 | 31,780 | 998 | We have a tree with N vertices numbered 1 to N. The i-th edge in the tree connects Vertex u_i and Vertex v_i, and its length is w_i. Your objective is to paint each vertex in the tree white or black (it is fine to paint all vertices the same color) so that the following condition is satisfied: * For any two vertices painted in the same color, the distance between them is an even number. Find a coloring of the vertices that satisfies the condition and print it. It can be proved that at least one such coloring exists under the constraints of this problem. | import sys
input = sys.stdin.readline
def main():
N = int(input())
A = []
for _ in range(N - 1):
A.append([int(x) for x in input().split()])
answer = [-1] * N
for a in A:
if a[2] % 2 == 0:
if answer[a[0] - 1] != -1:
answer[a[1] - 1] = answer[a[0] - 1]
elif answer[a[1] - 1] != -1:
answer[a[0] - 1] = answer[a[1] - 1]
else:
answer[a[0] - 1] = 1
answer[a[1] - 1] = 1
else:
if answer[a[0] - 1] == 1:
answer[a[1] - 1] = 0
elif answer[a[0] - 1] == 0:
answer[a[1] - 1] = 1
elif answer[a[1] - 1] != 1:
answer[a[0] - 1] = 0
elif answer[a[1] - 1] != 0:
answer[a[0] - 1] = 1
else:
answer[a[0] - 1] = 1
answer[a[1] - 1] = 0
print('\n'.join(map(str, answer)))
if __name__ == '__main__':
main()
| s824644218 | Accepted | 901 | 63,556 | 575 | import collections
N = int(input())
UVW = [[int(x) for x in input().split()] for _ in range(N - 1)]
ans = [-1] * N
A = [[] for _ in range(N)]
for u, v, w in UVW:
A[u - 1].append([v - 1, w % 2])
A[v - 1].append([u - 1, w % 2])
q = collections.deque()
q.append(0)
ans[0] = 0
while q:
c = q.popleft()
for ni, nd in A[c]:
if ans[ni] != -1:
continue
if nd == 0:
ans[ni] = ans[c]
else:
ans[ni] = ans[c] ^ 1
q.append(ni)
if sum(ans) == 0:
ans[0] = 1
for a in ans:
print(a)
|
s676951730 | p03997 | u460129720 | 2,000 | 262,144 | Wrong Answer | 19 | 3,060 | 70 | You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid. | a = int(input())
b = int(input())
h = int(input())
print((a + b)*h /2) | s398354220 | Accepted | 17 | 2,940 | 75 | a = int(input())
b = int(input())
h = int(input())
print(int((a + b)*h /2)) |
s158221033 | p02694 | u951985579 | 2,000 | 1,048,576 | Wrong Answer | 28 | 9,000 | 108 | Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time? | X = int(input())
total = 100
count = 0
while total <= X:
total += total//100
count += 1
print(count) | s440554512 | Accepted | 31 | 9,152 | 108 | X = int(input())
total = 100
count = 0
while total < X:
total += total//100
count += 1
print(count)
|
s522971982 | p03555 | u727787724 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 111 | You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise. | a=list(input())
b=list(input())
if a[0]==b[2] and a[1]==b[1] and a[2]==b[2]:
print('YES')
else:
print('NO') | s463032941 | Accepted | 17 | 2,940 | 112 | a=list(input())
b=list(input())
if a[0]==b[2] and a[1]==b[1] and a[2]==b[0]:
print('YES')
else:
print('NO')
|
s819468653 | p03929 | u761320129 | 1,000 | 262,144 | Wrong Answer | 23 | 3,316 | 300 | Snuke has a very long calendar. It has a grid with $n$ rows and $7$ columns. One day, he noticed that calendar has a following regularity. * The cell at the $i$-th row and $j$-th column contains the integer $7i+j-7$. A good sub-grid is a $3 \times 3$ sub-grid and the sum of integers in this sub-grid mod $11$ is $k$. How many good sub-grid are there? Write a program and help him. | N,K = map(int,input().split())
if N < 3:
print(0)
exit()
mem = [0] * 11
for i in range(11):
for j in range(2,7):
n = i*7 + j
if (9*n)%11 == K: mem[i] += 1
ans = sum(mem) * (N//11)
ans -= mem[0]
if N%11 == 0:
ans -= mem[-1]
else:
ans += sum(mem[:N%11])
print(ans) | s531095523 | Accepted | 17 | 3,064 | 315 | N,K = map(int,input().split())
if N<3:
print(0)
exit()
table = [[0]*5 for i in range(11)]
for i in range(11):
for j in range(5):
t = 9*(j+2 + 7*(i+1))
table[i][j] = t%11
c = [0]
for row in table:
c.append(c[-1] + row.count(K))
d,m = divmod(N-2, 11)
ans = c[m] + c[-1]*d
print(ans) |
s915706311 | p03779 | u652892331 | 2,000 | 262,144 | Wrong Answer | 27 | 2,940 | 108 | There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0. During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right. That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i. The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible. Find the earliest possible time to reach coordinate X. | x = int(input())
for t in range(1000000000):
if 2 * x < t * (t + 1):
print(t + 1)
break | s656663518 | Accepted | 26 | 2,940 | 105 | x = int(input())
for t in range(1000000000):
if 2 * x <= t * (t + 1):
print(t)
break |
s080211508 | p03435 | u057993957 | 2,000 | 262,144 | Wrong Answer | 151 | 12,480 | 387 | We have a 3 \times 3 grid. A number c_{i, j} is written in the square (i, j), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left. According to Takahashi, there are six integers a_1, a_2, a_3, b_1, b_2, b_3 whose values are fixed, and the number written in the square (i, j) is equal to a_i + b_j. Determine if he is correct. | import numpy as np
c = np.array([list(map(int, input().split())) for i in range(3)])
flag = True
for a1 in range(c.max()+1):
b1, b2, b3 = c[0, :] - a1
if (c[1, 0] - b1 != c[1, 1] - b2) or (c[1, 2] - b3 == c[1, 0] - b1):
flag = False
break
