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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s016034031
|
p03761
|
u548545174
| 2,000 | 262,144 |
Wrong Answer
| 25 | 3,572 | 312 |
Snuke loves "paper cutting": he cuts out characters from a newspaper headline and rearranges them to form another string. He will receive a headline which contains one of the strings S_1,...,S_n tomorrow. He is excited and already thinking of what string he will create. Since he does not know the string on the headline yet, he is interested in strings that can be created regardless of which string the headline contains. Find the longest string that can be created regardless of which string among S_1,...,S_n the headline contains. If there are multiple such strings, find the lexicographically smallest one among them.
|
import copy
n = int(input())
S = [list(input()) for _ in range(n)]
common_list = copy.deepcopy(S[0])
for s in S:
print(s, common_list)
for i in common_list:
if i not in s:
common_list.remove(i)
else:
s.remove(i)
print("".join(sorted(common_list)))
|
s232855903
|
Accepted
| 33 | 3,956 | 340 |
import string
alphabets = string.ascii_lowercase
n = int(input())
S = [input() for _ in range(n)]
counter = {}
for alphabet in alphabets:
alphabet_count = [s.count(alphabet) for s in S]
counter[alphabet] = min(alphabet_count)
ans = ""
for alphabet, count in counter.items():
ans += alphabet*count
print("".join(sorted(ans)))
|
s320036512
|
p03699
|
u547167033
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,064 | 260 |
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)]
dp=[[0]*10 for i in range(n+1)]
for i in range(1,n+1):
for j in range(10):
dp[i][(j+s[i-1])%10]=max(dp[i-1][(j+s[i-1])%10],dp[i-1][j]+s[i-1])
ans=0
for i in range(1,10):
ans=max(ans,dp[n][i])
print(ans)
|
s348224590
|
Accepted
| 18 | 3,064 | 276 |
n=int(input())
s=[int(input()) for i in range(n)]
dp=[[-10**9]*10 for i in range(n+1)]
dp[0][0]=0
for i in range(1,n+1):
for j in range(10):
dp[i][(j+s[i-1])%10]=max(dp[i-1][(j+s[i-1])%10],dp[i-1][j]+s[i-1])
ans=0
for i in range(1,10):
ans=max(ans,dp[n][i])
print(ans)
|
s347812726
|
p03997
|
u940652437
| 2,000 | 262,144 |
Wrong Answer
| 31 | 9,032 | 63 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a=int(input())
b=int(input())
h=int(input())
print((a+b)*h/0.5)
|
s081649812
|
Accepted
| 30 | 9,096 | 66 |
a=int(input())
b=int(input())
h=int(input())
print(int((a+b)*h/2))
|
s820897929
|
p03360
|
u350093546
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 86 |
There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations?
|
a=list(map(int,input().split()))
n=int(input())
x=max(a)*2**n
print(x+sum(a)-max(a)*2)
|
s827538137
|
Accepted
| 17 | 2,940 | 87 |
a=list(map(int,input().split()))
n=int(input())
x=max(a)*(2**n)
print(x+sum(a)-max(a))
|
s602582389
|
p03457
|
u606523772
| 2,000 | 262,144 |
Wrong Answer
| 472 | 27,380 | 421 |
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
N = int(input())
point = [0, 0]
t_list = []
ans = "Yes"
t = list(map(int, input().split()))
dis = abs(point[0]-t[1])+abs(point[1]-t[2])
t_list.append(t)
if t[0] < dis:
ans = "No"
point = [t[1], t[2]]
for i in range(N-1):
t = list(map(int, input().split()))
dis = abs(point[0]-t[1])+abs(point[1]-t[2])
if t_list[-1][0]-t[0] < dis:
ans = "No"
point = [t[1], t[2]]
t_list.append(t)
print(ans)
|
s053286715
|
Accepted
| 490 | 27,380 | 471 |
N = int(input())
point = [0, 0]
t_list = []
ans = "Yes"
t = list(map(int, input().split()))
dis = abs(point[0]-t[1])+abs(point[1]-t[2])
t_list.append(t)
if t[0] < dis or t[0]%2!=dis%2:
ans = "No"
point = [t[1], t[2]]
for i in range(N-1):
t = list(map(int, input().split()))
dis = abs(point[0]-t[1])+abs(point[1]-t[2])
if t[0]-t_list[-1][0] < dis or (t[0]-t_list[-1][0])%2!=dis%2:
ans = "No"
point = [t[1], t[2]]
t_list.append(t)
print(ans)
|
s393445450
|
p03351
|
u328755070
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 148 |
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a, b, c, d = list(map(int, input().split()))
if abs(a - c) <= d or (abs(a - c) <= d and a <= b and b <= c):
print('Yes')
else:
print('No')
|
s374917355
|
Accepted
| 17 | 2,940 | 146 |
a, b, c, d = list(map(int, input().split()))
if abs(a - c) <= d or (abs(a - b) <= d and abs(b - c) <= d):
print('Yes')
else:
print('No')
|
s827386971
|
p02614
|
u698868214
| 1,000 | 1,048,576 |
Wrong Answer
| 52 | 9,108 | 456 |
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())
board = [list(input()) for _ in range(H)]
ans = 0
for i in range(2**H):
part1 = []
part1 += board
for j in range(H):
if (i>>j)&1:
part1[j] = ["*"]*W
for k in range(2**W):
part2 = []
part2 += part1
for l in range(W):
if (k>>l)&1:
for n in range(H):
part2[n][l] = "*"
cnt = 0
for m in part2:
cnt += m.count("#")
if cnt == K:
ans += 1
print(ans)
|
s559504552
|
Accepted
| 63 | 9,048 | 316 |
H,W,K = map(int,input().split())
c = [list(input()) for _ in range(H)]
ans = 0
for I in range(2**H):
for J in range(2**W):
black = 0
for i in range(H):
for j in range(W):
if (I>>i)&1 == 0 and (J>>j)&1 == 0 and c[i][j] == "#":
black += 1
if black == K:
ans += 1
print(ans)
|
s870099392
|
p03168
|
u060938295
| 2,000 | 1,048,576 |
Wrong Answer
| 1,360 | 14,676 | 366 |
Let N be a positive odd number. There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i. Taro has tossed all the N coins. Find the probability of having more heads than tails.
|
import sys
import numpy as np
sys.setrecursionlimit(10 ** 9)
#def input():
# return sys.stdin.readline()[:-1]
N = int(input())
P = list(map(float,input().split()))
dp = np.zeros(N+1, dtype=float)
dp[0] = 1
for c in P:
dp = np.append(dp[:-1] * (1-c), 0) + np.append(0, dp[:-1] * c)
print(dp)
ans = np.sum(dp[N//2+1:])
print(ans)
|
s655799945
|
Accepted
| 282 | 12,504 | 364 |
import sys
import numpy as np
sys.setrecursionlimit(10 ** 9)
def input():
return sys.stdin.readline()[:-1]
N = int(input())
P = list(map(float,input().split()))
dp = np.zeros(N+1, dtype=float)
dp[0] = 1
for c in P:
dp = np.append(dp[:-1] * (1-c), 0) + np.append(0, dp[:-1] * c)
# print(dp)
ans = np.sum(dp[N//2+1:])
print(ans)
|
s440298129
|
p03635
|
u649558044
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 43 |
In _K-city_ , there are n streets running east-west, and m streets running north-south. Each street running east-west and each street running north-south cross each other. We will call the smallest area that is surrounded by four streets a block. How many blocks there are in K-city?
|
s = input()
print(s[0]+str(len(s)-2)+s[-1])
|
s440204968
|
Accepted
| 17 | 2,940 | 60 |
n, m = map(int, input().split(' '))
print((n - 1) * (m - 1))
|
s352064704
|
p02612
|
u204437779
| 2,000 | 1,048,576 |
Wrong Answer
| 30 | 9,144 | 36 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n = int(input())
t = n%1000
print(t)
|
s404069757
|
Accepted
| 29 | 9,168 | 108 |
n = int(input())
s = (n-(n%1000))/1000
if 1000*s == n:
print(0)
else:
t = int(1000*(s+1) - n)
print(t)
|
s390901586
|
p03433
|
u409064224
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 114 |
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())
while n < 0:
n -= 500
n += 500
if n <= a:
print("Yes")
else:
print("No")
|
s819595131
|
Accepted
| 17 | 2,940 | 92 |
n = int(input())
a = int(input())
n = n % 500
if n <= a:
print("Yes")
else:
print("No")
|
s205795111
|
p02389
|
u839788696
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,572 | 43 |
Write a program which calculates the area and perimeter of a given rectangle.
|
a,b = map(int, input().split())
print(a*b)
|
s812543019
|
Accepted
| 20 | 5,580 | 64 |
a,b = map(int, input().split())
m= a*b
l = 2*a + 2*b
print(m,l)
|
s669297380
|
p03659
|
u021548497
| 2,000 | 262,144 |
Wrong Answer
| 149 | 24,812 | 188 |
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|.
|
n = int(input())
a = [int(x) for x in input().split()]
ans = pow(10, 15)
key = sum(a)
p = key
for i in range(n):
p += a[i]
if ans > abs(key-2*p):
ans = abs(key-2*p)
print(ans)
|
s645930380
|
Accepted
| 146 | 24,832 | 188 |
n = int(input())
a = [int(x) for x in input().split()]
ans = pow(10, 15)
key = sum(a)
p = 0
for i in range(n-1):
p += a[i]
if ans > abs(key-2*p):
ans = abs(key-2*p)
print(ans)
|
s119631892
|
p03371
|
u552201227
| 2,000 | 262,144 |
Wrong Answer
| 117 | 3,060 | 154 |
"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively. Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.
|
a,b,c,x,y=map(int,input().split())
ans=a*x+b*y
for i in range(1,max(x,y)+1):
ab=i*2
abc=ab*c+a*max(0,x-1)+b*max(0,y-1)
ans = min(ans,abc)
print(ans)
|
s737668131
|
Accepted
| 120 | 3,060 | 155 |
a,b,c,x,y=map(int,input().split())
ans=a*x+b*y
for i in range(1,max(x,y)+1):
ab=i*2
abc=ab*c+a*max(0,x-i)+b*max(0,y-i)
ans = min(ans,abc)
print(ans)
|
s970678773
|
p03846
|
u595716769
| 2,000 | 262,144 |
Wrong Answer
| 117 | 13,880 | 489 |
There are N people, conveniently numbered 1 through N. They were standing in a row yesterday, but now they are unsure of the order in which they were standing. However, each person remembered the following fact: the absolute difference of the number of the people who were standing to the left of that person, and the number of the people who were standing to the right of that person. According to their reports, the difference above for person i is A_i. Based on these reports, find the number of the possible orders in which they were standing. Since it can be extremely large, print the answer modulo 10^9+7. Note that the reports may be incorrect and thus there may be no consistent order. In such a case, print 0.
