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 chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces. Find the number of chocolate pieces prepared at the beginning of the camp. | n = int(input())
d, x = [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 K operations? | 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 multiplication of the scores in the last K exams, including the exam in the graded term. M-kun scored A_i in the exam at the end of the i-th term. For each i such that K+1 \leq i \leq N, determine whether his grade for the i-th term is **strictly** greater than the grade for the (i-1)-th term. | 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 uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name. | 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 directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate. | 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 of the phrases once, in some order. | 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 sequence with length k, modulo 10^9+7. | 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 = y*(n+1-q-k+3)//(k-1)
| 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)%m)
x = x*(n+1-k)*inv[k+1]%m
y = y*(n+2-q-k)*inv[k]%m
|
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 operations. If it is, show one possible sequence of operations that achieves it. | # 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 column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it. | 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.readline().split()))
inf = float('inf')
mod = 10**9 + 7
s = rr()
print(s.replace('8', '7', 1))
| 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.readline().split()))
inf = float('inf')
mod = 10**9 + 7
s = rr()
print(s.replace('7', '8', 1))
|
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 b_i points of happiness. * C: Do homework at home. Gain c_i points of happiness. As Taro gets bored easily, he cannot do the same activities for two or more consecutive days. Find the maximum possible total points of happiness that Taro gains. | 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:
print(i, j, k, dp[i][j], hpns[i+1][k])
dp[i+1][j] = max(dp[i][k] + hpns[i+1][j], dp[i+1][j])
solve()
ans = 0
for i in range(3):
ans = max(ans, dp[N-1][i])
print(ans)
| 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])
dp[i+1][2] = hpns[i+1][2] + max(dp[i][0], dp[i][1])
solve()
ans = 0
for i in range(3):
ans = max(ans, dp[N-1][i])
print(ans)
|
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 of the phrases once, in some order. | 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, j}= `#`, and to make Square (i, j) white when s_{i, j}= `.`. However, since he is not a good painter, he can only choose two squares that are horizontally or vertically adjacent and paint those squares black, for some number of times (possibly zero). He may choose squares that are already painted black, in which case the color of those squares remain black. Determine if square1001 can achieve his objective. | 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] == '.' else 0
checker.append(num)
if i != H - 1:
num = 1 if lst[i + 1][j] == '.' else 0
checker.append(num)
if j != 0:
num = 1 if lst[i][j - 1] == '.' else 0
checker.append(num)
if j != W - 1:
num = 1 if lst[i][j + 1] == '.' else 0
checker.append(num)
if len(checker) == sum(checker) and len(checker) != 0:
print(False)
exit()
print(True) | 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] == '.' else 0
checker.append(num)
if i != H - 1:
num = 1 if lst[i + 1][j] == '.' else 0
checker.append(num)
if j != 0:
num = 1 if lst[i][j - 1] == '.' else 0
checker.append(num)
if j != W - 1:
num = 1 if lst[i][j + 1] == '.' else 0
checker.append(num)
if len(checker) == sum(checker) and len(checker) != 0:
print('No')
exit()
print('Yes') |
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 |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N. | 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 visits, but for the rest of the trip she will have to travel by road. If she visits the towns in the order that minimizes the distance traveled by road, what will that distance be? | 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])
def dfs(c, las, d):
global res
if c == R+1:
if res > d:
res = d
return
for i in range(1,R):
if used[i] != False:
used[i] = True
if las == -1:
dfs(c+1,i,0)
else:
dfs(c+1, i, d+dist[r[las]][r[i]])
used[i] = False
# global res
# for p in permutations(r):
# res = s
N, M, R = (int(i) for i in input().split())
r = read_int_list()
r = [x - 1 for x in r]
used = [bool]*9
res = float('inf')
dist = [[float("inf") for i in range(201)] for i in range((201))]
for i in range(M):
a, b, c = (int(i) for i in input().split())
a -= 1
b -= 1
dist[a][b] = c
dist[b][a] = c
floyd_warshall()
dfs(1,-1,0)
print(res) | 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][j]:
d[i][j] = dd
n, m, r = read_int_list()
v = read_int_list()
v = [x - 1 for x in v]
d = [[float('inf')] * n for i in range(n)]
for i in range(n):
d[i][i] = 0
for i in range(m):
a, b, c = read_int_list()
a -= 1
b -= 1
d[a][b] = c
d[b][a] = c
floyd_warshall()
res = float('inf')
for p in permutations(v):
s = 0
for i in range(r - 1):
