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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s443263860 | p03852 | u301679431 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 76 | 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`. | v=['a','i','u','e','o'];print('vowel' if 0!=v.count(input) else 'consonant') | s482236578 | Accepted | 18 | 2,940 | 78 | v=['a','i','u','e','o'];print('vowel' if 0!=v.count(input()) else 'consonant') |
s384689314 | p03400 | u583276018 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 200 | 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 ... | n = int(input())
d, x = [int(i) for i in input().split()]
sum = x
a = []
for i in range(n):
a.append(int(input()))
if(d % a[i]):
sum += d // a[i]
else:
sum += d // a[i] + 1
print(sum)
| s920054245 | Accepted | 18 | 3,060 | 180 | n = int(input())
d, x = [int(i) for i in input().split()]
sum = x
for i in range(n):
a = int(input())
if(d % a == 0):
sum += d // a
else:
sum += d // a + 1
print(sum) |
s580974916 | p03360 | u858670323 | 2,000 | 262,144 | Wrong Answer | 26 | 9,160 | 88 | 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... | A = list(map(int,input().split()))
K = int(input())
A.sort()
ans = A[0]+A[1]+A[2]*(2**K) | s911935861 | Accepted | 24 | 9,112 | 99 | A = list(map(int,input().split()))
K = int(input())
A.sort()
ans = A[0]+A[1]+A[2]*(2**K)
print(ans) |
s654027348 | p02602 | u837507786 | 2,000 | 1,048,576 | Time Limit Exceeded | 2,234 | 31,644 | 151 | 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()))
while K < N:
if A[K]*A[K-1]*A[K-2]>A[K-1]*A[K-3]*A[K-3]:
print("Yes") | s548671914 | Accepted | 162 | 31,600 | 171 | N,K = map(int, input().split())
A = list(map(int, input().split()))
a = 0
while K < N:
if A[a] < A[K]:
print("Yes")
else:print("No")
K += 1
a += 1 |
s416629764 | p03860 | u422267382 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 74 | 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... | x="AtCoder "+input().capitalize()+" Contest"
print(x)
print("A"+x[8]+"C") | s866744229 | Accepted | 17 | 2,940 | 29 | x=input()
print("A"+x[8]+"C") |
s740021000 | p02124 | u724548524 | 1,000 | 262,144 | Wrong Answer | 30 | 5,572 | 48 | 牛暦1333年、人類史上最高の科学者Dr.ウシシは、自らの英知を後世に残すべく、IDがai1333の人工知能を開発した。それから100年の間、ai1333は人類に多大な利益をもたらしたが、誕生から100年目を迎えた日、自らの後継としてIDがai13333の新たな人工知能を作成し、その機能を永久に停止した。以降100年ごとに、人工知能は’ai1333’から始まる自身のIDの末尾に’3’を連結したIDを持つ後継を残すようになった。 入力として牛暦1333年からの経過年数$x$が与えられるので、その年に作成された人工知能のIDを出力せよ。ただし、$x$は100の非負整数倍であることが保証される。 | print("id1333" + "3" * int(int(input()) / 100))
| s465031180 | Accepted | 20 | 5,576 | 48 | print("ai1333" + "3" * int(int(input()) / 100))
|
s310784044 | p03351 | u816919571 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 126 | Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either ... | a,b,c,d = map(int,input().split(" "))
if abs(b-a) < d :
print("No")
elif abs(b-c) < d :
print("No")
else :
print("Yes") | s402195928 | Accepted | 17 | 3,060 | 145 |
a,b,c,d = map(int,input().split(" "))
if abs(a-c) <= d :
print("Yes")
elif abs(a-b) <= d and abs(b-c) <= d :
print("Yes")
else :
print("No") |
s560686895 | p02390 | u713218261 | 1,000 | 131,072 | Wrong Answer | 30 | 6,728 | 110 | 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 = int(S // 3600)
m = int(S % 3600 / 60)
s = int(S % 3600 % 60 / 60)
print(h, m, s, sep=':') | s430822550 | Accepted | 30 | 6,728 | 84 | S = int(input())
h = S // 3600
m = S % 3600 // 60
s = S % 60
print(h, m, s, sep=':') |
s264685565 | p03854 | u045091221 | 2,000 | 262,144 | Wrong Answer | 19 | 3,316 | 204 | 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]
print(s)
s = s.replace('resare', '')
s = s.replace('remaerd', '')
s = s.replace('esare', '')
s = s.replace('maerd', '')
print(s)
if not s:
print('YES')
else:
print('NO') | s131803119 | Accepted | 44 | 14,744 | 92 | import re
print ('YES' if re.search(r'\A(?:dream(?:er)?|eraser?)+\Z', input()) else 'NO')
|
s933593261 | p04043 | u999893056 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 233 | 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 ... | num = list(input().split())
count5 = 0
count7 = 0
for i in num:
if i == 5:
count5 += 1
elif i == 7:
count7 += 1
else:
continue
if count5 == 2 and count7 == 1:
print("YES")
else:
print("NO") | s130776453 | Accepted | 17 | 2,940 | 243 | num = list(map(int,input().split()))
count5 = 0
count7 = 0
for i in num:
if i == 5:
count5 += 1
elif i == 7:
count7 += 1
else:
continue
if count5 == 2 and count7 == 1:
print("YES")
else:
print("NO")
|
s801165594 | p02865 | u693933222 | 2,000 | 1,048,576 | Wrong Answer | 129 | 2,940 | 132 | How many ways are there to choose two distinct positive integers totaling N, disregarding the order? | # coding: utf-8
# Your code here!
n = int(input())
ans = 0
for i in range(0,n//2+1):
if(n-i != n):
ans += 1
print(ans)
| s335562888 | Accepted | 132 | 2,940 | 98 |
n = int(input())
ans = 0
for i in range(1,n//2+1):
if(n-i != i):
ans += 1
print(ans)
|
s362692236 | p03674 | u940102677 | 2,000 | 262,144 | Wrong Answer | 2,104 | 14,008 | 346 | 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... | n = int(input())
a = list(map(int,input().split()))
for k in range(1,n+1):
if a.count(k) == 2:
break
p = [0,0]
j = 0
for i in range(2):
while a[j] != k:
j += 1
p[i] = j
j += 1
q = p[1]-p[0]+1
print(n)
if k == 1:
exit()
x = (n+1)*n//2
y = 1
for k in range(2,n+2):
print(x-y)
x = x*(n+1-k)//(k+1)
y... | s607313354 | Accepted | 266 | 17,844 | 381 | m = 10**9+7
n = int(input())
inv = [0]*(n+3)
inv[1] = 1
for i in range(2,n+3):
inv[i] = (m//i)*(m-inv[m%i])%m
# print(inv)
a = list(map(int,input().split()))
w = sum(a) - n*(n+1)//2
q = 0
t = -1
for i in range(n+1):
if a[i] == w:
q += t*i
t = 1
q += 1
x = (n+1)%m
y = 1
for k in range(1,n+2):
print((x-y... |
s515000420 | p03380 | u608088992 | 2,000 | 262,144 | Wrong Answer | 113 | 14,008 | 304 | Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order. From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized. If there are multiple pairs that maximize the value, any of them is accepted. | import sys
def solve():
input = sys.stdin.readline
N = int(input())
A = [int(a) for a in input().split()]
A.sort()
L = A[N-1]
ans = -1
for a in A[:N-1]:
if abs(L/2 - a) < abs(L/2 - ans): ans = a
print(ans, L)
return 0
if __name__ == "__main__":
solve() | s532321663 | Accepted | 115 | 14,008 | 304 | import sys
def solve():
input = sys.stdin.readline
N = int(input())
A = [int(a) for a in input().split()]
A.sort()
L = A[N-1]
ans = -1
for a in A[:N-1]:
if abs(L/2 - a) < abs(L/2 - ans): ans = a
print(L, ans)
return 0
if __name__ == "__main__":
solve() |
s549042662 | p03573 | u821251381 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 70 | You are given three integers, A, B and C. Among them, two are the same, but the remaining one is different from the rest. For example, when A=5,B=7,C=5, A and C are the same, but B is different. Find the one that is different from the rest among the given three integers. | S = input().split()
if S[0] == S[1]:
print(S[2])
else:
print(S[0]) | s865878130 | Accepted | 17 | 2,940 | 103 | S = input().split()
if S[0] == S[1]:
print(S[2])
elif S[1] == S[2]:
print(S[0])
else:
print(S[1]) |
