wrong_submission_id stringlengths 10 10 | problem_id stringlengths 6 6 | user_id stringlengths 10 10 | time_limit 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 10 | acc_status stringclasses 1
value | acc_cpu_time float64 10 27.8k | acc_memory float64 2.94k 960k | acc_code_size int64 19 14.9k | acc_code stringlengths 19 14.9k |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s983923449 | p03698 | u519939795 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 159 | You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different. | import sys
s=input()
for i in range(len(s)+1):
for j in range(len(s)+1):
if s[i]==s[j]:
print('no')
sys.exit()
print('yes') | s962151455 | Accepted | 17 | 2,940 | 134 | #ABC063.B
S = input()
cnt = 0
for i in S:
if S.count(i) != 1:
cnt +=1
if cnt == 0:
print('yes')
else:
print('no')
|
s446686574 | p03836 | u848173626 | 2,000 | 262,144 | Wrong Answer | 19 | 3,064 | 530 | 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 ... | from sys import stdin
SX, SY, TX, TY = map(int, stdin.readline().rstrip().split())
# S->T
print('U' * (TY - SY), end='')
print('R' * (TX - SX), end='')
# T->S
print('D' * (TY - SY), end='')
print('L' * (TX - SX), end='')
print('L', end='')
print('U' * (TY - SY + 1), end='')
print('R' * (TX - SX + 1), end='')
pr... | s576167548 | Accepted | 17 | 3,064 | 522 | from sys import stdin
SX, SY, TX, TY = map(int, stdin.readline().rstrip().split())
# S->T
print('U' * (TY - SY), end='')
print('R' * (TX - SX), end='')
# T->S
print('D' * (TY - SY), end='')
print('L' * (TX - SX), end='')
print('L', end='')
print('U' * (TY - SY + 1), end='')
print('R' * (TX - SX + 1), end='')
pr... |
s017670845 | p02262 | u286589639 | 6,000 | 131,072 | Wrong Answer | 20 | 5,600 | 454 | Shell Sort is a generalization of [Insertion Sort](http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_1_A) to arrange a list of $n$ elements $A$. 1 insertionSort(A, n, g) 2 for i = g to n-1 3 v = A[i] 4 j = i - g 5 while j >= 0 && A[j] > v 6 A[j+... | def insertion_sort(A ,N, g):
for i in range(g, N):
v = A[i]
j = i - g
while j>=0 and A[j]>v:
A[j+g] = A[j]
j = j - g
A[j+g] = v
def shell_sort(A, N):
cnt = 0
m = N//2
G = [i for i in reversed(range(1, m, 2))]
for g in G:
insertion_sort(A, N, g)
N = int(input())
A = [int(input()) for i in range(... | s128961482 | Accepted | 22,860 | 45,516 | 497 | def insertion_sort(A ,N, g):
global cnt
for i in range(g, N):
v = A[i]
j = i - g
while j>=0 and A[j]>v:
A[j+g] = A[j]
j = j - g
cnt += 1
A[j+g] = v
def shell_sort(A, N):
global cnt
cnt = 0
G = []
h = 1
while h<=len(A):
G.append(h)
h = 3*h+1
G.reverse()
m = len(G)
print(m)
print(' '.jo... |
s839819775 | p02390 | u586792237 | 1,000 | 131,072 | Wrong Answer | 20 | 5,580 | 138 | Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively. | x = int(input())
h = x // 3600
m = (x - (h*3600))//60
s = x-((h*3600)+(m*60))
print(h, ':', m, ':', s)
print(h, ':', m, ':', s, sep='')
| s985973005 | Accepted | 20 | 5,588 | 114 | x = int(input())
h = x // 3600
m = (x - (h*3600))//60
s = x-((h*3600)+(m*60))
print(h, ':', m, ':', s, sep='')
|
s560826771 | p02578 | u625864724 | 2,000 | 1,048,576 | Wrong Answer | 2,206 | 32,188 | 284 | N persons are standing in a row. The height of the i-th person from the front is A_i. We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person: Condition: Nobody in front of the person is taller than the person. Here, the height of a ... | n = int(input())
lst = list(map(int,input().split()))
ans = 0
for i in range(n):
tall = 0
for j in range(i):
if (lst[j] > tall):
tall = lst[j]
print (i, tall)
if (lst[i] < tall):
sa = tall - lst[i]
print(i,sa)
ans = ans + sa
lst[i] = tall
print(ans)
| s883303435 | Accepted | 161 | 32,144 | 210 | n = int(input())
lst = list(map(int,input().split()))
ans = 0
for i in range(n - 1):
if (lst[i] > lst[i + 1]):
sa = lst[i] - lst[i + 1]
ans = ans + sa
lst[i + 1] = lst[i]
print(ans)
|
s946480299 | p03360 | u597047658 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 224 | There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after... | import heapq
A = list(map(int, input().split()))
A = [-a for a in A]
K = int(input())
heapq.heapify(A)
print(A)
for i in range(K):
max_val = heapq.heappop(A)
heapq.heappush(A, 2*max_val)
print(sum([-a for a in A]))
| s237517291 | Accepted | 17 | 3,060 | 215 | import heapq
A = list(map(int, input().split()))
A = [-a for a in A]
K = int(input())
heapq.heapify(A)
for i in range(K):
max_val = heapq.heappop(A)
heapq.heappush(A, 2*max_val)
print(sum([-a for a in A]))
|
s951949203 | p01102 | u196653484 | 8,000 | 262,144 | Wrong Answer | 30 | 5,584 | 414 | The programming contest named _Concours de Programmation Comtemporaine Interuniversitaire_ (CPCI) has a judging system similar to that of ICPC; contestants have to submit correct outputs for two different inputs to be accepted as a correct solution. Each of the submissions should include the program that generated the ... | results=[]
while True:
a=input()
if a==".":
break
b=input()
count=0
for i,j in zip(a,b):
if(a!=b):
count+=1
if count==0:
results.append(0)
elif count==1:
results.append(1)
else:
results.append(2)
for i in results:
if i==0:
p... | s622750504 | Accepted | 30 | 5,596 | 1,019 | results=[]
while True:
a=input().split("\"")
if a[0]==".":
break
b=input().split("\"")
count1=0
count2=0
if len(a) != len(b):
results.append(2)
else:
for i in range(len(a)):
if(i%2==1):
if len(a[i]) != len(b[i]):
count1+... |
s787192434 | p03377 | u513081876 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 98 | 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 and X <= A+B:
print('Yes')
else:
print('No') | s962694027 | Accepted | 17 | 2,940 | 98 | A, B, X = map(int, input().split())
if A <= X and X <= A+B:
print('YES')
else:
print('NO') |
s208362457 | p03612 | u955125992 | 2,000 | 262,144 | Wrong Answer | 51 | 14,008 | 127 | You are given a permutation p_1,p_2,...,p_N consisting of 1,2,..,N. You can perform the following operation any number of times (possibly zero): Operation: Swap two **adjacent** elements in the permutation. You want to have p_i ≠ i for all 1≤i≤N. Find the minimum required number of operations to achieve this. | n = int(input())
p = list(map(int, input().split()))
ans = 0
for i in range(n):
if i == p[i]:
ans += 1
print(ans) | s342132612 | Accepted | 73 | 14,004 | 206 | n = int(input())
p = list(map(int, input().split()))
ans = 0
for i in range(n-1):
if i+1 == p[i]:
p[i], p[i+1] = p[i+1], p[i]
ans += 1
if p[n-1] == n:
ans += 1
print(ans)
|
s202800417 | p03457 | u727551259 | 2,000 | 262,144 | Wrong Answer | 450 | 11,816 | 478 | AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1... | n = int(input())
ti,xi,yi = [0],[0],[0]
ans = 'NO'
for ni in range(n):
t,x,y = [int(x) for x in input().split()]
ti.append(t)
xi.append(x)
yi.append(y)
for i in range(n):
dt = ti[i+1] - ti[i]
dx = xi[i+1] - xi[i]
dy = yi[i+1] - yi[i]
if dt < abs(dx)+abs(dy):
break
odev1 = (xi[i]+yi[i])%2
odev2 =... | s044920149 | Accepted | 441 | 11,728 | 478 | n = int(input())
ti,xi,yi = [0],[0],[0]
ans = 'No'
for ni in range(n):
t,x,y = [int(x) for x in input().split()]
ti.append(t)
xi.append(x)
yi.append(y)
for i in range(n):
dt = ti[i+1] - ti[i]
dx = xi[i+1] - xi[i]
dy = yi[i+1] - yi[i]
if dt < abs(dx)+abs(dy):
break
odev1 = (xi[i]+yi[i])%2
odev2 =... |
s060851670 | p03637 | u239375815 | 2,000 | 262,144 | Wrong Answer | 75 | 14,252 | 233 | We have a sequence of length N, a = (a_1, a_2, ..., a_N). Each a_i is a positive integer. Snuke's objective is to permute the element in a so that the following condition is satisfied: * For each 1 ≤ i ≤ N - 1, the product of a_i and a_{i + 1} is a multiple of 4. Determine whether Snuke can achieve his objective. | n = int(input())
