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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s625405979 | p03214 | u528388170 | 2,525 | 1,048,576 | Wrong Answer | 17 | 3,060 | 165 | Niwango-kun is an employee of Dwango Co., Ltd. One day, he is asked to generate a thumbnail from a video a user submitted. To generate a thumbnail, he needs to select a frame of the video according to the following procedure: * Get an integer N and N integers a_0, a_1, ..., a_{N-1} as inputs. N denotes the number of the frames of the video, and each a_i denotes the representation of the i-th frame of the video. * Select t-th frame whose representation a_t is nearest to the average of all frame representations. * If there are multiple such frames, select the frame with the smallest index. Find the index t of the frame he should select to generate a thumbnail. | n=int(input())
a=[int(i) for i in input().split()]
s=sum(a)
l=len(a)
x=s/l
ans=100
for cnt,i in enumerate(a):
y=abs(i-x)
if y<ans:
ans=cnt
print(ans) | s571899059 | Accepted | 17 | 3,064 | 173 | n=int(input())
a=[int(i) for i in input().split()]
s=sum(a)
l=len(a)
x=s/l
z=100
for cnt,i in enumerate(a):
y=abs(i-x)
if y<z:
ans=cnt
z=y
print(ans) |
s899402041 | p04012 | u516447519 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 123 | 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. | w = str(input())
for i in range(len(w)):
if w.count(w[i]) % 2 ==0:
print('Yes')
else:
print('No')
| s401078989 | Accepted | 17 | 2,940 | 165 | w = str(input())
count = int()
for i in range(len(w)):
if w.count(w[i]) % 2 == 0:
count += 1
if count == len(w):
print('Yes')
else:
print('No') |
s934711844 | p03359 | u468972478 | 2,000 | 262,144 | Wrong Answer | 23 | 9,152 | 75 | In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year. For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year. In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For example, 5-5 is Takahashi. How many days from 2018-1-1 through 2018-a-b are Takahashi? | a, b = map(int, input().split())
if a <= b:
print(a + 1)
else:
print(a) | s966741399 | Accepted | 29 | 9,152 | 61 | a, b = map(int, input().split())
print(a - 1 if a > b else a) |
s969431959 | p03545 | u061127257 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 415 | Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted. | num = list(map(int,list(input())))
for i in range(len(num) ** 2):
op = []
for j in range(len(num)):
if (i >> j) & 1:
op.append("+")
else:
op.append("-")
sum = num[0]
for k in range(1,len(num)):
if op[k] == "+":
sum += num[k]
elif op[k] == "-":
sum -= num[k]
if sum == 7:
print("{}{}{}{}{}{}{}=7".format(num[0],op[0],num[1],op[1],num[2],op[2],num[3]))
break | s861330065 | Accepted | 17 | 3,064 | 419 | num = list(map(int,list(input())))
for i in range(len(num) ** 2):
op = []
for j in range(len(num)):
if (i >> j) & 1:
op.append("+")
else:
op.append("-")
sum = num[0]
for k in range(len(num)-1):
if op[k] == "+":
sum += num[k+1]
elif op[k] == "-":
sum -= num[k+1]
if sum == 7:
print("{}{}{}{}{}{}{}=7".format(num[0],op[0],num[1],op[1],num[2],op[2],num[3]))
break |
s744660471 | p03448 | u256833330 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 153 | 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 _ in range(4)]
print(len([1 for i in range(a+1) for j in range(b+1) if (x-500*a-100*b)%50==0 and 0<=(x-500*a-100*b)//50 <=c]))
| s238287649 | Accepted | 18 | 2,940 | 153 | a,b,c,x=[int(input()) for _ in range(4)]
print(len([1 for i in range(a+1) for j in range(b+1) if (x-500*i-100*j)%50==0 and 0<=(x-500*i-100*j)//50 <=c]))
|
s601318781 | p03469 | u911793272 | 2,000 | 262,144 | Wrong Answer | 30 | 8,944 | 31 | On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date 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. | S = input()
print("2018"+S[:4]) | s817568402 | Accepted | 27 | 8,984 | 32 | S = input()
print("2018"+S[4:])
|
s158539794 | p03795 | u136843617 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 44 | Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y. | x = int(input())
print(x*1500 - (x//15)*200) | s067724392 | Accepted | 17 | 2,940 | 106 | def solve():
N = int(input())
print(N*800 - (N//15) * 200)
if __name__ == '__main__':
solve() |
s088444273 | p02418 | u100813820 | 1,000 | 131,072 | Wrong Answer | 20 | 7,508 | 9 | Write a program which finds a pattern $p$ in a ring shaped text $s$. | import re | s704773730 | Accepted | 50 | 7,632 | 969 | # 18-Character-Ring.py
# Input
# ????????????????????? p ????????????????????????
# Output
# Constraints
# 1???p????????????s????????????100
# Sample Input 1
# vanceknowledgetoad
# advance
# Sample Output 1
# Yes
# Sample Input 2
# vanceknowledgetoad
# advanced
# Sample Output 2
# No
import re
pattern=[]
s=input()
p=input()
s= s+s
# for p2 in list(p):
# pattern.append(p2)
# pattern.append(".*")
# pattern = "".join(pattern)
# print(pattern)
matchOB = re.search(p, s)
if matchOB:
print("Yes")
else:
print("No") |
s313788032 | p02646 | u731807761 | 2,000 | 1,048,576 | Wrong Answer | 112 | 27,152 | 2,157 | Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally. | import numpy as np
import sys
#import copy
def xnxn(n=0, flag="v"):
"""
args:
int n: a number of lows to read
example:
input 1
retrun 1
input 1 2 3
return [1,2,3]
input 1 2 3
4
return [[1,2,3],
[4]]
"""
if n == 0:
temp = list(map(int, input().split()))
if len(temp) == 1:
if flag == "l":
return temp
else:
return temp[0]
elif len(temp) > 1:
return temp
else:
temp = [list(map(int, input().split())) for _ in range(n)]
return temp
def sp_xnxn(n=0):
"""
args:
int n: a number of lows to read
example
input 1
retrun [1]
input 123
return [1,2,3]
input 123
4
return [[1,2,3],
[4]]
"""
if n == 0:
return [int(k) for k in xsxs()]
else:
return [[int(k) for k in xsxs()] for _ in range(n)]
def xsxs(n=0):
"""
args:
int n: a number of lows to read
example:
input AA
retrun AA
input A BBB CC
return ["A","BBB","CC"]
input A BB CCC
D
return [["A","BBB","CC"],
["D"]]
"""
if n == 0:
temp = list(input().split())
if len(temp) == 1:
return temp[0]
elif len(temp) > 1:
return temp
else:
return [list(input().split()) for _ in range(n)]
def sp_xsxs(n=0):
"""
args:
int n: a number of lows to read
example:
input A
retrun ["A"]
input ABC
return ["A","B","C"]
input ABC
D
return [["A","B","C"],
["D"]]
"""
if n == 0:
return [s for s in xsxs()]
else:
return [[s for s in xsxs()] for _ in range(n)]
a, v = xnxn()
b, w = xnxn()
t = xnxn()
if v - w <= 0:
print("No")
elif abs(b - a) / (v - w) <= t:
print("Yes")
else:
print("No")
| s035041849 | Accepted | 111 | 27,004 | 1,426 | import numpy as np
import sys
# import copy
def xnxn(n=0, flag="v"):
"""
args:
int n: a number of lows to read
example:
input 1
retrun 1
input 1 2 3
return [1,2,3]
input 1 2 3
4
return [[1,2,3],
[4]]
"""
if n == 0:
temp = list(map(int, input().split()))
if len(temp) == 1:
if flag == "l":
return temp
else:
return temp[0]
elif len(temp) > 1:
return temp
else:
temp = [list(map(int, input().split())) for _ in range(n)]
return temp
a, v = xnxn()
b, w = xnxn()
t = xnxn()
# if b - a > 0:
# b = min(b + w, 10 ** 9)
# a = min(a + v, 10 ** 9)
# if b == a:
# print("YES")
# sys.exit()
# elif b - a < 0:
# b = max(b - w, 10 ** 9)
# a = max(a - v, 10 ** 9)
# if b == a:
# print("YES")
# sys.exit()
# print("NO")
if b - a > 0:
# b_f = min(b+t*w, 10 ** 9)
# a_f = min(a + t * v, 10 ** 9)
b_f = b+t*w
a_f = a+t*v
if b_f > a_f:
print("NO")
else:
print("YES")
else:
# b_f = max(b-t*w, -10 ** 9)
# a_f = max(a-t*v, -10 ** 9)
b_f = b-t*w
a_f = a-t*v
if b_f < a_f:
print("NO")
else:
print("YES")
|
s571466470 | p02396 | u328199937 | 1,000 | 131,072 | Wrong Answer | 130 | 5,568 | 125 | In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem. | num = 1
while True:
a = input()
if a == '0':
break
print('case ' + str(num) + ':' + str(a))
num += 1
| s952431910 | Accepted | 130 | 5,564 | 126 | num = 1
while True:
a = input()
if a == '0':
break
print('Case ' + str(num) + ': ' + str(a))
num += 1
|
s419018785 | p03160 | u681444474 | 2,000 | 1,048,576 | Wrong Answer | 153 | 14,680 | 332 | 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()))
def dp_solver(x):
H = H_
dp=[float('inf')]*N
dp[0]=0
dp[1]=abs(H[1]-H[0])
for i in range(N-2):
dp[i+2] = min(dp[i+2],dp[i]+abs(H[i+2]-H[i]))
dp[i+2] = min(dp[i+2],dp[i+1]+abs(H[i+2]-H[i+1]))
print(dp)
return dp[x-1]
print(dp_solver(N)) | s120961646 | Accepted | 121 | 20,716 | 208 | # coding: utf-8
n = int(input())
h = list(map(int,input().split()))
dp = [0] * n
dp[1] = abs(h[0]-h[1])
for i in range(2, n):
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]) |
s485277962 | p03449 | u064408584 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 336 | We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j). You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down. The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them. At most how many candies can you collect when you choose the best way to travel? | def C_Candies():
N=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
print(a)
print(b)
count=0
for i in range(N):
if sum(a[i+1:N])<sum(b[i:N-1]):
break
count = count +1
print(count)
print(sum(a[0:count+1])+sum(b[count:N]))
C_Candies() | s914308825 | Accepted | 17 | 3,064 | 268 | def C_Candies():