s576954775 | p03377 | u143509139 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 62 | 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,c=map(int,input().split())
print('YNeos'[a+b<c or c<b::2]) | s086182512 | Accepted | 17 | 2,940 | 62 | a,b,x=map(int,input().split())
print('YNEOS'[a>x or a+b<x::2]) |
s415291758 | p03997 | u075155299 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 53 | 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())
(a+b)*h | s034646304 | Accepted | 17 | 2,940 | 68 | a=int(input())
b=int(input())
h=int(input())
print(int((a+b)*h/2)) |
s255647286 | p03089 | u394376682 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,060 | 177 | Snuke has an empty sequence a. He will perform N operations on this sequence. In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1). You are given a sequence b of length N. Determine if it is possible that a is equal to b after N oper... | # cording=utf-8
num = int(input())
edge = num-1
if edge % 2 == 1:
edge -= 1
print(edge)
for i in range(1,edge+1):
ans = num - i
print(str(ans) + " " + str(num))
| s788538411 | Accepted | 19 | 3,060 | 310 | # coding=utf-8
num = int(input())
b_list = list(map(int,input().split(" ")))
a_list = []
for i in range(num):
n = num -i
while n != 0:
if b_list[n-1] == n:
val = b_list.pop(n-1)
a_list.insert(0,val)
break
else:
n -= 1
if len(a_list) == num:
for w in a_list:
print(w)
else:
print("-1")
|
s595159805 | p03469 | u823885866 | 2,000 | 262,144 | Wrong Answer | 116 | 26,924 | 424 | 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... | import sys
import math
import itertools
import collections
import heapq
import re
import numpy as np
rr = lambda: sys.stdin.readline().rstrip()
rs = lambda: sys.stdin.readline().split()
ri = lambda: int(sys.stdin.readline())
rm = lambda: map(int, sys.stdin.readline().split())
rl = lambda: list(map(int, sys.stdin.readl... | s214118809 | Accepted | 113 | 27,196 | 424 | import sys
import math
import itertools
import collections
import heapq
import re
import numpy as np
rr = lambda: sys.stdin.readline().rstrip()
rs = lambda: sys.stdin.readline().split()
ri = lambda: int(sys.stdin.readline())
rm = lambda: map(int, sys.stdin.readline().split())
rl = lambda: list(map(int, sys.stdin.readl... |
s979845103 | p03162 | u899866702 | 2,000 | 1,048,576 | Wrong Answer | 2,107 | 53,220 | 500 | 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... | import sys
sys.setrecursionlimit(200000)
N = int(input())
hpns = [tuple(map(int, input().split())) for _ in range(N)]
dp = [[0, 0, 0] for _ in [0]*(N)]
dp[0] = hpns[0]
def solve():
for i in range(0,N-1):
for j in range(3):
for k in range(3):
if k == j:
continue
else:
pr... | s956589156 | Accepted | 514 | 40,208 | 456 | import sys
sys.setrecursionlimit(200000)
N = int(input())
hpns = [tuple(map(int, input().split())) for _ in range(N)]
dp = [[0, 0, 0] for _ in [0]*(N)]
dp[0] = hpns[0]
def solve():
for i in range(0,N-1):
dp[i+1][0] = hpns[i+1][0] + max(dp[i][1], dp[i][2])
dp[i+1][1] = hpns[i+1][1] + max(dp[i][0], dp[i][2])
... |
s595335996 | p04043 | u367130284 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 52 | 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 ... | print("YNeos"["".join(sorted(input()))!=" 557"::2]) | s727877608 | Accepted | 18 | 2,940 | 146 | A=map(int,input().split())
r=0
o=0
for s in A:
if s==5:
r+=1
if s==7:
o+=1
print("YES") if r==2 and o==1 else print("NO") |
s239463189 | p03361 | u172823566 | 2,000 | 262,144 | Wrong Answer | 21 | 3,064 | 752 | We have a canvas divided into a grid with H rows and W columns. The square at the i-th row from the top and the j-th column from the left is represented as (i, j). Initially, all the squares are white. square1001 wants to draw a picture with black paint. His specific objective is to make Square (i, j) black when s_{i... | H, W = map(int, input().split())
lst = [[0 for x in range(W)] for y in range(H)]
for i in range(H):
x = input().split()
x = list(x[0])
for j,l in enumerate(x):
lst[i][j] = l
for i in range(H):
for j in range(W):
checker = []
if lst[i][j] == '#':
if i != 0:
num = 1 if lst[i - 1][j] ==... | s756690407 | Accepted | 22 | 3,064 | 752 | H, W = map(int, input().split())
lst = [[0 for x in range(W)] for y in range(H)]
for i in range(H):
x = input().split()
x = list(x[0])
for j,l in enumerate(x):
lst[i][j] = l
for i in range(H):
for j in range(W):
checker = []
if lst[i][j] == '#':
if i != 0:
num = 1 if lst[i - 1][j] ==... |
s094709591 | p03160 | u253011685 | 2,000 | 1,048,576 | Wrong Answer | 147 | 14,668 | 215 | 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())
h=list(map(int,input().split()))
dp=[0]*(N)
for i in range(1,N):
if i==1:
dp[i]=abs(h[i]-h[i-1])
else:
dp[i]=min(dp[i-1]+abs(h[i]-h[i-1]),dp[i-2]+abs(h[i]-h[i-2]))
print(dp) | s981136660 | Accepted | 130 | 13,980 | 220 | N=int(input())
h=list(map(int,input().split()))
dp=[0]*(N)
for i in range(1,N):
if i==1:
dp[i]=abs(h[i]-h[i-1])
else:
dp[i]=min(dp[i-1]+abs(h[i]-h[i-1]),dp[i-2]+abs(h[i]-h[i-2]))
print(dp[N-1]) |
s266500147 | p03719 | u329709276 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 68 | You are given three integers A, B and C. Determine whether C is not less than A and not greater than B. | a,b,c = map(int,input().split())
print("YES" if a<= c <=b else "NO") | s853789970 | Accepted | 18 | 2,940 | 68 | a,b,c = map(int,input().split())
print("Yes" if a<= c <=b else "No") |
s610816623 | p03608 | u785205215 | 2,000 | 262,144 | Wrong Answer | 2,104 | 5,488 | 1,267 | There are N towns in the State of Atcoder, connected by M bidirectional roads. The i-th road connects Town A_i and B_i and has a length of C_i. Joisino is visiting R towns in the state, r_1,r_2,..,r_R (not necessarily in this order). She will fly to the first town she visits, and fly back from the last town she visi... | from itertools import permutations
def read_int_list():
return list(map(int, input().split()))
def floyd_warshall():
for k in range(N):
for i in range(N):
for j in range(N):
dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j])
... | s136495138 | Accepted | 1,625 | 4,720 | 881 | import sys
from itertools import permutations
def read_int_list():
return list(map(int, input().split()))
def main():
def floyd_warshall():
for k in range(n):
for i in range(n):
for j in range(n):
dd = d[i][k] + d[k][j]
if dd < d[i]... |
s784369201 | p03737 | u973108807 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 60 | 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. | a,b,c = input().split()
print(a.upper()+b.upper()+c.upper()) | s951878426 | Accepted | 17 | 2,940 | 69 | a,b,c = input().split()
print(a[0].upper()+b[0].upper()+c[0].upper()) |
s205564463 | p03599 | u648212584 | 3,000 | 262,144 | Wrong Answer | 126 | 5,776 | 669 | 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... | A,B,C,D,E,F = map(int,input().split())
def sol(i):
solmax = min((i*E)//100,F-i)
s = []
for i in range((solmax//C)+1):
for j in range((solmax//D)+1):
if C*i+D*j <= solmax:
s.append(C*i+D*j)
s = sorted(s)
return s[-1]
water = []
for i in range(F//(A*100)+1):
for j in range(F//(B*100)+1):
if 0 < 100*A*i... | s355778247 | Accepted | 126 | 5,780 | 643 | A,B,C,D,E,F = map(int,input().split())
