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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s941325843 | p03455 | u409254176 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 30 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | a,b = map(int,input().split()) | s624571190 | Accepted | 27 | 9,156 | 77 | a,b=map(int,input().split())
if a*b %2==0:
print("Even")
else:
print("Odd") |
s084784060 | p02402 | u227984374 | 1,000 | 131,072 | Wrong Answer | 20 | 5,580 | 82 | 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()))
print(max(A), min(A), sum(A))
| s638652854 | Accepted | 20 | 6,580 | 82 | n = int(input())
A = list(map(int,input().split()))
print(min(A), max(A), sum(A))
|
s683046950 | p03836 | u619458041 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 411 | Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him. | import sys
def main():
input = sys.stdin.readline
X = list(map(int, input().split()))
diffx = X[2] - X[0]
diffy = X[3] - X[1]
f1 = 'U' * diffy + 'R' * diffx
b1 = 'D' * diffy + 'L' * diffx
f2 = 'L' + 'U' * (diffy + 1) + 'R' * (diffx + 1) + 'D'
b2 = 'R' + 'D' * (diffy + 1) + 'L' * (diffx + 1) + 'U'
return f1 + b2 + f2 + b2
if __name__ == '__main__':
print(main())
| s544929622 | Accepted | 17 | 3,064 | 411 | import sys
def main():
input = sys.stdin.readline
X = list(map(int, input().split()))
diffx = X[2] - X[0]
diffy = X[3] - X[1]
f1 = 'U' * diffy + 'R' * diffx
b1 = 'D' * diffy + 'L' * diffx
f2 = 'L' + 'U' * (diffy + 1) + 'R' * (diffx + 1) + 'D'
b2 = 'R' + 'D' * (diffy + 1) + 'L' * (diffx + 1) + 'U'
return f1 + b1 + f2 + b2
if __name__ == '__main__':
print(main())
|
s019037973 | p03861 | u265506056 | 2,000 | 262,144 | Wrong Answer | 31 | 9,048 | 83 | You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x? | a,b,x=map(int,input().split())
A=a//x
B=b//x
ans=B-A
if a==0:
ans+=1
print(ans) | s752828033 | Accepted | 27 | 9,056 | 85 | a,b,x=map(int,input().split())
A=a//x
B=b//x
ans=B-A
if a%x==0:
ans+=1
print(ans) |
s166771302 | p04043 | u816265237 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 121 | Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order. | ln = sorted(list(map(int, input().split())))
if ln[0] == ln[1] ==5 and ln[2] ==7:
print('Yes')
else:
print('No')
| s628783639 | Accepted | 18 | 2,940 | 121 | ln = sorted(list(map(int, input().split())))
if ln[0] == ln[1] ==5 and ln[2] ==7:
print('YES')
else:
print('NO')
|
s169190380 | p02389 | u389610071 | 1,000 | 131,072 | Wrong Answer | 20 | 7,512 | 70 | Write a program which calculates the area and perimeter of a given rectangle. | nums = input().split()
a = int(nums[0])
b = int(nums[1])
print(a * b) | s400296287 | Accepted | 20 | 7,704 | 85 | nums = input().split()
a = int(nums[0])
b = int(nums[1])
print(a * b, 2 * a + 2 * b) |
s759089488 | p03711 | u118147328 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 196 | Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group. | x,y = map(int, input().split())
risuto1 = [1,3,5,7,8,10,12]
risuto2 = [4,6,9,11]
risuto3 = [2]
if x & y in risuto1:
print("yes")
elif x & y in risuto2:
print("yes")
else:
print("no") | s613399385 | Accepted | 17 | 3,064 | 222 | x,y = map(int, input().split())
risuto1 = [1,3,5,7,8,10,12]
risuto2 = [4,6,9,11]
risuto3 = [2]
if x in risuto1 and y in risuto1:
print("Yes")
elif x in risuto2 and y in risuto2:
print("Yes")
else:
print("No") |