if (c[2, 0] - b1 != c[2, 1] - b2) or (c[2, 2] - b3 == c[2, 0] - b1):
flag = False
break
print("Yes" if flag else "No") | s461108508 | Accepted | 150 | 14,524 | 389 | import numpy as np
c = np.array([list(map(int, input().split())) for i in range(3)])
flag = True
for a1 in range(c.max()+1):
b1, b2, b3 = c[0, :] - a1
if (c[1, 0] - b1 != c[1, 1] - b2) or (c[1, 2] - b3 != c[1, 0] - b1):
flag = False
break
if (c[2, 0] - b1 != c[2, 1] - b2) or (c[2, 2] - b3 != c[2, 0] - b1):
flag = False
break
print("Yes" if flag else "No") |
s958932692 | p02392 | u365686736 | 1,000 | 131,072 | Wrong Answer | 20 | 5,580 | 76 | Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No". | a,b,c = map(int, input().split())
if a < b < c:
print("Yes")
print("No") | s954853915 | Accepted | 20 | 7,648 | 76 | a,b,c = map(int, input().split())
if a < b < c:print("Yes")
else:print("No") |
s591328649 | p03386 | u818283165 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 234 | Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers. | a,b,k = map(int,input().split())
ans = []
for i in range(k):
if a+i > b:
break
ans.append(a+i)
for i in range(k):
if b-i < a:
break
ans.append(b-i)
ans.sort()
ans = set(ans)
for i in ans:
print(i) | s288834631 | Accepted | 19 | 3,060 | 240 | a,b,k = map(int,input().split())
ans = []
for i in range(k):
if a+i > b:
break
ans.append(a+i)
for i in range(k):
if b-i < a:
break
ans.append(b-i)
ans = set(ans)
ans =sorted(ans)
for i in ans:
print(i) |
s309587527 | p03048 | u050708958 | 2,000 | 1,048,576 | Wrong Answer | 2,104 | 2,940 | 222 | Snuke has come to a store that sells boxes containing balls. The store sells the following three kinds of boxes: * Red boxes, each containing R red balls * Green boxes, each containing G green balls * Blue boxes, each containing B blue balls Snuke wants to get a total of exactly N balls by buying r red boxes, g green boxes and b blue boxes. How many triples of non-negative integers (r,g,b) achieve this? | r,g,b,n = [int(i) for i in input().split()]
ans = 0
for i in range(n):
for j in range(n):
s = i * r + j * g
if s > n:
continue
if (n - s) % b == 0:
ans += 1
print(ans)
| s944677491 | Accepted | 1,927 | 2,940 | 223 | r,g,b,n = [int(i) for i in input().split()]
ans = 0
for i in range(n+1):
for j in range(n+1):
s = i * r + j * g
if s > n:
break
if (n - s) % b == 0:
ans += 1
print(ans)
|
s946529074 | p03024 | u516272298 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 77 | Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility. | s = str(input())
if s.count("o") >= 8:
print("YES")
else:
print("NO") | s544569841 | Accepted | 17 | 2,940 | 87 | s = str(input())
if 15-len(s)+s.count("o") >= 8:
print("YES")
else:
print("NO") |
s234894929 | p02615 | u221272125 | 2,000 | 1,048,576 | Wrong Answer | 149 | 31,532 | 233 | Quickly after finishing the tutorial of the online game _ATChat_ , you have decided to visit a particular place with N-1 players who happen to be there. These N players, including you, are numbered 1 through N, and the **friendliness** of Player i is A_i. The N players will arrive at the place one by one in some order. To make sure nobody gets lost, you have set the following rule: players who have already arrived there should form a circle, and a player who has just arrived there should cut into the circle somewhere. When each player, except the first one to arrive, arrives at the place, the player gets **comfort** equal to the smaller of the friendliness of the clockwise adjacent player and that of the counter-clockwise adjacent player. The first player to arrive there gets the comfort of 0. What is the maximum total comfort the N players can get by optimally choosing the order of arrivals and the positions in the circle to cut into? | N = int(input())
A = list(map(int,input().split()))
A.sort()
if N == 2:
print(A[1])
quit()
a = A.pop()
b = A.pop()
c = A.pop()
B = [c,b,a]
ans = b + c
i = 0
while A:
a = A.pop()
ans += b
b = c
c = a
print(ans) | s899651674 | Accepted | 143 | 31,436 | 197 | N = int(input())
A = list(map(int,input().split()))
A.sort()
k = N - 2
a = A.pop()
ans = a
n = k // 2
for i in range(n):
a = A.pop()
ans += 2*a
if k - 2*n == 1:
ans += A[-1]
print(ans) |
s719133337 | p02742 | u430336181 | 2,000 | 1,048,576 | Wrong Answer | 28 | 9,156 | 276 | We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move: | H, W = map(int, input().split())
H_ = H % 2
W_ = W % 2
if H_ == 0:
print(H/2*W//1)
else:
h = H // 2 + 1
h_ = H // 2
if W_ == 0:
print((h + h_)*W/2//1)
elif W_ == 1:
w = W // 2+1
w_ = W // 2
print((h * w + h_ * w_)//1)
| s594654942 | Accepted | 28 | 9,188 | 351 | H, W = map(int, input().split())
H_ = H % 2
W_ = W % 2
if H == 1 or W == 1:
print(1)
elif H_ == 0:
print(str(round(H/2*W//1)))
elif H_ == 1:
h = H // 2 + 1
h_ = H // 2
if W_ == 0:
print(str(round((h + h_)*W/2)))
elif W_ == 1:
w = W // 2+1
w_ = W // 2
print(str(round(h * w + h_ * w_)))
|
s594247974 | p03433 | u292810930 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 90 | E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins. | N = int(input())
A = int(input())
if N % 500 <= A:
print('YES')
else:
print('NO')
| s185070067 | Accepted | 17 | 2,940 | 90 | N = int(input())
A = int(input())
if N % 500 <= A:
print('Yes')
else:
print('No')
|