s += d[p[i]][p[i + 1]]
if s < res:
res = s
print(res)
main() |
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 beaker. * Operation 3: Put C grams of sugar into the beaker. * Operation 4: Put D grams of sugar into the beaker. In our experimental environment, E grams of sugar can dissolve into 100 grams of water. Snuke will make sugar water with the highest possible density. The beaker can contain at most F grams of substances (water and sugar combined), and there must not be any undissolved sugar in the beaker. Find the mass of the sugar water Snuke will make, and the mass of sugar dissolved in it. If there is more than one candidate, any of them will be accepted. We remind you that the sugar water that contains a grams of water and b grams of sugar is \frac{100b}{a + b} percent. Also, in this problem, pure water that does not contain any sugar is regarded as 0 percent density sugar water. | 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+100*B*j < F:
water.append(100*A*i+100*B*j)
water = list(set(water))
suger = []
for i in range(len(water)):
suger.append(sol(water[i]))
print(water)
print(suger)
con = 0
ws = 0
s = 0
for k in range(len(water)):
if con <= suger[k]/(water[k]+suger[k]):
con = suger[k]/(water[k]+suger[k])
ws = (water[k]+suger[k])
s = suger[k]
print(ws,s) | 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+100*B*j < F:
water.append(100*A*i+100*B*j)
water = list(set(water))
suger = []
for i in range(len(water)):
suger.append(sol(water[i]))
con = 0
ws = 0
s = 0
for k in range(len(water)):
if con <= suger[k]/(water[k]+suger[k]):
con = suger[k]/(water[k]+suger[k])
ws = (water[k]+suger[k])
s = suger[k]
print(ws,s) |
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 2 * d_1 \times d_{10} = m Takahashi wants more Product Days, and he made a new calendar called Takahashi Calendar where a year consists of M month from Month 1 to Month M, and each month consists of D days from Day 1 to Day D. In Takahashi Calendar, how many Product Days does a year have? | 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 j < len(s) and j < len(t):
if slist[j] == tlist[j]:
continue
elif slist[j] < tlist[j]:
break
else:
result = 'No'
break
elif len(s) > len(t):
result = 'No'
break
print(result) | 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):
if slist[j] == tlist[j]:
continue
elif slist[j] < tlist[j]:
break
else:
result = 'No'
break
elif len(s) > len(t):
result = 'No'
break
print(result) |
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 LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
N, K = LI()
print(N, K)
if N % 2 == 0:
return K <= N*2
else:
return K <= N//2 +1
if main():
print('YES')
else:
print('NO')
| 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 LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
N, K = LI()
if N % 2 == 0:
return K <= N/2
else:
return K <= N//2 +1
if main():
print('YES')
else:
print('NO')
|
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 have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have? | 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 satisfy the following binary search tree property: * Let $x$ be a node in a binary search tree. If $y$ is a node in the left subtree of $x$, then $y.key \leq x.key$. If $y$ is a node in the right subtree of $x$, then $x.key \leq y.key$. The following figure shows an example of the binary search tree. For example, keys of nodes which belong to the left sub-tree of the node containing 80 are less than or equal to 80, and keys of nodes which belong to the right sub-tree are more than or equal to 80. The binary search tree property allows us to print out all the keys in the tree in sorted order by an inorder tree walk. A binary search tree should be implemented in such a way that the binary search tree property continues to hold after modifications by insertions and deletions. A binary search tree can be represented by a linked data structure in which each node is an object. In addition to a key field and satellite data, each node contains fields _left_ , _right_ , and _p_ that point to the nodes corresponding to its left child, its right child, and its parent, respectively. To insert a new value $v$ into a binary search tree $T$, we can use the procedure insert as shown in the following pseudo code. The insert procedure is passed a node $z$ for which $z.key = v$, $z.left = NIL$, and $z.right = NIL$. The procedure modifies $T$ and some of the fields of $z$ in such a way that $z$ is inserted into an appropriate position in the tree. 1 insert(T, z) 2 y = NIL // parent of x 3 x = 'the root of T' 4 while x ≠ NIL 5 y = x // set the parent 6 if z.key < x.key 7 x = x.left // move to the left child 8 else 9 x = x.right // move to the right child 10 z.p = y 11 12 if y == NIL // T is empty 13 'the root of T' = z 14 else if z.key < y.key 15 y.left = z // z is the left child of y 16 else 17 y.right = z // z is the right child of y Write a program which performs the following operations to a binary search tree $T$. * insert $k$: Insert a node containing $k$ as key into $T$. * print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively. You should use the above pseudo code to implement the insert operation. $T$ is empty at the initial state. | 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:
y = x
if z.key < x.key:
x = x.left
else:
x = x.right
z.parent = y
if y == None:
self.root = z
elif z.key < y.key:
y.left = z
else:
y.right = z
def preorder(self, node):
print(" {}".format(node.key), end = "")
if node.left:
self.preorder(node.left)
if node.right:
self.preorder(node.right)
def inorder(self, node):
if node.left:
self.preorder(node.left)
print(" {}".format(node.key), end = "")
if node.right:
self.preorder(node.right)
def out(self):
self.inorder(self.root)
print('')
self.preorder(self.root)
print('')
tree = Tree()
N = int(input())
for _ in range(N):
order = input().split()
if order[0] == 'insert':
tree.insert(int(order[1]))
else:
tree.out() | 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:
y = x
if z.key < x.key:
x = x.left
else:
x = x.right
z.parent = y
if y == None:
self.root = z
elif z.key < y.key:
y.left = z
else:
y.right = z
def preorder(self, node):
print(" {}".format(node.key), end = "")
if node.left:
self.preorder(node.left)
if node.right:
self.preorder(node.right)
def inorder(self, node):
if node.left:
self.inorder(node.left)
print(" {}".format(node.key), end = "")
if node.right:
self.inorder(node.right)
def out(self):
self.inorder(self.root)
print('')
self.preorder(self.root)
print('')
tree = Tree()
N = int(input())
for _ in range(N):
order = input().split()
if order[0] == 'insert':
tree.insert(int(order[1]))
else:
tree.out() |
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 of the phrases once, in some order. | 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 remaining cards. Here, N is odd, which guarantees that at least one card can be kept. Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck. | 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] == 1:
i += 1
elif l_1[i][1] >= 2:
l_1[i][1] -= 1
while True:
if i + k <= len(l_1) - 1:
if l_1[i + k][1] >= 2:
l_1[i + k][1] -= 1
ans += 2
break
else:
k += 1
else:
ans += 2
break
print(n-ans) | 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 the squares in that column black. At least how many operations do you need in order to have N or more black squares in the grid? It is guaranteed that, under the conditions in Constraints, having N or more black squares is always possible by performing some number of operations. | 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 of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts. Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition: * For each of the N kinds of doughnuts, make at least one doughnut of that kind. At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition. | 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
else:
pass
print(ans) | 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:
ans += 1
else:
pass
print(ans) |
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 can have the same number of cookies. | 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 : orange * Rating 2800-3199 : red Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not. Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i. Find the minimum and maximum possible numbers of different colors of the users. | 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 person. Find the numbers of cookies Takahashi and Aoki respectively have after performing K operations in total. | 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
B -= A_plus
A += A_plus
print(A,B)
| 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_plus
A += A_plus
print(A,B)
|
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 integer a multiple of 4? | 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, \ldots, i + K. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N. | 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 above for the first time? | 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 of the phrases once, in some order. | 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 ** 2 + p ** 2 + q ** 2 + i * p + p * q + i * q == n:
count += 1
print(count)
count = 0 | 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:
NList[i ** 2 + p ** 2 + q ** 2 + i * p + p * q + i * q - 1] += 1
for n_count in NList:
print(n_count)
|
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==0 and x % 500!=0:
f = 10-c
print(f*100)
elif x % 10 ==0 and x % 100!=0:
j =10-i
print("g", g)
print(g*100+j*10)
elif x % 10!=0:
k = int(str(x)[3])
l = 9 - i
m = 10-k
print(k)
print(m)
print(g*100+l*10+m)
f(853)
# print("b",b)
# print("c",c)
# print("h",h)
# print("i",i)
# print("j",j)
| 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")
elif y % 100==0 and y % 500!=0:
f = 10-c
print(f*100)
elif y % 10 ==0 and y % 100!=0:
j =10-i
print(g*100+j*10)
elif y % 10!=0:
k = int(str(y)[3])
l = 9 - i
m = 10-k
print(g*100+l*10+m)
y = input()
x = f(int(y))
|
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 person includes the stool. Find the minimum total height of the stools needed to meet this goal. | 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 highest score of a problem among the ones solved by him. Determine the set of problems that should be solved. | 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 i+1!=z:print(i+1)
break
if temp==x:
for i in range(i):print(i+1)
break
#%%submit!