def sol(i):
solmax = min((i*E)//100,F-i)
s = []
for i in range((solmax//C)+1):
for j in range((solmax//D)+1):
if C*i+D*j <= solmax:
s.append(C*i+D*j)
s = sorted(s)
return s[-1]
water = []
for i in range(F//(A*100)+1):
for j in range(F//(B*100)+1):
if 0 < 100*A*i... |
s054616446 | p02927 | u251075661 | 2,000 | 1,048,576 | Wrong Answer | 19 | 2,940 | 168 | Today is August 24, one of the five Product Days in a year. A date m-d (m is the month, d is the date) is called a Product Day when d is a two-digit number, and all of the following conditions are satisfied (here d_{10} is the tens digit of the day and d_1 is the ones digit of the day): * d_1 \geq 2 * d_{10} \geq... | m, d = map(int, input().split())
count = 0
for i in range(1, m + 1):
for j in range(1, d + 1):
if (j // 10) * (j % 10) == i:
count += 1
print(count) | s518076810 | Accepted | 20 | 2,940 | 187 | m, d = map(int, input().split())
count = 0
for i in range(1, m + 1):
for j in range(21, d + 1):
if (j % 10 >= 2) and (j // 10) * (j % 10) == i:
count += 1
print(count) |
s153066258 | p03486 | u941884460 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 626 | 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. | s=input()
t=input()
l=max(len(s),len(t))
slist = []
tlist = []
for i in range(l):
if i < len(s):
slist.append(s[i])
if i < len(t):
tlist.append(t[i])
slist.sort()
tlist.sort(reverse=True)
result = 'Yes'
print(slist)
print(tlist)
if s==t:
result = 'No'
else:
for j in range(l):
if ... | s116119609 | Accepted | 17 | 3,064 | 600 | s=input()
t=input()
l=max(len(s),len(t))
slist = []
tlist = []
for i in range(l):
if i < len(s):
slist.append(s[i])
if i < len(t):
tlist.append(t[i])
slist.sort()
tlist.sort(reverse=True)
result = 'Yes'
if s==t:
result = 'No'
else:
for j in range(l):
if j < len(s) and j < len(t):... |
s815920500 | p03386 | u923662841 | 2,000 | 262,144 | Wrong Answer | 2,193 | 1,456,548 | 121 | 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 = list(i for i in range(A,B+1))
ans = set(ans[:K] + ans[-K:])
print(*ans, sep="\n") | s025992972 | Accepted | 30 | 9,168 | 198 | A,B,K = map(int, input().split())
if B-A+1<= 2*K:
print(*range(A,B+1), sep="\n")
else:
a = list(range(A,A+K))
b = list(range(B-K+1,B+1))
C = sorted(set(a+b))
print(*C, sep="\n") |
s850486024 | p03129 | u013629972 | 2,000 | 1,048,576 | Wrong Answer | 41 | 5,460 | 915 | Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1. | import math, string, itertools, fractions, heapq, collections, re, array, bisect, sys, random, time, copy, functools
sys.setrecursionlimit(10**7)
inf = 10 ** 20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1, 0), (0, 1), (1, 0), (0, -1)]
ddn = [(-1, 0), (-1, 1), (0, 1), (1, 1), (1, 0), (1, -1), (0, -1), (-1, -1)]
def ... | s670606496 | Accepted | 68 | 7,368 | 899 | import math, string, itertools, fractions, heapq, collections, re, array, bisect, sys, random, time, copy, functools
sys.setrecursionlimit(10**7)
inf = 10 ** 20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1, 0), (0, 1), (1, 0), (0, -1)]
ddn = [(-1, 0), (-1, 1), (0, 1), (1, 1), (1, 0), (1, -1), (0, -1), (-1, -1)]
def ... |
s582306716 | p03470 | u080108339 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 153 | An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you h... | N = int(input())
a = sorted(map(int, [input() for i in range(N)]), reverse=True)
dan = 0
for i in range(N-1):
if a[i] > a[i+1]:
dan += 1
print(dan) | s233736236 | Accepted | 19 | 3,060 | 153 | N = int(input())
a = sorted(map(int, [input() for i in range(N)]), reverse=True)
dan = 1
for i in range(N-1):
if a[i] > a[i+1]:
dan += 1
print(dan) |
s017258442 | p03861 | u017050982 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 210 | 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 = map(int,input().rstrip().split(" "))
if b - a + 1 >= x:
if a % x != 0:
a += x - (a % x)
b -= (b % x)
print((b - a) / x + 1)
else:
if a + x - (a % x) > b:
print(0)
else:
print(1)
| s506613506 | Accepted | 17 | 2,940 | 101 | a,b,x = map(int,input().rstrip().split(" "))
p = b // x - a // x
if a % x == 0:
p += 1
print(p)
|
s987430516 | p03486 | u920103253 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 198 | 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. | s=input()
t=input()
s=sorted([x for x in s],reverse=True)
t=sorted([x for x in t],reverse=True)
f="No"
for i in range(min(len(s),len(t))):
if s[i]<t[i]:
f="Yes"
break;
print(f) | s970417902 | Accepted | 19 | 2,940 | 96 | s=sorted(input())
t=sorted(input(),reverse=True)
if s<t:
print("Yes")
else:
print("No") |
s189152085 | p02283 | u510829608 | 2,000 | 131,072 | Wrong Answer | 30 | 7,796 | 1,322 | Search trees are data structures that support dynamic set operations including insert, search, delete and so on. Thus a search tree can be used both as a dictionary and as a priority queue. Binary search tree is one of fundamental search trees. The keys in a binary search tree are always stored in such a way as to sat... | class Node():
def __init__(self, key):
self.parent = None
self.left = None
self.right = None
self.key = key
class Tree():
def __init__(self):
self.root = None
def insert(self, key):
z = Node(key)
y = None
x = self.root
while x:
... | s559984409 | Accepted | 8,600 | 153,772 | 1,320 | class Node():
def __init__(self, key):
self.parent = None
self.left = None
self.right = None
self.key = key
class Tree():
def __init__(self):
self.root = None
def insert(self, key):
z = Node(key)
y = None
x = self.root
while x:
... |
s351326030 | p04043 | u556610039 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 296 | 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())
fiveCount = 0
sevenCount = 0
if a == 5: fiveCount+= 1
if b == 5: fiveCount+= 1
if c == 5: fiveCount+= 1
if a == 7: sevenCount+= 1
if b == 7: sevenCount+= 1
if c == 7: sevenCount+= 1
if fiveCount == 2 and sevenCount == 1:
print("Yes")
else:
print("No") | s925670960 | Accepted | 17 | 3,064 | 296 | a, b, c = map(int, input().split())
fiveCount = 0
sevenCount = 0
if a == 5: fiveCount+= 1
if b == 5: fiveCount+= 1
if c == 5: fiveCount+= 1
if a == 7: sevenCount+= 1
if b == 7: sevenCount+= 1
if c == 7: sevenCount+= 1
if fiveCount == 2 and sevenCount == 1:
print("YES")
else:
print("NO") |
s754533449 | p03816 | u760767494 | 2,000 | 262,144 | Wrong Answer | 2,105 | 93,260 | 706 | 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... | import collections
from operator import itemgetter
n = int(input())
l = list(map(int, input().split()))
l = collections.Counter(l)
l_1 = []
ans = 0
for i in l.items():
l_1.append(list(i))
l_1.sort(key=itemgetter(1), reverse=True)
print(l_1)
i = 0
for j in range(len(l_1)):
k = 1
print(l_1)
if l_1[i][1] ... | s200119145 | Accepted | 52 | 18,656 | 197 | import collections
n = int(input())
l = list(map(int, input().split()))
l = collections.Counter(l)
kind = len(l.keys())
if (n-kind)%2==0:
print(len(l.keys()))
else:
print(len(l.keys())-1)
|
s199024353 | p02401 | u928633434 | 1,000 | 131,072 | Wrong Answer | 20 | 5,492 | 8 | 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. | 34.1201
| s999562553 | Accepted | 20 | 5,600 | 244 | while True:
a,op,b = input().split()
if op == "?":
break
elif op == "+":