s485426260 | p03997 | u847033024 | 2,000 | 262,144 | Wrong Answer | 27 | 9,100 | 73 | You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid. | a = int(input())
b = int(input())
c = int(input())
print((a + b) / 2 * c) | s203740251 | Accepted | 25 | 9,056 | 78 | a = int(input())
b = int(input())
c = int(input())
print(int((a + b) / 2 * c)) |
s797948278 | p03761 | u117348081 | 2,000 | 262,144 | Wrong Answer | 22 | 3,316 | 499 | Snuke loves "paper cutting": he cuts out characters from a newspaper headline and rearranges them to form another string. He will receive a headline which contains one of the strings S_1,...,S_n tomorrow. He is excited and already thinking of what string he will create. Since he does not know the string on the headline yet, he is interested in strings that can be created regardless of which string the headline contains. Find the longest string that can be created regardless of which string among S_1,...,S_n the headline contains. If there are multiple such strings, find the lexicographically smallest one among them. | from collections import defaultdict
n = int(input())
s = []
t = []
a = defaultdict(int)
for i in range(n):
d = defaultdict(int)
st = list(input())
st.sort()
for j in range(len(st)):
d[st[j]]+=1
if i==0:
t = set(st)
for j in t:
a[j] = d[j]
else:
u = set(st)
t = t & u
for j in t:
a[j] = min(a[j],d[j])
s.append(st)
ans = ""
for i in t:
for j in range(a[i]):
ans = ans + i
print(ans) | s667199753 | Accepted | 22 | 3,316 | 503 | from collections import defaultdict
n = int(input())
s = []
t = []
a = defaultdict(int)
for i in range(n):
d = defaultdict(int)
st = list(input())
st.sort()
for j in range(len(st)):
d[st[j]]+=1
if i==0:
t = set(st)
for j in t:
a[j] = d[j]
else:
u = set(st)
t = t & u
for j in t:
a[j] = min(a[j],d[j])
ans = ""
t = list(t)
t.sort()
for i in t:
for j in range(a[i]):
ans = ans + i
print(ans) |
s733651099 | p03455 | u474464800 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 89 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | a, b = map(int, input().split())
if (a*b) % 2 == 0:
print("Odd")
else:
print("Even") | s470284348 | Accepted | 18 | 2,940 | 89 | a, b = map(int, input().split())
if (a*b) % 2 == 0:
print("Even")
else:
print("Odd")
|
s605110788 | p03544 | u941753895 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 122 | It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers. You are given an integer N. Find the N-th Lucas number. Here, the i-th Lucas number L_i is defined as follows: * L_0=2 * L_1=1 * L_i=L_{i-1}+L_{i-2} (i≥2) | l=[]
l.append(2)
l.append(1)
for i in range(84):
l.append(l[i]+l[i+1])
N=int(input())
print(l[N-1]) | s507053935 | Accepted | 17 | 2,940 | 120 | l=[]
l.append(2)
l.append(1)
for i in range(85):
l.append(l[i]+l[i+1])
N=int(input())
print(l[N]) |
s330591840 | p02614 | u935642171 | 1,000 | 1,048,576 | Wrong Answer | 30 | 9,188 | 525 | We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is `.`, and black if c_{i,j} is `#`. Consider doing the following operation: * Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns. You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices. | H,W,K = map(int,input().split())
s = []
black = 0
row = [0]*(H+1)
col = [0]*(W+1)
for i in range(H):
s.append(list(input()))
black += s[i].count('#')
row[i+1] += s[i].count('#')
for i in range(W):
for j in range(H):
if s[j][i]=='#':
col[i+1] += 1
count = 0
for y,i in enumerate(row):
if black - i == K:
count += 1
for z,j in enumerate(col):
if z==0 and y>=1:
continue
if black - (i+j) == K:
if i+j!=0:
count += 1
if K in col[1:]:
count += col[1:].count(K)
print(count) | s539973515 | Accepted | 68 | 9,124 | 316 | H,W,K = map(int,input().split())
s = [list(input()) for i in range(H)]
ans = 0
for i in range(1<<H):
for j in range(1<<W):
cnt = 0
for h in range(H):
for w in range(W):
if (i>>h)&1 == 1 and (j>>w)&1==1:
if s[h][w]=='#':
cnt += 1
if cnt==K:
ans += 1
print(ans) |
s620920062 | p03759 | u791664126 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 58 | Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful. | a,b,c=map(int,input().split());print('YNeos'[b-a!=c-b::2]) | s342417886 | Accepted | 17 | 2,940 | 59 | a,b,c=map(int,input().split());print('YNEOS'[b-a!=c-b::2])
|
s492358339 | p00008 | u776559258 | 1,000 | 131,072 | Wrong Answer | 140 | 7,544 | 218 | Write a program which reads an integer n and identifies the number of combinations of a, b, c and d (0 ≤ a, b, c, d ≤ 9) which meet the following equality: a + b + c + d = n For example, for n = 35, we have 4 different combinations of (a, b, c, d): (8, 9, 9, 9), (9, 8, 9, 9), (9, 9, 8, 9), and (9, 9, 9, 8). | import itertools
def combination(n):
c=0
for i in itertools.product(range(10),repeat=4):
(a,b,c,d)=i
if a+b+c+d==n:
c+=1
return c
while True:
try:
n=int(input())
combination(n)
except EOFError:
break | s137984907 | Accepted | 140 | 7,580 | 225 | import itertools
def combination(n):
m=0
for i in itertools.product(range(10),repeat=4):
(a,b,c,d)=i
if a+b+c+d==n:
m+=1
return m
while True:
try:
n=int(input())
print(combination(n))
except EOFError:
break |
s100877310 | p02646 | u345778634 | 2,000 | 1,048,576 | Wrong Answer | 22 | 9,180 | 215 | Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally. | A, V = map(int, input().split())
B, W = map(int, input().split())