N=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
count=0
for i in range(N):
if count < sum(a[0:i+1])+sum(b[i:N+1]):
count =sum(a[0:i+1])+sum(b[i:N+1])
print(count)
C_Candies() |
s815768737 | p03943 | u778348725 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 281 | Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students. | a,b,c = (int(x) for x in input().split())
half_pack = int((a+b+c)/2)
if(a == half_pack) and ((b+c) == half_pack):
print("YES")
elif(b == half_pack) and ((a+c) == half_pack):
print("YES")
elif(c == half_pack) and ((b+a) == half_pack):
print("YES")
else:
print("NO")
| s973476636 | Accepted | 18 | 3,060 | 281 | a,b,c = (int(x) for x in input().split())
half_pack = int((a+b+c)/2)
if(a == half_pack) and ((b+c) == half_pack):
print("Yes")
elif(b == half_pack) and ((a+c) == half_pack):
print("Yes")
elif(c == half_pack) and ((b+a) == half_pack):
print("Yes")
else:
print("No")
|
s327935801 | p02833 | u903005414 | 2,000 | 1,048,576 | Wrong Answer | 322 | 22,284 | 248 | For an integer n not less than 0, let us define f(n) as follows: * f(n) = 1 (if n < 2) * f(n) = n f(n-2) (if n \geq 2) Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N). | import sys
import numpy as np
N = int(input())
if N % 2 != 0:
print(0)
sys.exit()
N = N // 10 * 10
ans = 0
for i in range(1, 19):
if N < 10**i:
break
ans += N // 10**i
ans += N // 50
ans -= N // 200
print('ans', ans)
| s834407029 | Accepted | 17 | 2,940 | 140 | import sys
N = int(input())
if N % 2 != 0:
print(0)
sys.exit()
N //= 2
ans = 0
while N:
ans += N // 5
N //= 5
print(ans)
|
s002640022 | p02601 | u514334797 | 2,000 | 1,048,576 | Wrong Answer | 28 | 9,144 | 212 | M-kun has the following three cards: * A red card with the integer A. * A green card with the integer B. * A blue card with the integer C. He is a genius magician who can do the following operation at most K times: * Choose one of the three cards and multiply the written integer by 2. His magic is successful if both of the following conditions are satisfied after the operations: * The integer on the green card is **strictly** greater than the integer on the red card. * The integer on the blue card is **strictly** greater than the integer on the green card. Determine whether the magic can be successful. | # B>G>R
R,G,B = map(int,input().split())
K = int(input())
cnt = 0
while G<R:
G *= 2
cnt += 1
while not B>G:
B *= 2
cnt += 1
print(cnt)
if cnt<=K:
print("Yes")
else:
print("No") | s218364874 | Accepted | 30 | 8,968 | 201 | # B>G>R
R,G,B = map(int,input().split())
K = int(input())
cnt = 0
while not G>R:
G *= 2
cnt += 1
while not B>G:
B *= 2
cnt += 1
if cnt<=K:
print("Yes")
else:
print("No") |
s056738773 | p03556 | u403355272 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 39 | Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer. | x = int(input())
print((int(x ** 0.5))) | s863248015 | Accepted | 17 | 2,940 | 44 | x = int(input())
print((int(x ** 0.5)) ** 2) |
s872221593 | p03861 | u279229189 | 2,000 | 262,144 | Wrong Answer | 19 | 3,060 | 102 | 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? | v = input().split(" ")
a = int(v[0])
b = int(v[1])
x = int(v[2])
fb = b/x
fa = (a-1)/x
print(fb-fa) | s696385584 | Accepted | 17 | 3,060 | 112 | v = input().split(" ")
a = int(v[0])
b = int(v[1])
x = int(v[2])
fb = b//x
fa = (a-1)//x
print(max(fb-fa, 0)) |
s472175945 | p03455 | u985376351 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 96 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | a, b = map(int, input().split())
if a%2==0 and b%2==0:
print('Even')
else:
print('Odd') | s098882286 | Accepted | 19 | 2,940 | 92 | a, b = map(int, input().split())
c = a*b
if c%2==0:
print('Even')
else:
print('Odd') |
s449925884 | p02665 | u135847648 | 2,000 | 1,048,576 | Wrong Answer | 2,211 | 180,596 | 462 | Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1. | def main():
n = int(input())
A = list(map(int, input().split()))
if A[0] != 0:
print(-1)
exit()
tree = [2 ** i for i in range(n + 1)]
tree[-1] = A[-1]
for i in range(n):
if A[i] > tree[i]:
print(-1)
exit()
for i in range(n - 1, 0, -1):
cnt = A[i] + tree[i + 1]
tree[i] = min(tree[i], cnt)
#print(tree)
print(sum(tree))
if __name__ == "__main__":
main()
| s333461392 | Accepted | 84 | 20,024 | 822 | def main():
n = int(input())
A = list(map(int, input().split()))
B = [0] * (n+1)
if n == 0:
x = A[0]
if x == 1:
print(1)
else:
print(-1)
exit()
# A0 + B0 = 1
now_s = sum(A)
s = now_s
for i in range(n+1):
if i == 0:
B[i] = 1 - A[i]
else:
now_s -= A[i]
B[i] = min(now_s, 2 * B[i - 1] - A[i])
#print(A,B,now_s)
if B[i] < 0:
print(-1)
exit()
print(sum(B) + s)
if __name__ == "__main__":
main()
|
s426795358 | p03644 | u408375121 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 63 | Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times. | N = int(input())
i = 0
while 2 ** i <= N:
i += 1
print(i - 1) | s636688243 | Accepted | 17 | 2,940 | 71 | N = int(input())
i = 0
while 2 ** i <= N:
i += 1
print(2 ** (i - 1))
|
s610344384 | p03337 | u413165887 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 136 | You are given two integers A and B. Find the largest value among A+B, A-B and A \times B. | n,*a=map(int,open(0).read().split());x=s=l=c=0
for r,b in enumerate(a):
x^=b;s+=b
while x<s:x^=a[l];s-=a[l];l+=1
c+=r-l+1
print(c) | s293672836 | Accepted | 17 | 2,940 | 58 | a, b = map(int, input().split())
print(max([a*b,a-b,a+b])) |
s457167108 | p02694 | u022215787 | 2,000 | 1,048,576 | Wrong Answer | 24 | 9,020 | 84 | 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())
c = 100
r = 0
while c <= x:
c = int(c*1.01)
r += 1
print(r) | s336294559 | Accepted | 22 | 9,168 | 88 | x = int(input())
c = 100
r = 0
while c < x:
c = c + int(c*0.01)
r += 1
print(r) |
s235830411 | p03944 | u804358525 | 2,000 | 262,144 | Wrong Answer | 69 | 3,188 | 729 | There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white. Snuke plotted N points into the rectangle. The coordinate of the i-th (1 ≦ i ≦ N) point was (x_i, y_i). Then, he created an integer sequence a of length N, and for each 1 ≦ i ≦ N, he painted some region within the rectangle black, as follows: * If a_i = 1, he painted the region satisfying x < x_i within the rectangle. * If a_i = 2, he painted the region satisfying x > x_i within the rectangle. * If a_i = 3, he painted the region satisfying y < y_i within the rectangle. * If a_i = 4, he painted the region satisfying y > y_i within the rectangle. Find the area of the white region within the rectangle after he finished painting. | # -*- coding: utf-8 -*-
w,h,n = map(int, input().split())
print(w,h)
area = [[0 for i in range(w)] for j in range(h)]
for i in range(n):
a,b,c = map(int, input().split())
if(c == 1):
for j in range(h):
for k in range(a):
area[j][k] = 1
if (c == 2):
for j in range(h):
for k in range(a,w):
area[j][k] = 1
if(c == 3):
for j in range(b):
for k in range(w):
area[j][k] = 1
if (c == 4):
for j in range(b,h):
for k in range(w):
area[j][k] = 1
counter = 0
for i in range(h):
for j in range(w):
if(area[i][j] == 0):
counter += 1
print(counter) | s002799309 | Accepted | 74 | 3,064 | 718 | # -*- coding: utf-8 -*-
w,h,n = map(int, input().split())
area = [[0 for i in range(w)] for j in range(h)]
for i in range(n):
a,b,c = map(int, input().split())
if(c == 1):
for j in range(h):
for k in range(a):
area[j][k] = 1
if (c == 2):
for j in range(h):
for k in range(a,w):
area[j][k] = 1
if(c == 3):
for j in range(b):
for k in range(w):
area[j][k] = 1
if (c == 4):
for j in range(b,h):
for k in range(w):
area[j][k] = 1
counter = 0
for i in range(h):
for j in range(w):
if(area[i][j] == 0):
counter += 1
print(counter) |
s342839867 | p02613 | u504573674 | 2,000 | 1,048,576 | Wrong Answer | 159 | 9,124 | 328 | 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. | t = int(input())
c0 = 0
c1 = 0
c2 = 0
c3 = 0
for i in range(t):
s = str(input())
if s == 'AC':
c0 += 1
elif s == 'WA':
c1 += 1
elif s == 'TLE':
c2 += 1
elif s == 'RE':
c3 += 1
print('AC X ' + str(c0))
print('WA X ' + str(c1))
print('TLE X ' + str(c2))
print('RE X ' + str(c3)) | s301551905 | Accepted | 160 | 9,096 | 328 | t = int(input())
c0 = 0
c1 = 0
c2 = 0
c3 = 0
for i in range(t):
s = str(input())
if s == 'AC':
c0 += 1
elif s == 'WA':
c1 += 1
elif s == 'TLE':
c2 += 1
elif s == 'RE':
c3 += 1
print('AC x ' + str(c0))
print('WA x ' + str(c1))
print('TLE x ' + str(c2))
print('RE x ' + str(c3)) |
s851072259 | p02396 | u313089641 | 1,000 | 131,072 | Wrong Answer | 80 | 7,960 | 154 | In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem. | num = []
while True:
x = int(input())
if x == 0:
break
num.append(x)
for i, n in enumerate(num):
print('Case{} :{}'.format(i, n)) | s551582103 | Accepted | 70 | 8,124 | 163 | num = []
while True:
x = int(input())
if x == 0:
break
num.append(x)
for i, n in enumerate(num, start=1):
print('Case {}: {}'.format(i, n)) |
s712872434 | p03699 | u652656291 | 2,000 | 262,144 | Wrong Answer | 19 | 3,060 | 202 | You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well. However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade? | n = int(input())
s =[int(input()) for i in range(n)]
s.sort()
s_sum =sum(s)
if s_sum % 10 != 0:
print(s_sum)
else:
for i in range(n):