s495506017 | p03160 | u698919163 | 2,000 | 1,048,576 | Wrong Answer | 140 | 13,980 | 333 | There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N. | n = int(input())
h = list(map(int,input().split()))
cost1 = 0
cost2 = 0
cost = [0]
cost.append(abs(h[0]-h[1]))
print(cost)
for flog in range(2,len(h)):
cost1 = cost[flog-2] + abs(h[flog] - h[flog-2])
cost2 = cost[flog-1] + abs(h[flog] - h[flog-1])
cost.append(min(cost1,cost2))
#print(cost)
print(cost[-1]) | s016381032 | Accepted | 131 | 13,928 | 280 | N = int(input())
h = list(map(int,input().split()))
dp = [0]*N
dp[1] = abs(h[1]-h[0])
for i in range(2,N):
dp[i] = min(dp[i-1]+abs(h[i]-h[i-1]),dp[i-2]+abs(h[i]-h[i-2]))
print(dp[-1]) |
s193559912 | p00001 | u741801763 | 1,000 | 131,072 | Wrong Answer | 20 | 7,628 | 223 | There is a data which provides heights (in meter) of mountains. The data is only for ten mountains. Write a program which prints heights of the top three mountains in descending order. | if __name__=="__main__":
dataset = []
for i in range(10):
a = int(input())
dataset.append(a)
print(dataset)
for j in range(3):
print(max(dataset))
dataset.remove(max(dataset)) | s287354820 | Accepted | 20 | 7,760 | 204 | if __name__=="__main__":
dataset = []
for i in range(10):
a = int(input())
dataset.append(a)
for j in range(3):
print(max(dataset))
dataset.remove(max(dataset)) |
s266589839 | p03861 | u620157187 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 129 | You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x? | a, b, x = map(int, input().split())
max_i = b//x
if a%x == 0:
min_i = a//x
else:
min_i = a//x+1
print(max_i-min_i) | s212201650 | Accepted | 17 | 2,940 | 131 | a, b, x = map(int, input().split())
max_i = b//x
if a%x == 0:
min_i = a//x
else:
min_i = a//x+1
print(max_i-min_i+1) |
s916863260 | p03139 | u969190727 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 61 | We conducted a survey on newspaper subscriptions. More specifically, we asked each of the N respondents the following two questions: * Question 1: Are you subscribing to Newspaper X? * Question 2: Are you subscribing to Newspaper Y? As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2. What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y? Write a program to answer this question. | n,a,b=map(int,input().split())
if a>b:
a,b=b,a
print(b,n-a) | s455844118 | Accepted | 17 | 2,940 | 61 | n,a,b=map(int,input().split())
print(min(a,b),max(0,(a+b)-n)) |
s814732220 | p03130 | u167908302 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 198 | There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads. Determine if we can visit all the towns by traversing each of the roads exactly once. | #coding:utf-8
ans = [0] * 4
for i in range(3):
a, b = map(int, input().split())
ans[a-1] += 1
ans[b-1] += 1
for i in range(4):
if ans[i] >= 3:
print("No")
exit()
print("Yes")
| s132888375 | Accepted | 17 | 2,940 | 189 | #coding:utf-8
ans = [0] * 4
for i in range(3):
a, b = map(int, input().split())
ans[a-1] += 1
ans[b-1] += 1
for i in ans:
if i >= 3:
print("NO")
exit()
print("YES")
|
s678734094 | p03549 | u829249049 | 2,000 | 262,144 | Wrong Answer | 167 | 24,172 | 252 | Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`. When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases. Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds. Now, he goes through the following process: * Submit the code. * Wait until the code finishes execution on all the cases. * If the code fails to correctly solve some of the M cases, submit it again. * Repeat until the code correctly solve all the cases in one submission. Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer). | N,M=map(int,input().split())