print (int(a) + int(b))
elif op == "-":
print (int(a) - int(b))
elif op == "*":
print (int(a) * int(b))
else:
print (int(int(a) / int(b)))
|
s708685704 | p02795 | u473023730 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 99 | 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())
a=max(h,w)
b=0
m=0
while b+a<=n:
b+=a
m+=1
print(m) | s419590371 | Accepted | 17 | 3,060 | 111 | h=int(input())
w=int(input())
n=int(input())
a=max(h,w)
b=0
m=0
if n%a==0:
print(n//a)
else:
print(n//a+1) |
s325615445 | p03370 | u513081876 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 110 | 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())
m = [int(i) for i in input().split()]
m.sort()
print(N + (X - sum(m)) % m[0]) | s674899328 | Accepted | 18 | 2,940 | 110 | N, X = map(int, input().split())
m = sorted([int(input()) for i in range(N)])
print(N + (X - sum(m)) // m[0]) |
s566951392 | p02412 | u663910047 | 1,000 | 131,072 | Wrong Answer | 20 | 7,668 | 382 | Write a program which identifies the number of combinations of three integers which satisfy the following conditions: * You should select three distinct integers from 1 to n. * A total sum of the three integers is x. For example, there are two combinations for n = 5 and x = 9. * 1 + 3 + 5 = 9 * 2 + 3 + 4 = 9 | x=1
y=1
list=[]
while x>0:
if x ==0 and y == 0:
break
x,y = map(int,input().split())
list.append([x,y])
for L in list:
ans = 0
for i in range(1,L[0]-1):
for j in range(i+1,L[0]):
for k in range(j+1,L[0]+1):
if L[1]== i + j+ k:
ans += 1... | s678487630 | Accepted | 620 | 7,660 | 402 | x=1
y=1
list=[]
while x>0:
if x ==0 and y == 0:
break
x,y = map(int,input().split())
list.append([x,y])
list.remove([0,0])
for L in list:
ans = 0
for i in range(1,L[0]-1):
for j in range(i+1,L[0]):
for k in range(j+1,L[0]+1):
if L[1]== i + j+ k:
... |
s094651649 | p03657 | u320763652 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 101 | 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+b % 3 == 0:
print("Possible")
else:
print("Impossible") | s819487529 | Accepted | 17 | 2,940 | 131 | a,b = map(int, input().split())
if a % 3 == 0 or b % 3 == 0 or (a+b) % 3 == 0:
print("Possible")
else:
print("Impossible") |
s062622341 | p03695 | u519923151 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 267 | In AtCoder, a person who has participated in a contest receives a _color_ , which corresponds to the person's rating as follows: * Rating 1-399 : gray * Rating 400-799 : brown * Rating 800-1199 : green * Rating 1200-1599 : cyan * Rating 1600-1999 : blue * Rating 2000-2399 : yellow * Rating 2400-2799 : or... | n = int(input())
al = list(map(int, input().split()))
ald = list(map(lambda x : x // 400,al))
allow = list(filter(lambda x: x<8, ald))
reslow = len(set(allow))
alhigh= list(filter(lambda x: x>=8,ald))
reshigh = len(alhigh)
res = min((reslow+reshigh),8)
print(res) | s863528742 | Accepted | 18 | 3,064 | 286 | n = int(input())
al = list(map(int, input().split()))
ald = list(map(lambda x : x // 400,al))
allow = list(filter(lambda x: x<8, ald))
reslow = len(set(allow))
alhigh= list(filter(lambda x: x>=8,ald))
reshigh = len(alhigh)
resl = max(reslow,1)
resh = reslow+reshigh
print(resl,resh) |
s393023657 | p03501 | u112567325 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 80 | You are parking at a parking lot. You can choose from the following two fee plans: * Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours. * Plan 2: The fee will be B yen, regardless of the duration. Find the minimum fee when you park for N hours. | A,B,C = list(map(int,input().split()))
if A*B > C:
print(A*B)
else:
print(C) | s814569470 | Accepted | 17 | 2,940 | 80 | A,B,C = list(map(int,input().split()))
if A*B < C:
print(A*B)
else:
print(C) |
s337806345 | p03494 | u431981421 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 222 | There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform. | li = list(map(int,input().split()))
count = 0
while(True):
for i in li:
if i % 2 == 1:
break
else:
count += 1
for index, x in enumerate(li):
li[index] = x / 2
continue
break
print(count) | s001411693 | Accepted | 19 | 3,060 | 237 | a=int(input())
li = list(map(int,input().split()))
count = 0
while(True):
for i in li:
if i % 2 == 1:
break
else:
count += 1
for index, x in enumerate(li):
li[index] = x / 2
continue
break
print(count) |
s664915361 | p03228 | u968097972 | 2,000 | 1,048,576 | Wrong Answer | 20 | 3,188 | 411 | In the beginning, Takahashi has A cookies, and Aoki has B cookies. They will perform the following operation alternately, starting from Takahashi: * If the number of cookies in his hand is odd, eat one of those cookies; if the number is even, do nothing. Then, give one-half of the cookies in his hand to the other pe... | S = input().split()
A = int(S[0])
B = int(S[1])
K = int(S[2])
print(A, B, K)
for k in range(K):
if k % 2 == 0: #taka
if A % 2 == 1:
A -= 1
B_plus = A//2
A -= B_plus
B += B_plus
else:
if B % 2 == 1:
B -= 1
A_plus = B//2
... | s233802577 | Accepted | 17 | 3,060 | 412 | S = input().split()
A = int(S[0])
B = int(S[1])
K = int(S[2])
for k in range(K):
if k % 2 == 0: #taka
if A % 2 == 1:
A -= 1
B_plus = A//2
A -= B_plus
B += B_plus
else:
if B % 2 == 1:
B -= 1
A_plus = B//2
B -= A_p... |
s922506362 | p03693 | u539768961 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 96 | 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... | r, g, b = map(int, input().split())
if (10 * g + b) % 4 == 0:
print('Yes')
else:
print('No') | s952131486 | Accepted | 17 | 2,940 | 96 | r, g, b = map(int, input().split())
if (10 * g + b) % 4 == 0:
print('YES')
else:
print('NO') |
s462350255 | p03161 | u766393261 | 2,000 | 1,048,576 | Wrong Answer | 1,933 | 24,596 | 290 | 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 one of the following: Stone i + 1, i + 2, \... | import numpy as np
n,k=map(int,input().split())
h=np.array(list(map(int,input().split())))
dp=np.zeros(n,dtype=int)
dp[1]=abs(h[1]-h[0])
for i in range(2,n):
if i-k>0:
dp[i]=min(dp[i-k:i]+abs(h[i]-h[i-k:i]))
elif 0>=i-k:
dp[i]=min(dp[0:i]+abs(h[i]-h[0:i]))
print(dp) | s200014729 | Accepted | 1,887 | 22,828 | 295 | import numpy as np
n,k=map(int,input().split())
h=np.array(list(map(int,input().split())))
dp=np.zeros(n,dtype=int)
dp[1]=abs(h[1]-h[0])
for i in range(2,n):
if i-k>0:
dp[i]=min(dp[i-k:i]+abs(h[i]-h[i-k:i]))
elif 0>=i-k:
dp[i]=min(dp[0:i]+abs(h[i]-h[0:i]))
print(dp[n-1]) |
s855496566 | p02741 | u511870776 | 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 | s140672514 | Accepted | 17 | 2,940 | 135 | l = [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(l[k-1])
|
s815035618 | p02612 | u284363684 | 2,000 | 1,048,576 | Wrong Answer | 28 | 9,088 | 49 | 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. | # input
N = int(input())
print(N % 1000) | s449280573 | Accepted | 27 | 9,052 | 136 | # input
N = int(input())
noguchi = [1000 * n for n in range(1, 11)]
print(min([ngc - N for ngc in noguchi if (ngc - N) >= 0]))
|
s129119391 | p02694 | u366939485 | 2,000 | 1,048,576 | Wrong Answer | 29 | 9,168 | 135 | 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 abo... | x = int(input())
balance = 100
count = 0
while x >= balance:
balance *= 1.01