|
n = int(input())
L = [int(i) for i in input().split()]
L.sort()
print(L)
def out_detect(n,L):
flag = 0
if n%2 == 0:
for i in range(n):
if L[i] != -(-(i+1)//2)*2-1:
flag = 1
break
else:
if L[0] != 0:
flag = 1
return (flag)
for i in range(1,n):
if L[i] != (-(-i//2)*2):
flag = 1
break
return (flag)
if out_detect(n,L) == 1:
print(0)
else:
if n%2 == 0:
print(pow(2,n//2))
else:
print(pow(2,(n-1)//2))
|
s725564396
|
Accepted
| 104 | 13,880 | 505 |
n = int(input())
L = [int(i) for i in input().split()]
L.sort()
def out_detect(n,L):
flag = 0
if n%2 == 0:
for i in range(n):
if L[i] != -(-(i+1)//2)*2-1:
flag = 1
break
else:
if L[0] != 0:
flag = 1
return (flag)
for i in range(1,n):
if L[i] != (-(-i//2)*2):
flag = 1
break
return (flag)
if out_detect(n,L) == 1:
print(0)
else:
if n%2 == 0:
out = pow(2,n//2)
else:
out = pow(2,(n-1)//2)
print(out%(pow(10,9)+7))
|
s195086327
|
p03760
|
u772180901
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 173 |
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions. You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
|
o = input()
e = input()
ans = ""
m = min(len(o),len(e))
for i in range(m):
ans += o[i]
ans += e[i]
if o < e:
ans += e[-1]
elif e < o:
ans += o[-1]
print(ans)
|
s387722842
|
Accepted
| 18 | 3,064 | 193 |
o = input()
e = input()
ans = ""
m = min(len(o),len(e))
for i in range(m):
ans += o[i]
ans += e[i]
if len(o) < len(e):
ans += e[-1]
elif len(e) < len(o):
ans += o[-1]
print(ans)
|
s740052256
|
p03739
|
u966740656
| 2,000 | 262,144 |
Wrong Answer
| 70 | 14,644 | 632 |
You are given an integer sequence of length N. The i-th term in the sequence is a_i. In one operation, you can select a term and either increment or decrement it by one. At least how many operations are necessary to satisfy the following conditions? * For every i (1≤i≤n), the sum of the terms from the 1-st through i-th term is not zero. * For every i (1≤i≤n-1), the sign of the sum of the terms from the 1-st through i-th term, is different from the sign of the sum of the terms from the 1-st through (i+1)-th term.
|
def c(ints):
for i in range(len(ints)):
if ints[i] != 0:
sig = 1 if ints[i] > 0 else -1
total = ints[i]
mov = i
j = i
break
if i == len(ints) - 1:
return i + 1
for i_ in ints[j+1:]:
tmp = total + i_
if tmp == 0:
mov +=1
tmp = -sig
elif sig * tmp > 0:
mov += abs(tmp) + 1
tmp = -sig
sig *= -1
total = tmp
return mov
_ = input()
inp = input()
inp = inp.split(' ')
inp = [int(i_) for i_ in inp]
c(inp)
|
s018560458
|
Accepted
| 97 | 14,468 | 707 |
def c(ints):
sig = -1
sig_ = +1
total = 0
total_ = 0
mov = 0
mov_ = 0
for i_ in ints:
tmp = total + i_
tmp_ = total_ + i_
if tmp == 0:
mov +=1
tmp = -sig
elif sig * tmp > 0:
mov += abs(tmp) + 1
tmp = -sig
if tmp_ == 0:
mov_ +=1
tmp_ = -sig_
elif sig_ * tmp_ > 0:
mov_ += abs(tmp_) + 1
tmp_ = -sig_
sig *= -1
total = tmp
sig_ *= -1
total_ = tmp_
return min(mov, mov_)
_ = input()
inp = input()
inp = inp.split(' ')
inp = [int(i_) for i_ in inp]
print(c(inp))
|
s293282779
|
p02678
|
u207241407
| 2,000 | 1,048,576 |
Wrong Answer
| 2,207 | 66,220 | 936 |
There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.
|
import numpy as np
N, M = map(int, input().split())
ab = np.array([list(map(int, input().split())) for _ in range(M)])
dp = [[0, float("inf")] for _ in range(N + 1)]
tmp = {1}
while tmp:
for i in list(tmp):
abi = ab[np.any(ab == i, axis=1)]
ab = np.delete(ab, np.where(ab == i)[0], axis=0)
for a, b in abi:
if a == 1:
dp[b][0] = 1
dp[b][1] = 1
tmp.add(b)
elif b == 1:
dp[a][0] = 1
dp[a][1] = 1
tmp.add(a)
else:
if dp[a][1] > dp[b][1] + 1:
dp[a][0] = b
dp[a][1] = dp[b][1] + 1
if dp[b][1] > dp[a][1] + 1:
dp[b][0] = a
dp[b][1] = dp[a][1] + 1
tmp.add(a)
tmp.add(b)
tmp.remove(i)
for ans in dp[2 : N + 1]:
print(ans[0])
|
s379229747
|
Accepted
| 448 | 34,940 | 704 |
import sys
from collections import deque
def main():
n, m = map(int, sys.stdin.readline().split())
to = [[] for _ in range(n + 1)]
for _ in range(m):
a, b = map(int, sys.stdin.readline().split())
to[a].append(b)
to[b].append(a)
INF = float("inf")
dist = [INF] * (n + 1)
dist[1] = 0
pre = [0] * (n + 1)
que = deque([])
que.append(1)
while que:
p = que.popleft()
for i in to[p]:
if dist[i] > dist[p] + 1:
dist[i] = dist[p] + 1
pre[i] = p
que.append(i)
print("Yes")
for i in range(2, n + 1):
print(pre[i])
if __name__ == "__main__":
main()
|
s396651067
|
p03251
|
u589432040
| 2,000 | 1,048,576 |
Wrong Answer
| 20 | 3,060 | 211 |
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.
|
N, M, X, Y = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
for Z in range(X+1, Y+1):
if max(x)<Z and min(y)>=Z:
print("No War")
else:
print("War")
|
s816851057
|
Accepted
| 18 | 3,060 | 231 |
N, M, X, Y = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
for Z in range(X+1, Y+1):
if max(x)<Z and min(y)>=Z:
war = "No War"
break
else:
war = "War"
print(war)
|
s061724829
|
p02388
|
u186524656
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,724 | 23 |
Write a program which calculates the cube of a given integer x.
|
print(2**3)
print(3**3)
|
s058622011
|
Accepted
| 30 | 6,724 | 28 |
x = int(input())
print(x**3)
|
s194436113
|
p03478
|
u243312682
| 2,000 | 262,144 |
Wrong Answer
| 48 | 3,580 | 362 |
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
def main():
inp = list(map(int, input().split()))
n = inp[0]
a = inp[1]
b = inp[2]
n_sum = 0
res = 0
for i in range(1, n+1):
n_list = list(map(int, list(str(i))))
n_sum = sum(n_list)
print(i, n_sum)
if a <= n_sum and n_sum <= b:
res += i
print(res)
if __name__ == '__main__':
main()
|
s935063708
|
Accepted
| 35 | 3,064 | 338 |
def main():
inp = list(map(int, input().split()))
n = inp[0]
a = inp[1]
b = inp[2]
n_sum = 0
res = 0
for i in range(1, n+1):
n_list = list(map(int, list(str(i))))
n_sum = sum(n_list)
if a <= n_sum and n_sum <= b:
res += i
print(res)
if __name__ == '__main__':
main()
|
s699440706
|
p03739
|
u665038048
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 472 |
You are given an integer sequence of length N. The i-th term in the sequence is a_i. In one operation, you can select a term and either increment or decrement it by one. At least how many operations are necessary to satisfy the following conditions? * For every i (1≤i≤n), the sum of the terms from the 1-st through i-th term is not zero. * For every i (1≤i≤n-1), the sign of the sum of the terms from the 1-st through i-th term, is different from the sign of the sum of the terms from the 1-st through (i+1)-th term.
|
a = list(map(int, input().split()))
t1, s1 = 0, 0
for i, x in enumerate(a):
s1 += x
if i % 2 == 0:
if s1 < 1:
t1 += (1-s1)
s1 = 1
else:
if s1 > -1:
t1 += (s1+1)
s1 = -1
t2, s2 = 0, 0
for i, x in enumerate(a):
s2 += x
if i % 2 == 0:
if s2 > -1:
t2 += (s2+1)
s2 = -1
else:
if s2 < 1:
t2 += (1-s2)
s2 = 1
print(min(t1, t2))
|
s200093085
|
Accepted
| 119 | 14,332 | 489 |
n = int(input())
a = list(map(int, input().split()))
t1, s1 = 0, 0
for i, x in enumerate(a):
s1 += x
if i % 2 == 0:
if s1 < 1:
t1 += (1-s1)
s1 = 1
else:
if s1 > -1:
t1 += (s1+1)
s1 = -1
t2, s2 = 0, 0
for i, x in enumerate(a):
s2 += x
if i % 2 == 0:
if s2 > -1:
t2 += (s2+1)
s2 = -1
else:
if s2 < 1:
t2 += (1-s2)
s2 = 1
print(min(t1, t2))
|
s117691539
|
p03478
|
u923659712
| 2,000 | 262,144 |
Wrong Answer
| 39 | 2,940 | 161 |
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
n,a,b=map(int,input().split())
oo=0
for i in range(1,n+1):
f=str(i)
l=0
for j in range(len(f)):
l+=int(f[j])
if a<=l<=b:
oo+=l
print(oo)
|
s577591651
|
Accepted
| 37 | 3,060 | 156 |
n,a,b=map(int,input().split())
oo=0
for i in range(1,n+1):
f=str(i)
l=0
for j in range(len(f)):
l+=int(f[j])
if a<=l<=b:
oo+=i
print(oo)
|
s405160629
|
p02694
|
u861886710
| 2,000 | 1,048,576 |
Wrong Answer
| 25 | 9,064 | 105 |
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?