a = list(map(int,input().split()))
a_n4 = a_4 = [x for x in a if x%4!=0]
a_4 = [x for x in a if x%4==0]
a_2 = [x for x in a if x%4!=0 and x%2==0]
print("Yes "if (len(a_n4)-1-(int(len(a_2)/2))<=len(a_4)) else "No")
| s865007864 | Accepted | 60 | 15,020 | 278 | N = int(input())
a = list(map(int, input().split()))
c = [0,0,0,0]
for x in a:
c[x % 4] += 1
# if (c[1]==0) and (c[3]==0):
# print('Yes')
if c[1] + c[3] < c[0] + 1:
print('Yes')
elif (c[1] + c[3] == c[0] + 1)and(c[2] == 0):
print('Yes')
else:
print('No')
|
s398507147 | p03449 | u905203728 | 2,000 | 262,144 | Wrong Answer | 19 | 3,064 | 308 | We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j). You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down. The square (i, j) contains A_{i, j} candies. You ... | n=int(input())
box=[list(map(int,input().split())) for i in range(2)]
count=box[0][0]
point=-1
add=0
for j in range(n):
for i in range(n):
if add<=i:
count +=box[1][i]
else:
count +=box[0][i]
point=max(point,count)
count=box[0][0]
add +=1
print(point) | s547577492 | Accepted | 20 | 3,064 | 310 | n=int(input())
box=[list(map(int,input().split())) for i in range(2)]
count=box[0][0]
point=-1
add=0
for j in range(n):
for i in range(n):
if add<=i:
count +=box[1][i]
else:
count +=box[0][i+1]
point=max(point,count)
count=box[0][0]
add +=1
print(point) |
s014761454 | p02602 | u038216098 | 2,000 | 1,048,576 | Wrong Answer | 2,206 | 31,548 | 234 | M-kun is a student in Aoki High School, where a year is divided into N terms. There is an exam at the end of each term. According to the scores in those exams, a student is given a grade for each term, as follows: * For the first through (K-1)-th terms: not given. * For each of the K-th through N-th terms: the m... | N,K=map(int,input().split())
A=list(map(int,input().split()))
point=[1]*N
for i in range(K-1,N):
for j in range(K):
point[i]*=A[i-j]
for i in range(N-K-2,N-1):
if(point[i]>=point[i+1]):
print("No")
else:
print("Yes") | s066100657 | Accepted | 147 | 31,756 | 155 | N,K=map(int,input().split())
A=list(map(int,input().split()))
point=[1]*N
for i in range(N-K):
if(A[K+i]>A[i]):
print("Yes")
else:
print("No")
|
s348347758 | p03844 | u533039576 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 13 | Joisino wants to evaluate the formula "A op B". Here, A and B are integers, and the binary operator op is either `+` or `-`. Your task is to evaluate the formula instead of her. | eval(input()) | s285055337 | Accepted | 17 | 2,940 | 21 | print(eval(input()))
|
s828260923 | p02795 | u575101291 | 2,000 | 1,048,576 | Wrong Answer | 20 | 3,316 | 104 | We have a grid with H rows and W columns, where all the squares are initially white. You will perform some number of painting operations on the grid. In one operation, you can do one of the following two actions: * Choose one row, then paint all the squares in that row black. * Choose one column, then paint all t... | h = int(input())
w = int(input())
n = int(input())
if h < w:
t = n / w
else:
t = n / h
print(int(t)) | s090362995 | Accepted | 17 | 3,060 | 123 | import math
h = int(input())
w = int(input())
n = int(input())
if h < w:
t = n / w
else:
t = n / h
print(math.ceil(t)) |
s035435965 | p03860 | u331464808 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 40 | Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppe... | s = input()
print("A" + str(s[0]) + "C") | s521662914 | Accepted | 17 | 2,940 | 53 | a,x,c=map(str,input().split())
print('A'+ x[0] + 'C') |
s891786161 | p03674 | u075012704 | 2,000 | 262,144 | Wrong Answer | 2,109 | 32,380 | 561 | You are given an integer sequence of length n+1, a_1,a_2,...,a_{n+1}, which consists of the n integers 1,...,n. It is known that each of the n integers 1,...,n appears at least once in this sequence. For each integer k=1,...,n+1, find the number of the different subsequences (not necessarily contiguous) of the given s... | from statistics import mode
from scipy.misc import comb
N = int(input())
A = list(map(int, input().split()))
MOD = 10 ** 9
C = mode(A)
C_indexes = [i for i, x in enumerate(A) if x == C]
outside = C_indexes[0] + len(A) - C_indexes[1] - 1
for k in range(1, len(A)+1):
if k > outside + 1:
print(comb((N+1)... | s744882268 | Accepted | 305 | 31,324 | 796 | N = int(input())
A = list(map(int, input().split()))
MOD = 10 ** 9 + 7
duplicate_x = None
checked = set()
for a in A:
if a in checked:
duplicate_x = a
checked.add(a)
x_l_index = A.index(duplicate_x)
x_r_index = N + 1 - A[::-1].index(duplicate_x) - 1
factorial = [1, 1]
inverse = [1, 1]
invere_base... |
s460658553 | p02261 | u294922877 | 1,000 | 131,072 | Wrong Answer | 20 | 5,612 | 1,199 | 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 Bubbl... | def bubble_sort(n, key=0):
c = 0
r = [x if isinstance(x, list) else [x] for x in n[:]]
l = len(r) - 1
for i in range(0, l):
for j in range(l, i, -1):
if r[j][key] < r[j-1][key]:
r[j], r[j-1] = r[j-1], r[j]
c += 1
r = [x[0] if len(x) == 1 else x for... | s960394481 | Accepted | 20 | 5,616 | 1,201 | def bubble_sort(n, key=0):
c = 0
r = [x if isinstance(x, list) else [x] for x in n[:]]
l = len(r) - 1
for i in range(0, l):
for j in range(l, i, -1):
if r[j][key] < r[j-1][key]:
r[j], r[j-1] = r[j-1], r[j]
c += 1
r = [x[0] if len(x) == 1 else x for... |
s941733337 | p03548 | u756988562 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 197 | We have a long seat of width X centimeters. There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters. We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two... | input_lines = input().split(" ")
X = int(input_lines[0])
Y = int(input_lines[1])
Z = int(input_lines[2])
judge = X//(Y)
if judge*(Y+Z)>=Z:
print(str(X//(Y+Z)-1))
else:
print(str(X//(Y+Z))) | s047206028 | Accepted | 17 | 2,940 | 202 | input_lines = input().split(" ")
X = int(input_lines[0])
Y = int(input_lines[1])
Z = int(input_lines[2])
judge = X//(Y+Z)
if judge*(Y+Z)+Z>X:
print(str(X//(Y+Z)-1))
else:
print(str(X//(Y+Z)))
|
s925754360 | p03369 | u948911484 | 2,000 | 262,144 | Wrong Answer | 25 | 8,884 | 29 | 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 ... | print(700+input().count("o")) | s870511309 | Accepted | 24 | 9,028 | 41 | print([7,8,9,10][input().count("o")]*100) |
s426114435 | p02397 | u042882066 | 1,000 | 131,072 | Wrong Answer | 60 | 7,652 | 109 | Write a program which reads two integers x and y, and prints them in ascending order. | while True:
x, y = map(int, input().split(" "))
if x == 0 and y == 0:
break
print(x, y) | s986515807 | Accepted | 60 | 7,612 | 181 | # -*- coding: utf-8 -*-
while True:
x, y = map(int, input().split(" "))
if x == 0 and y == 0:
break
elif x <= y:
print(x, y)
elif x >= y:
print(y, x) |
s844267679 | p03007 | u606045429 | 2,000 | 1,048,576 | Wrong Answer | 232 | 13,964 | 617 | There are N integers, A_1, A_2, ..., A_N, written on a blackboard. We will repeat the following operation N-1 times so that we have only one integer on the blackboard. * Choose two integers x and y on the blackboard and erase these two integers. Then, write a new integer x-y. Find the maximum possible value of the... | N = int(input())
A = [int(i) for i in input().split()]
A.sort()
L = [a for a in A if a <= 0]
R = [a for a in A if a > 0]
if L != [] and R != []:
low = L[0]
for r in R[:-1]:
print(low, r)
low -= r
up = R[-1]
for l in L[1:]:
print(up, l)