if s[i] % 10 != 0:
s[i] = 0
break
print(s_sum) | s193690804 | Accepted | 18 | 2,940 | 152 | n = int(input())
s = [0]
for i in range(n):
s.append(int(input()))
m=sum(s)
s.sort()
for x in s:
if (m-x)%10!=0:
print(m-x)
exit()
print(0)
|
s556371526 | p02399 | u264450287 | 1,000 | 131,072 | Wrong Answer | 20 | 5,600 | 75 | Write a program which reads two integers a and b, and calculates the following values: * a ÷ b: d (in integer) * remainder of a ÷ b: r (in integer) * a ÷ b: f (in real number) | a,b = map(int,input().split())
d = a // b
r = a % b
f = a / b
print(d,r,f)
| s682062103 | Accepted | 20 | 5,608 | 92 | a,b = map(int,input().split())
d = a // b
r = a % b
f = a / b
print(d,r,"{:.5f}".format(f))
|
s982003374 | p03543 | u093492951 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 85 | We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**? | N = input()
if N[0] == N[1] and N[1] == N[2]:
print("yes")
else:
print("no") | s451830393 | Accepted | 17 | 2,940 | 122 | N = input()
if (N[0] == N[1] and N[1] == N[2]) or (N[1] == N[2] and N[2] == N[3]):
print("Yes")
else:
print("No") |
s160710332 | p02659 | u945405878 | 2,000 | 1,048,576 | Wrong Answer | 22 | 9,096 | 116 | Compute A \times B, truncate its fractional part, and print the result as an integer. | a, b = input().split(" ")
A, B = int(a), float(b)
B100 = int(round(B * 100))
C = int(round(A * B100 / 100))
print(C) | s498029663 | Accepted | 20 | 9,100 | 110 | a, b = input().split(" ")
A, B = int(a), float(b)
B100 = round(B * 100)
C = A * B100
ans = C // 100
print(ans) |
s359888698 | p03827 | u902151549 | 2,000 | 262,144 | Wrong Answer | 19 | 3,184 | 210 | You have an integer variable x. Initially, x=0. Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`. Find the maximum value taken by x during the operations (including before the first operation, and after the last operation). | # coding: utf-8
import time
import re
import math
def main():
n=int(input())
s=input()
x=0
mx=0
for a in s:
if(a=='I'):x+=1
elif(a=='D'):x-=1
mx=max([mx,x])
main() | s781972256 | Accepted | 19 | 3,184 | 223 | # coding: utf-8
import time
import re
import math
def main():
n=int(input())
s=input()
x=0
mx=0
for a in s:
if(a=='I'):x+=1
elif(a=='D'):x-=1
mx=max([mx,x])
print(mx)
main() |
s215440508 | p02259 | u253463900 | 1,000 | 131,072 | Wrong Answer | 30 | 7,484 | 404 | Write a program of the Bubble Sort algorithm which sorts a sequence _A_ in ascending order. The algorithm should be based on the following pseudocode: BubbleSort(A) 1 for i = 0 to A.length-1 2 for j = A.length-1 downto i+1 3 if A[j] < A[j-1] 4 swap A[j] and A[j-1] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 4 of the pseudocode. | n = int(input())
s = input()
A = list(map(int,s.split()))
flag = 1
while(flag == 1):
for k in range(0,len(A)-1):
print(A[k],end=" ")
print(A[len(A)-1])
flag = 0
for j in range(n-1,0,-1):
if A[j] < A[j-1]:
tmp = A[j]
A[j] = A[j-1]
A[j-1] = tmp
flag = 1
for k in range(0,len(A)-1):
print(A[k],end=" ")
print(A[len(A)-1]) | s217929547 | Accepted | 60 | 7,616 | 362 | n = int(input())
s = input()
A = list(map(int,s.split()))
flag = 1
cnt = 0
while(flag == 1):
flag = 0
for j in range(n-1,0,-1):
if A[j] < A[j-1]:
tmp = A[j]
A[j] = A[j-1]
A[j-1] = tmp
flag = 1
cnt += 1
for k in range(0,len(A)-1):
print(A[k],end=" ")
print(A[len(A)-1])
print (cnt) |
s780981360 | p03480 | u846150137 | 2,000 | 262,144 | Wrong Answer | 2,104 | 9,952 | 430 | You are given a string S consisting of `0` and `1`. Find the maximum integer K not greater than |S| such that we can turn all the characters of S into `0` by repeating the following operation some number of times. * Choose a contiguous segment [l,r] in S whose length is at least K (that is, r-l+1\geq K must be satisfied). For each integer i such that l\leq i\leq r, do the following: if S_i is `0`, replace it with `1`; if S_i is `1`, replace it with `0`. | l=input()
def hntn(a,s,t):
e=s+t
n="".join([str((int(i)+1) % 2) for i in a[s:e]])
return a[:s]+ n + a[e:]
for i in range(len(l),0,-1):
c=[]
for w in range(len(l),i-1,-1):
d='0'*len(l)
for j in range(2*(len(l)-i+1)):
b='0'*(len(l)-i+1-len(str(format(j, 'b'))))+str(format(j, 'b'))
for v in range(len(b)):
if b[v]=='1':
d=hntn(d,v,w)
c.append(d)
if l in c:
break
print(i) | s262283351 | Accepted | 66 | 3,188 | 114 | l=input()
n=len(l)
x = n
for i in range(1,n):
if l[i-1] != l[i]:
nx = max(i, n-i)
x = min(x,nx)
print(x) |
s261019402 | p03637 | u494209838 | 2,000 | 262,144 | Wrong Answer | 307 | 23,072 | 397 | We have a sequence of length N, a = (a_1, a_2, ..., a_N). Each a_i is a positive integer. Snuke's objective is to permute the element in a so that the following condition is satisfied: * For each 1 ≤ i ≤ N - 1, the product of a_i and a_{i + 1} is a multiple of 4. Determine whether Snuke can achieve his objective. | # -*- coding: utf-8 -*-
import numpy as np
N = np.int64(input())
a = np.array([int(x) for x in input().split()])
mod_a = np.mod(a, 4)
count4 = (mod_a==0).sum()
count2 = (mod_a==2).sum()
odd = ((mod_a==1) | (mod_a==3)).sum()
if N % 2 == 0:
if count4 >= odd:
print('yes')
else:
print('no')
else:
if count4 >= odd - 1:
print('yes')
else:
print('no')
| s824488720 | Accepted | 363 | 23,072 | 397 | # -*- coding: utf-8 -*-
import numpy as np
N = np.int64(input())
a = np.array([int(x) for x in input().split()])
mod_a = np.mod(a, 4)
count4 = (mod_a==0).sum()
count2 = (mod_a==2).sum()
odd = ((mod_a==1) | (mod_a==3)).sum()
if N % 2 == 0:
if count4 >= odd:
print('Yes')
else:
print('No')
else:
if count4 >= odd - 1:
print('Yes')
else:
print('No')
|
s060437673 | p02747 | u854405453 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 63 | A Hitachi string is a concatenation of one or more copies of the string `hi`. For example, `hi` and `hihi` are Hitachi strings, while `ha` and `hii` are not. Given a string S, determine whether S is a Hitachi string. | s = input()
s.replace("hi", "")
print("Yes" if s=="" else "No") | s105989124 | Accepted | 17 | 2,940 | 65 | s = input()
s=s.replace("hi", "")
print("Yes" if s=="" else "No") |
s652809155 | p03545 | u543954314 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 217 | Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted. | e = (1,-1)
a,b,c,d = map(int, list(input()))
for i in e:
for j in e:
for k in e:
if a+b*i+c*j+d*k == 7:
break
dic = {1:"+",-1:"-"}
l = [a,dic[i],b,dic[j],c,dic[k],d]
print("".join(map(str,l))+"=7") | s747649345 | Accepted | 18 | 3,064 | 231 | from itertools import product
e = (1,-1)
a,b,c,d = map(int, list(input()))
for i,j,k in product(e,repeat=3):
if a+b*i+c*j+d*k == 7:
break
dic = {1:"+",-1:"-"}
l = [a,dic[i],b,dic[j],c,dic[k],d]
print("".join(map(str,l))+"=7") |
s343605031 | p03699 | u769870836 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 206 | You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well. However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade? | n=int(input())
l=sorted([int(input()) for i in range(n)])
if sum(l)%10!=0:
print(sum(l))
else:
ans=sum(l)
for i in range(n):
ans-=l[-1-i]
if ans%10!=0:
print(ans)
quit()
print(0) | s088928685 | Accepted | 18 | 3,060 | 206 | n=int(input())
l=sorted([int(input()) for i in range(n)])
if sum(l)%10!=0:
print(sum(l))
else:
ans=sum(l)
for i in range(n):
if l[i]%10!=0:
ans-=l[i]
print(ans)
quit()
print(0) |
s843106658 | p02240 | u424457654 | 1,000 | 131,072 | Wrong Answer | 30 | 5,604 | 778 | Write a program which reads relations in a SNS (Social Network Service), and judges that given pairs of users are reachable each other through the network. | def dfs(r, c):
global color
global g
s = [r]
color[r] = c
while len(s) > 0:
u = s.pop()
for u in g[u]:
v = color[u]
if v is None:
color[u] = c
s.append(u)
def assigncolor():
global g
global color
global n
c = 1
for i in range(n):
if color[i] == None:
dfs(i, c)
c += 1
n, m = list(map(int, input().split()))
g = [[] for i in range(n)]
for i in range(m):
s, t = list(map(int, input().split()))
g[s].append(t)
g[t].append(s)
color = [None for i in range(n)]
assigncolor()
q = int(input())
for i in range(q):
if color[s] == color[t]:
print("yes")
else:
print("no") | s045476713 | Accepted | 710 | 23,388 | 821 | def dfs(r, c):
global color
global g
s = [r]
color[r] = c
while len(s) > 0:
u = s.pop()
for u in g[u]:
v = color[u]
if v is None:
color[u] = c
s.append(u)
def assigncolor():
global g
global color
global n
c = 1
for i in range(n):
if color[i] == None:
dfs(i, c)
c += 1
n, m = list(map(int, input().split()))
g = [[] for i in range(n)]
for i in range(m):
s, t = list(map(int, input().split()))
g[s].append(t)
g[t].append(s)
color = [None for i in range(n)]
assigncolor()
q = int(input())
for i in range(q):
s, t = list(map(int, input().split()))
if color[s] == color[t]:
print("yes")
else:
print("no") |
s902398302 | p03545 | u327532412 | 2,000 | 262,144 | Wrong Answer | 19 | 3,064 | 282 | Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted. | *A, = list(input())
s = ['+', '-']
for i in range(2**3):
formula = A[0]
for j in range(3):
if i >> j & 1:
formula += s[0]
else:
formula += s[1]
formula += A[j+1]
if eval(formula) == 7:
print(formula)
exit() | s742159238 | Accepted | 29 | 9,064 | 258 | ABCD = list(input())
ope = ['+', '-']
for i in range(2**3):