T = int(input())
if W >= V:
print("No")
else:
L = abs(B - A)
if T * (V - W) >= L :
print("Yes")
else:
print("No") | s479722991 | Accepted | 25 | 9,180 | 215 | A, V = map(int, input().split())
B, W = map(int, input().split())
T = int(input())
if W >= V:
print("NO")
else:
L = abs(B - A)
if T * (V - W) >= L :
print("YES")
else:
print("NO") |
s260536498 | p02854 | u189487046 | 2,000 | 1,048,576 | Wrong Answer | 244 | 32,712 | 213 | Takahashi, who works at DISCO, is standing before an iron bar. The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters. Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length. However, this may not be possible as is, so he will do the following operations some number of times **before** he does the cut: * Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan). * Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen. Find the minimum amount of money needed before cutting the bar into two parts with the same length. | n = int(input())
A = list(map(int, input().split()))
ca = [0]
for i in range(n):
ca.append(A[i]+ca[i])
print(ca)
ans = 2020202020
for i in range(n):
ans = min(ans, abs(ca[i]-(ca[-1]-ca[i])))
print(ans)
| s703265257 | Accepted | 216 | 26,764 | 203 | n = int(input())
A = list(map(int, input().split()))
ca = [0]
for i in range(n):
ca.append(A[i]+ca[i])
ans = 2020202020
for i in range(n):
ans = min(ans, abs(ca[i]-(ca[-1]-ca[i])))
print(ans)
|
s118253056 | p03836 | u557235596 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 426 | Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him. | def main():
sx, sy, tx, ty = map(int, input().split())
result = "U" * (ty - sy)
result += "R" * (tx - sx)
result += "D" * (ty - sy)
result += "L" * (tx - sx)
result += "L"
result += "U" * (ty - sy + 1)
result += "R" * (tx - sx + 1)
result += "D"
result += "U"
result += "D" * (ty - sy + 1)
result += "L" * (tx - sx + 1)
result += "R"
if __name__ == "__main__":
main()
| s844321642 | Accepted | 17 | 3,060 | 444 | def main():
sx, sy, tx, ty = map(int, input().split())
result = "U" * (ty - sy)
result += "R" * (tx - sx)
result += "D" * (ty - sy)
result += "L" * (tx - sx)
result += "L"
result += "U" * (ty - sy + 1)
result += "R" * (tx - sx + 1)
result += "D"
result += "R"
result += "D" * (ty - sy + 1)
result += "L" * (tx - sx + 1)
result += "U"
print(result)
if __name__ == "__main__":
main()
|
s570232104 | p03379 | u223904637 | 2,000 | 262,144 | Wrong Answer | 152 | 25,228 | 249 | When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N. | import bisect
n=int(input())
l=list(map(int,input().split()))
l.sort()
if n==2:
print(l[1],l[0])
exit()
m=max(l)
k=m//2
if m%2==1:
k+=1
i=bisect.bisect_left(l,k)
if abs(l[i]-k)<=abs(l[i+1]-k):
print(m,l[i])
else:
print(m,l[i+1])
| s087542558 | Accepted | 323 | 25,220 | 178 | n=int(input())
li = list(map(int,input().split()))
s=sorted(li)
a=s[round(n/2)-1]
b=s[round(n/2)]
for i in range(n):
if li[i]<a+1:
print(b)
else:
print(a) |
s382987765 | p03565 | u896741788 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 279 | E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One more thing he found is a sheet of paper with the following facts written on it: * Condition 1: The string S contains a string T as a contiguous substring. * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1. Print the string S. If such a string does not exist, print `UNRESTORABLE`. | s,y=(input() for i in range(2))
ans="{"*len(s);pa=ans
for i in range(len(s)-len(y)):
for j in range(len(y)):
if s[i+j]!='?' and s[i+j]!=y[j]:break
else:ans=min(ans,s[:i].replace("?","a")+y+s[i+len(y):].replace("?","a"))
if pa==ans:print("UNRESTORABLE")
else:print(ans) | s711355304 | Accepted | 27 | 9,084 | 319 | s,t=input(),input()
ans='z'*len(s)
for i in range(len(s)-len(t)+1):
pre=s[:i].replace("?",'a')
for x,y in zip(list(s[i:]),list(t)):
if x!="?"and x!=y:break
pre+=y
else:
pre+=s[i+len(t):].replace("?",'a')
ans=min(ans,pre)
print(ans if ans!='z'*len(s) else 'UNRESTORABLE') |
s864275080 | p03131 | u593567568 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 118 | Snuke has one biscuit and zero Japanese yen (the currency) in his pocket. He will perform the following operations exactly K times in total, in the order he likes: * Hit his pocket, which magically increases the number of biscuits by one. * Exchange A biscuits to 1 yen. * Exchange 1 yen to B biscuits. Find the maximum possible number of biscuits in Snuke's pocket after K operations. | K,A,B = map(int,input().split())
ans = K + 1
t = ((K+1) // (A+1)) * B
t += (K+1) % (A+1)
ans = max(ans,t)
print(ans)
| s096134804 | Accepted | 17 | 2,940 | 153 | K,A,B = map(int,input().split())
if K < A+1 or B <= A+2:
print(K+1)
exit(0)
K -= A - 1
ans = A
p,q = divmod(K,2)
ans += (B-A) * p + q
print(ans)
|
s826495728 | p03852 | u116348130 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 139 | Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: `a`, `e`, `i`, `o` and `u`. | S = input()
sc = ["eraser", "erase", "dreamer", "dream"]
for i in sc:
S = S.replace(i, "")
if S:
print("NO")
else:
print("YES") | s273722360 | Accepted | 17 | 2,940 | 82 | c = input()
p = "aeiou"
if c in p:
print("vowel")
else:
print("consonant") |
s181103346 | p02401 | u078042885 | 1,000 | 131,072 | Wrong Answer | 30 | 7,412 | 63 | Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part. | while 1:
s=input()
if '?' in s:break
print(eval(s)) | s842937176 | Accepted | 20 | 7,404 | 68 | while 1:
s=input()
if '?' in s:break
print(int(eval(s))) |
s146198442 | p03760 | u319690708 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 664 | Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions. You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password. | from sys import stdin
# import itertools as itr
A = str(input())
B = str(input())
A = list(A)
B = list(B)
AL = len(A)
BL = len(B)
print(AL)
b = []
for i in range(BL):
b.append(A[i] + B[i])
if AL -BL == 1:
b.append(A[-1])
print(",".join(b).replace(",", ""))
# print(C)
| s904150179 | Accepted | 17 | 3,060 | 227 | A = str(input())
B = str(input())
A = list(A)
B = list(B)
AL = len(A)
BL = len(B)
b = []
for i in range(BL):
b.append(A[i] + B[i])
if AL -BL == 1:
# b.append(A[-1])
b.append(A[-1])
print(",".join(b).replace(",", "")) |
s399068425 | p03814 | u637824361 | 2,000 | 262,144 | Wrong Answer | 17 | 3,500 | 58 | Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s). Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`. | s = input()
A = s.find('a')
Z = s.rfind('z')
print(Z-A+1)
| s455441033 | Accepted | 17 | 3,512 | 57 | s = input()
A = s.find('A')
Z = s.rfind('Z')
print(Z-A+1) |
s491707959 | p02880 | u544587633 | 2,000 | 1,048,576 | Wrong Answer | 19 | 2,940 | 230 | Having learned the multiplication table, Takahashi can multiply two integers between 1 and 9 (inclusive) together. Given an integer N, determine whether N can be represented as the product of two integers between 1 and 9. If it can, print `Yes`; if it cannot, print `No`. | import sys
for line in sys.stdin:
N = int(line)
found=False
for i in range(1, 10, 1):
if N % i == 0:
found = True
break
if found:
print('Yes')
else:
print('No') | s380899194 | Accepted | 19 | 3,060 | 217 | import sys
import math
l = []
for i in range(1, 10, 1):
for j in range(1,10,1):
l.append(i*j)
for line in sys.stdin:
N = int(line)
if N in l:
print('Yes')
else:
print('No')
|
s314545038 | p03485 | u197300260 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 95 | You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer. | a,b=map(int,input().split(" "))
print(a)
print(b)
c=a+b
d=c%2
e=c//2
if d==1:
e=e+1
print(e) | s626218542 | Accepted | 17 | 2,940 | 331 |
# Python 3rd Try
from math import ceil
def ceil2number(a, b):
result = 0
c = (a + b) / 2
result = ceil(c)
return result
if __name__ == "__main__":
answer = 0
a, b = map(int, input().split(' '))
answer = ceil2number(a, b)
print(answer)
|
s971800756 | p03611 | u361826811 | 2,000 | 262,144 | Wrong Answer | 177 | 21,420 | 329 | You are given an integer sequence of length N, a_1,a_2,...,a_N. For each 1≤i≤N, you have three choices: add 1 to a_i, subtract 1 from a_i or do nothing. After these operations, you select an integer X and count the number of i such that a_i=X. Maximize this count by making optimal choices. | import sys
import itertools
import numpy as np
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
N, *a = map(int, read().split())
A = np.array(a)
counter = np.bincount(A)
print(type(counter))
B = counter.copy()
B[1:] += counter[:-1]
B[:-1] += counter[1:]
print(B.max())
| s860743441 | Accepted | 177 | 21,408 | 308 | import sys
import itertools
import numpy as np
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
N, *a = map(int, read().split())
A = np.array(a)
counter = np.bincount(A)
B = counter.copy()
B[1:] += counter[:-1]
B[:-1] += counter[1:]
print(B.max())
|
s449233279 | p03624 | u397563544 | 2,000 | 262,144 | Wrong Answer | 48 | 4,664 | 288 | You are given a string S consisting of lowercase English letters. Find the lexicographically (alphabetically) smallest lowercase English letter that does not occur in S. If every lowercase English letter occurs in S, print `None` instead. | from collections import Counter
s = sorted(str(input()))
l = list(Counter(s))
print(l)
a = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
for i in range(0,len(l)):
if l[i]!=a[i]:
print(a[i])
break | s047546465 | Accepted | 33 | 3,960 | 181 | import string
x = input()
def func(x):
diff = set(string.ascii_lowercase)- set(x)
if diff:
print(sorted(diff)[0])
else:
print('None')
func(x) |
s710008664 | p02853 | u254634413 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,060 | 195 | We held two competitions: Coding Contest and Robot Maneuver. In each competition, the contestants taking the 3-rd, 2-nd, and 1-st places receive 100000, 200000, and 300000 yen (the currency of Japan), respectively. Furthermore, a contestant taking the first place in both competitions receives an additional 400000 yen. DISCO-Kun took the X-th place in Coding Contest and the Y-th place in Robot Maneuver. Find the total amount of money he earned. | X, Y = map(int, input().split())
prices = {1: 3e5, 2: 2e5, 3: 1e5}
ans = 0
if X < 4:
ans += prices[X]
if Y < 4:
ans += prices[Y]
if Y == 1 and X == 1:
ans += 4e5
print(ans)
| s349793146 | Accepted | 18 | 2,940 | 207 | X, Y = map(int, input().split())
prices = {1: 300000, 2: 200000, 3: 100000}
ans = 0
if X < 4:
ans += prices[X]
if Y < 4:
ans += prices[Y]
if Y == 1 and X == 1:
ans += 400000
print(ans)
|