up -= l
print(up, low)
... | s106425612 | Accepted | 307 | 24,016 | 746 | N = int(input())
A = [int(i) for i in input().split()]
A.sort()
L = [a for a in A if a <= 0]
R = [a for a in A if a > 0]
ans = []
if L != [] and R != []:
low = L[0]
for r in R[:-1]:
ans.append([low, r])
low -= r
up = R[-1]
for l in L[1:]:
ans.append([up, l])
up -= l
... |
s707729913 | p02407 | u853619096 | 1,000 | 131,072 | Wrong Answer | 20 | 7,524 | 81 | Write a program which reads a sequence and prints it in the reverse order. | n=input()
n=int(n)
a=input().split()
a=a[::-1]
for i in a:
print(" ".join(a)) | s541308233 | Accepted | 20 | 7,636 | 76 | n=int(input())
a=list(map(str,input().split()))
a=a[::-1]
print(" ".join(a)) |
s201514092 | p02612 | u878473776 | 2,000 | 1,048,576 | Wrong Answer | 26 | 9,164 | 95 | 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 = int(input(""))
if N % 1000 == 0:
N1 = N // 1000
else:
N1 = N // 1000 + 1
print(N1) | s384324305 | Accepted | 31 | 9,124 | 106 | N = int(input(""))
if N % 1000 == 0:
N1 = N // 1000
else:
N1 = N // 1000 + 1
print(N1 * 1000 - N) |
s180342043 | p04044 | u022871813 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 132 | Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically sm... | n, l = map(int,input().split())
list = []
for i in range(n):
a = input()
list.append(a)
for i in sorted(list):
print(i)
| s377386699 | Accepted | 17 | 3,060 | 125 | n, l = map(int,input().split())
list = []
for i in range(n):
a = input()
list.append(a)
print(''.join(sorted(list)))
|
s881254886 | p03574 | u637551956 | 2,000 | 262,144 | Wrong Answer | 24 | 3,188 | 1,667 | You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square con... | H,W = map(int, input().split())
S = [list(input()) for i in range(H)]
S=[[0 if j == "." else "#" for j in i] for i in S]
for i in range(H):
for j in range(W):
if S[i][j]=='#':
try:S[i-1][j-1]
except IndexError:pass
else:
if i-1... | s863312092 | Accepted | 25 | 3,572 | 1,692 | H,W = map(int, input().split())
S = [list(input()) for i in range(H)]
S=[[0 if j == "." else "#" for j in i] for i in S]
for i in range(H):
for j in range(W):
if S[i][j]=='#':
try:S[i-1][j-1]
except IndexError:pass
else:
if i-1... |
s873417098 | p02409 | u092736322 | 1,000 | 131,072 | Wrong Answer | 20 | 5,608 | 409 | You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room. For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth fl... | n=int(input())
S=[]
for i in range(1200):
S.append(0)
for i in range(n):
k=input().split()
b=int(k[0])
f=int(k[1])
r=int(k[2])
v=int(k[3])
S[(b-1)*30+(f-1)*10+r-1]+=v
for i in range(4):
for j in range(3):
msg=""
for t in range(9):
msg+=str(S[i*30+j*10+t])+" ... | s965441481 | Accepted | 20 | 5,604 | 393 | n=int(input())
S=[]
for i in range(1200):
S.append(0)
for i in range(n):
k=input().split()
b=int(k[0])
f=int(k[1])
r=int(k[2])
v=int(k[3])
S[(b-1)*30+(f-1)*10+r-1]+=v
for i in range(4):
for j in range(3):
msg=""
for t in range(10):
msg+=" "+str(S[i*30+j*10+t... |
s397470814 | p03644 | u865413330 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 210 | Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can... | n = int(input())
count = 0
currCount = 0
for i in range(n):
currCount = 0
while i % 2 != 0:
i = int(i / 2)
currCount += 1
if count < currCount:
count = currCount
print(count) | s956216876 | Accepted | 17 | 3,060 | 296 | n = int(input())
count = 0
currCount = 0
ans = 0
for i in range(n+1):
currCount = 0
currNum = i
while (i % 2) == 0:
if i == 0:
break
i = int(i / 2)
currCount += 1
if count <= currCount:
count = currCount
ans = currNum
print(ans) |
s177911106 | p03597 | u717626627 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 49 | We have an N \times N square grid. We will paint each square in the grid either black or white. If we paint exactly A squares white, how many squares will be painted black? | a = int(input())
b = int(input())
print(a^2 - b) | s734445527 | Accepted | 17 | 2,940 | 49 | a = int(input())
b = int(input())
print(a*a - b) |
s749504771 | p03474 | u056486147 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 326 | The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom. | a, b = map(int, input().split())
s = list(input())
print(s)
print(len(s), a+b+1)
flag = True
if len(s) == a+b+1:
if s[a] == "-":
s.pop(a)
for i in s:
if i == "-":
flag = False
else:
flag = False
else:
flag = False
if flag:
print("Yes")
else:
print(... | s068371991 | Accepted | 18 | 2,940 | 319 | import sys
a, b = map(int, input().split())
s = list(input())
if len(s) == a+b+1:
if s[a] == "-":
s.pop(a)
for i in s:
if i == "-":
print("No")
sys.exit()
else:
print("No")
sys.exit()
else:
print("No")
sys.exit()
print("Yes")
|
s927786355 | p03370 | u253952966 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 141 | Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams o... | n, x = map(int, input().split())
mn = []
for _ in range(n):
mi = int(input())
mn.append(mi)
mn.sort()
x -= sum(mn)
print(n + (x % mn[0])) | s725284889 | Accepted | 18 | 2,940 | 142 | n, x = map(int, input().split())
mn = []
for _ in range(n):
mi = int(input())
mn.append(mi)
mn.sort()
x -= sum(mn)
print(n + (x // mn[0])) |
s445656750 | p03485 | u374146618 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 66 | 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 = [int(x) for x in input().split()]
x = (a+b)/2
print(x//1+1) | s456911962 | Accepted | 17 | 2,940 | 110 | a, b = [int(x) for x in input().split()]
x = (a+b)/2
if x%1==0:
print(int(x))
else:
print(int(x//1+1)) |
s034344095 | p03456 | u762474234 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 172 | 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 = input().split()
c = int(a + b)
square_root = math.sqrt(c)
# print(c/square_root, square_root)
print('YES') if square_root.is_integer() else print('NO')
| s669032642 | Accepted | 17 | 2,940 | 172 | import math
a, b = input().split()
c = int(a + b)
square_root = math.sqrt(c)
# print(c/square_root, square_root)
print('Yes') if square_root.is_integer() else print('No')
|
s337216561 | p04043 | u518958552 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 221 | 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 ... | a, b, c = map(int,input().split())
if a == 5 and b == 7 and c == 5:
print("Yes")
elif a == 5 and b == 5 and c == 7:
print("Yes")
elif a == 7 and b == 5 and c == 5:
print("Yes")
else:
print("No") | s240913636 | Accepted | 17 | 2,940 | 132 | a, b, c = map(int,input().split())
d = [a, b, c]
if d.count(5) == 2 and d.count(7) == 1:
print("YES")
else:
print("NO") |
s880785767 | p03943 | u842696304 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 126 | 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. Not... | x = list(input().split())
x.sort()
if int(x[0]) + int(x[1]) == int(x[2]):
print('YES')
else:
print('FALSE')
print(x)
| s341185995 | Accepted | 17 | 2,940 | 119 | a, b, c = map(int, input().split())
if a + b == c or b + c == a or c + a == b:
print("Yes")
else:
print("No")
|
s466541235 | p02288 | u357267874 | 2,000 | 131,072 | Wrong Answer | 30 | 5,604 | 773 | A binary heap which satisfies max-heap property is called max-heap. In a max- heap, for every node $i$ other than the root, $A Write a program which reads an array and constructs a max-heap from the array based on the following pseudo code. $maxHeapify(A, i)$ move the value of $A[i]$ down to leaves to make a sub-tr... | def get_left_index(A, i):
if len(A) > 2 * i:
return 2 * i
else:
return None
def get_right_index(A, i):
if len(A) > 2 * i + 1:
return 2 * i + 1