p=(1/2)**M
listP=[p]
listP+=[0]*(10**6)
sumP=[0]*(10**6)
sumP[0]=p
ans=0
time=1900*M+100*(N-M)
for i in range(10**5):
ans+=time*listP[i]*(i+1)
sumP[i+1]=sumP[i]+p*(1-sumP[i])
listP[i+1]=p*(1-sumP[i])
print(int(ans)) | s072294848 | Accepted | 17 | 2,940 | 75 | N,M=map(int,input().split())
time=1900*M+100*(N-M)
ans=time*2**M
print(ans) |
s708587049 | p02578 | u551967750 | 2,000 | 1,048,576 | Wrong Answer | 154 | 31,988 | 187 | N persons are standing in a row. The height of the i-th person from the front is A_i. We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person: Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool. Find the minimum total height of the stools needed to meet this goal. | n = input()
l = [int(i) for i in input().split()]
max = 0
sum = 0
print(l)
for i in range(1,len(l)):
if l[i-1] > max:
max = l[i-1]
if l[i] < max:
sum += max - l[i]
print(sum) | s124391673 | Accepted | 138 | 32,156 | 178 | n = input()
l = [int(i) for i in input().split()]
max = 0
sum = 0
for i in range(1,len(l)):
if l[i-1] > max:
max = l[i-1]
if l[i] < max:
sum += max - l[i]
print(sum) |
s267794873 | p03338 | u598009172 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 167 | You are given a string S of length N consisting of lowercase English letters. We will cut this string at one position into two strings X and Y. Here, we would like to maximize the number of different letters contained in both X and Y. Find the largest possible number of different letters contained in both X and Y when we cut the string at the optimal position. | N = int(input())
S = str(input())
ans=0
for i in range(N):
X = S[:i]
Y = S[i+1:]
Z = set(X) & set(Y)
if len(Z) > ans :
ans = len(Z)
print(ans)
| s606665488 | Accepted | 18 | 3,060 | 168 | N = int(input())
S = str(input())
ans=0
for i in range(1,N-1):
X = S[:i]
Y = S[i:]
Z = set(X) & set(Y)
if len(Z) > ans :
ans = len(Z)
print(ans) |
s282161260 | p03095 | u802772880 | 2,000 | 1,048,576 | Wrong Answer | 22 | 3,572 | 104 | You are given a string S of length N. Among its subsequences, count the ones such that all characters are different, modulo 10^9+7. Two subsequences are considered different if their characters come from different positions in the string, even if they are the same as strings. Here, a subsequence of a string is a concatenation of **one or more** characters from the string without changing the order. | n=int(input())
s=input()
m=list(set(s))
ans=1
for i in m:
ans*=s.count(i)-1
print((ans-1)%(10**9+7)) | s241278190 | Accepted | 22 | 3,444 | 104 | n=int(input())
s=input()
m=list(set(s))
ans=1
for i in m:
ans*=s.count(i)+1
print((ans-1)%(10**9+7)) |
s469561909 | p04029 | u055687574 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 37 | There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total? | n = int(input())
print(sum(range(n))) | s489307804 | Accepted | 17 | 2,940 | 40 | n = int(input())
print(sum(range(n+1)))
|
s618309458 | p03502 | u363407238 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 142 | An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10. Given an integer N, determine whether it is a Harshad number. | n = input()
digits_sum = sum(map(int, n.split()))
print(n)
print(digits_sum)
if int(n) == digits_sum:
print('YES')
else:
print('NO')