f = ABCD[0]
for j in range(3):
if i >> j & 1:
f += '+'
else:
f += '-'
f += ABCD[j+1]
if eval(f) == 7:
print(f+"=7")
exit() |
s756251267 | p02406 | u299231628 | 1,000 | 131,072 | Wrong Answer | 20 | 5,584 | 155 | In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; } | n = int(input())
i = 1
a = ''
while i <= n:
if i % 3 == 0:
a += ' ' + str(i)
if i % 10 == 3:
a += ' ' + str(i)
i += 1
print(a)
| s881545463 | Accepted | 30 | 5,980 | 326 | n = int(input())
i = 1
l = []
while i <= n:
x = i
if x % 3 == 0: l.append(i)
elif x % 10 == 3: l.append(i)
else:
x /= 10
x = int(x)
while x:
if x % 10 == 3:
l.append(i)
break
x /= 10
x = int(x)
i += 1
print('',*l)
|
s897914647 | p03943 | u646336933 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 106 | Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students. | a, b, c = map(int, input().split())
if a+b==c or b+c==a or c+a==b:
print("YES")
else:
print("NO")
| s435364952 | Accepted | 17 | 2,940 | 106 | a, b, c = map(int, input().split())
if a+b==c or b+c==a or c+a==b:
print("Yes")
else:
print("No")
|
s300086778 | p03501 | u740047492 | 2,000 | 262,144 | Wrong Answer | 18 | 3,188 | 79 | 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. | n, a, b = map(int, input().split())
if (n*a) > b:
print(n*a)
else:
print(b) | s421349627 | Accepted | 17 | 2,940 | 79 | n, a, b = map(int, input().split())
if (n*a) < b:
print(n*a)
else:
print(b) |
s295972450 | p03251 | u111508936 | 2,000 | 1,048,576 | Wrong Answer | 863 | 21,524 | 352 | Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out. | import numpy as np
import math
# K = int(input())
# N, K = map(int, input().split())
# A = list(map(int, input().split()))
N, M, X, Y = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
maxx = max(x)
miny = min(y)
if miny > maxx and maxx < Y and miny < X:
print('No War')
else:
print('War')
| s317721934 | Accepted | 149 | 12,508 | 371 | import numpy as np
import math
# K = int(input())
# N, K = map(int, input().split())
# A = list(map(int, input().split()))
N, M, X, Y = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
maxx = max(x)
miny = min(y)
maxxx = max(maxx, X)
minyy = min(miny, Y)
if maxxx < minyy:
print('No War')
else:
print('War')
|
s642667350 | p03845 | u519939795 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 161 | Joisino is about to compete in the final round of a certain programming competition. In this contest, there are N problems, numbered 1 through N. Joisino knows that it takes her T_i seconds to solve problem i(1≦i≦N). Also, there are M kinds of drinks offered to the contestants, numbered 1 through M. If Joisino takes drink i(1≦i≦M), her brain will be stimulated and the time it takes for her to solve problem P_i will become X_i seconds. It does not affect the time to solve the other problems. A contestant is allowed to take exactly one of the drinks before the start of the contest. For each drink, Joisino wants to know how many seconds it takes her to solve all the problems if she takes that drink. Here, assume that the time it takes her to solve all the problems is equal to the sum of the time it takes for her to solve individual problems. Your task is to write a program to calculate it instead of her. | n = int(input())
T = list(map(int, input().split()))
s = sum(T)
m = int(input())
for _ in range(m):
p, x = map(int, input().split())
sub = -x + T[p - 1]
| s367084143 | Accepted | 18 | 3,060 | 179 | n = int(input())
T = list(map(int, input().split()))
s = sum(T)
m = int(input())
for _ in range(m):
p, x = map(int, input().split())
sub = -x + T[p - 1]
print(s - sub) |
s609109006 | p02417 | u874395007 | 1,000 | 131,072 | Wrong Answer | 20 | 5,556 | 211 | Write a program which counts and reports the number of each alphabetical letter. Ignore the case of characters. | import sys
data = sys.stdin.read()
letters = 'abcdefghijklmnopqrstuvwxyz'
for letter in letters:
count = 0
for d in data:
if letter == d:
count += 1
print(f'{letter} : {count}')
| s054347869 | Accepted | 20 | 5,564 | 219 | import sys
data = sys.stdin.read()
letters = 'abcdefghijklmnopqrstuvwxyz'
for letter in letters:
count = 0
for d in data:
if letter == d.lower():
count += 1
print(f'{letter} : {count}')
|
s561993514 | p03836 | u480847874 | 2,000 | 262,144 | Wrong Answer | 18 | 3,064 | 504 | 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 m():
sx, sy, tx, ty = map(int, input().split())
X = tx-sx
Y = ty-sy
ans = ""
for _ in range(X):
ans += "R"
for _ in range(Y):
ans += "U"
for _ in range(X):
ans += "L"
for _ in range(Y):
ans += "D"
ans += "D"
for _ in range(X):
ans += "R"
ans += "R"
ans += "U"
for _ in range(Y):
ans += "U"
ans += "U"
ans += "L" * (X+1)
ans += "D" * (Y+1)
ans += "R"
return ans
print(m()) | s591375295 | Accepted | 17 | 3,064 | 455 | def m():
sx, sy, tx, ty = map(int, input().split())
X = tx-sx
Y = ty-sy
ans = ""
ans += "U" * Y
ans += "R" * X
ans += "D" * Y
ans += "L" * X
ans += "L"
ans += "U" * (Y+1)
ans += "R" * (X+1)
ans += "D"
ans += "R"
ans += "D" * (Y+1)
ans += "L" * (X+1)
ans += "U"
return ans
print(m()) |
s028794460 | p03485 | u564060397 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 91 | You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer. | a,b=map(int,input().split())
if (a+b)%2==0:
print((a+b)/2)
else:
print((a+b)//2+1) | s160718447 | Accepted | 17 | 2,940 | 49 | a,b=map(int,input().split())
print(-(-(a+b)//2)) |
s340616000 | p03486 | u420522278 | 2,000 | 262,144 | Wrong Answer | 18 | 3,064 | 202 | 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()
sl = sorted(s)
tl = sorted(t)
out = 'No'
for i in range(0,min(len(s), len(t))):
if sl[i] < tl[-1*i]:
out = 'Yes'
break
elif sl[i] > tl[-1*i]:
break
print(out) | s029371668 | Accepted | 17 | 2,940 | 130 | s = input()
t = input()
sl = ''.join(sorted(s))
tl = ''.join(reversed(sorted(t)))
if sl < tl:
print('Yes')
else:
print('No') |
s312010741 | p02602 | u996506712 | 2,000 | 1,048,576 | Wrong Answer | 168 | 31,600 | 162 | 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()))
print(a)
for i in range(n-k):
if a[k+i]>a[i]:
print('Yes')
else:
print('No') | s512842190 | Accepted | 144 | 31,524 | 153 | n,k=map(int,input().split())
a=list(map(int,input().split()))
for i in range(n-k):
if a[k+i]>a[i]:
print('Yes')
else:
print('No') |
s399869989 | p03433 | u725044506 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 96 | 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())
a = N % 500
if a <= A:
print('YES')
else:
print('NO') | s328268044 | Accepted | 17 | 2,940 | 96 | N = int(input())
A = int(input())
a = N % 500
if a <= A:
print('Yes')
else:
print('No') |
s567573114 | p03352 | u064408584 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 126 | 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())
a=0
for i in range(300):
for j in range(11):
if i**j>x: break
if a<i**j: a=i**j
print(a) | s709342699 | Accepted | 17 | 2,940 | 128 | x=int(input())
a=0
for i in range(300):
for j in range(2,11):
if i**j>x: break
if a<i**j: a=i**j
print(a) |
s294888575 | p02697 | u614875193 | 2,000 | 1,048,576 | Wrong Answer | 69 | 9,280 | 66 | You are going to hold a competition of one-to-one game called AtCoder Janken. _(Janken is the Japanese name for Rock-paper-scissors.)_ N players will participate in this competition, and they are given distinct integers from 1 through N. The arena has M playing fields for two players. You need to assign each playing field two distinct integers between 1 and N (inclusive). You cannot assign the same integer to multiple playing fields. The competition consists of N rounds, each of which proceeds as follows: * For each player, if there is a playing field that is assigned the player's integer, the player goes to that field and fight the other player who comes there. * Then, each player adds 1 to its integer. If it becomes N+1, change it to 1. You want to ensure that no player fights the same opponent more than once during the N rounds. Print an assignment of integers to the playing fields satisfying this condition. It can be proved that such an assignment always exists under the constraints given. | N,M=map(int,input().split())
for i in range(M):
print(1+i,N-i) | s599988063 | Accepted | 102 | 21,180 | 393 | N,M=map(int,input().split())
ans=[]
if N%2==0:
N-=1
half=N//2
if half%2==0:
for i in range(half//2):
ans.append((i+1,half-i))
for j in range(half//2):
ans.append((half+1+j,N-j))
else:
for i in range(half//2):
ans.append((i+1,half-i))
for j in range(half//2+1):
ans.append((half+1+j,N-j))
#print(ans)
for i in range(M):
print(*ans[i]) |
s998721280 | p03625 | u969708690 | 2,000 | 262,144 | Wrong Answer | 155 | 33,264 | 224 | We have N sticks with negligible thickness. The length of the i-th stick is A_i. Snuke wants to select four different sticks from these sticks and form a rectangle (including a square), using the sticks as its sides. Find the maximum possible area of the rectangle. | import collections
N=int(input())
L=list(map(int,input().split()))
c = collections.Counter(L)
A=c.keys()
B=c.values()
R = zip(A,B)
R=sorted(R,reverse=True)
a, b = zip(*R)
if b[0]>=4:
print(a[0]**2)
else:
print(a[0]*a[1]) | s022752133 | Accepted | 111 | 22,384 | 353 | import collections
N=int(input())
L=list(map(int,input().split()))