else:
return None
def max_heapfy(A, i):
left = get_left_index(A, i)
right = get_right_index(A, i)
largest = i
i... | s635196351 | Accepted | 1,200 | 64,160 | 756 | def get_left_index(A, i):
if len(A) > 2 * i:
return 2 * i
else:
return None
def get_right_index(A, i):
if len(A) > 2 * i + 1:
return 2 * i + 1
else:
return None
def max_heapfy(A, i):
left = get_left_index(A, i)
right = get_right_index(A, i)
largest = i
i... |
s772676751 | p02865 | u448751328 | 2,000 | 1,048,576 | Wrong Answer | 21 | 3,316 | 85 | How many ways are there to choose two distinct positive integers totaling N, disregarding the order? | N = int(input())
if N % 2 == 0:
ans = N / 2 - 1
else:
ans = (N -1) / 2
print(ans) | s020301845 | Accepted | 18 | 3,064 | 71 | N=int(input())
if N%2==0:
print(N//2-1)
else:
print((N+1)//2-1) |
s141688080 | p03543 | u565380863 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 4 | 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**? | 1234 | s663789335 | Accepted | 17 | 2,940 | 106 | #-*- coding: utf-8 -*-
N = input()
print('Yes' if N[0] == N[1] == N[2] or N[1] == N[2] == N[3] else 'No')
|
s848553784 | p03401 | u143492911 | 2,000 | 262,144 | Wrong Answer | 2,104 | 19,808 | 281 | There are N sightseeing spots on the x-axis, numbered 1, 2, ..., N. Spot i is at the point with coordinate A_i. It costs |a - b| yen (the currency of Japan) to travel from a point with coordinate a to another point with coordinate b along the axis. You planned a trip along the axis. In this plan, you first depart from... | n=int(input())
a=list(map(int,input().split()))
ans=[]
a.sort()
cnt=0
for i in range(n):
v=a[i]
a[i]=0
a.sort()
for j in range(1,n):
cnt+=a[j]-a[j-1]
print(cnt)
ans.append(cnt+(max(a)-min(a)))
cnt=0
a[i]=v
for i in ans:
print(i)
| s859810469 | Accepted | 216 | 14,048 | 223 | n=int(input())
a=list(map(int,input().split()))
a.insert(0,0)
a.insert(n+1,0)
total=0
for i in range(n+1):
total+=abs(a[i+1]-a[i])
for i in range(n):
print(total-abs(a[i+1]-a[i])-abs(a[i+2]-a[i+1])+abs(a[i+2]-a[i])) |
s700654759 | p03759 | u629709614 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 86 | 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())
if b-a == c-a:
print("YES")
else:
print("NO") | s812293212 | Accepted | 17 | 2,940 | 86 | a, b, c = map(int, input().split())
if b-a == c-b:
print("YES")
else:
print("NO") |
s335110353 | p03854 | u899395423 | 2,000 | 262,144 | Wrong Answer | 80 | 3,188 | 345 | You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`. | s = input()
s = s[::-1]
words = ["dream", "dreamer", "erase", "eraser"]
words = [w[::-1] for w in words]
lens = [len(w) for w in words]
while True:
for l, w in zip(lens, words):
if s[:l] == w:
s = s[l:]
break
else:
print("No")
break
if len(s) == 0:
p... | s496662948 | Accepted | 83 | 3,316 | 345 | s = input()
s = s[::-1]
words = ["dream", "dreamer", "erase", "eraser"]
words = [w[::-1] for w in words]
lens = [len(w) for w in words]
while True:
for l, w in zip(lens, words):
if s[:l] == w:
s = s[l:]
break
else:
print("NO")
break
if len(s) == 0:
p... |
s346474431 | p03657 | u717626627 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 128 | Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them ca... | a, b = map(int, input().split())
if a % 3 == 0 or b % 3 == 0 or a + b % 3 == 0:
print('Possible')
else:
print('Impossible') | s677877937 | Accepted | 17 | 2,940 | 130 | a, b = map(int, input().split())
if a % 3 == 0 or b % 3 == 0 or (a + b) % 3 == 0:
print('Possible')
else:
print('Impossible') |
s193860091 | p03637 | u219607170 | 2,000 | 262,144 | Wrong Answer | 69 | 15,020 | 208 | We have a sequence of length N, a = (a_1, a_2, ..., a_N). Each a_i is a positive integer. Snuke's objective is to permute the element in a so that the following condition is satisfied: * For each 1 ≤ i ≤ N - 1, the product of a_i and a_{i + 1} is a multiple of 4. Determine whether Snuke can achieve his objective. | N = int(input())
A = list(map(int, input().split()))
four, odd = 0, 0
for a in A:
if a%4 == 0:
four += 1
elif a%2 == 1:
odd += 1
result = 'No' if four > odd else 'Yes'
print('result')
| s653363903 | Accepted | 65 | 15,020 | 257 | N = int(input())
A = list(map(int, input().split()))
four, two, odd = 0, 0, 0
for a in A:
if a%4 == 0:
four += 1
elif a%2 == 0:
two += 1
else:
odd += 1
result = 'Yes' if four + 1 >= odd + bool(two) else 'No'
print(result) |
s132657025 | p03737 | u702786238 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 83 | You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words. | s = input()
s.upper().split(" ")
print("{}{}{}".format(s[0][0], s[1][0], s[2][0])) | s129069150 | Accepted | 17 | 2,940 | 88 | s = input()
s = s.upper().split(" ")
print("{}{}{}".format(s[0][0], s[1][0], s[2][0]))
|
s272715292 | p03737 | u363836311 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 95 | You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words. | s1,s2,s3=map(str, input().split())
S1=list(s1)
S2=list(s2)
S3=list(s3)
print(S1[0]+S2[0]+S3[0]) | s918920961 | Accepted | 17 | 2,940 | 135 | s1,s2,s3=map(str, input().split())
S1=list(s1)
S2=list(s2)
S3=list(s3)
print(chr(ord(S1[0])-32)+chr(ord(S2[0])-32)+chr(ord(S3[0])-32))
|
s557740880 | p03657 | u798260206 | 2,000 | 262,144 | Wrong Answer | 26 | 9,164 | 168 | Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them ca... | a,b = map(int,input().split())
Flag = "Impossible"
if a%3==0:
Flag = "Possible"
elif b%3 ==0:
Flag="Possible"
elif (a+b)%3:
Flag = "Possible"
print(Flag)
| s695192039 | Accepted | 22 | 9,132 | 172 | a,b = map(int,input().split())
Flag = "Impossible"
if a%3==0:
Flag = "Possible"
elif b%3 ==0:
Flag="Possible"
elif (a+b)%3==0:
Flag = "Possible"
print(Flag)
|
s509203331 | p03475 | u581187895 | 3,000 | 262,144 | Wrong Answer | 113 | 3,188 | 548 | A railroad running from west to east in Atcoder Kingdom is now complete. There are N stations on the railroad, numbered 1 through N from west to east. Tomorrow, the opening ceremony of the railroad will take place. On this railroad, for each integer i such that 1≤i≤N-1, there will be trains that run from Station i t... |
import math
N = int(input())
CSF = [list(map(int, input().split())) for _ in range(N-1)]
for i in range(N):
t = 0
for j in range(i, N-1):
c, s, f = CSF[j]
if t <= s:
t = s+c
else:
t += math.ceil(t / f) * f + c
print(t) | s846190009 | Accepted | 88 | 3,188 | 547 |
import math
N = int(input())
CSF = [list(map(int, input().split())) for _ in range(N-1)]
for i in range(N):
t = 0
for j in range(i, N-1):
c, s, f = CSF[j]
if t <= s:
t = s+c
else:
t = math.ceil(t / f) * f + c
print(t) |
s504075519 | p03645 | u257974487 | 2,000 | 262,144 | Wrong Answer | 2,106 | 52,620 | 374 | 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 Snuk... | N, M = map(int,input().split())
P = [list(map(int,input().split())) for _ in range(M)]
P.sort()
Q = []
s = 0
ans = ("IMPOSSIBLE")
for i in range (M):
if P[i][0] == 1:
Q.append(P[i][1])
else:
s = i - 1
break
if s > 0:
for j in range(s, M):
if P[j][1] == N:
if P[j][... | s301061127 | Accepted | 1,174 | 52,616 | 639 | N, M = map(int,input().split())
P = [list(map(int,input().split())) for _ in range(M)]
P.sort()
q = -1
ans = ("IMPOSSIBLE")
for i in range (M):
if P[i][0] == 1:
p = P[i][1]
if q < 0:
if i > 0:
q = i - 1
else:
q = 0
for j in range(q, M):... |
s552954041 | p03416 | u201234972 | 2,000 | 262,144 | Wrong Answer | 111 | 3,740 | 143 | Find the number of _palindromic numbers_ among the integers between A and B (inclusive). Here, a palindromic number is a positive integer whose string representation in base 10 (without leading zeros) reads the same forward and backward. | A,B = map(int,input().split())