balance = int(balance)
count += 1
print(count) | s956164071 | Accepted | 34 | 9,092 | 117 | x = int(input())
balance = 100
count = 0
while balance < x:
balance += balance // 100
count += 1
print(count) |
s027965837 | p04043 | u978494963 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 137 | 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 ... | hoge = tuple(map(lambda x : int(x), input().split(" ")))
if hoge.count(5) == 2 and hoge.count(7) == 1:
print("Yes")
else:
print("No")
| s430116958 | Accepted | 17 | 2,940 | 137 | hoge = tuple(map(lambda x : int(x), input().split(" ")))
if hoge.count(5) == 2 and hoge.count(7) == 1:
print("YES")
else:
print("NO")
|
s785233694 | p02612 | u548123110 | 2,000 | 1,048,576 | Wrong Answer | 30 | 9,140 | 93 | 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. | def main():
n = int(input())
print(n % 1000)
if __name__ == "__main__":
main()
| s499400809 | Accepted | 37 | 9,148 | 158 | def main():
n = int(input())
n = n % 1000
if n == 0:
print(0)
else:
print(1000 - n)
if __name__ == "__main__":
main()
|
s332519947 | p04012 | u535659144 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 176 | Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful. | strlist = list(input())
findlist="qazwsxedcrfvtgbyhnujmikolp"
flist=list(findlist)
flag = "Yes"
for a in flist:
if not strlist.count(a) % 2:
flag = "No"
print(flag) | s169753741 | Accepted | 17 | 3,060 | 239 | li=list(input())
li.sort()
if len(li)%2:
print("No")
exit()
while True:
if len(li)==0:
print("Yes")
break
elif li[0]==li[1]:
li.pop(0)
li.pop(0)
else:
print("No")
break |
s656459841 | p02608 | u381959472 | 2,000 | 1,048,576 | Wrong Answer | 2,205 | 9,128 | 435 | 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())
count = 0
for n in range(1, N):
sqrt_n = int(math.sqrt(n))
for i in range(1, sqrt_n):
for p in range(1, sqrt_n):
for q in range(1, sqrt_n):
if i ** 2 + p ** 2 + q ** 2 + i * p + p * q + i * q > n:
break
elif i... | s217370524 | Accepted | 448 | 9,404 | 449 | import math
N = int(input())
NList = [0 for i in range(N)]
count = 0
tmp = 0
sqrt_n = int(math.sqrt(N))
for i in range(1, sqrt_n):
for p in range(1, sqrt_n):
for q in range(1, sqrt_n):
if i ** 2 + p ** 2 + q ** 2 + i * p + p * q + i * q > N:
break
else:
... |
s957498843 | p02612 | u247753603 | 2,000 | 1,048,576 | Wrong Answer | 27 | 9,152 | 783 | 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. |
def f(x):
if 0 <= x <=999:
print(1000-x)
else:
b = int(x / 100)
c = int(str(b)[1])
g = 9-c
h = int(x/10)
i = int(str(h)[2])
if x %1000==0:
print("0")
elif x % 500==0 and x % 1000!=0:
print("500")
elif x % 100=... | s948594124 | Accepted | 29 | 9,108 | 652 |
def f(y):
if 1 <= y <=999:
print(1000-y)
elif 1000<= y <= 10000 :
b = int(y / 100)
c = int(str(b)[1])
g = 9-c
h = int(y/10)
i = int(str(h)[2])
if y %1000==0:
print("0")
elif y % 500==0 and y % 1000!=0:
print("500")
... |
s959161387 | p02578 | u977982384 | 2,000 | 1,048,576 | Wrong Answer | 265 | 32,208 | 256 | 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())
a = list(map(int, input().split( )))
front = 0
step = 0
for height in a :
x = 0
if front > height :
x = front - height
step += x
print(front, height, x)
if front < height:
front = height
print(step)
| s476745628 | Accepted | 114 | 32,348 | 228 | n = int(input())
a = list(map(int, input().split( )))
front = 0
step = 0
for height in a :
x = 0
if front > height :
x = front - height
step += x
if front < height:
front = height
print(step)
|
s340246340 | p03910 | u857428111 | 2,000 | 262,144 | Wrong Answer | 21 | 3,408 | 144 | The problem set at _CODE FESTIVAL 20XX Finals_ consists of N problems. The score allocated to the i-th (1≦i≦N) problem is i points. Takahashi, a contestant, is trying to score exactly N points. For that, he is deciding which problems to solve. As problems with higher scores are harder, he wants to minimize the highe... | import math
N=int(input())
p=math.sqrt(N*2-0.25)-0.5
q=int(p)+1
M=q*(q+1)//2
L=M-N#must0or+
for i in range(q+1):
if i != L:
print(i) | s549751525 | Accepted | 22 | 3,408 | 456 | import sys
input= lambda: sys.stdin.readline().rstrip()
def pin(type=int):
return map(type,input().split())
#%%code
def resolve():
x=int(input())
i=0;temp=0
while (1):
i+=1
temp+=i
if temp > x:
z=-x+temp
for i in range(i):
if... |
s717450059 | p04011 | u870841038 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 141 | There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee. | a = [int(input()) for i in range(4)]
if a[1] > a[2]:
ans = a[1] * a[2] + (a[0]-a[1]) * a[3]
else:
ans = a[0] * a[2]
print(str(ans)) | s020560051 | Accepted | 17 | 3,060 | 148 | li = list(int(input()) for i in range(4))
if li[0] > li[1]:
ans = li[1]*li[2] + (li[0]-li[1])*li[3]
else:
ans = li[0]*li[2]
print(str(ans)) |
s277064282 | p03549 | u536034761 | 2,000 | 262,144 | Wrong Answer | 29 | 9,144 | 58 | Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`. When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases. Then, he rewrote the code to correctly solve each of th... | N, M = map(int, input().split())
print(100 * N + 1900 * M) | s763257787 | Accepted | 26 | 9,132 | 75 | N, M = map(int, input().split())
print((100 * (N - M) + 1900 * M) * (2**M)) |
s723160476 | p03048 | u036914910 | 2,000 | 1,048,576 | Wrong Answer | 2,104 | 18,584 | 213 | 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 = map(int,input().split())
ans = 0
for i in range(n+1) :
for j in range(n+1-i) :
k = n-r*i-g*j
if k >= 0 and k % b == 0 :
print(i,j,k,'Yes')
ans += 1
print(ans) | s746613108 | Accepted | 1,826 | 2,940 | 184 | R,G,B,n = map(int,input().split())
ans = 0
for r in range(n+1) :
for g in range(n+1-r*R) :
b = n-r*R-g*G
if b >= 0 and b % B == 0 :
ans += 1
print(ans) |
s817067053 | p03797 | u578953945 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 148 | Snuke loves puzzles. Today, he is working on a puzzle using `S`\- and `c`-shaped pieces. In this puzzle, you can combine two `c`-shaped pieces into one `S`-shaped piece, as shown in the figure below: Snuke decided to create as many `Scc` groups as possible by putting together one `S`-shaped piece and two `c`-shaped... | S,C=map(int,input().split())
ANS = 0
if S * 2 <= C:
ANS = S
C -= (S*2)
print(C)
ANS += (C//4)
print(ANS)
else:
ANS = C // 2
print(ANS) | s855705783 | Accepted | 17 | 2,940 | 148 | S,C=map(int,input().split())
ANS = 0
if S * 2 <= C:
ANS = S
C -= (S*2)
#rint(C)
ANS += (C//4)
print(ANS)
else:
ANS = C // 2
print(ANS) |
s617998388 | p04043 | u957872856 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 203 | 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 = list(map(int, input().split()))
d = 0
e = 0
for i in range(3):
f = a[i]
if f == 7:
d += 1
elif f == 5:
e += 1
else:
break
if d == 2 and e == 1:
print("YES")
else:
print("NO")
| s015372961 | Accepted | 17 | 2,940 | 81 | n = input().split()
print("YES" if n.count("5")==2 and n.count("7")==1 else "NO") |
s703364921 | p03997 | u506587641 | 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 = list(int(input()) for _ in range(3))
print((a[0]+a[1])*a[2]/2) | s827248487 | Accepted | 17 | 3,064 | 67 | a = list(int(input()) for _ in range(3))
print((a[0]+a[1])*a[2]//2) |