resolve() |
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 those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds. Now, he goes through the following process: * Submit the code. * Wait until the code finishes execution on all the cases. * If the code fails to correctly solve some of the M cases, submit it again. * Repeat until the code correctly solve all the cases in one submission. Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer). | 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 green boxes and b blue boxes. How many triples of non-negative integers (r,g,b) achieve this? | 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 pieces. Find the maximum number of `Scc` groups that can be created when Snuke has N `S`-shaped pieces and M `c`-shaped pieces. | 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 of the phrases once, in some order. | 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 must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him. | def 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[2:], t[:2])
t[0] -= 1
t[3] += 1
route_2 = 'L' + search_route(t[:2], t[2:]) + 'D'
t[2] += 1
t[1] -= 1
route_2 = route + 'R' + search_route(t[2:], t[:2]) + 'U'
print(route+route_2)
if __name__ == '__main__':
main()
| 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()))
route = search_route(t[:2], t[2:]) + search_route(t[2:], t[:2])
tmp = t[:]
t[0] -= 1
t[3] += 1
route_2 = 'L' + search_route(t[:2], t[2:]) + 'D'
t = tmp
t[1] -= 1
t[2] += 1
route_2 = route_2 + 'R' + search_route(t[2:], t[:2]) + 'U'
print(route+route_2)
if __name__ == '__main__':
main()
|
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 cost of 1. It is guaranteed that there is no toll gate in Square 0, Square X and Square N. Find the minimum cost incurred before reaching the goal. | 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]:
cost_r += 1
j += 1
if j >= len(a):
break
else:
continue
print(cost_l, cost_r)
print(min(cost_l, cost_r))
| 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]:
cost_r += 1
j += 1
if j >= len(a):
break
else:
continue
print(min(cost_l, cost_r))
|
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 the point with coordinate 0, then visit the N spots in the order they are numbered, and finally return to the point with coordinate 0. However, something came up just before the trip, and you no longer have enough time to visit all the N spots, so you decided to choose some i and cancel the visit to Spot i. You will visit the remaining spots as planned in the order they are numbered. You will also depart from and return to the point with coordinate 0 at the beginning and the end, as planned. For each i = 1, 2, ..., N, find the total cost of travel during the trip when the visit to Spot i is canceled. | 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 operations. If it is, show one possible sequence of operations that achieves it. | 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:
print(a)
else:
print(-1) |
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. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache. | 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__":
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__":
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 are neither two adjacent dogs nor two adjacent monkeys. How many such arrangements there are? Find the count modulo 10^9+7 (since animals cannot understand numbers larger than that). Here, dogs and monkeys are both distinguishable. Also, two arrangements that result from reversing each other are distinguished. | # -*- 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 + 1):
ans *= i
ans %= MOD
return ans
def solve():
n, m = [int(x) for x in input().split()]
n, m = min(n,m), max(n,m)
x = m - n
if x > 1:
print(0)
elif x == 1:
print((fmod(n) * fmod(m)*2)%MOD)
else:
print((fmod(n) * fmod(m))%MOD)
t = 1
# t = int(input())
for case in range(1,t+1):
ans = solve()
"""
azyxwvutsrqponmlkjihgfedcb
"""
| 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 + 1):
ans *= i
ans %= MOD
return ans
def solve():
n, m = [int(x) for x in input().split()]
n, m = min(n,m), max(n,m)
x = m - n
if x > 1:
print(0)
elif x == 1:
print((fmod(n) * fmod(m))%MOD)
else:
print((fmod(n) * fmod(m)*2)%MOD)
t = 1
# t = int(input())
for case in range(1,t+1):
ans = solve()
"""
azyxwvutsrqponmlkjihgfedcb
"""
|
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 at the same time must all have different colors. The one who eats the last apple from the tree will be declared winner. If both you and Lunlun play optimally, which will win? | 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 green boxes and b blue boxes. How many triples of non-negative integers (r,g,b) achieve this? | 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
print(count)
|
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 |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N. | 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]))
print(cost_min_list[n-1]) | 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_list[i - 2]))
print(cost_min_list[n - 1]) |