ans = 0
for x in range(A,B+1):
x = str(x)
print(x)
if x[:1] == x[:2:-1]:
ans += 1
print(ans) | s508432199 | Accepted | 65 | 2,940 | 130 | A,B = map(int,input().split())
ans = 0
for x in range(A,B+1):
x = str(x)
if x[:2] == x[:2:-1]:
ans += 1
print(ans) |
s372490203 | p02608 | u111652094 | 2,000 | 1,048,576 | Wrong Answer | 2,031 | 9,216 | 548 | Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N). | import math
N=int(input())
for i in range(1,N+1):
ans=0
if i<=5:
print(0)
else:
ma=int(math.sqrt(i)/6)
for x in range(1,ma):
for y in range(x,ma):
for z in range(y,ma):
if x**2+y**2+z**2+x*y+y*z+z*x==i:
if x=... | s022472906 | Accepted | 198 | 9,348 | 474 | import math
N=int(input())
anslist=[0]*(N+1)
ma=int(math.sqrt(N))+1
for x in range(1,ma):
for y in range(x,ma):
for z in range(y,ma):
n=x**2+y**2+z**2+x*y+y*z+z*x
if n <=N:
if x==y and y==z:
anslist[n]+=1
elif x==y or y==z or z... |
s933019076 | p03599 | u136090046 | 3,000 | 262,144 | Wrong Answer | 17 | 3,064 | 413 | Snuke is making sugar water in a beaker. Initially, the beaker is empty. Snuke can perform the following four types of operations any number of times. He may choose not to perform some types of operations. * Operation 1: Pour 100A grams of water into the beaker. * Operation 2: Pour 100B grams of water into the bea... |
ABCDEF = input()
split_ABCDEF = ABCDEF.split()
int_split_ABCDEF = [int(i) for i in split_ABCDEF]
A = int_split_ABCDEF[0]*100
B = int_split_ABCDEF[1]*100
C = int_split_ABCDEF[2]
D = int_split_ABCDEF[3]
E = int_split_ABCDEF[4]
F = int_split_ABCDEF[5]
if A > B:
sugar = int((B / 100) * E)
print("{} {}".format(sug... | s864454293 | Accepted | 340 | 39,648 | 840 | def solve():
A, B, C, D, E, F = map(int, input().split())
memo = {}
for a in range(31):
for b in range(31):
for c in range(101):
for d in range(101):
w = (100 * A) * a + (100 * B) * b
s = C * c + D * d
if w == 0... |
s354467169 | p03080 | u317587743 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 120 | 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 = list(input())
r = s.count('R')
b = s.count('B')
if r > b:
print('YES')
else:
print('NO') | s635286887 | Accepted | 17 | 2,940 | 120 | N = int(input())
s = list(input())
r = s.count('R')
b = s.count('B')
if r > b:
print('Yes')
else:
print('No') |
s054437622 | p02659 | u297651868 | 2,000 | 1,048,576 | Wrong Answer | 23 | 9,164 | 81 | Compute A \times B, truncate its fractional part, and print the result as an integer. | a,b=map(float,input().split())
left=a//100*b
right=a%100*b
print(int(left+right)) | s944073948 | Accepted | 27 | 9,752 | 74 | from decimal import Decimal
a,b=map(Decimal,input().split())
print(a*b//1) |
s983429924 | p02613 | u165436807 | 2,000 | 1,048,576 | Wrong Answer | 150 | 16,024 | 214 | 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`,... | n = int(input())
s = [0]*n
for i in range(n):
s[i] = input()
ac = s.count('AC')
wa = s.count('WA')
tle = s.count('TLE')
re = s.count('RE')
print('AC x',ac)
print('WA x',wa)
print('TLE x',tle)
print('Re x',re)
| s569081540 | Accepted | 147 | 16,088 | 214 | n = int(input())
s = [0]*n
for i in range(n):
s[i] = input()
ac = s.count('AC')
wa = s.count('WA')
tle = s.count('TLE')
re = s.count('RE')
print('AC x',ac)
print('WA x',wa)
print('TLE x',tle)
print('RE x',re)
|
s685373556 | p03471 | u616522759 | 2,000 | 262,144 | Wrong Answer | 2,104 | 3,060 | 328 | The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situat... | N, Y = map(int, input().split())
ans = '-1 -1 -1'
for i in range(N + 1):
if ans != '-1 -1 -1':
break
for j in range(N + 1):
for k in range(N + 1):
if Y == 10000 * i + 5000 * j + 1000 * k \
and i + j + k <= N:
ans = str(i) + ' ' + str(j) + ' ' + str(k)
pr... | s967599352 | Accepted | 1,365 | 3,060 | 299 | N, Y = map(int, input().split())
ans = '-1 -1 -1'
for i in range(N + 1):
for j in range(N + 1):
if Y == 10000 * i + 5000 * j + 1000 * (N - i - j)\
and i + j + (N - i - j) <= N\
and N - i - j >= 0:
ans = str(i) + ' ' + str(j) + ' ' + str(N - i - j)
print(ans) |
s425499607 | p02389 | u435300817 | 1,000 | 131,072 | Wrong Answer | 20 | 7,592 | 111 | Write a program which calculates the area and perimeter of a given rectangle. | input_height_width = input()
height, width = [int(x) for x in input_height_width.split()]
print(height * width) | s098961731 | Accepted | 20 | 7,644 | 112 | values = input()
height, width = [int(x) for x in values.split()]
print(height * width, height * 2 + width * 2) |
s790941918 | p02392 | u192877548 | 1,000 | 131,072 | Wrong Answer | 30 | 7,636 | 146 | Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No". | # coding: utf-8
a,b,c = map(int,input().split())
if a < b:
if b < c:
print("YES")
else:
print("NO")
else:
print("NO") | s233321890 | Accepted | 20 | 7,640 | 156 | # coding: utf-8
a,b,c = map(int,input().split())
if a < b and a < c:
if b < c:
print("Yes")
else:
print("No")
else:
print("No") |
s696765142 | p02388 | u630518143 | 1,000 | 131,072 | Wrong Answer | 20 | 5,596 | 322 | Write a program which calculates the cube of a given integer x. | x = int(input("整数xを入力してください。（0<=x<=100）："))
while not(0<=x and x<=100):
print("0<=x<=100を満たす整数を入力してください。")
x = int(input("整数xを入力してください。（0<=x<=100）："))
x = x ** 3
print(str(x) + "の３乗は" + str(x) + "です。")
| s925347446 | Accepted | 20 | 5,576 | 35 | x = int(input())
x = x**3
print(x)
|
s216104514 | p02578 | u724088399 | 2,000 | 1,048,576 | Wrong Answer | 29 | 9,176 | 155 | N persons are standing in a row. The height of the i-th person from the front is A_i. We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person: Condition: Nobody in front of the person is taller than the person. Here, the height of a ... | A = list(map(int,input().split()))
sm = 0
now = A[0]
for num in A:
if num < now:
sm += now-num
elif num > now:
now = num
print(sm)
| s079866888 | Accepted | 104 | 32,220 | 172 | N = int(input())
A = list(map(int,input().split()))
sm = 0
now = A[0]
for num in A:
if num < now:
sm += now-num
elif num > now:
now = num
print(sm)
|
s508245986 | p03478 | u840310460 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 262 | Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive). | N, A, B = [int(i) for i in input().split()]
def func_10(x):
ans = 0
while x/10 != 0:
ans += x % 10
x = x//10
return ans
ans = 0
for i in (1, N+1):
if A <= func_10(i) <= B:
ans += i
print(ans) | s995689448 | Accepted | 29 | 2,940 | 235 | N, A, B = [int(i) for i in input().split()]
def func_de(n):
ans = 0
while n/10 != 0:
ans += n % 10
n //= 10
return ans
ANS = 0
for i in range(N+1):
if A <= func_de(i) <= B:
ANS += i
print(ANS) |
s447342192 | p02390 | u822442916 | 1,000 | 131,072 | Wrong Answer | 20 | 7,680 | 73 | Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively. | S=int(input())
h=S/3600
k=S//3600
m=k/60
s=k-m
print("%d:%d:%d" %(h,m,s)) | s704594958 | Accepted | 40 | 7,684 | 73 | S=int(input())
h=S/3600
k=S%3600
m=k/60
s=k%60
print("%d:%d:%d" %(h,m,s)) |
s373533578 | p03455 | u747602774 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 119 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | import math
s=list(input())
X=int(s[0]+s[1])
root=math.sqrt(X)
a=root//1
if a**2!=X:
print('No')
else:
print('Yes') | s381465032 | Accepted | 17 | 2,940 | 81 | x,y = map(int,input().split())
if x*y%2==0:
print('Even')
else:
print('Odd') |
s145951453 | p03861 | u633450100 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 115 | You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x? | a,b,x = [int(i) for i in input().split()]