|
import math
X = int(input())
A = 100
i = 0
while X >= A:
A = math.floor(A * 1.01)
i += 1
print(i)
|
s796232325
|
Accepted
| 24 | 9,116 | 104 |
import math
X = int(input())
A = 100
i = 0
while X > A:
A = math.floor(A * 1.01)
i += 1
print(i)
|
s295803036
|
p03910
|
u156743806
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 62 |
The problem set at _CODE FESTIVAL 20XX Finals_ consists of N problems. The score allocated to the i-th (1≦i≦N) problem is i points. Takahashi, a contestant, is trying to score exactly N points. For that, he is deciding which problems to solve. As problems with higher scores are harder, he wants to minimize the highest score of a problem among the ones solved by him. Determine the set of problems that should be solved.
|
import math
N=int(input())
print(int(math.sqrt(N*2-0.25)+0.5))
|
s554752257
|
Accepted
| 21 | 3,408 | 202 |
import math
N=int(input())
M=int(math.sqrt(N*2-0.25)+0.5)
left = M*(M-1) // 2
sum_=0
for i in range(1,M-(N-left)):
print(i)
sum_ += i
for i in range(M-(N-left),M):
print(i+1)
sum_ += i+1
|
s136831187
|
p03456
|
u110656379
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 111 |
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.
|
import math
a,b=input().split()
c=int(a+b)
d=math.sqrt(c)
if type(d)=="int":
print("Yes")
else:
print("No")
|
s654661389
|
Accepted
| 17 | 2,940 | 109 |
import math
a,b=input().split()
c=int(a+b)
d=int(math.sqrt(c))
if d**2==c:
print("Yes")
else:
print("No")
|
s933610440
|
p02255
|
u777299405
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,720 | 240 |
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.
|
n = int(input())
a = list(map(int, input().split()))
for i in range(1, len(a)):
# print(''.join(map(str, a)))
v = a[i]
j = i - 1
while j >= 0 and a[j] > v:
a[j + 1] = a[j]
j -= 1
a[j + 1] = v
print(*a)
|
s319026649
|
Accepted
| 40 | 6,756 | 217 |
n = int(input())
a = list(map(int, input().split()))
for i in range(1, len(a)):
print(*a)
v = a[i]
j = i - 1
while j >= 0 and a[j] > v:
a[j + 1] = a[j]
j -= 1
a[j + 1] = v
print(*a)
|
s042238532
|
p03493
|
u898042052
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 74 |
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
input_list = [int(i) for i in input().split()]
print(input_list.count(1))
|
s921401139
|
Accepted
| 17 | 2,940 | 74 |
args = input()
lst = [int(x) for x in list(str(args))]
print(lst.count(1))
|
s610114571
|
p02409
|
u217069758
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,604 | 301 |
You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room. For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth floor at building b. If v is negative, it means that −v persons left. Assume that initially no person lives in the building.
|
n = int(input())
house = [[[0 for i in range(10)] for j in range(3)] for k in range(4)]
for i in range(n):
b,f,r,v = map(int, input().split())
house[b - 1][f - 1][r - 1] += v
for k in range(4):
for j in range(3):
print(str(house[k][j]))
if k != 3:
print('#' * 20)
|
s229247117
|
Accepted
| 20 | 5,608 | 375 |
n = int(input())
oh = [[[0 for i in range(10)] for j in range(3)] for k in range(4)]
for tr in range(n):
b, f, r, v = map(int, input().split())
oh[b - 1][f - 1][r - 1] += v
for k in range(4):
for j in range(3):
tmp = ''
for i in range(10):
tmp += (' ' + str(oh[k][j][i]))
print(tmp)
if k < 3:
print('#' * 20)
|
s137607098
|
p03860
|
u020176853
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 52 |
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
#21:27
s = list(input().upper())
print("A"+s[0]+"C")
|
s555678381
|
Accepted
| 17 | 2,940 | 153 |
#21:27
abc = list(map(str, input().split()))
prex = list(abc[0].upper())
x = list(abc[1].upper())
posx = list(abc[2].upper())
print(prex[0]+x[0]+posx[0])
|
s363648748
|
p03549
|
u871303155
| 2,000 | 262,144 |
Wrong Answer
| 24 | 3,060 | 214 |
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`. When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases. Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds. Now, he goes through the following process: * Submit the code. * Wait until the code finishes execution on all the cases. * If the code fails to correctly solve some of the M cases, submit it again. * Repeat until the code correctly solve all the cases in one submission. Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
|
import math
N, M = map(int, input().split())
t = 100 * (N - M) + 1900 * M
ans = 0
first_p = (1/2) ** M
p = first_p
for i in range(1, 10001):
ans += i * t * p
p = p * (1 - first_p)
print(math.ceil(ans))
|
s362737209
|
Accepted
| 18 | 3,060 | 111 |
import math
N, M = map(int, input().split())
t = 100 * (N - M) + 1900 * M
ans = t * (2 ** M)
print(int(ans))
|
s042079341
|
p02614
|
u887152994
| 1,000 | 1,048,576 |
Wrong Answer
| 79 | 9,216 | 789 |
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.
|
import itertools
H,W,K= map(int,input().split())
List=[list(input()) for i in range(H)]
nums=[]
total=0
for i in range(1,H+W+1):
nums.append(i)
for q in range(0,H+W+1):
hoge=[]
for balls in itertools.combinations(nums, q):
hoge.append(balls)
for p in range(0,len(hoge)):
List_sub=List
for i in range(1,H+1):
if i in hoge[p]:
for j in range(0,W):
List_sub[i-1][j]="x"
for i in range(H+1,H+W+1):
if i in hoge[p]:
for j in range(0,H):
List_sub[j][i-H-1]="x"
s=0
for i in range(0,H):
for j in range(0,W):
if List_sub[i][j]=="#":
s+=1
if s== K:
total+=1
print(total)
|
s093408936
|
Accepted
| 165 | 9,452 | 804 |
import itertools
import copy
H,W,K= map(int,input().split())
List=[list(input()) for i in range(H)]
nums=[]
total=0
for i in range(1,H+W+1):
nums.append(i)
for q in range(0,H+W+1):
hoge=[]
for balls in itertools.combinations(nums, q):
hoge.append(balls)
for p in range(0,len(hoge)):
Listo=copy.deepcopy(List)
for i in range(1,H+1):
if i in hoge[p]:
for j in range(0,W):
Listo[i-1][j]="x"
for i in range(H+1,H+W+1):
if i in hoge[p]:
for j in range(0,H):
Listo[j][i-H-1]="x"
s=0
for i in range(0,H):
for j in range(0,W):
if Listo[i][j]=="#":
s+=1
if s== K:
total+=1
print(total)
|
s243513451
|
p03696
|
u279493135
| 2,000 | 262,144 |
Wrong Answer
| 295 | 21,516 | 796 |
You are given a string S of length N consisting of `(` and `)`. Your task is to insert some number of `(` and `)` into S to obtain a _correct bracket sequence_. Here, a correct bracket sequence is defined as follows: * `()` is a correct bracket sequence. * If X is a correct bracket sequence, the concatenation of `(`, X and `)` in this order is also a correct bracket sequence. * If X and Y are correct bracket sequences, the concatenation of X and Y in this order is also a correct bracket sequence. * Every correct bracket sequence can be derived from the rules above. Find the shortest correct bracket sequence that can be obtained. If there is more than one such sequence, find the lexicographically smallest one.
|
import sys, re
from collections import deque, defaultdict, Counter
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians, atan, degrees
from itertools import permutations, combinations, product, accumulate
from operator import itemgetter, mul
from copy import deepcopy
from string import ascii_lowercase, ascii_uppercase, digits
from fractions import gcd
from bisect import bisect
import numpy as np
def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
mod = 10 ** 9 + 7
N = INT()
S = input()
T = S.replace("()", "")
l = 0
r = 0
for x in T:
if x == ")":
l += 1
else:
r += 1
print("("*l + S + ")"*r)
|
s342581653
|
Accepted
| 17 | 3,060 | 166 |
N = int(input())
S = input()
l = 0
r = 0
for s in S:
if s == "(":
r += 1
else:
if r >= 1:
r -= 1
else:
l += 1
print("("*l + S + ")"*r)
|
s335497286
|
p02612
|
u489247698
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 9,148 | 44 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
A = N % 1000
print(int(A))
|
s386734125
|
Accepted
| 31 | 9,144 | 80 |
N = int(input())
if N%1000==0:
A = 0
else:
A =1000-( N % 1000)
print(int(A))
|
s820192796
|
p03943
|
u272522520
| 2,000 | 262,144 |
Wrong Answer
| 20 | 2,940 | 115 |
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.
|
s = list(map(int,input().split()))
SUM = s[0]+s[1]+s[2]
if SUM%2 ==0:
print("YES")
else:
print("NO")
|
s950201357
|
Accepted
| 17 | 3,060 | 156 |
a,b,c = list(map(int, input().split()))
if a == b+c:
print("Yes")
elif b == a+c:
print("Yes")
elif c == a+b:
print("Yes")
else:
print("No")
|
s989767066
|
p03448
|
u382303205
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 191 |
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())
count = 0
for i in range(a+1):
for j in range(b+1):
if x <= i*500 + j*100 + c*50:
count += 1
print(count)
|
s235122814
|
Accepted
| 18 | 3,064 | 214 |
a=int(input())
b=int(input())
c=int(input())
x=int(input())
count = 0
for i in range(a+1):
for j in range(b+1):
if x >= i*500 + j*100 and x <= i*500 + j*100 + c*50:
count += 1
print(count)
|
s325532931
|
p03563
|
u450883456
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 46 |
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.
|
a = int(input())
b = int(input())
print(2*a-b)
|
s590487605
|
Accepted
| 17 | 2,940 | 46 |
a = int(input())
b = int(input())
print(2*b-a)
|
s203935887
|
p03469
|
u500279510
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 43 |
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.strip('2017')))
|
s383847437
|
Accepted
| 17 | 2,940 | 44 |
S = input()
print('2018'+(S.lstrip('2017')))
|
s151608021
|
p03447
|
u064408584
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 286 |
You went shopping to buy cakes and donuts with X yen (the currency of Japan). First, you bought one cake for A yen at a cake shop. Then, you bought as many donuts as possible for B yen each, at a donut shop. How much do you have left after shopping?