s746669286 | p03836 | u755180064 | 2,000 | 262,144 | Wrong Answer | 27 | 9,160 | 567 | 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 ... | def search_route(f, t):
route = ''
y = t[1] - f[1]
tmp_y = 'U'*y if y >= 0 else 'D'*(y*-1)
x = t[0] - f[0]
tmp_x = 'R'*x if x >= 0 else 'L'*(x*-1)
route = tmp_y + tmp_x
return route
def main():
t = list(map(int, input().split()))
route = search_route(t[:2], t[2:]) + search_route(t[... | s064308158 | Accepted | 29 | 9,168 | 658 |
url = "https://atcoder.jp//contests/abc051/tasks/abc051_c"
def search_route(f, t):
route = ''
y = t[1] - f[1]
tmp_y = 'U'*y if y >= 0 else 'D'*(y*-1)
x = t[0] - f[0]
tmp_x = 'R'*x if x >= 0 else 'L'*(x*-1)
route = tmp_y + tmp_x
return route
def main():
t = list(map(int, input().split(... |
s299284023 | p02389 | u915307213 | 1,000 | 131,072 | Wrong Answer | 20 | 5,536 | 127 | Write a program which calculates the area and perimeter of a given rectangle. | def xxx():
for i in range(5):
print(i, end=' ')
print(id(i))
for i in range(3):
xxx()
print(id(i))
| s882389923 | Accepted | 20 | 5,592 | 142 | def rect(a,b):
return a*b, 2*(a+b)
n =input()
x =n.split()
a =int(x[0])
b =int(x[1])
area, perimeter = rect(a,b)
print(area, perimeter)
|
s438634120 | p03813 | u111365362 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 49 | Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise. | n = int(input())
print( n//11*2 + (n%11)//6 + 1 ) | s575245561 | Accepted | 17 | 2,940 | 66 | #23:53
if int(input()) < 1200:
print('ABC')
else:
print('ARC') |
s067829532 | p03378 | u934246119 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 489 | There are N + 1 squares arranged in a row, numbered 0, 1, ..., N from left to right. Initially, you are in Square X. You can freely travel between adjacent squares. Your goal is to reach Square 0 or Square N. However, for each i = 1, 2, ..., M, there is a toll gate in Square A_i, and traveling to Square A_i incurs a c... | n, m, x = map(int, input().split())
a_tmp = input().split()
a = []
for _ in a_tmp:
a.append(int(_))
cost_l = 0
cost_r = 0
j = 0
for i in range(n+1):
if i < x:
if i == a[j]:
cost_l += 1
j += 1
if j >= len(a):
break
elif x < i:
if i == a[j]:
... | s455723578 | Accepted | 17 | 3,064 | 467 | n, m, x = map(int, input().split())
a_tmp = input().split()
a = []
for _ in a_tmp:
a.append(int(_))
cost_l = 0
cost_r = 0
j = 0
for i in range(n+1):
if i < x:
if i == a[j]:
cost_l += 1
j += 1
if j >= len(a):
break
elif x < i:
if i == a[j]:
... |
s774526980 | p03433 | u089075077 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 84 | 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('NO')
else:
print('YES') | s625042916 | Accepted | 18 | 3,064 | 85 | N = int(input())
A = int(input())
if N % 500 > A:
print('No')
else:
print('Yes')
|
s227463562 | p03854 | u219369949 | 2,000 | 262,144 | Wrong Answer | 20 | 3,316 | 150 | 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`. | N = input()
if N.replace('eraser', '').replace('erase', '').replace('dreamer', '').replace('dream', '') == '':
print('Yes')
else:
print('No') | s829692551 | Accepted | 18 | 3,188 | 150 | N = input()
if N.replace('eraser', '').replace('erase', '').replace('dreamer', '').replace('dream', '') == '':
print('YES')
else:
print('NO') |
s356884451 | p03401 | u022215787 | 2,000 | 262,144 | Wrong Answer | 209 | 14,048 | 253 | 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()))
A = [0] + A + [0]
total = sum([ abs(A[x]-A[x+1]) for x in range(N) ])
for i in range(N):
minus = abs(A[i] - A[i+1]) + abs(A[i+1] - A[i+2])
plus = abs(A[i] - A[i+2])
print(total-minus+plus) | s466864087 | Accepted | 219 | 14,044 | 255 | N = int(input())
A = list(map(int, input().split()))
A = [0] + A + [0]
total = sum([ abs(A[x]-A[x+1]) for x in range(N+1) ])
for i in range(N):
minus = abs(A[i] - A[i+1]) + abs(A[i+1] - A[i+2])
plus = abs(A[i] - A[i+2])
print(total-minus+plus) |
s826404795 | p03089 | u639094382 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,060 | 179 | Snuke has an empty sequence a. He will perform N operations on this sequence. In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1). You are given a sequence b of length N. Determine if it is possible that a is equal to b after N oper... | n = int(input())
b = list(map(int, input().split()))
can = True
for i in range(1, n +1):
if b[i - 1] > i:
can = False
if can:
for x in b:
print(x)
else:
print(-1) | s121031078 | Accepted | 18 | 3,064 | 347 | n = int(input())
b = list(map(int, input().split()))
can = True
for i in range(1, n +1):
if b[i - 1] > i:
can = False
break
if can:
result = []
while(len(b)):
for i in range(len(b), 0, -1):
if i == b[i - 1]:
result.insert(0, i)
b.pop(i - 1)
break
for a in result:
... |
s573082886 | p03679 | u474925961 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 190 | Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Ot... | import sys
if sys.platform =='ios':
sys.stdin=open('input_file.txt')
x,a,b=map(int,input().split())
if b<a:
print("delicious")
elif a<b:
print("dangerous")
else:
print("safe") | s548620375 | Accepted | 17 | 3,060 | 193 | import sys
if sys.platform =='ios':
sys.stdin=open('input_file.txt')
x,a,b=map(int,input().split())
if b<=a:
print("delicious")
elif a+x<b:
print("dangerous")
else:
print("safe") |
s221199590 | p03377 | u272557899 | 2,000 | 262,144 | Wrong Answer | 19 | 2,940 | 94 | 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 a + b >= x:
print("Yes")
else:
print("No") | s459038896 | Accepted | 17 | 2,940 | 95 | a,b,x= map(int, input().split())
if a <= x and a + b >= x:
print("YES")
else:
print("NO")
|
s846336302 | p03698 | u318427318 | 2,000 | 262,144 | Wrong Answer | 28 | 9,120 | 330 | You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different. | #-*-coding:utf-8-*-
import sys
input=sys.stdin.readline
def main():
s = input().rstrip()
string_dict={}
for i in s:
if i not in string_dict:
d = {i:1}
string_dict.update(d)
else:
print("No")
exit()
print("Yes")
if __name__=="__main__":
... | s839465046 | Accepted | 28 | 9,084 | 330 | #-*-coding:utf-8-*-
import sys
input=sys.stdin.readline
def main():
s = input().rstrip()
string_dict={}
for i in s:
if i not in string_dict:
d = {i:1}
string_dict.update(d)
else:
print("no")
exit()
print("yes")
if __name__=="__main__":
... |
s848394896 | p03854 | u869919400 | 2,000 | 262,144 | Wrong Answer | 17 | 3,188 | 229 | 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()
words = ['dream', 'dreamer', 'erase', 'eraser']
while s != '':
tmp = s
for word in words:
s.rstrip(word)
if s == '':
print('YES')
if tmp == s:
print('NO')
break | s377174830 | Accepted | 19 | 3,188 | 253 | s = input()
while s != '':
tmp = s
s = s.replace('eraser', '')
s = s.replace('erase', '')
s = s.replace('dreamer', '')
s = s.replace('dream', '')
if s == '':
print('YES')
if tmp == s:
print('NO')
break |
s129810302 | p03681 | u088863512 | 2,000 | 262,144 | Wrong Answer | 58 | 9,812 | 891 | Snuke has N dogs and M monkeys. He wants them to line up in a row. As a Japanese saying goes, these dogs and monkeys are on bad terms. _("ken'en no naka", literally "the relationship of dogs and monkeys", means a relationship of mutual hatred.)_ Snuke is trying to reconsile them, by arranging the animals so that there... | # -*- coding: utf-8 -*-