if (b-a) % x == 0:
print((b-a) // x)
else:
print((b-a) // x + 1)
| s312378485 | Accepted | 17 | 2,940 | 127 | a,b,x = [int(i) for i in input().split()]
if a % x == 0:
A = a // x
else:
A = a // x + 1
B = b // x
print(B - A + 1)
|
s644017821 | p03457 | u871867619 | 2,000 | 262,144 | Wrong Answer | 700 | 29,584 | 523 | AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1... | import numpy as np
N = int(input())
travel_list = [np.array([0, 0, 0])]
add_list = [np.array([int(i) for i in input().split()]) for i in range(N)]
travel_list.extend(add_list)
def is_travel(arr1, arr2):
dt, dx, dy = abs(arr2 - arr1)
if dt <= dx + dy:
return False
elif dt % 2 != (dx + dy) % 2:
... | s045849113 | Accepted | 1,406 | 29,564 | 539 | import numpy as np
N = int(input())
travel_list = [np.array([0, 0, 0])]
add_list = [np.array([int(i) for i in input().split()]) for i in range(N)]
travel_list.extend(add_list)
def is_travel(arr1, arr2):
dt, dx, dy = abs(arr2 - arr1)
if dt < dx + dy:
return False
elif dt % 2 != (dx + dy) % 2:
... |
s729307755 | p03612 | u513434790 | 2,000 | 262,144 | Wrong Answer | 168 | 14,008 | 428 | You are given a permutation p_1,p_2,...,p_N consisting of 1,2,..,N. You can perform the following operation any number of times (possibly zero): Operation: Swap two **adjacent** elements in the permutation. You want to have p_i ≠ i for all 1≤i≤N. Find the minimum required number of operations to achieve this. | N = int(input())
p = tuple(map(int, input().split()))
a = list(p)
b = list(p)
c = 0
for i in range(N):
if i+1 == a[i] and i == N-1:
c += 1
elif i+1 == a[i]:
a[i], a[i+1] = a[i+1], a[i]
c += 1
cc = 0
for i in range(N)[::-1]:
if i+1 == b[i] and i == 0:
cc += 1
elif... | s372502019 | Accepted | 130 | 14,132 | 411 | N = int(input())
p = tuple(map(int, input().split()))
a = list(p)
b = list(p)
c = 0
for i in range(N):
if i+1 == a[i] and i == N-1:
c += 1
elif i+1 == a[i]:
a[i], a[i+1] = a[i+1], a[i]
c += 1
cc = 0
for i in range(N)[::-1]:
if i+1 == b[i] and i == 0:
cc += 1
elif... |
s083368030 | p02865 | u929585607 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 22 | How many ways are there to choose two distinct positive integers totaling N, disregarding the order? | print(int(input())//2) | s902358224 | Accepted | 17 | 2,940 | 26 | print((int(input())-1)//2) |
s560147228 | p03049 | u228223940 | 2,000 | 1,048,576 | Wrong Answer | 38 | 3,828 | 561 | Snuke has N strings. The i-th string is s_i. Let us concatenate these strings into one string after arranging them in some order. Find the maximum possible number of occurrences of `AB` in the resulting string. | import re
N = int(input())
s = [input() for _ in range(N)]
tmp = 0
print(s[1],s[1][-1:],s[1][0:1])
B_A = 0
B_any = 0
any_A = 0
for i in range(N):
tmp = tmp + s[i].count("AB")
if s[i][0:1] == "B":
if s[i][-1:] == "A":
B_A = B_A + 1
else:
B_any = B_any + 1
elif s[i... | s467543949 | Accepted | 38 | 3,956 | 469 | import re
N = int(input())
s = [input() for _ in range(N)]
tmp = 0
B_A = 0
B_any = 0
any_A = 0
for i in range(N):
tmp = tmp + s[i].count("AB")
if s[i][0:1] == "B":
if s[i][-1:] == "A":
B_A = B_A + 1
else:
B_any = B_any + 1
elif s[i][-1:] == "A":
any_A = an... |
s577738630 | p03693 | u896741788 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 75 | AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this i... | a,s,d=map(int,input().split())
print("YES") if 10*s+d%4==0 else print("NO") | s069314628 | Accepted | 17 | 2,940 | 77 | a,s,d=map(int,input().split())
print("YES") if (10*s+d)%4==0 else print("NO") |
s803813019 | p02845 | u535236942 | 2,000 | 1,048,576 | Wrong Answer | 2,108 | 14,396 | 201 | N people are standing in a queue, numbered 1, 2, 3, ..., N from front to back. Each person wears a hat, which is red, blue, or green. The person numbered i says: * "In front of me, exactly A_i people are wearing hats with the same color as mine." Assuming that all these statements are correct, find the number of p... | n = int(input())
a = list(map(int,input().split()))
ans = 1
for i in range(n):
if a[:i].count(a[i]) == 2:
ans = ans*3%1000000007
if a[:i].count(a[i]) == 1:
ans = ans*2%1000000007
print(ans) | s089389363 | Accepted | 194 | 14,396 | 412 | n = int(input())
a = list(map(int,input().split()))
mod = 1000000007
ans = 1
l = []
for i in range(n):
if a[i] == 0:
l.append(0)
ans = ans*(4-len(l))%mod
else:
id = -1
count = 0
for j in range(len(l)):
if l[j] == a[i] - 1:
count += 1
id = j
if id == -1:
ans = 0
... |
s578341497 | p03457 | u335973735 | 2,000 | 262,144 | Wrong Answer | 453 | 39,456 | 482 | AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1... | N = int(input())
z_t = []*N
t0 = 0
x0 = 0
y0 = 0
ans = 'NO'
for i in range(N):
s = input().rstrip().split()
z_t.append(s)
print(z_t)
for i in range(N):
dt = int(z_t[i][0]) - t0
dx = int(z_t[i][1]) - x0
dy = int(z_t[i][2]) - y0
extra_t = dt - (dx+dy)
if extra_t >= 0 and extra_t%2 == 0:
... | s168546270 | Accepted | 412 | 32,144 | 478 | N = int(input())
z_t = []*N
t0 = 0
x0 = 0
y0 = 0
ans = 'No'
for i in range(N):
s = input().rstrip().split()
z_t.append(s)
for i in range(N):
dt = int(z_t[i][0]) - t0
dx = abs(int(z_t[i][1]) - x0)
dy = abs(int(z_t[i][2]) - y0)
extra_t = dt - (dx+dy)
if extra_t >= 0 and extra_t%2 == 0:
... |
s498857940 | p03696 | u580225095 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 250 | You are given a string S of length N consisting of `(` and `)`. Your task is to insert some number of `(` and `)` into S to obtain a _correct bracket sequence_. Here, a correct bracket sequence is defined as follows: * `()` is a correct bracket sequence. * If X is a correct bracket sequence, the concatenation of... | # -*- coding: utf-8 -*-
x = int(input())
y = input()
z = y
while z.find("()") != -1:
z = z.replace("()","")
a = z.count("(")
b = z.count(")")
ans = z
for i in range(a):
ans = ans + ")"
for i in range(b):
ans = "(" + ans
print(ans)
| s543302317 | Accepted | 17 | 3,060 | 250 | # -*- coding: utf-8 -*-
x = int(input())
y = input()
z = y
while z.find("()") != -1:
z = z.replace("()","")
a = z.count("(")
b = z.count(")")
ans = y
for i in range(a):
ans = ans + ")"
for i in range(b):
ans = "(" + ans
print(ans)
|
s098490102 | p03547 | u202877219 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 104 | In programming, hexadecimal notation is often used. In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively. In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`,... | a,b = map(str, input().split())
if a>b:
print (">")
elif b<a:
print ("<")
else:
print ("=") | s085563504 | Accepted | 17 | 2,940 | 108 | x,y = map(str, input().split())
if x>y:
print (">")
elif x < y:
print ("<")
else:
print ("=")
|
s692279587 | p03471 | u350049649 | 2,000 | 262,144 | Wrong Answer | 765 | 3,060 | 202 | The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situat... | N,Y=map(int,input().split())
a1=-1
a2=-1
a3=-1
for a in range(N+1):
for b in range(N-a,N+1):
c=N-a-b
if a*10000+b*5000+c*1000==Y:
a1=a
a2=b
a3=c
break
print(a1,a2,a3) | s520515793 | Accepted | 759 | 3,060 | 200 | N,Y=map(int,input().split())
a1=-1
a2=-1
a3=-1
for a in range(N+1):
for b in range(N-a+1):
c=N-a-b
if a*10000+b*5000+c*1000==Y:
a1=a
a2=b
a3=c
break
print(a1,a2,a3) |
s844743716 | p03168 | u589303510 | 2,000 | 1,048,576 | Wrong Answer | 1,627 | 222,760 | 463 | Let N be a positive odd number. There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i. Taro has tossed all the N coins. Find the probability of having more heads than tails. | def coins(n,prob):