s821547343 | p03455 | u826785572 | 2,000 | 262,144 | Wrong Answer | 23 | 3,316 | 102 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | a, b = map(int, input().split())
x = a * b
if x // 2 == 0:
print('Even')
else:
print('Odd')
| s174507113 | Accepted | 17 | 2,940 | 101 | a, b = map(int, input().split())
x = a * b
if x % 2 == 0:
print('Even')
else:
print('Odd')
|
s639796048 | p02613 | u656330453 | 2,000 | 1,048,576 | Wrong Answer | 144 | 16,220 | 279 | Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format. | n = int(input())
a = [input() for i in range(n)]
ac = 0
wa = 0
tle = 0
re = 0
for s in a:
if a=='AC':
ac+=1
if a=='WA':
wa+=1
if a=='TLE':
tle+=1
if a=='RE':
re+=1
print(f'AC × {ac}')
print(f'WA × {wa}')
print(f'TLE × {tle}')
print(f'RE × {re}')
| s789100058 | Accepted | 142 | 16,540 | 194 | from collections import Counter
n = int(input())
s = [input() for _ in range(n)]
counter = Counter(s)
output = ['AC', 'WA', 'TLE', 'RE']
for key in output:
print(f'{key} x {counter[key]}') |
s346676735 | p02255 | u908651435 | 1,000 | 131,072 | Wrong Answer | 20 | 5,532 | 48 | Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step. | a=input()
n=input().split()
s=n.sort()
print(s)
| s389559883 | Accepted | 30 | 5,976 | 363 | def show (nums):
for i in range(len(nums)):
if i!=len(nums)-1:
print(nums[i],end=' ')
else :
print(nums[i])
n=int(input())
nums=list(map(int,input().split()))
show(nums)
for i in range(1,n):
v=nums[i]
j=i-1
while (j>=0 and nums[j]>v):
nums[j+1]=nums[j]
j-=1
nums[j+1]=v
show(nums)
|
s257173331 | p03946 | u007550226 | 2,000 | 262,144 | Wrong Answer | 456 | 15,060 | 604 | There are N towns located in a line, conveniently numbered 1 through N. Takahashi the merchant is going on a travel from town 1 to town N, buying and selling apples. Takahashi will begin the travel at town 1, with no apple in his possession. The actions that can be performed during the travel are as follows: * _Move_ : When at town i (i < N), move to town i + 1. * _Merchandise_ : Buy or sell an arbitrary number of apples at the current town. Here, it is assumed that one apple can always be bought and sold for A_i yen (the currency of Japan) at town i (1 ≦ i ≦ N), where A_i are distinct integers. Also, you can assume that he has an infinite supply of money. For some reason, there is a constraint on merchandising apple during the travel: the sum of the number of apples bought and the number of apples sold during the whole travel, must be at most T. (Note that a single apple can be counted in both.) During the travel, Takahashi will perform actions so that the _profit_ of the travel is maximized. Here, the profit of the travel is the amount of money that is gained by selling apples, minus the amount of money that is spent on buying apples. Note that we are not interested in apples in his possession at the end of the travel. Aoki, a business rival of Takahashi, wants to trouble Takahashi by manipulating the market price of apples. Prior to the beginning of Takahashi's travel, Aoki can change A_i into another arbitrary non-negative integer A_i' for any town i, any number of times. The cost of performing this operation is |A_i - A_i'|. After performing this operation, different towns may have equal values of A_i. Aoki's objective is to decrease Takahashi's expected profit by at least 1 yen. Find the minimum total cost to achieve it. You may assume that Takahashi's expected profit is initially at least 1 yen. | n,t= map(int,input().split())
a= list(map(int,input().split()))
mp,mprof = 999999999,0
bid = []
bsus = []
sid = []
nc=0
for i in range(n):
cp = a[i]-mp
if cp<0:
nc += min(len(bid),len(sid))
mp = a[i]
sid=[]
bid=[]
bsus=[i]
elif cp>0:
if mprof <= cp:
bid += bsus
bsus = []
if mprof < cp:
mprof,sid,nc = cp,[i],0
else:sid.append(i)
else:bsus.append(i)
print(a[i],nc,mprof,bid,sid,bsus)
else:
if len(bid)!=0 and len(sid)!=0:
nc += min(len(bid),len(sid))
print(nc) | s653341232 | Accepted | 122 | 15,060 | 566 | n,t= map(int,input().split())
a= list(map(int,input().split()))
mp,mprof = 999999999,0
bid = []
bsus = []
sid = []
nc=0
for i in range(n):
cp = a[i]-mp
if cp<0:
nc += min(len(bid),len(sid))
mp = a[i]
sid=[]
bid=[]
bsus=[i]
elif cp>0:
if mprof <= cp:
bid += bsus
bsus = []
if mprof < cp:
mprof,sid,nc = cp,[i],0
else:sid.append(i)
else:bsus.append(i)
else:
if len(bid)!=0 and len(sid)!=0:
nc += min(len(bid),len(sid))
print(nc) |
s753020397 | p03964 | u540290227 | 2,000 | 262,144 | Wrong Answer | 27 | 3,060 | 154 | AtCoDeer the deer is seeing a quick report of election results on TV. Two candidates are standing for the election: Takahashi and Aoki. The report shows the ratio of the current numbers of votes the two candidates have obtained, but not the actual numbers of votes. AtCoDeer has checked the report N times, and when he checked it for the i-th (1≦i≦N) time, the ratio was T_i:A_i. It is known that each candidate had at least one vote when he checked the report for the first time. Find the minimum possible total number of votes obtained by the two candidates when he checked the report for the N-th time. It can be assumed that the number of votes obtained by each candidate never decreases. | N=int(input())
t=a=1
for i in range(N):
T,A=map(int,input().split())
num=max(-(-t//T),-(-a//A))
t,a=num*T,num*A
print(t, a)
print(t + a)