|
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 sum(a[i+1:N])<sum(b[i:N-1]):
break
count = count +1
print(sum(a[0:count])+sum(b[count-1:N+1]))
C_Candies()
|
s768935669
|
Accepted
| 17 | 2,940 | 61 |
X =int(input())
A=int(input())
B=int(input())
print((X-A)%B)
|
s884786231
|
p03605
|
u235066013
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 72 |
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
N=str(input())
if N[0]==9 or N[1]==9:
print('Yes')
else:
print('NO')
|
s792430692
|
Accepted
| 17 | 2,940 | 76 |
N=str(input())
if N[0]=='9' or N[1]=='9':
print('Yes')
else:
print('No')
|
s273793875
|
p03485
|
u275488119
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 49 |
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
a, b = map(int, input().split())
print((a+b+1)/2)
|
s794924822
|
Accepted
| 57 | 3,060 | 54 |
a, b = map(int, input().split())
print(int((a+b+1)/2))
|
s184345604
|
p02850
|
u948524308
| 2,000 | 1,048,576 |
Wrong Answer
| 2,105 | 34,032 | 473 |
Given is a tree G with N vertices. The vertices are numbered 1 through N, and the i-th edge connects Vertex a_i and Vertex b_i. Consider painting the edges in G with some number of colors. We want to paint them so that, for each vertex, the colors of the edges incident to that vertex are all different. Among the colorings satisfying the condition above, construct one that uses the minimum number of colors.
|
N=int(input())
R=[[] for i in range(N)]
C=[[] for i in range(N)]
a=[]
b=[]
L=[]
for i in range(N-1):
ai,bi=map(int,input().split())
a.append(ai-1)
b.append(bi-1)
R[a[i]].append(b[i])
L.append(len(R[a[i]]))
K=max(L)
if len(R[0])!=K:
K+=1
print(K)
for i in range(N-1):
for k in range(1,K+1):
if not k in C[a[i]]:
print(k)
C[a[i]].append(k)
C[b[i]].append(k)
break
else:continue
|
s235671010
|
Accepted
| 820 | 61,508 | 627 |
N=int(input())
G=[[] for i in range(N)]
a=[]
b=[]
for i in range(N-1):
ai,bi=map(int,input().split())
a.append(ai-1)
b.append(bi-1)
G[ai-1].append(bi-1)
G[bi-1].append(ai-1)
k=[0]*N
from collections import deque
q=deque([0])
visited=[0]*N
ans={}
p=[-1]*N
kmx=0
while q:
x=q.popleft()
visited[x]=1
cnt=1
for i in G[x]:
if visited[i]!=0:continue
if p[x]==cnt:
cnt+=1
ans[(x,i)]=cnt
ans[(i,x)]=cnt
p[i] = cnt
cnt += 1
kmx=max(kmx,cnt-1)
q.append(i)
print(kmx)
for i in range(N-1):
print(ans[(a[i],b[i])])
|
s393924071
|
p03155
|
u477320129
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 74 |
It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board. The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns. How many ways are there to choose where to put the notice so that it completely covers exactly HW squares?
|
N = int(input())
H = int(input())
W = int(input())
print((N-H+1)+(N-W+1))
|
s754761053
|
Accepted
| 17 | 2,940 | 74 |
N = int(input())
H = int(input())
W = int(input())
print((N-H+1)*(N-W+1))
|
s619387958
|
p03555
|
u171366497
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 68 |
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
a=input()
b=input()
if a==b[::-1]:
print('Yes')
else:print('No')
|
s167618505
|
Accepted
| 18 | 2,940 | 68 |
a=input()
b=input()
if a==b[::-1]:
print('YES')
else:print('NO')
|
s476644796
|
p02393
|
u406434162
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,592 | 202 |
Write a program which reads three integers, and prints them in ascending order.
|
a,b,c = map(int,input().split())
if a > b:
x = a
a = b
b = a
elif b > c:
x = b
b = c
c = x
elif a > b:
x = a
a = b
b = x
print(a,b,c,end = " ")
|
s176802801
|
Accepted
| 20 | 5,596 | 214 |
a,b,c = map(int,input().split())
if a > b:
x = a
a = b
b = x
if b > c:
x = b
b = c
c = x
if a > b:
x = a
a = b
b = x
print(a,b,c,sep = " ")
|
s026264519
|
p03814
|
u506587641
| 2,000 | 262,144 |
Wrong Answer
| 64 | 11,256 | 171 |
Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s). Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`.
|
s = str(input())
Alst = []
Zlst = []
for i in range(len(s)):
if s[i]=='A':
Alst.append(i)
elif s[i]=='Z':
Zlst.append(i)
print(max(Zlst)-min(Alst))
|
s201795865
|
Accepted
| 66 | 11,204 | 173 |
s = str(input())
Alst = []
Zlst = []
for i in range(len(s)):
if s[i]=='A':
Alst.append(i)
elif s[i]=='Z':
Zlst.append(i)
print(max(Zlst)-min(Alst)+1)
|
s312584276
|
p03759
|
u854093727
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 89 |
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful.
|
a,b,c = map(int,input().split())
if b-a == c-b :
print("Yes")
else:
print("No")
|
s988040313
|
Accepted
| 17 | 2,940 | 89 |
a,b,c = map(int,input().split())
if b-a == c-b :
print("YES")
else:
print("NO")
|
s439751374
|
p02927
|
u268792407
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,060 | 197 |
Today is August 24, one of the five Product Days in a year. A date m-d (m is the month, d is the date) is called a Product Day when d is a two-digit number, and all of the following conditions are satisfied (here d_{10} is the tens digit of the day and d_1 is the ones digit of the day): * d_1 \geq 2 * d_{10} \geq 2 * d_1 \times d_{10} = m Takahashi wants more Product Days, and he made a new calendar called Takahashi Calendar where a year consists of M month from Month 1 to Month M, and each month consists of D days from Day 1 to Day D. In Takahashi Calendar, how many Product Days does a year have?
|
m,d=map(int,input().split())
month = [i for i in range(1,m+1)]
ct=0
if d<10:
print(0)
exit()
for i in range(11,d+1):
a=int(str(i)[0])
b=int(str(i)[1])
if a*b in month:
ct+=1
print(ct)
|
s420156332
|
Accepted
| 17 | 3,060 | 216 |
m,d=map(int,input().split())
month = [i for i in range(1,m+1)]
ct=0
if d<10:
print(0)
exit()
for i in range(11,d+1):
a=int(str(i)[0])
b=int(str(i)[1])
if a*b in month and a>=2 and b>=2:
ct+=1
print(ct)
|
s652343646
|
p03578
|
u858136677
| 2,000 | 262,144 |
Wrong Answer
| 2,105 | 36,956 | 309 |
Rng is preparing a problem set for a qualification round of CODEFESTIVAL. He has N candidates of problems. The difficulty of the i-th candidate is D_i. There must be M problems in the problem set, and the difficulty of the i-th problem must be T_i. Here, one candidate of a problem cannot be used as multiple problems. Determine whether Rng can complete the problem set without creating new candidates of problems.
|
n = int(input())
d = list(map(int,input().split()))
m = int(input())
t = list(map(int,input().split()))
for i in range(n):
for k in range(len(d)):
print(len(d))
if len(t) == 0:
print('YES')
break
elif t[0] == d[k]:
del d[k]
del t[0]
break
if len(t) == 0:
print('YES')
else:
print('NO')
|
s095656850
|
Accepted
| 284 | 57,056 | 316 |
import collections
n = int(input())
d = list(map(int,input().split()))
m = int(input())
t = list(map(int,input().split()))
dcounter = collections.Counter(d)
tcounter = collections.Counter(t)
check = True
for i in tcounter.items():
if dcounter[i[0]] < i[1]:
check = False
if check:
print('YES')
else:
print('NO')
|
s588831951
|
p03193
|
u391731808
| 2,000 | 1,048,576 |
Wrong Answer
| 21 | 3,188 | 125 |
There are N rectangular plate materials made of special metal called AtCoder Alloy. The dimensions of the i-th material are A_i \times B_i (A_i vertically and B_i horizontally). Takahashi wants a rectangular plate made of AtCoder Alloy whose dimensions are exactly H \times W. He is trying to obtain such a plate by choosing one of the N materials and cutting it if necessary. When cutting a material, the cuts must be parallel to one of the sides of the material. Also, the materials have fixed directions and cannot be rotated. For example, a 5 \times 3 material cannot be used as a 3 \times 5 plate. Out of the N materials, how many can produce an H \times W plate if properly cut?
|
N,H,W = map(int,input().split())
AB = [list(map(int,input().split())) for _ in [0]*N]
print(sum(a<=H and b<=W for a,b in AB))
|
s281169994
|
Accepted
| 20 | 3,188 | 126 |
N,H,W = map(int,input().split())
AB = [list(map(int,input().split())) for _ in [0]*N]
print(sum(a>=H and b>=W for a,b in AB))
|
s085293342
|
p03457
|
u711539583
| 2,000 | 262,144 |
Wrong Answer
| 441 | 27,324 | 328 |
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
import sys
n = int(input())
txy = []
for i in range(n):
txy.append(list(map(int, input().split())))
pt = 0
py = 0
px = 0
for t, x, y in txy:
dt = t - pt
dx = x - px
dy = y - py
ax = abs(dx) % 2
ay = abs(dy) % 2
a = ax + ay % 2
if abs(dx) + abs(dy) > dt or a != dt % 2:
print("No")
sys.exit()
print("Yes")
|
s091335309
|
Accepted
| 432 | 27,324 | 331 |
import sys
n = int(input())
txy = []
for i in range(n):
txy.append(list(map(int, input().split())))
pt = 0
py = 0
px = 0
for t, x, y in txy:
dt = t - pt
dx = x - px
dy = y - py
ax = abs(dx) % 2
ay = abs(dy) % 2
a = (ax + ay) % 2
if abs(dx) + abs(dy) > dt or a != dt % 2:
print("No")
sys.exit()
print("Yes")
|
s499671005
|
p03155
|
u814986259
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 69 |
It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board. The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns. How many ways are there to choose where to put the notice so that it completely covers exactly HW squares?
|
N=int(input())
H=int(input())
W=int(input())
H+=1-N
W+=1-N
print(H*W)
|
s639416809
|
Accepted
| 30 | 9,096 | 67 |
N=int(input())
H=int(input())
W=int(input())
print((N-H+1)*(N-W+1))
|
s255637578
|
p03597
|
u539517139
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 42 |
We have an N \times N square grid. We will paint each square in the grid either black or white. If we paint exactly A squares white, how many squares will be painted black?
|
n=int(input())
a=int(input())
print(n*2-a)
|
s029122495
|
Accepted
| 17 | 2,940 | 44 |
n=int(input())
a=int(input())
print(n**2-a)
|
s162260308
|
p02612
|
u649840780
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 9,152 | 38 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n = int(input())
print(1000-(n//1000))
|
s237821796
|
Accepted
| 27 | 9,148 | 44 |
n = int(input())
print((1000-(n%1000))%1000)
|
s295182621
|
p03795
|
u754022296
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 61 |
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.