dp = [[0]*(n+1) for _ in range(n+1)]
dp[0][0] = 1
for i in range(1,n+1):
for j in range(i+1):
if j == 0:
dp[i][j] = dp[i-1][j]*prob[i-1]
else:
dp[i][j] = dp[i-1][j]*(1-prob[i-1])+dp[i-1][j-1]*prob[i-1]
total = 0
for ... | s254566404 | Accepted | 1,607 | 222,560 | 467 | def coins(n,prob):
dp = [[0]*(n+1) for _ in range(n+1)]
dp[0][0] = 1
for i in range(1,n+1):
for j in range(i+1):
if j == 0:
dp[i][j] = dp[i-1][j]*(1-prob[i-1])
else:
dp[i][j] = dp[i-1][j]*(1-prob[i-1])+dp[i-1][j-1]*prob[i-1]
total = 0
... |
s195933623 | p03847 | u786020649 | 2,000 | 262,144 | Wrong Answer | 30 | 9,188 | 441 | You are given a positive integer N. Find the number of the pairs of integers u and v (0≦u,v≦N) such that there exist two non-negative integers a and b satisfying a xor b=u and a+b=v. Here, xor denotes the bitwise exclusive OR. Since it can be extremely large, compute the answer modulo 10^9+7. | import sys
p=10**9+7
dthbit=lambda d,n: (n>>d)&1
def main(n):
dp=[[0 for _ in range(3)] for _ in range(64)]
dp[63][0]=1
for d in range(62,-1,-1):
b=dthbit(d,n)
s=dp[d+1][:]
dp[d][0]=dp[d+1][0]+(1^b)*dp[d+1][1] % p
dp[d][1]=b*dp[d+1][0]+dp[d+1][1] % p
dp[d][2]=(1+b)*... | s843596843 | Accepted | 29 | 9,176 | 408 | import sys
p=10**9+7
dthbit=lambda d,n: (n>>d)&1
def main(n):
dp=[[0 for _ in range(3)] for _ in range(64)]
dp[63][0]=1
for d in range(62,-1,-1):
b=dthbit(d,n)
s=dp[d+1][:]
dp[d][0]=dp[d+1][0]+(1^b)*dp[d+1][1] % p
dp[d][1]=b*dp[d+1][0]+dp[d+1][1] % p
dp[d][2]=(1+b)*... |
s261826424 | p03352 | u802772880 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,064 | 229 | You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2. | X=int(input())
l2=[1,2,4,8,16,32,64,128,256,512]
l3=[3,9,27,81,243,729]
l5=[5,25,125,625]
l6=[6,36,216]
l7=[7,49,343]
l10=[10,100,1000]
le=[n*n for n in range(11,32)]
l=l2+l3+l5+l6+l7+l10+le
m=max([i for i in l if i<=X])
print(m) | s148607886 | Accepted | 18 | 2,940 | 151 | X=int(input())
Y=1
for b in range(1,X):
for p in range(2,X):
if b**p<=X:
Y=max(Y,b**p)
else:
break
print(Y) |
s981577920 | p03069 | u528005130 | 2,000 | 1,048,576 | Wrong Answer | 78 | 12,836 | 455 | There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is `.`, and the stone is black if the character is `#`. Takahashi wants to change the colors of some stones to black or white so t... | # coding: utf-8
# Your code here!
N = int(input())
S = [x for x in input()]
cum_white = []
count = 0
for i, s in enumerate(S):
if s == '.':
count += 1
cum_white.append(count)
if len(cum_white) == 0 or len(cum_white) == N:
print(0)
else:
min_num = int(1e10)
for i, n_white in enumerate(... | s158393411 | Accepted | 180 | 12,396 | 385 | # coding: utf-8
# Your code here!
N = int(input())
S = ['.'] + [x for x in input()] + ['#']
cum_white = []
count = 0
for i, s in enumerate(S):
if s == '#':
count += 1
cum_white.append(count)
min_num = int(1e10)
for i, n_white in enumerate(cum_white):
num = n_white + (N + 1 - i - (cum_white[-1... |
s816832679 | p02393 | u731896389 | 1,000 | 131,072 | Wrong Answer | 20 | 7,304 | 36 | Write a program which reads three integers, and prints them in ascending order. | n= input().split()
n.sort()
print(n) | s281758227 | Accepted | 30 | 7,644 | 63 | a=list(map(int,input().split()))
a.sort()
print(a[0],a[1],a[2]) |
s919249252 | p03644 | u475675023 | 2,000 | 262,144 | Wrong Answer | 19 | 2,940 | 75 | Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can... | n=int(input())
for i in range(9):
if 2**i >= n:
print(2**i)
break | s048486483 | Accepted | 17 | 2,940 | 64 | n=int(input())
m=1
while 2**m<=n:
m+=1
else:
print(2**(m-1)) |
s946659324 | p03369 | u023229441 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 37 | 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 ... | print(700+(input().count("○"))*100) | s342247335 | Accepted | 17 | 2,940 | 36 | print(700+(input().count("o"))*100)
|
s792245876 | p03493 | u284363684 | 2,000 | 262,144 | Wrong Answer | 30 | 9,020 | 52 | Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble. | # input
S = list(input())
print(S.count(1)) | s127624880 | Accepted | 25 | 8,996 | 54 | # input
S = list(input())
print(S.count("1")) |
s880213233 | p03474 | u187516587 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 323 | The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom. | p=input().split()
A=int(p[0])
B=int(p[1])
S=input()
if S[A]=="-":
t=S.split('-')
if len(t)==2:
a,b=int(t[0]),int(t[1])
print(a)
if 10**(A-1)<=a<10**A and 10**(B-1)<=b<10**B:
print("Yes")
else:
print("No")
else:
print("No")
else:
print("No... | s891752581 | Accepted | 17 | 3,060 | 175 | p=input().split()
A=int(p[0])
B=int(p[1])
S=input()
if S[A]=="-":
t=S.split('-')
if len(t)==2:
print("Yes")
else:
print("No")
else:
print("No") |
s356867701 | p03407 | u077075933 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 68 | An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan. | A, B, C = map(int, input().split())
print("YES" if C<+A+B else "NO") | s986739614 | Accepted | 17 | 2,940 | 68 | A, B, C = map(int, input().split())
print("Yes" if C<=A+B else "No") |
s577240503 | p02601 | u005569385 | 2,000 | 1,048,576 | Wrong Answer | 30 | 9,136 | 293 | M-kun has the following three cards: * A red card with the integer A. * A green card with the integer B. * A blue card with the integer C. He is a genius magician who can do the following operation at most K times: * Choose one of the three cards and multiply the written integer by 2. His magic is successfu... | A,B,C = map(int,input().split())
K = int(input())
for i in range(0,K+1):
A = A*(2*i)
for j in range(0,K+1):
B = B*(2*j)
for h in range(0,K+1):
C = C*(2*h)
if (i+j+h) <= K and C>B>A:
print("Yes")
exit()
print("No") | s961739469 | Accepted | 31 | 9,192 | 252 | A,B,C = map(int,input().split())
K = int(input())
for i in range(0,K+1):
for j in range(0,K+1):
for h in range(0,K+1):
if A*(2**i)<B*(2**j)<C*(2**h) and (i+j+h)<=K:
print("Yes")
exit()
print("No") |
s360254824 | p03759 | u394482932 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 43 | 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. | for s in input().split():print(s[0],end='') | s822734946 | Accepted | 17 | 2,940 | 58 | a,b,c=map(int,input().split());print('YNEOS'[b-a!=c-b::2]) |
s862093787 | p03658 | u492030100 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 135 | Snuke has N sticks. The length of the i-th stick is l_i. Snuke is making a snake toy by joining K of the sticks together. The length of the toy is represented by the sum of the individual sticks that compose it. Find the maximum possible length of the toy. | N, K = map(int, input().split())
L = [l for l in range(N)]
L.sort(reverse=True)
ans = 0
for i in range(K):
ans += L[i]
print(ans)
| s041152239 | Accepted | 18 | 2,940 | 147 | N, K = map(int, input().split())
L = [int(l) for l in input().split()]
L.sort(reverse=True)
ans = 0
for i in range(K):
ans += L[i]
print(ans)
|
s575734686 | p02842 | u671211357 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,060 | 266 | 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).... | import math
N = int(input())
kari = N*100/108
hako = math.ceil(kari)
ans = math.floor(kari)
if hako*108/100==N or ans*108/100==N:
print(kari)
elif math.floor(hako*108/100)==N:
print(hako)
elif math.floor(ans*108/100)==N:
print(ans)
else:
print(":(") | s102884812 | Accepted | 17 | 3,060 | 378 | import math
N = int(input())
kari = N*100/108
hako = math.ceil(kari)
ans = math.floor(kari)
if hako*108/100==N or ans*108/100==N:
print(math.floor(kari))