| s940000070 | Accepted | 20 | 3,064 | 138 | N=int(input())
t=a=1
for i in range(N):
T,A=map(int,input().split())
num=max(-(-t//T),-(-a//A))
t,a=num*T,num*A
print(t + a)
|
s809850191 | p04029 | u546132222 | 2,000 | 262,144 | Wrong Answer | 25 | 9,068 | 158 | There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total? | # 043A
n =int(input())
ansewr = (n + 1) * n / 2
print(ansewr)
| s049694016 | Accepted | 24 | 9,148 | 159 | # 043A
n = int(input())
answer = (n + 1) * n // 2
print(answer)
|
s344803021 | p02385 | u126478680 | 1,000 | 131,072 | Wrong Answer | 30 | 5,604 | 1,487 | Write a program which reads the two dices constructed in the same way as [Dice I](description.jsp?id=ITP1_11_A), and determines whether these two dices are identical. You can roll a dice in the same way as [Dice I](description.jsp?id=ITP1_11_A), and if all integers observed from the six directions are the same as that of another dice, these dices can be considered as identical. | #! python3
class dice():
def __init__(self, arr):
self.top = arr[0]
self.south = arr[1]
self.east = arr[2]
self.west = arr[3]
self.north = arr[4]
self.bottom = arr[5]
def rotate(self, ope):
if ope == 'S':
self.top, self.north, self.bottom, self.south = self.north, self.bottom, self.south, self.top
elif ope == 'N':
self.top, self.south, self.bottom, self.north = self.south, self.bottom, self.north, self.top
elif ope == 'E':
self.top, self.west, self.bottom, self.east = self.west, self.bottom, self.east, self.top
elif ope == 'W':
self.top, self.east, self.bottom, self.west = self.east, self.bottom, self.west, self.top
def get_surfaces(self):
return [self.top, self.south, self.east, self.north, self.west, self.bottom]
dc1 = dice([int(x) for x in input().split(' ')])
dc2 = dice([int(x) for x in input().split(' ')])
judged = False
if len((dc1.get_surfaces() and dc2.get_surfaces())) == 6:
if dc2.top != dc1.top:
opes = ['S', 'E', 'S', 'E', 'S']
for op in opes:
dc2.rotate(op)
if dc2.top == dc1.top: break
for i in range(4):
if dc2.south == dc1.south:
break
dc2.rotate('R')
if (dc1.east == dc2.east) and (dc1.north == dc2.north) and (dc1.bottom == dc2.bottom):
judged = True
if judged:
print('Yes')
else:
print('No')
| s755200495 | Accepted | 20 | 5,612 | 1,709 | #! python3
class dice():
def __init__(self, arr):
self.top = arr[0]
self.south = arr[1]
self.east = arr[2]
self.west = arr[3]
self.north = arr[4]
self.bottom = arr[5]
def rotate(self, ope):
if ope == 'S':
self.top, self.north, self.bottom, self.south = self.north, self.bottom, self.south, self.top
elif ope == 'N':
self.top, self.south, self.bottom, self.north = self.south, self.bottom, self.north, self.top
elif ope == 'E':
self.top, self.west, self.bottom, self.east = self.west, self.bottom, self.east, self.top
elif ope == 'W':
self.top, self.east, self.bottom, self.west = self.east, self.bottom, self.west, self.top
elif ope == 'R': # clockwise
self.south, self.east, self.north, self.west = self.east, self.north, self.west, self.south
elif ope == 'L': # reversed clockwise
self.south, self.east, self.north, self.west = self.west, self.north, self.east, self.south
else:
pass
def get_surfaces(self):
return [self.top, self.south, self.east, self.north, self.west, self.bottom]
dc1 = dice([int(x) for x in input().split(' ')])
dc2 = dice([int(x) for x in input().split(' ')])
judged = False
if len((dc1.get_surfaces() and dc2.get_surfaces())) == 6:
opes = ['', 'S', 'E', 'S', 'E', 'S']
for op in opes:
dc2.rotate(op)
for i in range(4):
if dc1.get_surfaces() == dc2.get_surfaces():
judged = True
break
dc2.rotate('R')
if judged: break
if judged:
print('Yes')
else:
print('No')
|
s784598308 | p03729 | u853586331 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 97 | You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`. | a,b,c=map(str, input().split())
if a[-1]==b[0] and b[-1]==c[0]:
print ('Yes')
else:
print('No') | s303253259 | Accepted | 17 | 2,940 | 97 | a,b,c=map(str, input().split())
if a[-1]==b[0] and b[-1]==c[0]:
print ('YES')
else:
print('NO') |
s415721456 | p03048 | u677393869 | 2,000 | 1,048,576 | Wrong Answer | 2,104 | 3,472 | 208 | Snuke has come to a store that sells boxes containing balls. The store sells the following three kinds of boxes: * Red boxes, each containing R red balls * Green boxes, each containing G green balls * Blue boxes, each containing B blue balls Snuke wants to get a total of exactly N balls by buying r red boxes, g green boxes and b blue boxes. How many triples of non-negative integers (r,g,b) achieve this? | A,B,C,N=map(int, input().split())
count=0
for x in range(A+1):
for y in range(B+1):
for z in range(C+1):
if N==x+y+z:
count+=1
print(x,y,z)
print(count) | s412195330 | Accepted | 1,871 | 3,060 | 176 | A,B,C,N=map(int, input().split())
count=0
for x in range(N+1):
for y in range(N-x*A+1):
b=(N-x*A-y*B)
if b>=0 and b%C==0:
count+=1
print(count) |
s518222585 | p03171 | u858742833 | 2,000 | 1,048,576 | Wrong Answer | 2,115 | 117,116 | 490 | Taro and Jiro will play the following game against each other. Initially, they are given a sequence a = (a_1, a_2, \ldots, a_N). Until a becomes empty, the two players perform the following operation alternately, starting from Taro: * Remove the element at the beginning or the end of a. The player earns x points, where x is the removed element. Let X and Y be Taro's and Jiro's total score at the end of the game, respectively. Taro tries to maximize X - Y, while Jiro tries to minimize X - Y. Assuming that the two players play optimally, find the resulting value of X - Y. | import sys