|
n = int(input())
print(800*(min(n, 15)) + 600*(max(0, n-15)))
|
s342292460
|
Accepted
| 17 | 2,940 | 43 |
n = int(input())
print(800*n - 200*(n//15))
|
s507062874
|
p02678
|
u934868410
| 2,000 | 1,048,576 |
Wrong Answer
| 1,171 | 68,848 | 476 |
There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.
|
import queue
from collections import defaultdict
n,m = map(int,input().split())
g = defaultdict(set)
ans = [0]*n
for _ in range(m):
a,b = map(int,input().split())
a -= 1
b -= 1
g[a].add(b)
g[b].add(a)
visited = [False]*n
visited[0] = True
q = queue.Queue()
q.put(0)
while not q.empty():
v = q.get()
for p in g[v]:
if not visited[p]:
ans[p] = v
q.put(p)
visited[p] = True
for x in ans[1:]:
print(x+1)
|
s213480943
|
Accepted
| 1,131 | 68,864 | 489 |
import queue
from collections import defaultdict
n,m = map(int,input().split())
g = defaultdict(set)
ans = [0]*n
for _ in range(m):
a,b = map(int,input().split())
a -= 1
b -= 1
g[a].add(b)
g[b].add(a)
visited = [False]*n
visited[0] = True
q = queue.Queue()
q.put(0)
while not q.empty():
v = q.get()
for p in g[v]:
if not visited[p]:
ans[p] = v
q.put(p)
visited[p] = True
print('Yes')
for x in ans[1:]:
print(x+1)
|
s178652745
|
p04000
|
u952022797
| 3,000 | 262,144 |
Wrong Answer
| 2,130 | 232,816 | 686 |
We have a grid with H rows and W columns. At first, all cells were painted white. Snuke painted N of these cells. The i-th ( 1 \leq i \leq N ) cell he painted is the cell at the a_i-th row and b_i-th column. Compute the following: * For each integer j ( 0 \leq j \leq 9 ), how many subrectangles of size 3×3 of the grid contains exactly j black cells, after Snuke painted N cells?
|
# -*- coding: utf-8 -*-
import sys
def main():
H, W, N = map(int, input().split(" "))
AB = []
for _ in range(N):
a, b = map(int, input().split(" "))
AB.append((a, b))
ans = {}
for i, val in enumerate(AB):
a, b = val
a, b = a-1, b-1
for ba in range(max(0, a-2), min(H-2, a+1)):
for bb in range(max(0, b-2), min(W-2, b+1)):
if (ba, bb) in ans.keys():
ans[(ba, bb)] += 1
else:
ans[(ba, bb)] = 1
if bb == 3:
print("{} {}".format(a+1, b+1))
ans2 = [0] * 10
total = (H-2) * (W-2)
t = 0
print(ans)
for i in ans.values():
ans2[i] += 1
t += 1
ans2[0] = total - t
for i in range(10):
print(ans2[i])
if __name__ == "__main__":
main()
|
s492284171
|
Accepted
| 1,402 | 165,180 | 621 |
# -*- coding: utf-8 -*-
import sys
def main():
H, W, N = map(int, input().split(" "))
AB = []
for _ in range(N):
a, b = map(int, input().split(" "))
AB.append((a, b))
ans = {}
for i, val in enumerate(AB):
a, b = val
a, b = a-1, b-1
for ba in range(max(0, a-2), min(H-2, a+1)):
for bb in range(max(0, b-2), min(W-2, b+1)):
if (ba, bb) in ans.keys():
ans[(ba, bb)] += 1
else:
ans[(ba, bb)] = 1
ans2 = [0] * 10
total = (H-2) * (W-2)
t = 0
for i in ans.values():
ans2[i] += 1
t += 1
ans2[0] = total - t
for i in range(10):
print(ans2[i])
if __name__ == "__main__":
main()
|
s214357355
|
p03992
|
u859897687
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 32 |
This contest is `CODE FESTIVAL`. However, Mr. Takahashi always writes it `CODEFESTIVAL`, omitting the single space between `CODE` and `FESTIVAL`. So he has decided to make a program that puts the single space he omitted. You are given a string s with 12 letters. Output the string putting a single space between the first 4 letters and last 8 letters in the string s.
|
s=input()
print(s[:4]+' '+s[5:])
|
s439040838
|
Accepted
| 17 | 2,940 | 32 |
s=input()
print(s[:4]+' '+s[4:])
|
s728693767
|
p02402
|
u743390845
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,540 | 238 |
Write a program which reads a sequence of $n$ integers $a_i (i = 1, 2, ... n)$, and prints the minimum value, maximum value and sum of the sequence.
|
n= int(input())
l= list(map(int,input().split()))
s= 0
for i in range(n-1):
if(l[i+1]<l[0]):
l[0],l[i+1]= l[i+1],l[0]
for j in range(n-1):
if(l[n-1]<l[j]):
l[n-1],l[j]= l[j],l[n-1]
for k in range(n-1):
s+= l[i]
print(l[0],l[n-1],s)
|
s177226934
|
Accepted
| 20 | 8,656 | 226 |
n= int(input())
l= list(map(int,input().split()))
s= 0
for i in range(n):
if(l[i]<l[0]):
l[0],l[i]= l[i],l[0]
for j in range(n):
if(l[n-1]<l[j]):
l[n-1],l[j]= l[j],l[n-1]
for k in range(n):
s+= l[k]
print(l[0],l[n-1],s)
|
s665551000
|
p03386
|
u252828980
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 198 |
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
a,b,k = map(int,input().split())
lia,lib = [],[]
for i in range(a,min(a+k,b)):
lia.append(i)
for i in range(max(a,b-k+1),b+1):
lib.append(i)
lia.extend(lib)
for i in set(lia):
print(i)
|
s153927757
|
Accepted
| 18 | 3,064 | 213 |
a,b,k = map(int,input().split())
lia,lib = [],[]
for i in range(a,min(a+k,b)):
lia.append(i)
for i in range(max(a,b-k+1),b+1):
lib.append(i)
lia = list(set(lia+lib))
lia.sort()
for i in lia:
print(i)
|
s255458891
|
p03719
|
u168416324
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,080 | 83 |
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
a,b,c=map(int,input().split())
if b<=a and a<=c:
print("Yes")
else:
print("No")
|
s244930280
|
Accepted
| 28 | 9,120 | 116 |
def yaya(boo):
if boo:
print("Yes")
else:
print("No")
a,b,c=map(int,input().split())
yaya(a<=c and c<=b)
|
s866691206
|
p03971
|
u283929013
| 2,000 | 262,144 |
Wrong Answer
| 100 | 4,016 | 352 |
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these. Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests. * A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B. * An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students. A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these. Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
|
n,a,b = map(int,input().split())
S = input()
passed = 0
f_passed = 0
for s in S:
if s == "a":
if passed <= a + b:
print("Yes")
passed += 1
else:
print("No")
elif s == "b":
if passed <= a + b and f_passed <= b:
print("Yes")
passed += 1
f_passed += 1
else:
print("No")
else:
print("No")
|
s019740977
|
Accepted
| 104 | 4,016 | 355 |
n,a,b = map(int,input().split())
S = input()
passed = 0
f_passed = 0
for s in S:
if s == "a":
if passed < a + b:
print("Yes")
passed += 1
else:
print("No")
elif s == "b":
if passed < a + b and f_passed + 1 <= b:
print("Yes")
passed += 1
f_passed += 1
else:
print("No")
else:
print("No")
|
s884489225
|
p03455
|
u619128059
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 118 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
num = list(map(int,input().split()))
for m in num:
t = m % 2
if t == 0:
print("Even")
else:
print("Odd")
|
s700588895
|
Accepted
| 17 | 2,940 | 123 |
num = list(map(int,input().split()))
a = num[0]
b = num[1]
c = (a * b) % 2
if c == 0:
print("Even")
else:
print("Odd")
|
s998849713
|
p03816
|
u562446079
| 2,000 | 262,144 |
Wrong Answer
| 62 | 14,692 | 133 |
Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written. He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept. Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.
|
def Main():
n = int(input())
A = list(map(int, input().split()))
s = set(A)
print(len(s))
if __name__ == '__main__':
Main()
|
s406215743
|
Accepted
| 62 | 14,396 | 187 |
def Main():
n = int(input())
A = list(map(int, input().split()))
s = set(A)
if len(s)%2 == 0:
print(len(s) - 1)
else:
print(len(s))
if __name__ == '__main__':
Main()
|
s729745612
|
p03193
|
u457901067
| 2,000 | 1,048,576 |
Wrong Answer
| 21 | 3,060 | 160 |
There are N rectangular plate materials made of special metal called AtCoder Alloy. The dimensions of the i-th material are A_i \times B_i (A_i vertically and B_i horizontally). Takahashi wants a rectangular plate made of AtCoder Alloy whose dimensions are exactly H \times W. He is trying to obtain such a plate by choosing one of the N materials and cutting it if necessary. When cutting a material, the cuts must be parallel to one of the sides of the material. Also, the materials have fixed directions and cannot be rotated. For example, a 5 \times 3 material cannot be used as a 3 \times 5 plate. Out of the N materials, how many can produce an H \times W plate if properly cut?
|
from math import floor
N, H, W = map(int, input().split())
ans = 0
for i in range(N):
A, B = map(int, input().split())
ans += (A//H) * (B//W)
print(ans)
|
s666493799
|
Accepted
| 21 | 3,060 | 175 |
from math import floor
N, H, W = map(int, input().split())
ans = 0
for i in range(N):
A, B = map(int, input().split())
if (A//H) * (B//W) > 0:
ans += 1
print(ans)
|
s316571912
|
p02663
|
u409974118
| 2,000 | 1,048,576 |
Wrong Answer
| 23 | 9,080 | 345 |
In this problem, we use the 24-hour clock. Takahashi gets up exactly at the time H_1 : M_1 and goes to bed exactly at the time H_2 : M_2. (See Sample Inputs below for clarity.) He has decided to study for K consecutive minutes while he is up. What is the length of the period in which he can start studying?
|
T = input()
t = T.count("?")