c = collections.Counter(L)
A=list(c.keys())
B=list(c.values())
R=list()
for i in range(len(A)):
if B[i]>=2:
R.append([A[i],B[i]])
R=sorted(R,reverse=True)
if len(R)==0:
print(0)
exit()
if R[0][1]>=4:
print(R[0][0]**2)
elif len(R)<2:
print(0)
else:
print(R[0][0]*R[1][0]) |
s971580559 | p02393 | u058433718 | 1,000 | 131,072 | Wrong Answer | 20 | 5,600 | 229 | Write a program which reads three integers, and prints them in ascending order. | data = input()
nums = [int(i) for i in data.split()]
max_idx = len(nums)
for i in range(1, max_idx):
for j in range(max_idx-i):
if nums[j] > nums[j+1]:
nums[j], nums[j+1] = nums[j+1], nums[j]
print(nums) | s805199166 | Accepted | 20 | 5,608 | 257 | data = input()
nums = [int(i) for i in data.split()]
max_idx = len(nums)
for i in range(1, max_idx):
for j in range(max_idx-i):
if nums[j] > nums[j+1]:
nums[j], nums[j+1] = nums[j+1], nums[j]
print(' '.join([str(i) for i in nums])) |
s345364312 | p03693 | u532966492 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 57 | 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? | print("YNEOS"[int("".join(sorted(input()))[2:])%4!=0::2]) | s933296393 | Accepted | 17 | 2,940 | 111 | def main():
a, b, c = map(int, input().split())
print(['NO', 'YES'][(a*100+b*10+c) % 4 == 0])
main()
|
s505450425 | p02743 | u891945807 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 109 | Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold? | a,b,c = map(int,input().split())
res = (c-a-b)**2 - a*b
if res > 0 :
print('Yes')
else:
print('No') | s074224975 | Accepted | 17 | 2,940 | 126 | a,b,c = map(int,input().split())
res = (c-a-b)**2 - 4* a*b
if res > 0 and c-a-b > 0 :
print('Yes')
else:
print('No') |
s725771506 | p04012 | u408620326 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 101 | 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. | w=input().split()
ans='No'
for wd in w:
if w.count(wd)%2!=0:
break
else:
ans='Yes'
print(ans) | s411309948 | Accepted | 17 | 2,940 | 93 | w=input()
ans='No'
for wd in w:
if w.count(wd)%2!=0:
break
else:
ans='Yes'
print(ans) |
s955672619 | p03836 | u964299793 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 284 | Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him. | sx,sy,tx,ty=map(int,input().split())
print('U'*(ty-sy),end='')
print('R'*(tx-sx),end='')
print('D'*(ty-sy),end='')
print('L'*(tx-sx),end='')
print('L',end='')
print('U'*(ty-sy+1),end='')
print('R'*(tx-sx+1),end='')
print('D'*(ty-sy+1),end='')
print('L'*(tx-sx+1),end='')
print('U')
| s071814483 | Accepted | 17 | 3,064 | 320 | sx,sy,tx,ty=map(int,input().split())
print('U'*(ty-sy),end='')
print('R'*(tx-sx),end='')
print('D'*(ty-sy),end='')
print('L'*(tx-sx),end='')
print('L',end='')
print('U'*(ty-sy+1),end='')
print('R'*(tx-sx+1),end='')
print('D',end='')
print('R',end='')
print('D'*(ty-sy+1),end='')
print('L'*(tx-sx+1),end='')
print('U')
|
s412942740 | p03547 | u626468554 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 194 | In programming, hexadecimal notation is often used. In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively. In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`. When X and Y are seen as hexadecimal numbers, which is larger? | #n = int(input())
#n,k = map(int,input().split())
#x = list(map(int,input().split()))
a = list(input())
if a[0] > a[1]:
print(">")
elif a[0] == a[1]:
print("=")
else:
print("<")
| s477524012 | Accepted | 17 | 2,940 | 235 | #n = int(input())
#n,k = map(int,input().split())
#x = list(map(int,input().split()))
a = list(input().split())
li = ["A","B","C","D","E","F"]
if a[0] > a[1]:
print(">")
elif a[0] == a[1]:
print("=")
else:
print("<")
|
s937807910 | p03502 | u320098990 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 205 | An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10. Given an integer N, determine whether it is a Harshad number. | N = int(input())
f = 0
N_dammy = N
while N_dammy>0:
f += N_dammy % 10
N_dammy = int(N_dammy/10)
print("f:",f)
print("N_dammy:",N_dammy)
if N%f==0:
print("Yes")
else:
print("No") | s556920348 | Accepted | 17 | 2,940 | 156 | N = int(input())
f = 0
N_dammy = N
while N_dammy>0:
f += N_dammy % 10
N_dammy = int(N_dammy/10)
if N%f==0:
print("Yes")
else:
print("No") |
s896097873 | p04029 | u852790844 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 33 | There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total? | n = int(input())
print(n*(n+1)/2) | s915580984 | Accepted | 17 | 2,940 | 39 | n = int(input())
print(int(n*(n+1)/2))
|
s473102075 | p03569 | u539517139 | 2,000 | 262,144 | Wrong Answer | 60 | 3,316 | 182 | We have a string s consisting of lowercase English letters. Snuke can perform the following operation repeatedly: * Insert a letter `x` to any position in s of his choice, including the beginning and end of s. Snuke's objective is to turn s into a palindrome. Determine whether the objective is achievable. If it is achievable, find the minimum number of operations required. | s=input()
l=0
r=len(s)-1
x=0
while l<r:
if s[l]==s[r]:
l+=1
r-=1
elif s[l]=='x':
x+=1
l+=1
elif s[l]=='x':
x+=1
r-=1
else:
x=-1
break
print(x) | s571265590 | Accepted | 62 | 3,316 | 182 | s=input()
l=0
r=len(s)-1
x=0
while l<r:
if s[l]==s[r]:
l+=1
r-=1
elif s[l]=='x':
x+=1
l+=1
elif s[r]=='x':
x+=1
r-=1
else:
x=-1
break
print(x) |
s468859289 | p03434 | u201928947 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 181 | We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score. | n = int(input())
a = list(map(int,input().split()))
ans1 = 0
ans2 = 0
for i in range(n):
if i % 2 == 1:
ans1 += a[i]
else:
ans2 += a[i]
print(max(ans1,ans2)) | s891212843 | Accepted | 17 | 3,060 | 199 | n = int(input())
a = sorted(list(map(int,input().split())),reverse = True)
ans1 = 0
ans2 = 0
for i in range(n):
if i % 2 == 0:
ans1 += a[i]
else:
ans2 += a[i]
print(ans1-ans2) |
s221216903 | p03721 | u146597538 | 2,000 | 262,144 | Wrong Answer | 464 | 34,320 | 267 | There is an empty array. The following N operations will be performed to insert integers into the array. In the i-th operation (1≤i≤N), b_i copies of an integer a_i are inserted into the array. Find the K-th smallest integer in the array after the N operations. For example, the 4-th smallest integer in the array \\{1,2,2,3,3,3\\} is 3. | n,k = map(int, input().split())
ab = [list(map(int, input().split())) for _ in range(n)]
s_ab = sorted(ab, key=lambda x: x[0])
cnt = 0
ans = -1
print(s_ab)
for i in s_ab:
cnt += i[1]
if cnt >= k:
print(cnt)
ans = i[0]
break
print(ans) | s169700573 | Accepted | 499 | 28,640 | 187 | n,k = map(int, input().split())
ab = sorted([list(map(int, input().split())) for _ in range(n)])
cnt = 0
for i in ab:
cnt += i[1]
if cnt >= k:
print(i[0])
exit(0) |
s780015286 | p03162 | u506858457 | 2,000 | 1,048,576 | Wrong Answer | 632 | 35,660 | 460 | 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. | N = int(input())
a, b, c = [0] * N, [0] * N, [0] * N
for i in range(N):
a[i], b[i], c[i] = map(int, input().split())
dp = [[0] * 3 for _ in range(N)]
print(dp)
dp[0] = [a[0], b[0], c[0]]
for i in range(N - 1):
dp[i + 1][0] = max(dp[i + 1][0], dp[i][1] + a[i + 1], dp[i][2] + a[i + 1])
dp[i + 1][1] = max(dp[i + 1][1], dp[i][0] + b[i + 1], dp[i][2] + b[i + 1])
dp[i + 1][2] = max(dp[i + 1][2], dp[i][0] + c[i + 1], dp[i][1] + c[i + 1])
print(max(dp[N - 1])) | s421625946 | Accepted | 993 | 47,348 | 407 | n=int(input())
happy=[list(map(int, input().split())) for i in range(n)]
dp=[[0 for j in range(3)] for i in range(n+1)]
for i in range(1,n+1):
for place in range(3):
for placeY in range(3):
if place==placeY:
continue
dp[i][place]=max(dp[i][place],
dp[i-1][placeY]+happy[i-1][place])
ans=0
for place in range(3):
ans=max(ans,dp[n][place])
print(ans) |
s882609039 | p03574 | u011062360 | 2,000 | 262,144 | Wrong Answer | 29 | 3,444 | 522 | 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())
map = a = [ list(input()) for _ in range(h) ]
d = [[1, 0], [0, 1], [1, 1], [-1, 0], [0, -1], [-1, -1], [1, -1], [-1, 1]]
for i in range(h):
for j in range(w):
cnt = 0
if map[i][j] == ".":
for dx, dy in d:
nx = j + dx
ny = i + dy
if 0 <= nx < w and 0 <= ny < h and map[ny][nx] == "#":
cnt += 1
if map[i][j] != "#":
map[i][j] = cnt
for i in range(h):
print(*map[i]) | s202475042 | Accepted | 28 | 3,444 | 529 | h, w = map(int, input().split())
map = a = [ list(input()) for _ in range(h) ]
d = [[1, 0], [0, 1], [1, 1], [-1, 0], [0, -1], [-1, -1], [1, -1], [-1, 1]]
for i in range(h):
for j in range(w):
cnt = 0
if map[i][j] == ".":
for dx, dy in d:
nx = j + dx
ny = i + dy
if 0 <= nx < w and 0 <= ny < h and map[ny][nx] == "#":
cnt += 1
if map[i][j] != "#":
map[i][j] = cnt
for i in range(h):
print(*map[i],sep="") |
s065777915 | p03657 | u350093546 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 112 | 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%3!=0 and b%3!=0 and (a+b)%3!=0:
print('Impossible')
else:
print('Posible') | s723919861 | Accepted | 17 | 2,940 | 114 | a,b=map(int,input().split())
if a%3!=0 and b%3!=0 and (a+b)%3!=0:
print('Impossible')
else:
print('Possible')
|
s093867624 | p03110 | u810066979 | 2,000 | 1,048,576 | Wrong Answer | 18 | 3,060 | 203 | Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total? | import sys