elif math.floor(hako*108/100)==N:
print(hako)
elif math.floor(ans*108/100)==N:
print(ans)
# print(hako)
# print(ans)
else:
print(":(") |
s193631754 | p03469 | u498575211 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 31 | 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 col... | S = input()
print("2018"+S[5:]) | s820371018 | Accepted | 17 | 2,940 | 31 | S = input()
print("2018"+S[4:]) |
s101841550 | p02613 | u377989038 | 2,000 | 1,048,576 | Wrong Answer | 143 | 16,612 | 213 | 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`,... | from collections import Counter as ct
n = int(input())
s = ct([input() for _ in range(n)])
print("AC x " + str(s["AC"]))
print("WA x " + str(s["wA"]))
print("TLE x " + str(s["TLE"]))
print("RE x " + str(s["RE"])) | s139659295 | Accepted | 143 | 16,508 | 213 | from collections import Counter as ct
n = int(input())
s = ct([input() for _ in range(n)])
print("AC x " + str(s["AC"]))
print("WA x " + str(s["WA"]))
print("TLE x " + str(s["TLE"]))
print("RE x " + str(s["RE"])) |
s147743229 | p03486 | u239528020 | 2,000 | 262,144 | Wrong Answer | 25 | 9,092 | 208 | 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. | #!/usr/bin/env python3
import sys
sys.setrecursionlimit(10**6)
s = list(str(input()))
t = str(input())
s.sort()
s = "".join(s)
ans = sorted([s, t])
if ans[0] == t:
print("No")
else:
print("Yes")
| s058066570 | Accepted | 30 | 9,072 | 263 | #!/usr/bin/env python3
import sys
sys.setrecursionlimit(10**6)
s = list(str(input()))
t = list(str(input()))
s.sort()
t.sort(reverse=True)
s = "".join(s)
t = "".join(t)
ans = sorted([s, t])
# print(ans)
if ans[0] == t:
print("No")
else:
print("Yes")
|
s606374258 | p04029 | u824147251 | 2,000 | 262,144 | Wrong Answer | 37 | 3,064 | 33 | 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? | n = int(input())
print((n+1)*n/2) | s378277987 | Accepted | 42 | 3,444 | 34 | n = int(input())
print((n+1)*n//2) |
s204546982 | p02255 | u734765925 | 1,000 | 131,072 | Wrong Answer | 20 | 7,628 | 229 | 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 ... | n = int(input())
A = list(map(int,input().split()))
temp = 0
for i in range(1,n):
temp = A[i]
j = i -1
while j >=0 and A[j] > temp :
A[j+1]=A[j]
j-=1
A[j+1] = temp
print(" ".join(map(str,A))) | s331689025 | Accepted | 30 | 7,660 | 256 | n = int(input())
A = list(map(int,input().split()))
temp = 0
print(" ".join(map(str,A)))
for i in range(1,n):
temp = A[i]
j = i -1
while j >=0 and A[j] > temp :
A[j+1]=A[j]
j-=1
A[j+1] = temp
print(" ".join(map(str,A))) |
s070782881 | p03048 | u902468164 | 2,000 | 1,048,576 | Wrong Answer | 890 | 3,060 | 163 | 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... | r,g,b,n=[int(i) for i in input().split(" ")]
res = 0
for i in range(0,n+1,r):
for j in range(i,n+1,g):
if j % b == 0:
res += 1
print(res) | s141532350 | Accepted | 958 | 2,940 | 169 | r,g,b,n=[int(i) for i in input().split(" ")]
res = 0
for i in range(0,n+1,r):
for j in range(i,n+1,g):
if (n-j) % b == 0:
res += 1
print(res)
|
s648067977 | p03457 | u780675733 | 2,000 | 262,144 | Wrong Answer | 435 | 27,380 | 422 | AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1... | N = int(input())
Input = []
for i in range(N):
Input.append(list(map(int,input().split())))
x = 0
y = 0
t = 0
for i in Input:
dif_x = abs(x - i[1])
dif_y = abs(y - i[2])
dif_t = abs(t - i[0])
if dif_x + dif_y <= dif_t and (dif_x + dif_y)%2 == dif_t%2:
x = i[1]
y = i[2]
t ... | s227703883 | Accepted | 442 | 27,380 | 394 | N = int(input())
Input = []
for i in range(N):
Input.append(list(map(int,input().split())))
x = 0
y = 0
t = 0
for i in Input:
dif_x = abs(x - i[1])
dif_y = abs(y - i[2])
dif_t = abs(t - i[0])
if dif_x + dif_y <= dif_t and (dif_x + dif_y)%2 == dif_t%2:
x = i[1]
y = i[2]
t ... |
s365086804 | p03643 | u301823349 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 118 | 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 = int(input())
p = 0
while True:
if N < 2**p:
print(2**(p-1))
break
else:
p=p+1
| s688066500 | Accepted | 17 | 2,940 | 28 | N = input()
print("ABC" + N) |
s936772673 | p03162 | u469953228 | 2,000 | 1,048,576 | Wrong Answer | 959 | 50,332 | 304 | Taro's summer vacation starts tomorrow, and he has decided to make plans for it now. The vacation consists of N days. For each i (1 \leq i \leq N), Taro will choose one of the following activities and do it on the i-th day: * A: Swim in the sea. Gain a_i points of happiness. * B: Catch bugs in the mountains. Gain... | N = int(input())
dp = [[0] * 3 for _ in range(N+1)]
A = [list(map(int,input().split())) for _ in range(N)]
for i in range(N):
for j in range(3):
for k in range(3):
if j == k:
continue
dp[i+1][k] = max(dp[i+1][k], dp[i][k] + A[i][j])
print(max(dp[N]))
| s996135608 | Accepted | 980 | 50,336 | 304 | N = int(input())
dp = [[0] * 3 for _ in range(N+1)]
A = [list(map(int,input().split())) for _ in range(N)]
for i in range(N):
for j in range(3):
for k in range(3):
if j == k:
continue
dp[i+1][k] = max(dp[i+1][k], dp[i][j] + A[i][k])
print(max(dp[N]))
|
s050432332 | p02608 | u388971072 | 2,000 | 1,048,576 | Wrong Answer | 31 | 9,244 | 243 | Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N). | N = int(input())
N_list = [0]*(N+1)
for i in range(1,11):
for j in range(1,11):
for k in range(1,11):
if(i**2+j**2+k**2+i*j+j*k+k*i<=N):
N_list[i**2+j**2+k**2+i*j+j*k+k*i]+=1
for i in N_list:
print(i) | s116003770 | Accepted | 1,027 | 9,292 | 250 | N = int(input())
N_list = [0]*(N+1)
for i in range(1,101):
for j in range(1,101):
for k in range(1,101):
if(i**2+j**2+k**2+i*j+j*k+k*i<=N):
N_list[i**2+j**2+k**2+i*j+j*k+k*i]+=1
for i in N_list[1:]:
print(i) |
s037132831 | p02393 | u255164080 | 1,000 | 131,072 | Wrong Answer | 30 | 6,720 | 46 | Write a program which reads three integers, and prints them in ascending order. | nums = input().split()
nums.sort()
print(nums) | s419016480 | Accepted | 30 | 6,724 | 224 | nums = input().split()
a = int(nums[0])
b = int(nums[1])
c = int(nums[2])
if a > b:
tmp = a
a = b
b = tmp
if b > c:
tmp = b
b = c
c = tmp
if a > b:
tmp = a
a = b
b = tmp
print(a, b, c) |
s739158398 | p02613 | u068844030 | 2,000 | 1,048,576 | Wrong Answer | 136 | 16,160 | 234 | 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`,... | n = int(input())
s = [input() for i in range(n)]
AC = s.count("AC")
WA = s.count("WA")
TLE = s.count("TLE")
RE = s.count("RE")
print("AC × " + str(AC))
print("WA × " + str(WA))
print("TLE × " + str(TLE))
print("RE × " + str(RE)) | s983307635 | Accepted | 146 | 16,052 | 293 | n = int(input())
s = [input() for i in range(n)]
AC = s.count("AC")
WA = s.count("WA")
TLE = s.count("TLE")
RE = s.count("RE")
AC2 = "AC x " + str(AC)
WA2 = "WA x " + str(WA)
TLE2 = "TLE x " + str(TLE)
RE2 = "RE x " + str(RE)
list = [AC2,WA2,TLE2,RE2]
list_n = "\n".join(list)
print(list_n) |
s152640402 | p03160 | u830054172 | 2,000 | 1,048,576 | Wrong Answer | 2,104 | 109,972 | 243 | There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of... | n = int(input())
a = list(map(int, input().split()))
dp = [0]*n
for i in range(n):
if i >= 2:
dp[i] = min(dp[i-1]+abs(a[i]-a[i-1]), dp[i-2]+abs(a[i]-a[i-2]))
elif i == 1:
dp[i] = a[i]-a[i-1]
print(dp)
print(dp[-1]) | s105303366 | Accepted | 181 | 13,928 | 264 | N = int(input())
h = list(map(int, input().split())) + [0]*100010
dp = [float("inf")]*100010
dp[0] = 0
for i in range(N):
dp[i+1] = min(dp[i+1], dp[i] + abs(h[i+1] - h[i]))
dp[i+2] = min(dp[i+2], dp[i] + abs(h[i+2] - h[i]))
# print(dp)
print(dp[N-1]) |
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