sys.setrecursionlimit(10 ** 6)
def main():
N = int(input())
A = list(map(int, input().split()))
print(A)
c = [[None] * (N + 1) for _ in range(N + 1)]
def dp(r, l, A, c):
if r == l:
return A[r]
if c[r][l] != None:
return c[r][l]
print(r, l)
t = max(-dp(r + 1, l, A, c) + A[r],
-dp(r, l - 1, A, c) + A[l])
c[r][l] = t
return t
print(dp(0, N - 1, A, c))
main()
| s056976821 | Accepted | 1,544 | 3,444 | 275 | def main():
N = int(input())
A = list(map(int, input().split()))
c = [0] * N
for r in reversed(range(N)):
ar = A[r]
c[r] = t = ar
for l in range(r + 1, N):
c[l] = t = max(ar - c[l], A[l] - t)
print(c[-1])
main()
|
s352881429 | p03486 | u503476415 | 2,000 | 262,144 | Wrong Answer | 31 | 3,824 | 505 | You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order. | import random
s_given = list(input())
t_given = list(input())
s_len = len(s_given)
t_len = len(t_given)
s_random = random.sample(s_given, s_len)
t_random = random.sample(t_given, t_len)
print(s_random, t_random)
for i in range(t_len):
if s_random[i] < t_random[i]:
print('Yes')
break
elif s_random[i] > t_random[i]:
print('No')
break
if i == t_len - 1:
print('No')
break
elif i == s_len - 1 :
print('Yes')
break
| s822649054 | Accepted | 17 | 3,060 | 546 | s_given = list(input())
t_given = list(input())
s_len = len(s_given)
t_len = len(t_given)
s_sorted = sorted(s_given)
t_sorted = sorted(t_given, reverse=True)
for i in range(t_len):
if s_sorted[i] < t_sorted[i]:
print('Yes')
break
elif s_sorted[i] > t_sorted[i]:
print('No')
break
if i == t_len - 1:
print('No')
break
elif i == s_len - 1 :
print('Yes')
break
|
s158883634 | p02744 | u445916016 | 2,000 | 1,048,576 | Wrong Answer | 988 | 440,040 | 336 | In this problem, we only consider strings consisting of lowercase English letters. Strings s and t are said to be **isomorphic** when the following conditions are satisfied: * |s| = |t| holds. * For every pair i, j, one of the following holds: * s_i = s_j and t_i = t_j. * s_i \neq s_j and t_i \neq t_j. For example, `abcac` and `zyxzx` are isomorphic, while `abcac` and `ppppp` are not. A string s is said to be in **normal form** when the following condition is satisfied: * For every string t that is isomorphic to s, s \leq t holds. Here \leq denotes lexicographic comparison. For example, `abcac` is in normal form, but `zyxzx` is not since it is isomorphic to `abcac`, which is lexicographically smaller than `zyxzx`. You are given an integer N. Print all strings of length N that are in normal form, in lexicographically ascending order. |
def main():
N = int(input())
li = 'abcdefghijklmnopqrstuvwxyz'
def f(k):
nonlocal li
if k == 1:
return ['a']
else:
temp = f(k-1)
return [t+char for t in temp for char in li[:k]]
result = f(N)
print(sorted(result))
if __name__ == '__main__':
main()
| s674350982 | Accepted | 455 | 13,892 | 546 |
def main():
N = int(input())
li = 'abcdefghijklmnopqrstuvwxyz'
def condition(t,char):
nonlocal li
if li.index(sorted(t)[-1]) + 1 >= li.index(char):
return True
else:
return False
def f(k):
nonlocal li
if k == 1:
return ['a']
else:
temp = f(k-1)
return [t+char for t in temp for char in li[:k] if condition(t, char)]
result = f(N)
for t in sorted(result):
print(t)
if __name__ == '__main__':
main()
|
s750108676 | p03623 | u732870425 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 78 | Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|. | x, a, b = map(int, input().split())
print("A" if abs(x-a) > abs(x-b) else "B") | s509233195 | Accepted | 17 | 2,940 | 78 | x, a, b = map(int, input().split())
print("A" if abs(x-a) < abs(x-b) else "B") |
s492863045 | p03469 | u435281580 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 335 | On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it. | def bills(count, value):
max10000 = value // 10000
for i in range(max10000 + 1):
max5000 = min(count - i, (value - i * 10000) // 5000)
for j in range(max5000 + 1):
k = (value - i * 10000 - j * 5000) // 1000
if i + j + k == count:
return i, j, k
return -1, -1, -1 | s692753374 | Accepted | 17 | 2,940 | 38 | print(input().replace("2017", "2018")) |
s010118934 | p03079 | u514299323 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 100 | You are given three integers A, B and C. Determine if there exists an equilateral triangle whose sides have lengths A, B and C. | A, B, C = map(int,input("test").split(" "))
if A == B == C:
print('Yes')
else:
print('No')
| s672524152 | Accepted | 17 | 2,940 | 94 | A, B, C = map(int,input("").split(" "))
if A == B == C:
print('Yes')
else:
print('No') |
s424975643 | p02842 | u540761833 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 131 | Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact. | N = int(input())
if N%1.08 == 0:
print(N//1.08)
elif int(N/1.08) != int((N+1)/1.08):
print(int((N+1)/1.08))
else:
print(':(') | s568224057 | Accepted | 18 | 2,940 | 125 | N = int(input())
x = int(N//1.08)
if int(x*1.08) == N:
print(x)
elif int((x+1)*1.08) == N:
print(x+1)
else:
print(':(') |
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