import math
import itertools
n=input()
sum=0
for i in range(int(n)):
x,u=list(input().split())
if u=="JPY":
sum+=int(x)
else :
b_money=float(x)*380000.0
sum+=int(b_money)
print(sum) | s562329209 | Accepted | 18 | 3,060 | 199 | import sys
import math
import itertools
n=input()
sum=0
for i in range(int(n)):
x,u=list(input().split())
if u=="JPY":
sum+=int(x)
else :
b_money=float(x)*380000.0
sum+=b_money
print(sum) |
s463192976 | p03672 | u773686010 | 2,000 | 262,144 | Wrong Answer | 30 | 9,100 | 174 | We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not. You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input. | N = str(input())
cnt = 0
while True:
N = N[0:-1]
cnt += 1
if len(N) % 2 == 0:
if N[0:int(len(N)/2)] == N[int(len(N)/2):]:
break
print(cnt)
| s042673138 | Accepted | 30 | 9,060 | 154 | N = str(input())
while True:
N = N[0:-1]
if len(N) % 2 == 0:
if N[0:int(len(N)/2)] == N[int(len(N)/2):]:
break
print(len(N)) |
s036456548 | p03369 | u517152997 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 291 | In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen. | S = input("in")
if S == "ooo" :
print('1000')
elif S == "oox":
print('900')
elif S == "oxo" :
print('900')
elif S == "xoo":
print('900')
elif S == "oxx":
print('800')
elif S == "xox":
print('800')
elif S == "xxo":
print('800')
elif S == "xxx":
print('700')
| s037311718 | Accepted | 17 | 2,940 | 276 | S = input()
if S == "ooo" :
print('1000')
elif S == "oox":
print('900')
elif S == "oxo" :
print('900')
elif S == "xoo":
print('900')
elif S == "oxx":
print('800')
elif S == "xox":
print('800')
elif S == "xxo":
print('800')
else:
print('700')
|
s084328002 | p03090 | u335278042 | 2,000 | 1,048,576 | Wrong Answer | 20 | 3,572 | 323 | You are given an integer N. Build an undirected graph with N vertices with indices 1 to N that satisfies the following two conditions: * The graph is simple and connected. * There exists an integer S such that, for every vertex, the sum of the indices of the vertices adjacent to that vertex is S. It can be proved that at least one such graph exists under the constraints of this problem. | #import collections as c
#ip = lambda : map(int, input().split())
#################
N = int(input())
S = N*(N+1)/2
r = []
a = []
b = []
f = 1
for i in range(N,0,-1):
if f%4==1 or f%4==0:
a.append(i)
f += 1
else:
b.append(i)
f += 1
for ia in a:
for ib in b:
print(ia,ib) | s077238722 | Accepted | 24 | 3,612 | 695 | #import collections as c
#ip = lambda : map(int, input().split())
#################
N = int(input())
if N%2==0:
print(N*(N-1)//2-N//2)
S = N + 1
for i in range(1,N):
for j in range(i+1,N+1):
if i!=j and i+j!=S:
print(i,j)
elif N%2==1:
print(N*(N-1)//2-(N-1)//2)
S = N
for i in range(1,N):
for j in range(i+1,N+1):
if i!=j and i+j!=S:
print(i,j)
# a = []
# b = []
# f = 1
# if f%4==1 or f%4==0:
# a.append(i)
# f += 1
# else:
# b.append(i)
# f += 1
# print(len(a)*len(b))
# for ia in a:
# for ib in b:
# print(ia,ib) |
s257742278 | p03798 | u633450100 | 2,000 | 262,144 | Wrong Answer | 20 | 3,316 | 45 | Snuke, who loves animals, built a zoo. There are N animals in this zoo. They are conveniently numbered 1 through N, and arranged in a circle. The animal numbered i (2≤i≤N-1) is adjacent to the animals numbered i-1 and i+1. Also, the animal numbered 1 is adjacent to the animals numbered 2 and N, and the animal numbered N is adjacent to the animals numbered N-1 and 1. There are two kinds of animals in this zoo: honest sheep that only speak the truth, and lying wolves that only tell lies. Snuke cannot tell the difference between these two species, and asked each animal the following question: "Are your neighbors of the same species?" The animal numbered i answered s_i. Here, if s_i is `o`, the animal said that the two neighboring animals are of the same species, and if s_i is `x`, the animal said that the two neighboring animals are of different species. More formally, a sheep answered `o` if the two neighboring animals are both sheep or both wolves, and answered `x` otherwise. Similarly, a wolf answered `x` if the two neighboring animals are both sheep or both wolves, and answered `o` otherwise. Snuke is wondering whether there is a valid assignment of species to the animals that is consistent with these responses. If there is such an assignment, show one such assignment. Otherwise, print `-1`. | N = int(input())
s = input()
print("WWSSSS") | s654466747 | Accepted | 236 | 3,740 | 4,892 | N = int(input())
s = str(input())
A = ''
def majikayo():
k = 0
if A[0] == 'S' and s[0] == 'o':
if A[-1] == 'S' and s[-1] == 'o':
if A[1] == 'S' and A[-2] == 'S':
k = 1
elif A[-1] == 'S' and s[-1] == 'x':
if A[1] == 'S' and A[-2] == 'W':
k = 1
elif A[-1] == 'W' and s[-1] == 'o':
if A[1] == 'W' and A[-2] == 'W':
k = 1
elif A[-1] == 'W' and s[-1] == 'x':
if A[1] == 'W' and A[-2] == 'S':
k = 1
elif A[0] == 'S' and s[0] == 'x':
if A[-1] == 'S' and s[-1] == 'o':
if A[1] == 'W' and A[-2] == 'S':
k = 1
elif A[-1] == 'S' and s[-1] == 'x':
if A[1] == 'W' and A[-2] == 'W':
k = 1
elif A[-1] == 'W' and s[-1] == 'o':
if A[1] == 'S' and A[-2] == 'W':
k = 1
elif A[-1] == 'W' and s[-1] == 'x':
if A[1] == 'S' and A[-2] == 'S':
k = 1
elif A[0] == 'W' and s[0] == 'o':
if A[-1] == 'S' and s[-1] == 'o':
if A[1] == 'W' and A[-2] == 'W':
k = 1
elif A[-1] == 'S' and s[-1] == 'x':
if A[1] == 'W' and A[-2] == 'S':
k = 1
elif A[-1] == 'W' and s[-1] == 'o':
if A[1] == 'S' and A[-2] == 'S':
k = 1
elif A[-1] == 'W' and s[-1] == 'x':
if A[1] == 'S' and A[-2] == 'W':
k = 1
elif A[0] == 'W' and s[0] == 'x':
if A[-1] == 'S' and s[-1] == 'o':
if A[1] == 'S' and A[-2] == 'W':
k = 1
elif A[-1] == 'S' and s[-1] == 'x':
if A[1] == 'S' and A[-2] == 'S':
k = 1
elif A[-1] == 'W' and s[-1] == 'o':
if A[1] == 'W' and A[-2] == 'S':
k = 1
elif A[-1] == 'W' and s[-1] == 'x':
if A[1] == 'W' and A[-2] == 'W':
k = 1
if k == 1:
return 1
A = 'SS'
for i in range(1,N-1):
if s[i] == 'o' and A[i] == 'S':
if A[i-1] == 'S':
A += 'S'
else:
A += 'W'
elif s[i] == 'o' and A[i] == 'W':
if A[i-1] == 'S':
A += 'W'
else:
A += 'S'
elif s[i] == 'x' and A[i] == 'S':
if A[i-1] == 'S':
A += 'W'
else:
A += 'S'
elif s[i] == 'x' and A[i] == 'W':
if A[i-1] == 'S':
A += 'S'
else:
A += 'W'
if majikayo() == 1:
print(A)
else:
A = 'SW'
for i in range(1,N-1):
if s[i] == 'o' and A[i] == 'S':
if A[i-1] == 'S':
A += 'S'
else:
A += 'W'
elif s[i] == 'o' and A[i] == 'W':
if A[i-1] == 'S':
A += 'W'
else:
A += 'S'
elif s[i] == 'x' and A[i] == 'S':
if A[i-1] == 'S':
A += 'W'
else:
A += 'S'
elif s[i] == 'x' and A[i] == 'W':
if A[i-1] == 'S':
A += 'S'
else:
A += 'W'
if majikayo() == 1:
print(A)
else:
k = 0
A = 'WS'
for i in range(1,N-1):
if s[i] == 'o' and A[i] == 'S':
if A[i-1] == 'S':
A += 'S'
else:
A += 'W'
elif s[i] == 'o' and A[i] == 'W':
if A[i-1] == 'S':
A += 'W'
else:
A += 'S'
elif s[i] == 'x' and A[i] == 'S':
if A[i-1] == 'S':
A += 'W'
else:
A += 'S'
elif s[i] == 'x' and A[i] == 'W':
if A[i-1] == 'S':
A += 'S'
else:
A += 'W'
majikayo()
if majikayo() == 1:
print(A)
else:
k = 0
A = 'WW'
for i in range(1,N-1):
if s[i] == 'o' and A[i] == 'S':
if A[i-1] == 'S':
A += 'S'
else:
A += 'W'
elif s[i] == 'o' and A[i] == 'W':
if A[i-1] == 'S':
A += 'W'
else:
A += 'S'
elif s[i] == 'x' and A[i] == 'S':
if A[i-1] == 'S':
A += 'W'
else:
A += 'S'
elif s[i] == 'x' and A[i] == 'W':
if A[i-1] == 'S':
A += 'S'
else:
A += 'W'
if majikayo() == 1:
print(A)
else:
print(-1)
|
s300406007 | p03339 | u057109575 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,060 | 137 | There are N people standing in a row from west to east. Each person is facing east or west. The directions of the people is given as a string S of length N. The i-th person from the west is facing east if S_i = `E`, and west if S_i = `W`. You will appoint one of the N people as the leader, then command the rest of them to face in the direction of the leader. Here, we do not care which direction the leader is facing. The people in the row hate to change their directions, so you would like to select the leader so that the number of people who have to change their directions is minimized. Find the minimum number of people who have to change their directions. | S = input()
cnt = S.count('E')
ans = cnt
for v in S:
if v == 'E':
cnt -= 1
else:
cnt += 1
ans = min(ans, cnt)
print(ans) | s893479917 | Accepted | 124 | 3,676 | 148 | N = input()
S = input()
cnt = S.count('E')
ans = cnt
for v in S:
if v == 'E':
cnt -= 1
else:
cnt += 1
ans = min(ans, cnt)
print(ans)
|
s879787974 | p03545 | u996996256 | 2,000 | 262,144 | Wrong Answer | 18 | 3,064 | 343 | Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted. | s = input()
n= 2**(len(s)-1)
ans = 'inf'
for i in range(n):
total = int(s[0])
str = s[0]
for j in range(1,len(s)):
if i>>(len(s)-j)&1:
total += int(s[j])
str += '+' + s[j]
else:
total -= int(s[j])
str += '-' + s[j]
if total == 7:
ans = str
print(str+'=7') | s736820188 | Accepted | 17 | 3,060 | 318 | s = input()
for i in range(8):
total = int(s[0])
str = s[0]
for j in range(1, 4):
if i >> (j - 1) & 1:
total += int(s[j])
str += '+' + s[j]
else:
total -= int(s[j])
str += '-' + s[j]
if total == 7:
print(str + '=7')
exit() |