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 and Brown received. | 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 the number by 1, paying D coins. You can perform these operations in arbitrary order and an arbitrary number of times. What is the minimum number of coins you need to reach N? **You have to solve T testcases.** | # 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
stack = [(n, d)] # remaining, cost
while stack:
rem, cost = stack.pop()
if rem <= 1:
ans = min(ans, cost)
continue
for y, x in ABC:
if cost+y < ans:
stack.append((rem//x, (rem % x)*d+cost+y))
stack.append((-(-rem//x), (-rem % x)*d+cost+y))
ret.append(str(ans))
return '\n'.join(ret)
if __name__ == "__main__":
print(solve(*(open(0).read().splitlines())))
| 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)]
while stack:
rem, cost = stack.pop()
if min(dp[0], dp[rem]) <= cost:
continue
dp[rem] = cost
dp[0] = min(dp[0], cost+rem*d)
if 0 < rem:
for x, y in ABC:
div, m = divmod(rem, x)
if div:
stack.append((div, d*m+cost+y))
if m:
stack.append((div+1, d*(x-m)+cost+y))
ret.append(str(dp[0]))
return '\n'.join(ret)
if __name__ == "__main__":
print(solve(*(open(0).read().splitlines())))
|
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 uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name. | 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 the center of town. However, now buildings are crammed side by side and people can not walk along exactly what the paper says in. Your task is to write a program which hunts for the treature on the paper. For simplicity, 1 step is equivalent to 1 meter. Input consists of several pairs of two integers d (the first integer) and t (the second integer) separated by a comma. Input ends with "0, 0". Your program should print the coordinate (x, y) of the end point. There is the treature where x meters to the east and y meters to the north from the center of town. You can assume that d ≤ 100 and -180 ≤ t ≤ 180\. | 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 operation once for each square contains a card: * For each of the following nine squares, flip the card in it if it exists: the target square itself and the eight squares that shares a corner or a side with the target square. It can be proved that, whether each card faces up or down after all the operations does not depend on the order the operations are performed. Find the number of cards that face down after all the operations. | 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 coin are different. | 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`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format. | 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":
li[3]+=1
for i in range(4):
if i==0:
print("AC x {}",li[0])
elif i==1:
print("WA x {}",li[1])
elif i==2:
print("TLR x {}",li[2])
elif i==3:
print("RE x {}",li[3])
if __name__=="__main__":
main() | 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":
li[3]+=1
for i in range(4):
if i==0:
print("AC x {}".format(li[0]))
elif i==1:
print("WA x {}".format(li[1]))
elif i==2:
print("TLE x {}".format(li[2]))
elif i==3:
print("RE x {}".format(li[3]))
if __name__=="__main__":
main() |
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 remaining cards. Here, N is odd, which guarantees that at least one card can be kept. Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck. | 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 containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process. | 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:
x = i + dx
y = j + dy
if x < 0 or x >= H:
continue
if y <= 0 or y >= W:
continue
if S[x][y] == '#':
continue
answer[x][y] += 1
#print(answer)
for i in answer:
print(*i,sep='') | 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:
x = i + dx
y = j + dy
if x < 0 or x >= H:
continue
if y < 0 or y >= W:
continue
if S[x][y] == '#':
continue
answer[x][y] += 1
#print(answer)
for i in answer:
print(*i,sep='') |
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 the vertex from which $v$ was discovered. This process continues until all the vertices that are reachable from the original source vertex have been discovered. If any undiscovered vertices remain, then one of them is selected as a new source and the search is repeated from that source. DFS timestamps each vertex as follows: * $d[v]$ records when $v$ is first discovered. * $f[v]$ records when the search finishes examining $v$’s adjacency list. Write a program which reads a directed graph $G = (V, E)$ and demonstrates DFS on the graph based on the following rules: * $G$ is given in an adjacency-list. Vertices are identified by IDs $1, 2,... n$ respectively. * IDs in the adjacency list are arranged in ascending order. * The program should report the discover time and the finish time for each vertex. * When there are several candidates to visit during DFS, the algorithm should select the vertex with the smallest ID. * The timestamp starts with 1. | 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] = 1
for j in range(i, n):
if (graph[i][j] == 1) & (g[j] == 0):
search(j)
time = time + 1
f[i] = time if f[i] == 0 else f[i]
search(0)
[print(i+1, d[i], f[i]) for i in range(n)]
| 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] = 1
for j in range(n):
if (graph[i][j] == 1) & (g[j] == 0):
search(j)
time = time + 1
f[i] = time
for i in range(n):
if g[i] == 0:
search(i)
[print(i+1, d[i], f[i]) for i in range(n)]
|
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