num =[]
for i in range(0,t+1):
st = T.replace("?" , "D", i)
for n in range(0,t+1):
ste = st.replace("?","P",n)
for q in range(0,t+1):
stes = ste.replace("?","D",q)
num.append(stes)
print(num)
s = stes.count("DD")
f = stes.count("PD")
ans = s + f
print(ans)
|
s889354031
|
Accepted
| 23 | 9,160 | 96 |
h,m,H,M,k = map(int,input().split())
t1 = 60*h + m
t2 = 60*H + M
ans = t2 - t1 - k
print(ans)
|
s802723525
|
p03729
|
u976162616
| 2,000 | 262,144 |
Wrong Answer
| 19 | 2,940 | 101 |
You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`.
|
A,B,C = input().split()
if A[0] == B[-1] and B[0] == C[0]:
print ("YES")
else:
print ("NO")
|
s213220105
|
Accepted
| 17 | 2,940 | 102 |
A,B,C = input().split()
if A[-1] == B[0] and B[-1] == C[0]:
print ("YES")
else:
print ("NO")
|
s660229648
|
p02646
|
u808817704
| 2,000 | 1,048,576 |
Wrong Answer
| 24 | 9,072 | 267 |
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
A, V = tuple(int(c) for c in input().split(' '))
B, W = tuple(int(c) for c in input().split(' '))
T = int(input())
delta = V - W
if A == B:
print('YES')
exit()
elif delta <= 0:
print('NO')
exit()
else:
d = abs(A-B)
time = d // delta + 1
print(time)
|
s860649964
|
Accepted
| 22 | 9,192 | 305 |
A, V = tuple(int(c) for c in input().split(' '))
B, W = tuple(int(c) for c in input().split(' '))
T = int(input())
delta = V - W
if A == B:
print('YES')
exit()
elif delta <= 0:
print('NO')
exit()
else:
d = abs(A-B)
time = d / delta
if time <= T:
print('YES')
else:
print('NO')
|
s062061789
|
p02972
|
u426649993
| 2,000 | 1,048,576 |
Wrong Answer
| 762 | 37,300 | 458 |
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
N = int(input())
a = list(map(int, str(input()).split()))
box = dict(zip(range(N, 0, -1), [0] * N))
output = list()
for i in box:
mul = N // i
total_ball = 0
for m in range(1,mul+1):
total_ball += box[m * i]
if (a[i-1] == 1 and total_ball % 2 == 0) or (a[i-1] == 0 and total_ball % 2 == 1) :
box[i] = 1
output.append(str(i))
else:
pass
if len(output) != 0:
print(' '.join(output))
else:
print(0)
|
s650563050
|
Accepted
| 1,500 | 76,984 | 547 |
from collections import OrderedDict
N = int(input())
a = list(map(int, str(input()).split()))
box = OrderedDict(zip(range(N, 0, -1), [0] * N))
output = list()
count = 0
for i in box:
mul = N // i
total_ball = 0
for m in range(1,mul+1):
total_ball += box[m * i]
if (a[i-1] == 1 and total_ball % 2 == 0) or (a[i-1] == 0 and total_ball % 2 == 1) :
box[i] = 1
count += 1
output.append(str(i))
else:
pass
if len(output) != 0:
print(count)
print(' '.join(output))
else:
print(0)
|
s124213923
|
p03545
|
u076564993
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 383 |
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.
|
t=list(map(int,list(input())))
t.append(-1)
op=[]
def dfs(x,i):
if i==4 or len(op)==3:
if x==7:
ans=''
for j in range(4):
if j==3:
ans+=str(t[j])
else:
ans+=str(t[j])+op[j]
print(ans)
return
i+=1
op.append('+')
dfs(x+t[i],i)
op.pop()
op.append('-')
dfs(x-t[i],i)
op.pop()
i-=1
return
dfs(t[0],0)
|
s114736679
|
Accepted
| 17 | 3,064 | 542 |
t = list(map(int, list(input())))
t.append(7)
op = []
def dfs(x, i):
if i == 4 or len(op) == 3:
if x == 7:
op.append('=')
ans = ''
for j in range(5):
if j == 4:
ans += str(t[j])
else:
ans += str(t[j]) + op[j]
print(ans)
exit()
return
i += 1
op.append('+')
dfs(x + t[i], i)
op.pop()
op.append('-')
dfs(x - t[i], i)
op.pop()
i -= 1
return
dfs(t[0], 0)
|
s576259996
|
p02408
|
u605451279
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,592 | 372 |
Taro is going to play a card game. However, now he has only n cards, even though there should be 52 cards (he has no Jokers). The 52 cards include 13 ranks of each of the four suits: spade, heart, club and diamond.
|
n = int(input())
x = input().split()
y = int(x[1])
if x[0] == "S":
for i in range(13):
if i == x[1]:
print("S",i)
if x[0] == "H":
for j in range(13):
if j == x[1]:
print("H",j)
if x[0] == "C":
for k in range(13):
if k == x[1]:
print("C",k)
if x[0] == "D":
for l in range(13):
if l == x[1]:
print("D",l)
|
s821067281
|
Accepted
| 20 | 5,604 | 535 |
n = int(input())
s = list()
h = list()
c = list()
d = list()
for i in range(n):
a = input().split()
p = a[0]
q = int(a[1])
if p == 'S':
s.append(q)
elif p == 'H':
h.append(q)
elif p == 'C':
c.append(q)
else:# p == 'D'
d.append(q)
x = 1
for j in range(13):
if x not in s:
print('S',x)
x += 1
x =1
for k in range(13):
if x not in h:
print('H',x)
x += 1
x = 1
for l in range(13):
if x not in c:
print('C',x)
x += 1
x =1
for m in range(13):
if x not in d:
print('D',x)
x += 1
|
s469924779
|
p03544
|
u218843509
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 142 |
It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers. You are given an integer N. Find the N-th Lucas number. Here, the i-th Lucas number L_i is defined as follows: * L_0=2 * L_1=1 * L_i=L_{i-1}+L_{i-2} (i≥2)
|
n = int(input())
if n == 1:
print(2)
elif n == 2:
print(1)
else:
l = [2, 1]
for _ in range(n - 2):
l.append(l[-1] + l[-2])
print(l[-1])
|
s481747854
|
Accepted
| 17 | 2,940 | 120 |
n = int(input())
if n == 1:
print(1)
else:
l = [2, 1]
for _ in range(n - 1):
l.append(l[-1] + l[-2])
print(l[-1])
|
s209463403
|
p03351
|
u802796197
| 2,000 | 1,048,576 |
Wrong Answer
| 26 | 9,132 | 223 |
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a, b, c, d = map(int, input().split())
print(a, b, c, d)
ab = abs(a - b)
bc = abs(b - c)
ac = abs(a - c)
if ab <= d and bc <= d:
print('Yes')
elif ab <= d or bc <= d or ac <= d:
print('Yes')
else:
print('No')
|
s507730099
|
Accepted
| 28 | 9,020 | 327 |
a, b, c, d = map(int, input().split())
#print(a, b, c, d)
distance_ab = abs(a - b)
distance_bc = abs(b - c)
distance_ac = abs(a - c)
if distance_ab <= d and distance_bc <= d:
print('Yes')
elif distance_ac <= d:
print('Yes')
else:
print('No')
|
s981036266
|
p03067
|
u194894739
| 2,000 | 1,048,576 |
Wrong Answer
| 23 | 2,940 | 134 |
There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.
|
a, b, c = map(int, input().split())
if a > c and c > b:
print('Yes')
elif b > c and c > b:
print('Yes')
else:
print('No')
|
s719401222
|
Accepted
| 17 | 2,940 | 134 |
a, b, c = map(int, input().split())
if a > c and c > b:
print('Yes')
elif b > c and c > a:
print('Yes')
else:
print('No')
|
s627897738
|
p03089
|
u535171899
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,064 | 257 |
Snuke has an empty sequence a. He will perform N operations on this sequence. In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1). You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
|
n = int(input())
B = [0]+list(map(int,input().split()))
ans_li = []
for i in range(n):
for j in range(len(B)-1,0,-1):
if j==B[j]:
ans_li.append(B.pop(j))
break
print(-1);exit()
print('\n'.join(map(str,ans_li[::-1])))
|
s703894594
|
Accepted
| 18 | 3,064 | 271 |
n = int(input())
B = [0]+list(map(int,input().split()))
ans_li = []
for i in range(n):
for j in range(len(B)-1,0,-1):
if j==B[j]:
ans_li.append(B.pop(j))
break
else:
print(-1);exit()
print('\n'.join(map(str,ans_li[::-1])))
|
s827461749
|
p03623
|
u612721349
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 89 |
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x=sorted(set(chr(97+i) for i in range(26))-set(input()))
print("None" if not x else x[0])
|
s266840285
|
Accepted
| 17 | 2,940 | 100 |
print((lambda x:"A" if abs(x[0]-x[2]) > abs(x[0]-x[1]) else "B")([int(i) for i in input().split()]))
|
s885726210
|
p00011
|
u075836834
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,728 | 150 |
Let's play Amidakuji. In the following example, there are five vertical lines and four horizontal lines. The horizontal lines can intersect (jump across) the vertical lines. In the starting points (top of the figure), numbers are assigned to vertical lines in ascending order from left to right. At the first step, 2 and 4 are swaped by the first horizontal line which connects second and fourth vertical lines (we call this operation (2, 4)). Likewise, we perform (3, 5), (1, 2) and (3, 4), then obtain "4 1 2 5 3" in the bottom. Your task is to write a program which reads the number of vertical lines w and configurations of horizontal lines and prints the final state of the Amidakuji. In the starting pints, numbers 1, 2, 3, ..., w are assigne to the vertical lines from left to right.
|
w=int(input())
n=int(input())
A=[int(i+1) for i in range(w)]
for i in range(n):
x,y=map(int,input().split(','))
A[x-1],A[y-1]=A[y-1],A[x-1]
print(A)
|
s254797223
|
Accepted
| 20 | 7,664 | 173 |
w=int(input())
n=int(input())
A=[int(i+1) for i in range(w)]
for i in range(n):
x,y=map(int,input().split(','))
A[x-1],A[y-1]=A[y-1],A[x-1]
for i in range(w):
print(A[i])
|
s661722157
|
p02865
|
u116233709
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 62 |
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
n=int(input())
if n%2==0:
print((n/2)-1)
else:
print(n//2)
|
s965167548
|
Accepted
| 17 | 2,940 | 103 |
n=int(input())
if n==1 or n==2:
print(0)
elif n%2==0:
print(int((n/2)-1))
else:
print(int(n//2))
|
s850500081
|
p03503
|
u533885955
| 2,000 | 262,144 |
Wrong Answer
| 124 | 3,192 | 706 |
Joisino is planning to open a shop in a shopping street. Each of the five weekdays is divided into two periods, the morning and the evening. For each of those ten periods, a shop must be either open during the whole period, or closed during the whole period. Naturally, a shop must be open during at least one of those periods. There are already N stores in the street, numbered 1 through N. You are given information of the business hours of those shops, F_{i,j,k}. If F_{i,j,k}=1, Shop i is open during Period k on Day j (this notation is explained below); if F_{i,j,k}=0, Shop i is closed during that period. Here, the days of the week are denoted as follows. Monday: Day 1, Tuesday: Day 2, Wednesday: Day 3, Thursday: Day 4, Friday: Day 5. Also, the morning is denoted as Period 1, and the afternoon is denoted as Period 2. Let c_i be the number of periods during which both Shop i and Joisino's shop are open. Then, the profit of Joisino's shop will be P_{1,c_1}+P_{2,c_2}+...+P_{N,c_N}. Find the maximum possible profit of Joisino's shop when she decides whether her shop is open during each period, making sure that it is open during at least one period.