s761891555 | p03456 | u867826040 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 99 | AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number. | n = int("".join(input().split()))
x = n**(1/2)
if x%1 != 0:
print("No")
else:
print(int(x)) | s858185467 | Accepted | 18 | 3,060 | 98 | n = int("".join(input().split()))
x = n**(1/2)
if x%1 != 0:
print("No")
else:
print("Yes") |
s126917146 | p03007 | u994988729 | 2,000 | 1,048,576 | Wrong Answer | 259 | 19,532 | 355 | There are N integers, A_1, A_2, ..., A_N, written on a blackboard. We will repeat the following operation N-1 times so that we have only one integer on the blackboard. * Choose two integers x and y on the blackboard and erase these two integers. Then, write a new integer x-y. Find the maximum possible value of the final integer on the blackboard and a sequence of operations that maximizes the final integer. | from collections import deque
n = int(input())
a = list(map(int, input().split()))
a.sort()
a = deque(a)
left = a.popleft()
right = a.pop()
ans = []
for i in a:
if i > 0:
ans.append((left, i))
left -= i
else:
ans.append((right, i))
right -= i
else:
ans.append((right, left))
for x, y in ans:
print(x, y)
| s229691525 | Accepted | 261 | 19,524 | 373 | from collections import deque
n = int(input())
a = list(map(int, input().split()))
a.sort()
a = deque(a)
left = a.popleft()
right = a.pop()
ans = []
for i in a:
if i > 0:
ans.append((left, i))
left -= i
else:
ans.append((right, i))
right -= i
else:
ans.append((right, left))
print(right-left)
for x, y in ans:
print(x, y)
|
s385306064 | p02259 | u905313459 | 1,000 | 131,072 | Time Limit Exceeded | 40,000 | 7,376 | 256 | Write a program of the Bubble Sort algorithm which sorts a sequence _A_ in ascending order. The algorithm should be based on the following pseudocode: BubbleSort(A) 1 for i = 0 to A.length-1 2 for j = A.length-1 downto i+1 3 if A[j] < A[j-1] 4 swap A[j] and A[j-1] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 4 of the pseudocode. | import sys
x = input()
y = input().split()
r = sorted(y)
c = 0
while y != r:
for i, j in enumerate(y[:-1]):
if y[i] > y[i+1]:
y[i], y[i+1] = y[i+1], y[i]
c += 1
else:
break
print(" ".join(y))
print(c) | s556346758 | Accepted | 20 | 7,704 | 231 | x = int(input())
y = input()
y = list(map(int, y.split()))
r = sorted(y)
c = 0
while y != r:
for i, j in enumerate(y[:-1]):
if y[i] > y[i+1]:
y[i], y[i+1] = y[i+1], y[i]
c += 1
print(*y)
print(c) |
s531122085 | p03563 | u580093517 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 37 | Takahashi is a user of a site that hosts programming contests. When a user competes in a contest, the _rating_ of the user (not necessarily an integer) changes according to the _performance_ of the user, as follows: * Let the current rating of the user be a. * Suppose that the performance of the user in the contest is b. * Then, the new rating of the user will be the avarage of a and b. For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000. Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest. Find the performance required to achieve it. | print(int(input())**2 - int(input())) | s571478375 | Accepted | 18 | 2,940 | 39 | print(-(int(input()) - 2*int(input()))) |
s068035689 | p03361 | u460468647 | 2,000 | 262,144 | Wrong Answer | 21 | 3,188 | 521 | 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())
array = []
for _ in range(h):
array.append(list(map(str,input())))
print(array)
for i in range(h):
array[i].insert(0,".")
array[i].append(".")
print(array)
array.insert(0,["."]*(w+2))
array.append(["."]*(w+2))
print(array)
for i in range(1,h+1):
for j in range(1,w+1):
if array[i][j]=="#" and array[i][j-1] == array[i][j+1] == array[i-1][j] == array[i+1][j] == ".":
print("No")
break
if i==h and j==w:
print("Yes")
break
else:
continue
break | s949220300 | Accepted | 18 | 3,064 | 313 | H,W=map(int,input().split())
S=["."+input()+"." for i in range(H)]
S=["."*(W+2)]+S+["."*(W+2)]
flag=0
for i in range(H):
for j in range(W):
if S[i][j]=="#":
if S[i-1][j]=="." and S[i+1][j]=="." and S[i][j-1]=="." and S[i][j+1]==".":
flag=1
print("Yes" if flag==0 else "No") |
s404224557 | p03827 | u214547877 | 2,000 | 262,144 | Wrong Answer | 30 | 9,160 | 118 | You have an integer variable x. Initially, x=0. Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`. Find the maximum value taken by x during the operations (including before the first operation, and after the last operation). | N = int(input())
S = input()
ans = 0
for i in range(N):
if S[i] == "I":
ans += 1
else:
ans -= 1
print(ans) | s952173437 | Accepted | 31 | 9,176 | 139 | N = int(input())
S = input()
ans = 0
x = 0
for i in range(N):
if S[i] == "I":
x += 1
else:
x -= 1
ans = max(ans,x)
print(ans) |
s090412570 | p03605 | u911612592 | 2,000 | 262,144 | Wrong Answer | 33 | 9,092 | 79 | It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N? | s = str(input())
if s[0] == 9 or s[1] == 9:
print("Yes")
else:
print("No") | s582958345 | Accepted | 29 | 9,016 | 53 | if "9" in input():
print("Yes")
else:
print("No") |
s958266433 | p03659 | u099566485 | 2,000 | 262,144 | Wrong Answer | 229 | 24,948 | 216 | Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer a_i written on it. They will share these cards. First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards. Here, both Snuke and Raccoon have to take at least one card. Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively. They would like to minimize |x-y|. Find the minimum possible value of |x-y|. | #067-C
n=int(input())
a=list(map(int,input().split()))
sa=[a[0] for i in range(n)]
for i in range(1,n):
sa[i]=sa[i-1]+a[i]
mm=float("INF")
for i in range(n):
mm=min(mm,abs(a[i]-(sa[n-1]-a[i])))
print(int(mm)) | s495045545 | Accepted | 231 | 24,812 | 220 | #067-C
n=int(input())
a=list(map(int,input().split()))
sa=[a[0] for i in range(n)]
for i in range(1,n):
sa[i]=sa[i-1]+a[i]
mm=float("INF")
for i in range(n-1):
mm=min(mm,abs(sa[i]-(sa[n-1]-sa[i])))
print(int(mm)) |
s311130182 | p03090 | u509911932 | 2,000 | 1,048,576 | Wrong Answer | 45 | 4,484 | 878 | You are given an integer N. Build an undirected graph with N vertices with indices 1 to N that satisfies the following two conditions: * The graph is simple and connected. * There exists an integer S such that, for every vertex, the sum of the indices of the vertices adjacent to that vertex is S. It can be proved that at least one such graph exists under the constraints of this problem. | """
234 23
134 14
124 14
123 23
"""
import copy
N = 100
initial = [list(range(1, N + 1)) for i in range(N)]
for i in range(N):
initial[i].pop(i)
# print(initial)
neighbors = copy.deepcopy(initial)
S = sum(neighbors[-1])
while True:
# print(S)
# print(neighbors)
for i, tyouten in enumerate(neighbors):
delta = sum(tyouten) - S
if delta in tyouten:
tyouten.remove(delta)
neighbors[delta - 1].remove(i + 1)
elif delta == 0:
continue
else:
S -= 1
neighbors = copy.deepcopy(initial)
break
else:
break
# print(neighbors)
ans = []
for i, tyouten in enumerate(neighbors):
for j in tyouten:
if j > i+1:
ans.append(str(i+1)+" "+str(j))
print(len(ans))
for s in ans:
print(s)
| s891760819 | Accepted | 43 | 4,356 | 887 | """
234 23
134 14
124 14
123 23
"""
import copy
N = int(input())
initial = [list(range(1, N + 1)) for i in range(N)]
for i in range(N):
initial[i].pop(i)
# print(initial)
neighbors = copy.deepcopy(initial)
S = sum(neighbors[-1])
while True:
# print(S)
# print(neighbors)
for i, tyouten in enumerate(neighbors):
delta = sum(tyouten) - S
if delta in tyouten:
tyouten.remove(delta)
neighbors[delta - 1].remove(i + 1)
elif delta == 0:
continue
else:
S -= 1
neighbors = copy.deepcopy(initial)
break
else:
break
# print(neighbors)
ans = []
for i, tyouten in enumerate(neighbors):
for j in tyouten:
if j > i+1:
ans.append(str(i+1)+" "+str(j))
print(len(ans))
for s in ans:
print(s)
|
s058053849 | p03024 | u814986259 | 2,000 | 1,048,576 | Wrong Answer | 28 | 9,004 | 68 | Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility. | S=input()
k=S.count("o")
if k>=8:
print("YES")
else:
print("NO") | s984335156 | Accepted | 31 | 9,012 | 87 | S=input()
l=len(S)
k=S.count("o")
K=15-l
if k+K>=8:
print("YES")
else:
print("NO")
|
s694052030 | p03369 | u728498511 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 39 | In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen. | print(700+100*list(input()).count("x")) | s729169628 | Accepted | 17 | 2,940 | 39 | print(700+100*list(input()).count("o")) |
s878851259 | p03455 | u082584636 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 124 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. |
int_list = list(map(int, input().split()))
if (((int_list[0] + int_list[1]) % 2) == 0):
print('Even')
else:
print('Odd') | s698105848 | Accepted | 17 | 2,940 | 125 |
int_list = list(map(int, input().split()))
if (((int_list[0]*int_list[1]) % 2) == 0):
print('Even')
else:
print('Odd')
|
s171184664 | p03679 | u461636820 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 137 | 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. | X, A, B = map(int, input().split())
if A - B <= X:
print('delicious')
elif B - A >= X:
print('dangerous')
else:
print('safe') | s810075976 | Accepted | 18 | 2,940 | 133 | X, A, B = map(int, input().split())
if A >= B:
print('delicious')