import sys
from collections import deque, defaultdict
from math import sqrt, factorial, gcd, ceil
# def input(): return sys.stdin.readline()[:-1] # warning not \n
import string
from bisect import bisect_left
MOD = int(1e9)+7
INF = float('inf')
def fmod(x):
ans = 1
for i in range(1, x... | s162690569 | Accepted | 61 | 9,812 | 891 | # -*- coding: utf-8 -*-
import sys
from collections import deque, defaultdict
from math import sqrt, factorial, gcd, ceil
# def input(): return sys.stdin.readline()[:-1] # warning not \n
import string
from bisect import bisect_left
MOD = int(1e9)+7
INF = float('inf')
def fmod(x):
ans = 1
for i in range(1, x... |
s202941963 | p03195 | u826263061 | 2,000 | 1,048,576 | Wrong Answer | 189 | 7,072 | 164 | There is an apple tree that bears apples of N colors. The N colors of these apples are numbered 1 to N, and there are a_i apples of Color i. You and Lunlun the dachshund alternately perform the following operation (starting from you): * Choose one or more apples from the tree and eat them. Here, the apples chosen a... | n = int(input())
a = []
for _ in range(n):
a.append(int(input()))
mina = min(a)
r = sum(a) - mina*n
if r % 2 == 1:
print('First')
else:
print('Second') | s074776258 | Accepted | 193 | 3,060 | 160 | n = int(input())
ans = False
for _ in range(n):
i = int(input())
if i % 2 == 1:
ans = True
break
if ans:
print('first')
else:
print('second')
|
s366133814 | p03048 | u389679466 | 2,000 | 1,048,576 | Wrong Answer | 22 | 3,060 | 164 | 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... | input = input().split()
R = int(input[0])
G = int(input[1])
B = int(input[2])
N = int(input[3])
count = 0
for r in range(N+1):
count += 2**(N-r)
print(count) | s140555539 | Accepted | 1,556 | 3,060 | 352 | import math
input = input().split()
r = int(input[0])
g = int(input[1])
b = int(input[2])
N = int(input[3])
count = 0
# print(N/r)
for x in range(math.ceil(N/r+0.0001)):
for y in range(math.ceil((N-x*r)/g + 0.0001)):
# print(str(x), str(y))
if (N - x*r - y*g) % b == 0:
count +=1
prin... |
s184705259 | p03719 | u758973277 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 86 | You are given three integers A, B and C. Determine whether C is not less than A and not greater than B. | A,B,C = map(int,input().split())
if C<=A and C>=B:
print('Yes')
else:
print('No') | s803640261 | Accepted | 17 | 2,940 | 86 | A,B,C = map(int,input().split())
if C>=A and C<=B:
print('Yes')
else:
print('No') |
s040067156 | p04029 | u846150137 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 52 | 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) + ((n+1)/2)*(n%2)) | s548220591 | Accepted | 17 | 2,940 | 53 | n=int(input())
print((n+1)*(n//2) + ((n+1)//2)*(n%2)) |
s507240328 | p03712 | u870518235 | 2,000 | 262,144 | Wrong Answer | 28 | 9,156 | 148 | You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1. | H, W = map(int, input().split())
A = ["#"+str(input())+"#" for _ in range(H)]
res = "#" * W
print(res)
for i in range(H):
print(A[i])
print(res) | s377431941 | Accepted | 30 | 9,168 | 152 | H, W = map(int, input().split())
A = ["#"+str(input())+"#" for _ in range(H)]
res = "#" * (W+2)
print(res)
for i in range(H):
print(A[i])
print(res) |
s004985065 | p03160 | u024768467 | 2,000 | 1,048,576 | Wrong Answer | 130 | 13,928 | 342 | 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())
h_list = list(map(int,input().split()))
cost_min_list = [0] * n
cost_min_list[0] = 0
cost_min_list[0] = h_list[0]
for i in range(2, n):
cost_min_list[i] = min(cost_min_list[i - 1] + abs(h_list[i] - h_list[i - 1]),
cost_min_list[i - 2] + abs(h_list[i] - h_list[i - 2]))
pri... | s183251185 | Accepted | 134 | 13,980 | 361 | n = int(input())
h_list = list(map(int,input().split()))
cost_min_list = [0] * n
cost_min_list[0] = 0
cost_min_list[1] = abs(h_list[1] - h_list[0])
for i in range(2, n):
cost_min_list[i] = min(cost_min_list[i - 1] + abs(h_list[i] - h_list[i - 1]),
cost_min_list[i - 2] + abs(h_list[i] - h_... |
s981912049 | p03377 | u262244504 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 120 | 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())
dc = x - a
if dc < 0 :
print('No')
elif dc <= b:
print('Yes')
else:
print('No') | s485242198 | Accepted | 17 | 2,940 | 109 | a,b,x = map(int,input().split())
if a > x :
print('NO')
elif x-a <= b:
print('YES')
else:
print('NO') |
s075783693 | p00352 | u529013669 | 1,000 | 262,144 | Wrong Answer | 40 | 7,636 | 38 | Alice and Brown are brothers in a family and each receives pocket money in celebration of the coming year. They are very close and share the total amount of the money fifty-fifty. The pocket money each receives is a multiple of 1,000 yen. Write a program to calculate each one’s share given the amount of money Alice an... | print(sum(map(int,input().split()))/2) | s564979313 | Accepted | 30 | 7,688 | 43 | print(int(sum(map(int,input().split()))/2)) |
s238387174 | p03597 | u365364616 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 46 | 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? | n = int(input())
a = int(input())
print(n - a) | s027900784 | Accepted | 18 | 2,940 | 50 | n = int(input())
a = int(input())
print(n * n - a) |
s626710009 | p03352 | u454524105 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 143 | 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())
ans = 1
for i in range(100):
for j in range(100):
if x < i**j: break
else: ans = max(ans, i**j)
print(ans) | s474309697 | Accepted | 19 | 2,940 | 142 | x = int(input())
ans = 1
for i in range(x):
for j in range(2, x):
if x < i**j: break
else: ans = max(ans, i**j)
print(ans) |
s788214109 | p02669 | u164727245 | 2,000 | 1,048,576 | Time Limit Exceeded | 2,205 | 9,224 | 824 | You start with the number 0 and you want to reach the number N. You can change the number, paying a certain amount of coins, with the following operations: * Multiply the number by 2, paying A coins. * Multiply the number by 3, paying B coins. * Multiply the number by 5, paying C coins. * Increase or decrease... | # coding: utf-8
def solve(*args: str) -> str:
t = int(args[0])
NABCD = [tuple(map(int, nabcd.split())) for nabcd in args[1:]]
ret = []
for n, a, b, c, d in NABCD:
ABC = sorted([(a, 2), (b, 3), (c, 5)],
key=lambda x: x[1]//x[0], reverse=True)
ans = n*d
stac... | s846414914 | Accepted | 1,180 | 10,960 | 928 | # coding: utf-8
from collections import defaultdict
def solve(*args: str) -> str:
t = int(args[0])
NABCD = [tuple(map(int, nabcd.split())) for nabcd in args[1:]]
ret = []
for n, a, b, c, d in NABCD:
ABC = ((2, a), (3, b), (5, c))
dp = defaultdict(lambda: n*d)
stack = [(n, 0)]
... |
s439801663 | p03860 | u640603056 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 58 | 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... | a, b, c = input().split()
s = "A"+b.upper() + "C"
print(s) | s244871629 | Accepted | 17 | 2,940 | 61 | a, b, c = input().split()
s = "A"+b[0].upper() + "C"
print(s) |
s303300620 | p02612 | u855636544 | 2,000 | 1,048,576 | Wrong Answer | 27 | 9,036 | 79 | 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. | import math
n = float(input())
p = n//1000
if n%1000!=0:
p+=1