|
N = int(input())
F = []
for i in range(N):
F.append(list(map(int,input().split())))
P = []
for i in range(N):
P.append(list(map(int,input().split())))
j = 1
MAX = -100000000
for j in range(1023):
Fcount = [0 for a in range(N)]
b = bin(j)
b.lstrip("0b")
blist = list(b)
L = len(blist)
for k in range(10-L):
blist.insert(0,"0")
m = 0
for m in range(10):
if blist[m] == "1":
o = 0
for o in range(N):
if F[o][m] == 1:
Fcount[o]+=1
else:
pass
kari = 0
for p in range(N):
kari+=P[p][Fcount[p]]
if MAX <= kari:
MAX = kari
print(MAX)
|
s310906545
|
Accepted
| 243 | 3,444 | 754 |
#C
from collections import deque
import bisect
import math
import heapq
import itertools
#import sys
#input = sys.stdin.readline
inf = float("inf")
mod = 10**9 + 7
#mod = 998244353
#list(map(int,input().split()))
N = int(input())
F = [list(map(int,input().split())) for i in range(N)]
P = [list(map(int,input().split())) for i in range(N)]
ans = -inf
for i in range(1024):
if i == 0:
continue
b = bin(i)
b = b.lstrip("0b")
while len(b) < 10:
b = "0"+b
kari = 0
for j in range(N):
f = F[j]
p = P[j]
C = 0
for k in range(10):
if b[k] == "1" and f[k] == 1:
C+=1
kari += p[C]
ans = max(ans,kari)
#print(b)
print(ans)
|
s119094425
|
p03067
|
u848882989
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 189 |
There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.
|
import sys
args = list(map(int, input().split()))
print(args)
if args[0] < args[1] and args[1] < args[2] or args[2] < args[1] and args[1] < args[0]:
print("yes")
else:
print("no")
|
s801216773
|
Accepted
| 17 | 2,940 | 203 |
import sys
args = list(map(int, input().split()))
if (args[0] < args[2] and args[2] < args[1]) or (args[1] < args[2] and args[2] < args[0]):
print("Yes")
else:
print("No")
|
s298966032
|
p03943
|
u396961814
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 120 |
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
X = [int(x) for x in input().split()]
Y = sorted(X)
if Y[0] + Y[1] == Y[2]:
print('YES')
else:
print('NO')
|
s541012534
|
Accepted
| 17 | 2,940 | 114 |
X = [int(x) for x in input().split()]
Y = sorted(X)
if Y[0] + Y[1] == Y[2]:
print('Yes')
else:
print('No')
|
s309817511
|
p03591
|
u601018334
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 75 |
Ringo is giving a present to Snuke. Ringo has found out that Snuke loves _yakiniku_ (a Japanese term meaning grilled meat. _yaki_ : grilled, _niku_ : meat). He supposes that Snuke likes grilled things starting with `YAKI` in Japanese, and does not like other things. You are given a string S representing the Japanese name of Ringo's present to Snuke. Determine whether S starts with `YAKI`.
|
s = input().split()
if s[0:4]=='YAKI' :
print('Yes')
else :
print('No')
|
s306629934
|
Accepted
| 17 | 2,940 | 67 |
s = input()
if s[0:4]=='YAKI' :
print('Yes')
else :
print('No')
|
s515306319
|
p03545
|
u929569377
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 659 |
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.
|
array = [int(i) for i in list(input())]
flag = False
fugou = ['+', '-']
for i in fugou:
if i == '+':
tmp = array[0] + array[1]
else:
tmp = array[0] - array[1]
for j in fugou:
if j == '+':
tmp2 = tmp + array[2]
else:
tmp2 = tmp - array[2]
for k in fugou:
if k == '+':
if tmp2 + array[3] == 7:
print(str(array[0])
+ i + str(array[1])
+ j + str(array[2])
+ k + str(array[3]))
flag = True
break
else:
if tmp2 - array[3] == 7:
print(str(array[0])
+ i + str(array[1])
+ j + str(array[2])
+ k + str(array[3]))
flag = True
break
if flag: break
if flag: break
|
s353035824
|
Accepted
| 18 | 3,064 | 631 |
N = list(map(int, list(input())))
flag = False
for i in range(2):
for j in range(2):
for k in range(2):
B = (N[1] if i == 0 else -N[1])
C = (N[2] if j == 0 else -N[2])
D = (N[3] if k == 0 else -N[3])
if sum([N[0], B, C, D]) == 7:
print(N[0],end='')
if i == 0:
print('+',end='')
else:
print('-',end='')
print(N[1],end='')
if j == 0:
print('+',end='')
else:
print('-',end='')
print(N[2],end='')
if k == 0:
print('+',end='')
else:
print('-',end='')
print(N[3],end='')
print('=7')
flag = True
break
if flag:
break
if flag:
break
|
s256069835
|
p03434
|
u354126779
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 140 |
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())
Aseq=input().split()
A=[]
for i in range(n):
A.append(int(Aseq[i]))
A.sort()
AA=A[0::2]
AB=A[1::2]
print(sum(AA)-sum(AB))
|
s350729561
|
Accepted
| 17 | 3,060 | 152 |
n=int(input())
Aseq=input().split()
A=[]
for i in range(n):
A.append(int(Aseq[i]))
A.sort(reverse=True)
AA=A[0::2]
AB=A[1::2]
print(sum(AA)-sum(AB))
|
s456458950
|
p03778
|
u626468554
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 183 |
AtCoDeer the deer found two rectangles lying on the table, each with height 1 and width W. If we consider the surface of the desk as a two-dimensional plane, the first rectangle covers the vertical range of AtCoDeer will move the second rectangle horizontally so that it connects with the first rectangle. Find the minimum distance it needs to be moved.
|
#n = int(input())
#n,k = map(int,input().split())
#x = list(map(int,input().split()))
w,a,b = map(int,input().split())
a = min(b-w-a,a-b-w)
if a <0:
print(0)
else:
print(a)
|
s901931461
|
Accepted
| 17 | 2,940 | 58 |
w,a,b = map(int,input().split())
print(max(b-w-a,a-b-w,0))
|
s274807919
|
p02255
|
u510829608
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,648 | 202 |
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.
|
N = int(input())
l = list(map(int, input().split()))
print(*l)
for i in range(N):
t = l[i]
j = i-1
while j >= 0 and l[j] > t:
l[j+1], l[j] = l[j], l[j+1]
j -= 1
print(*l)
|
s077040344
|
Accepted
| 20 | 8,024 | 284 |
N = int(input())
A = list(map(int, input().split()))
print(*A)
def insertionSort(A, N):
for i in range(1, N):
v = A[i]
j = i-1
while j>= 0 and A[j] > v:
A[j+1] = A[j]
j -= 1
A[j+1] = v
print(*A)
insertionSort(A, N)
|
s519617632
|
p02402
|
u623996423
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,588 | 98 |
Write a program which reads a sequence of $n$ integers $a_i (i = 1, 2, ... n)$, and prints the minimum value, maximum value and sum of the sequence.
|
n = int(input())
a = list(map(int, input().split()))
for i in range(n):
print(a[i], end=' ')
|
s734622738
|
Accepted
| 30 | 6,580 | 114 |
n = int(input())
a = list(map(int, input().split()))
print(min(a), end=' ')
print(max(a), end=' ')
print(sum(a))
|
s855043226
|
p03457
|
u703391033
| 2,000 | 262,144 |
Wrong Answer
| 459 | 34,632 | 304 |
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
N=int(input())
a=[list(map(int,input().split())) for _ in range(N)]
dp=[[0,0] for _ in range(N)]
for i in range(N):
t=a[i][0]-dp[i-1][1]
x=a[i][1]
y=a[i][2]
if t>=x+y and t%2==(x+y+dp[i-1][1])%2 and dp[i-1][0]==1:
dp[i][0]=1
dp[i][1]=a[i][0]
if dp[N-1][0]==1:print("Yes")
else:print("No")
|
s934087692
|
Accepted
| 497 | 37,816 | 324 |
N=int(input())
a=[list(map(int,input().split())) for _ in range(N)]
dp=[[0,0] for _ in range(N+1)]
f=True
for i in range(N):
t=a[i][0]-dp[i][1]
x=a[i][1]
y=a[i][2]
if t>=x+y-dp[i][0] and t%2==(x+y+dp[i][1])%2:
dp[i+1][0]=x+y
dp[i+1][1]=a[i][0]
else:
f=False
break
if f:print("Yes")
else:print("No")
|
s975410157
|
p02343
|
u890713354
| 3,000 | 131,072 |
Wrong Answer
| 20 | 5,604 | 1,060 |
Write a program which manipulates a disjoint set S = {S1, S2, . . . , Sk}. First of all, the program should read an integer n, then make a disjoint set where each element consists of 0, 1, ... n−1 respectively. Next, the program should read an integer q and manipulate the set for q queries. There are two kinds of queries for different operations: * unite(x, y): unites sets that contain x and y, say Sx and Sy, into a new set. * same(x, y): determine whether x and y are in the same set.