elif B - A <= X:
print('safe')
else:
print('dangerous') |
s741695582 | p02690 | u606523772 | 2,000 | 1,048,576 | Time Limit Exceeded | 2,248 | 27,832 | 196 | Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X. | X = int(input())
for i in range(-800, 800):
for j in range(-800, 800):
print(i**5 - j**5)
if i**5 - j**5 == X:
ans = [i, j]
break
print(ans) | s333255655 | Accepted | 544 | 9,044 | 189 | X = int(input())
for i in range(-500, 500):
for j in range(-500, 500):
if i**5 - j**5 == X:
ans = [str(i), str(j)]
break
print(" ".join(ans)) |
s863268295 | p03448 | u119983020 | 2,000 | 262,144 | Wrong Answer | 48 | 3,060 | 202 | 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 = int(input())
B = int(input())
C = int(input())
X = int(input())
cnt = 0
for a in range(A):
for b in range(B):
for c in range(C):
if X == 500*a + 100*b + 50*c:
cnt += 1
print(cnt) | s436676476 | Accepted | 51 | 3,060 | 208 | A = int(input())
B = int(input())
C = int(input())
X = int(input())
cnt = 0
for a in range(A+1):
for b in range(B+1):
for c in range(C+1):
if X == 500*a + 100*b + 50*c:
cnt += 1
print(cnt) |
s092883903 | p02614 | u382407432 | 1,000 | 1,048,576 | Wrong Answer | 27 | 9,204 | 361 | We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is `.`, and black if c_{i,j} is `#`. Consider doing the following operation: * Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns. You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices. | H, W, K = map(int,(input().split()))
c=[]
for _ in range(H):
c.append(input())
ans=0
for maskR in range(H):
for maskC in range(W):
black=0
for i in range(H):
for j in range(W):
if ((maskR >> i) & 1) == 0 and ((maskC >> j) & 1) == 0 and c[i][j] == '#':
black = black + 1
if black == K:
ans = ans + 1
print(ans)
| s706687583 | Accepted | 66 | 9,208 | 367 | H, W, K = map(int,(input().split()))
c = []
for _ in range(H):
c.append(input())
ans = 0
for maskR in range((1<<H)-1):
for maskC in range((1<<W)-1):
black=0
for i in range(H):
for j in range(W):
if ((maskR >> i) & 1) == 0 and ((maskC >> j) & 1) == 0 and c[i][j] == '#':
black += 1
if black == K:
ans += 1
print(ans)
|
s139530306 | p03711 | u320763652 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 196 | Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group. | a,b = map(int, input().split())
listb = [4,6,9,11]
tmp = a in listb
print(tmp)
if (a == 2 and b ==2):
print("YES")
elif (a in listb and b in listb):
print("YES")
else:
print("NO") | s268120638 | Accepted | 18 | 3,064 | 274 | a,b = map(int, input().split())
lista = [1,3,5,7,8,10,12]
listb = [4,6,9,11]
tmp = a in listb
# print(tmp)
if (a == 2 and b ==2):
print("Yes")
elif (a in listb and b in listb):
print("Yes")
elif (a in lista and b in lista):
print("Yes")
else:
print("No") |
s132208432 | p04012 | u131406102 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 351 | 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. | a = list(input())
a = sorted(a)
def count(a):
b =[]
for i in range(0,len(a)-2):
if a[i] != a[i+1]:
b.append( i+1)
return b
b = count(a)
b.append(len(a))
for i in reversed(range(0,len(b)-1)):
b[i+1] =b[i+1] - b[i]
d = 0
for i in range(0,len(b)):
d = d +b[i]%2
if d ==0:
print('yes')
else:
print('no') | s977999972 | Accepted | 18 | 3,064 | 351 | a = list(input())
a = sorted(a)
def count(a):
b =[]
for i in range(0,len(a)-2):
if a[i] != a[i+1]:
b.append( i+1)
return b
b = count(a)
b.append(len(a))
for i in reversed(range(0,len(b)-1)):
b[i+1] =b[i+1] - b[i]
d = 0
for i in range(0,len(b)):
d = d +b[i]%2
if d ==0:
print('Yes')
else:
print('No') |
s097319999 | p03943 | u870841038 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 148 | Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students. | li = list(map(int, input().split()))
if li[0]+li[1] == li[2] or li[1]+li[2] == li[0] or li[2]+li[0] == li[1]:
print('YES')
else:
print('NO') | s876762461 | Accepted | 17 | 2,940 | 148 | li = list(map(int, input().split()))
if li[0]+li[1] == li[2] or li[1]+li[2] == li[0] or li[2]+li[0] == li[1]:
print('Yes')
else:
print('No') |
s640434774 | p02613 | u002663801 | 2,000 | 1,048,576 | Wrong Answer | 147 | 9,216 | 358 | Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format. | n = int(input())
ac = 0
wa = 0
tle = 0
re = 0
for i in range(n):
s = input()
if s == "AC":
ac += 1
if s == "WA":
wa += 1
if s == "TLE":
tle += 1
if s == "RE":
re += 1
a1 = "AC × " + str(ac)
a2 = "WA × " + str(wa)
a3 = "TLE × " + str(tle)
a4 = "RE × " + str(re)
print(a1)
print(a2)
print(a3)
print(a4) | s164949046 | Accepted | 151 | 9,220 | 354 | n = int(input())
ac = 0
wa = 0
tle = 0
re = 0
for i in range(n):
s = input()
if s == "AC":
ac += 1
if s == "WA":
wa += 1
if s == "TLE":
tle += 1
if s == "RE":
re += 1
a1 = "AC x " + str(ac)
a2 = "WA x " + str(wa)
a3 = "TLE x " + str(tle)
a4 = "RE x " + str(re)
print(a1)
print(a2)
print(a3)
print(a4) |
s126435944 | p03738 | u533039576 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 253 | You are given two positive integers A and B. Compare the magnitudes of these numbers. | a = input()
b = input()
max_len = 105
if len(a) < max_len:
a = '0' * (max_len - len(a)) + a
if len(b) < max_len:
b = '0' * (max_len - len(b)) + b
if a > b:
ans = "GRATER"
elif a == b:
ans = "EQUAL"
else:
ans = "LESS"
print(ans)
| s216927799 | Accepted | 18 | 3,060 | 254 | a = input()
b = input()
max_len = 105
if len(a) < max_len:
a = '0' * (max_len - len(a)) + a
if len(b) < max_len:
b = '0' * (max_len - len(b)) + b
if a > b:
ans = "GREATER"
elif a == b:
ans = "EQUAL"
else:
ans = "LESS"
print(ans)
|
s498565793 | p03473 | u750389519 | 2,000 | 262,144 | Wrong Answer | 28 | 9,088 | 27 | How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December? | M=input()
M=int(M)
M=48-M
| s324034406 | Accepted | 29 | 8,836 | 32 | M=input()
M=int(M)
print(48-M)
|
s673410077 | p03943 | u027403702 | 2,000 | 262,144 | Wrong Answer | 19 | 3,060 | 105 | Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students. | a,b,c = input().split()
if a + b == c or a + c == b or b + c == a:
print("Yes")
else:
print("No") | s293109907 | Accepted | 18 | 2,940 | 115 | a,b,c = map(int, input().split())
if a + b == c or a + c == b or b + c == a:
print("Yes")
else:
print("No") |
s361958606 | p03861 | u978510477 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 58 | 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().split())
print(b // x - a // x) | s060716374 | Accepted | 17 | 2,940 | 64 | a, b, x = map(int, input().split())
print(b // x - (a - 1) // x) |
s029147917 | p03624 | u396961814 | 2,000 | 262,144 | Wrong Answer | 60 | 4,800 | 158 | You are given a string S consisting of lowercase English letters. Find the lexicographically (alphabetically) smallest lowercase English letter that does not occur in S. If every lowercase English letter occurs in S, print `None` instead. | s = list(sorted(input()))
ans = 'None'
for i in range(26):
if chr(97 + i) not in s:
ans = chr(97 + i)
print(ans)
break
print(ans) | s958961185 | Accepted | 60 | 4,800 | 140 | s = list(sorted(input()))
ans = 'None'
for i in range(26):
if chr(97 + i) not in s:
ans = chr(97 + i)
break
print(ans)
|
s471006461 | p02233 | u798803522 | 1,000 | 131,072 | Wrong Answer | 20 | 7,604 | 220 | Write a program which prints $n$-th fibonacci number for a given integer $n$. The $n$-th fibonacci number is defined by the following recursive formula: \begin{equation*} fib(n)= \left \\{ \begin{array}{ll} 1 & (n = 0) \\\ 1 & (n = 1) \\\ fib(n - 1) + fib(n - 2) & \\\ \end{array} \right. \end{equation*} | target = int(input())
fib = [0 for n in range(target + 1)]
for i in range(target):
if i == 0:
continue
elif i == 1:
fib[i] = 1
else:
fib[i] = fib[i - 1] + fib[i - 2]
print(fib[target]) | s960290085 | Accepted | 20 | 7,768 | 228 | target = int(input()) + 1
fib = [0 for n in range(target + 1)]
for i in range(target + 1):
if i == 0:
continue
elif i == 1:
fib[i] = 1
else:
fib[i] = fib[i - 1] + fib[i - 2]
print(fib[target]) |
s789377090 | p00014 | u252368621 | 1,000 | 131,072 | Wrong Answer | 40 | 7,612 | 83 | Write a program which computes the area of a shape represented by the following three lines: $y = x^2$ $y = 0$ $x = 600$ It is clear that the area is $72000000$, if you use an integral you learn in high school. On the other hand, we can obtain an approximative area of the shape by adding up areas of many rectangles in the shape as shown in the following figure: $f(x) = x^2$ The approximative area $s$ where the width of the rectangles is $d$ is: area of rectangle where its width is $d$ and height is $f(d)$ $+$ area of rectangle where its width is $d$ and height is $f(2d)$ $+$ area of rectangle where its width is $d$ and height is $f(3d)$ $+$ ... area of rectangle where its width is $d$ and height is $f(600 - d)$ The more we decrease $d$, the higer-precision value which is close to $72000000$ we could obtain. Your program should read the integer $d$ which is a divisor of $600$, and print the area $s$. | d=int(input())
sum=0
for i in range(int(600/d)):
sum+=(((i*d)**2)*d)
print(sum) | s876835198 | Accepted | 30 | 7,676 | 171 | while(True):
try:
d=int(input())
sum=0
for i in range(int(600/d)):
sum+=(((i*d)**2)*d)
print(sum)
except:
break |
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