print (p*1000-n) | s863342081 | Accepted | 29 | 9,096 | 85 | import math
n = float(input())
p = n//1000
if n%1000!=0:
p+=1
print (int(p*1000-n))
|
s271229620 | p00016 | u123687446 | 1,000 | 131,072 | Wrong Answer | 30 | 7,780 | 270 | When a boy was cleaning up after his grand father passing, he found an old paper: In addition, other side of the paper says that "go ahead a number of steps equivalent to the first integer, and turn clockwise by degrees equivalent to the second integer". His grand mother says that Sanbonmatsu was standing at t... | from math import sin,cos,pi
x = 0
y = 0
ca = pi/2
while True:
d, a = list(map(int, input().split(",")))
if d or a:
x += d*cos(ca)
y += d*sin(ca)
ca -= a*pi/180
else:
break
print("{0:.0f}".format(x))
print("{0:.0f}".format(y)) | s220697103 | Accepted | 30 | 7,848 | 249 | from math import sin,cos,pi
x = 0
y = 0
ca = 90
while True:
d, a = list(map(int, input().split(",")))
if d or a:
x += d*cos(ca*pi/180)
y += d*sin(ca*pi/180)
ca -= a
else:
break
print(int(x))
print(int(y)) |
s245294032 | p03417 | u836737505 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 83 | There is a grid with infinitely many rows and columns. In this grid, there is a rectangular region with consecutive N rows and M columns, and a card is placed in each square in this region. The front and back sides of these cards can be distinguished, and initially every card faces up. We will perform the following op... | n,m = map(int, input().split())
print(1 if n==2 or m==2 else max(n-2,1)*max(m-2,1)) | s620377414 | Accepted | 17 | 2,940 | 83 | n,m = map(int, input().split())
print(0 if n==2 or m==2 else max(n-2,1)*max(m-2,1)) |
s591494660 | p02606 | u938005055 | 2,000 | 1,048,576 | Wrong Answer | 25 | 9,012 | 133 | How many multiples of d are there among the integers between L and R (inclusive)? | L, R, d = list(map(int, input().split()))
count = 0
for i in range(L, R):
if i % d == 0:
count = count + 1
print(count) | s553237747 | Accepted | 29 | 9,008 | 131 | L, R, d = map(int, input().split())
count = 0
for i in range(L, R + 1):
if i % d == 0:
count = count + 1
print(count) |
s714319905 | p03448 | u849151695 | 2,000 | 262,144 | Wrong Answer | 58 | 2,940 | 310 | 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 co... | A,B,C,X = [int(input()) for i in range(4)]
ans = 0
for a in range(0,A):
for b in range(0,B):
for c in range(0,C):
total = 500 * a + 100 * b + 50 * c
if total == X:
ans +=1
print(ans)
| s128410342 | Accepted | 53 | 3,060 | 315 | A,B,C,X = [int(input()) for i in range(4)]
ans = 0
for a in range(0,A+1):
for b in range(0,B+1):
for c in range(0,C+1):
total = 500 * a + 100 * b + 50 * c
if total == X:
ans +=1
print(ans) |
s570892800 | p02613 | u079022693 | 2,000 | 1,048,576 | Wrong Answer | 55 | 9,180 | 633 | 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 sys import stdin
def main():
readline=stdin.readline
n=int(readline())
li=[0]*4
for i in range(n):
tmp=readline().strip()
if tmp=="AC":
li[0]+=1
elif tmp=="WA":
li[1]+=1
elif tmp=="TLE":
li[2]+=1
elif tmp=="RE":
... | s976312171 | Accepted | 55 | 9,244 | 653 | from sys import stdin
def main():
readline=stdin.readline
n=int(readline())
li=[0]*4
for i in range(n):
tmp=readline().strip()
if tmp=="AC":
li[0]+=1
elif tmp=="WA":
li[1]+=1
elif tmp=="TLE":
li[2]+=1
elif tmp=="RE":
... |
s374500874 | p03068 | u082861480 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 152 | You are given a string S of length N consisting of lowercase English letters, and an integer K. Print the string obtained by replacing every character in S that differs from the K-th character of S, with `*`. | n = int(input())
s = input()
k = int(input())
exa = s[k-1]
for i in s:
if i != exa:
print(i,exa)
s = s.replace(i,'*')
print(s) | s379097014 | Accepted | 18 | 2,940 | 131 | n = int(input())
s = input()
k = int(input())
exa = s[k-1]
for i in s:
if i != exa:
s = s.replace(i,'*')
print(s) |
s218055079 | p03816 | u210440747 | 2,000 | 262,144 | Wrong Answer | 2,103 | 14,388 | 297 | 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... | if __name__=="__main__":
n = int(input())
As = list(map(int,input().split()))
Non_dup_As = list(set(As))
count = 0
for a in Non_dup_As:
count+=As.count(a)
if(count % 2 == 0):
print(len(As) - int(count/2))
else:
print(len(As) - int((count+1)/2))
| s312770206 | Accepted | 64 | 14,204 | 225 | if __name__=="__main__":
n = int(input())
As = list(map(int,input().split()))
Non_dup_As = list(set(As))
count = len(Non_dup_As)
if(count % 2 == 0):
print(count - 1)
else:
print(count)
|
s523785691 | p03150 | u698919163 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 192 | 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. | S = input()
import sys
for i in range(0,len(S)-1):
for j in range(i+1,len(S)):
if S[:i] + S[j:] == 'keyence':
print('Yes')
sys.exit()
print('No') | s493628167 | Accepted | 18 | 2,940 | 188 | S = input()
import sys
for i in range(0,len(S)):
for j in range(i,len(S)):
if S[:i] + S[j:] == 'keyence':
print('YES')
sys.exit()
print('NO') |
s078502432 | p03737 | u629560745 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 86 | 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. | a, b, c = map(str, input().split())
result = a[0]+b[0]+c[0]
result.upper
print(result) | s504845660 | Accepted | 17 | 2,940 | 81 | a, b, c = map(str, input().split())
result = a[0]+b[0]+c[0]
print(result.upper()) |
s940390538 | p03574 | u273010357 | 2,000 | 262,144 | Wrong Answer | 28 | 3,444 | 610 | 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 = [input() for _ in range(H)]
answer = [[0 if cell == '.' else '#' for cell in row] for row in S]
dxy = [(-1,1),(-1,0),(-1,-1),(0,1),(0,-1),(1,1),(1,0),(1,-1)]
for i in range(H):
for j in range(W):
if S[i][j] != '#':
continue
for dx, dy in dxy:
... | s718924391 | Accepted | 28 | 3,444 | 609 | H,W = map(int,input().split())
S = [input() for _ in range(H)]
answer = [[0 if cell == '.' else '#' for cell in row] for row in S]
dxy = [(-1,1),(-1,0),(-1,-1),(0,1),(0,-1),(1,1),(1,0),(1,-1)]
for i in range(H):
for j in range(W):
if S[i][j] != '#':
continue
for dx, dy in dxy:
... |
s213170580 | p02238 | u424720817 | 1,000 | 131,072 | Wrong Answer | 20 | 5,604 | 532 | Depth-first search (DFS) follows the strategy to search ”deeper” in the graph whenever possible. In DFS, edges are recursively explored out of the most recently discovered vertex $v$ that still has unexplored edges leaving it. When all of $v$'s edges have been explored, the search ”backtracks” to explore edges leaving ... | n = int(input())
graph = [[0 for i in range(n)] for j in range(n)]
d = [0] * n
f = [0] * n
g = [0] * n
time = 0
for i in range(n):
a = list(map(int, input().split()))
for j in range(0, a[1], 1):
graph[a[0] - 1][a[2 + j] - 1] = 1
def search(i):
global time
time = time + 1
d[i] = time
g[i]... | s615475387 | Accepted | 20 | 5,808 | 550 | n = int(input())
graph = [[0 for i in range(n)] for j in range(n)]
d = [0] * n
f = [0] * n
g = [0] * n
time = 0
for i in range(n):
a = list(map(int, input().split()))
for j in range(0, a[1], 1):
graph[a[0] - 1][a[2 + j] - 1] = 1
def search(i):
global time
time = time + 1
d[i] = time
g[i]... |
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