|
p=[]
rank=[]
answer=[]
def makeset(x):
p[x] = x
rank[x] = 0
def union(x,y):
link(findset(x),findset(y))
def link(x,y):
if rank[x] > rank[y]:
p[y] = x
else:
p[x] = y
if rank[x] == rank[y]:
rank[y] = rank[y] + 1
def findset(x):
if x != p[x]:
p[x] = findset(p[x])
return p[x]
t,s = map(int,input().split())
for i in range(t):
p.append(None)
rank.append(None)
while True:
try:
o,x,y = map(int,input().split())
except:
break
else:
if o == 0:
if p[x] == None:
makeset(x)
if p[y] == None:
makeset(y)
union(x,y)
print(p[x])
print(rank[x])
else:
if p[x] == None:
makeset(x)
if p[y] == None:
makeset(y)
if findset(x) == findset(y):
answer.append(1)
else:
answer.append(0)
for i in range(len(answer)):
print(answer[i])
|
s123519931
|
Accepted
| 590 | 6,972 | 1,009 |
p=[]
rank=[]
answer=[]
def makeset(x):
p[x] = x
rank[x] = 0
def union(x,y):
link(findset(x),findset(y))
def link(x,y):
if rank[x] > rank[y]:
p[y] = x
else:
p[x] = y
if rank[x] == rank[y]:
rank[y] = rank[y] + 1
def findset(x):
if x != p[x]:
p[x] = findset(p[x])
return p[x]
t,s = map(int,input().split())
for i in range(t):
p.append(None)
rank.append(None)
while True:
try:
o,x,y = map(int,input().split())
except:
break
else:
if o == 0:
if p[x] == None:
makeset(x)
if p[y] == None:
makeset(y)
union(x,y)
else:
if p[x] == None:
makeset(x)
if p[y] == None:
makeset(y)
if findset(x) == findset(y):
answer.append(1)
else:
answer.append(0)
for i in range(len(answer)):
print(answer[i])
|
s843277819
|
p02972
|
u900848911
| 2,000 | 1,048,576 |
Wrong Answer
| 744 | 7,504 | 431 |
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
N = int(input())
a = list(int(x) for x in input().split())
b = [0] * N
index = N // 2
for i in range(index,N):
b[i] = a[i]
count = 0
for i in range(index):
for j in range(index-i-1+(index-i),N,index-i):
count += a[j]
if count % 2 != a[i]:
b[i] = 1
count = 0
for i in range(N):
if b[i] == 1:
count += 1
print(count)
for i in range(N):
print(b[i],end=' ')
print()
|
s626078306
|
Accepted
| 626 | 14,092 | 718 |
N = int(input())
arr = list(int(x) for x in input().split())
flag = [-1] * N
ans = []
for i in range(N-1,-1,-1):
sum = 0
for j in range(i,N,i+1):
if j == i:
continue
else:
if flag[j] == 1:
sum += 1
if sum % 2 == arr[i]:
flag[i] = 0
else:
flag[i] = 1
ans.append(i+1)
num = len(ans)
if num == 0:
print(num)
else:
print(num)
print(*ans)
|
s457871168
|
p02412
|
u104171359
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,784 | 758 |
Write a program which identifies the number of combinations of three integers which satisfy the following conditions: * You should select three distinct integers from 1 to n. * A total sum of the three integers is x. For example, there are two combinations for n = 5 and x = 9. * 1 + 3 + 5 = 9 * 2 + 3 + 4 = 9
|
#!usr/bin/env python3
import sys
def subset_sum(numbers, target, partial=[]):
s = sum(partial)
if s == target:
if len(partial) != 3:
return None
print(*partial)
if s >= target:
return None
for i in range(len(numbers)):
n = numbers[i]
remaining = numbers[i+1:]
if len(partial) == 3:
break
subset_sum(remaining, target, partial + [n])
def program_runner():
while True:
lst = [int(num) for num in sys.stdin.readline().split()]
if lst[0] == lst[1] == 0:
break
num_list = [num for num in range(1, lst[0]+1)]
subset_sum(num_list, lst[1])
def main():
program_runner()
if __name__ == '__main__':
main()
|
s734100842
|
Accepted
| 310 | 7,636 | 433 |
#!/usr/local/env python3
import sys
def main():
while True:
count = 0
n, r = map(int, sys.stdin.readline().split())
if n == r == 0:
break
for n1 in range(1, n-1):
for n2 in range(n1+1, n):
for n3 in range(n2+1, n+1):
if n1 + n2 + n3 == r:
count += 1
print(count)
if __name__ == '__main__':
main()
|
s354161948
|
p03053
|
u830592648
| 1,000 | 1,048,576 |
Wrong Answer
| 1,073 | 248,032 | 788 |
You are given a grid of squares with H horizontal rows and W vertical columns, where each square is painted white or black. HW characters from A_{11} to A_{HW} represent the colors of the squares. A_{ij} is `#` if the square at the i-th row from the top and the j-th column from the left is black, and A_{ij} is `.` if that square is white. We will repeatedly perform the following operation until all the squares are black: * Every white square that shares a side with a black square, becomes black. Find the number of operations that will be performed. The initial grid has at least one black square.
|
from collections import deque
H,W=map(int, input().split())
S=[]
S.append(list(['X']*(W+2)))
for h in range(H):
S.append(['X']+list(input())+['X'])
S.append(list(['X']*(W+2)))
q=deque([])
d=[(1,0),(-1,0),(0,1),(0,-1)]
cnt_n=0
def add_q(S):
for h in range(1,H+1):
for w in range(1,W+1):
if S[h][w]=='#':
for dy,dx in d:
q.append([h+dy,w+dx])
return S,q
def chnge_are(S,q):
while q:
y,x = q.popleft()
S[y][x]='#'
return S
def res_chk(S,cnt_n):
res=0
for h in range(H+2):
if S[h].count('.')!=0:
res=1
if res==1:
cnt_n+=1
S,q=add_q(S)
S=chnge_are(S,q)
return res_chk(S,cnt_n)
else:
return cnt_n
res_chk(S,cnt_n)
|
s527649760
|
Accepted
| 681 | 52,740 | 302 |
from scipy.ndimage import distance_transform_cdt
H,W=map(int, input().split())
S=list(list(input()) for h in range(H))
for h in range(H):
for w in range(W):
if S[h][w]=="#":
S[h][w]=0
else:
S[h][w]=1
print(distance_transform_cdt(S, metric='taxicab').max())
|
s973848734
|
p02612
|
u080885857
| 2,000 | 1,048,576 |
Wrong Answer
| 25 | 9,016 | 67 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
import math
n=int(input())
total = math.ceil(n/1000)
print(total-n)
|
s553572083
|
Accepted
| 29 | 9,008 | 75 |
import math
n=int(input())
total = math.ceil(n/1000)
print((total)*1000-n)
|
s226999693
|
p03827
|
u732743460
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 139 |
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())
x=0
s=input()
ans=0
for i in range(n):
if s[i]=="I":
x=x+1
else:
x=x-1
ans=max(ans,x)
print(x)
|
s939512001
|
Accepted
| 18 | 2,940 | 141 |
n = int(input())
x=0
s=input()
ans=0
for i in range(n):
if s[i]=="I":
x=x+1
else:
x=x-1
ans=max(ans,x)
print(ans)
|
s916604338
|
p03795
|
u670961163
| 2,000 | 262,144 |
Wrong Answer
| 28 | 9,140 | 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.
|
n = int(input())
print(800*n - 200 * n// 15)
|
s262097358
|
Accepted
| 25 | 8,932 | 47 |
n = int(input())
print(800*n - 200 * (n// 15))
|
s973610323
|
p02606
|
u946648678
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 9,152 | 120 |
How many multiples of d are there among the integers between L and R (inclusive)?
|
l, r, d = [int(x) for x in input().split()]
count = 0
for i in range(l, r+1):
if not l%d:
count += 1
print(count)
|
s052079447
|
Accepted
| 23 | 9,152 | 120 |
l, r, d = [int(x) for x in input().split()]
count = 0
for i in range(l, r+1):
if not i%d:
count += 1
print(count)
|
s032241953
|
p03828
|
u550943777
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 401 |
You are given an integer N. Find the number of the positive divisors of N!, modulo 10^9+7.
|
N = int(input())
mod = 10**9+7
arr = [1 for i in range(N+1)]
for i in range(2,N+1):
if arr[i] > 0:
j = 2*i
while(j<N+1):
arr[j] = 0
j += i
prime = []
for i in range(2,N):
if arr[i] != 0:
prime.append(i)
count = 1
for n in prime:
tmp = n
i = 0
while(tmp < N+1):
i += N//tmp
tmp *= n
count *= (i+1)
print(count%mod)
|
s060249801
|
Accepted
| 19 | 3,064 | 403 |
N = int(input())
mod = 10**9+7
arr = [1 for i in range(N+1)]
for i in range(2,N+1):
if arr[i] > 0:
j = 2*i
while(j<N+1):
arr[j] = 0
j += i
prime = []
for i in range(2,N+1):
if arr[i] != 0:
prime.append(i)
count = 1
for n in prime:
tmp = n
i = 0
while(tmp < N+1):
i += N//tmp
tmp *= n
count *= (i+1)
print(count%mod)
|
s590181043
|
p02833
|
u940139461
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 353 |
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).
|
N = int(input())
# c = 0
# while True:
# a = N // 5
# b = N % 5
# if b != 0:
# N = b
# else:
# N = a
# c += N
# if N < 5:
# break
# print(int(c / 2))
n = 1
ans = 0
while 5 ** n < N:
a = (N // (5 ** n))
b = (N % (10 ** n))
if b == 0:
a = int(a / 2)
ans += a
n += 1
print(ans)
|
s955190483
|
Accepted
| 17 | 2,940 | 229 |
import math
N = int(input())
def main(N):
if N % 2 == 1:
print(0)
return
n = 1
ans = 0
N //= 2
while N:
N //= 5
ans += N
print(ans)
if __name__ == "__main__":
main(N)
|
s661728298
|
p03386
|
u595375942
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 132 |
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
a,b,k=map(int,input().split())
L=set([i for i in range(a,min(a+k,b))]+[i for i in range(max(b-k+1,a),b+1)])
for a in L:
print(a)
|
s631333716
|
Accepted
| 19 | 3,060 | 99 |
a,b,k=map(int,input().split())
R=range(a,b+1)
for x in sorted(set(R[:k])|set(R[-k:])):
print(x)
|
s847451262
|
p03658
|
u878138257
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 114 |
Snuke has N sticks. The length of the i-th stick is l_i. Snuke is making a snake toy by joining K of the sticks together. The length of the toy is represented by the sum of the individual sticks that compose it. Find the maximum possible length of the toy.
|
n,k = map(int, input().split())
listA = list(map(int, input().split()))
listA.sort()
listB = listA[:k]
sum(listB)
|
s689095752
|
Accepted
| 18 | 2,940 | 134 |
n,k = map(int, input().split())
listA = list(map(int, input().split()))
listA.sort(reverse=True)
listB = listA[:k]
print(sum(listB))
|
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