| s425527366 | Accepted | 17 | 2,940 | 113 | n = input()
digit_sum = sum(map(int, list(n)))
if int(n) % digit_sum == 0:
print('Yes')
else:
print('No') |
s200552925 | p03139 | u163320134 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 109 | We conducted a survey on newspaper subscriptions. More specifically, we asked each of the N respondents the following two questions: * Question 1: Are you subscribing to Newspaper X? * Question 2: Are you subscribing to Newspaper Y? As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2. What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y? Write a program to answer this question. | n,x,y=map(int,input().split())
if (x+y)>n:
mn=(x+y)-n
else:
mn=0
mx=max(x,y)
print('{} {}'.format(mx,mn)) | s134708215 | Accepted | 17 | 3,060 | 109 | n,x,y=map(int,input().split())
if (x+y)>n:
mn=(x+y)-n
else:
mn=0
mx=min(x,y)
print('{} {}'.format(mx,mn)) |
s111908327 | p03729 | u280512618 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 83 | 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`. | s=input().split()
print("YES" if s[0][-1]==s[0][0] and s[1][-1]==s[2][0] else 'NO') | s784802787 | Accepted | 17 | 2,940 | 84 | s=input().split()
print("YES" if s[0][-1]==s[1][0] and s[1][-1]==s[2][0] else 'NO')
|
s033458628 | p03861 | u427344224 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 151 | You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x? | a, b, x = map(int, input().split())
def cal(n, x):
if a == 0:
return 0
else:
return n / x + 1
print(str(cal(b, x) - cal(a, x))) | s935240840 | Accepted | 17 | 2,940 | 77 | a, b, x = map(int, input().split())
A = (a - 1) // x
B = b // x
print(B - A)
|
s862840145 | p01713 | u352394527 | 2,000 | 131,072 | Wrong Answer | 20 | 5,532 | 775 | 日本では防災研究が盛んに行われており，近年その重要性がますます増している． 避難経路の評価も重要な研究のひとつである． 今回は直線状の通路の安全評価を行う． 通路は W 個のユニットに分けられており，一方の端のユニットからもう一方の端のユニットまで 0, 1, 2, … , W-1 の番号がつけられている． 通路内の各ユニットには，入口の扉，出口の扉，防火扉のいずれか1つが存在する． 入り口の扉，出口の扉，防火扉はそれぞれ通路内に複数個存在しうる． この問題では時刻 t=0 で火災が発生したと想定する． それにより，通路の外部にいて避難しようとしている人々が入口の扉を通じて通路へ入り，より安全な場所へ出るために出口の扉へ脱出しようとするものとする． 避難中のそれぞれの人は単位時刻ごとに 1 つのユニットを移動するか，今のユニットに留まることができる． すなわち，時刻 t にある人がユニット i にいたとするとき，その人は時刻 t+1 ではユニット i-1, i, i+1 の3つの数字のうち 0 以上 W-1 以下であるものを選択し，その番号のユニットへ移動することができる． 防火扉があるユニットは，ある一定時刻以降になると完全に遮断されてしまい，避難中の人々はそのユニットに立ち入りできなくなる．また，そのユニット内に居た人々もそこから他のユニットに移動できなくなってしまう． この問題における目的は，それぞれの扉の情報が与えられるので，避難中の人々が最適に行動した時に最大で何人が出口の扉へたどり着けるか計算することである． 通路の情報がW個の整数a_iで与えられる． * a_i = 0のとき，i 番目のユニットが出口の扉であることをあらわす． * a_i < 0のとき，i 番目のユニットが防火扉により時間 |a_i| 以降出入りできなくなることを表す． * a_i > 0のとき，時刻 t=0, 1, 2, … , a_{i}-1 のそれぞれにおいて，ちょうど一人の人が i 番目のユニットに現れる．時刻 t に現れた人は，時刻 t+1 以降から移動を開始する． なお，1つのユニットに複数の人々が存在してもかまわない． 出口の扉へたどり着ける最大の人数を求めよ． | def search_left(ind):
doors = []
for j in range(ind - 1, -1, -1):
if alst[j] < 0:
doors.append((j, -alst[j]))
if alst[j] == 0:
ret = alst[ind]
for pos, lim in doors:
ret = min(ret, lim - (ind - pos))
ret = max(0, ret)
return ret
return 0
def search_right(ind):
doors = []
for j in range(ind + 1, w):
if alst[j] < 0:
doors.append((j, -alst[j]))
if alst[j] == 0:
ret = alst[ind]
for pos, lim in doors:
ret = min(ret, lim - (pos - ind))
ret = max(0, ret)
return ret
return 0
def main():
w = int(input())
alst = list(map(int, input().split()))
ans = 0
for ind, a in enumerate(alst):
if a > 0:
ans += max(search_left(ind), search_right(ind))
print(ans)
| s314936449 | Accepted | 320 | 20,404 | 513 | w = int(input())
alst = list(map(int, input().split()))
INF = 10 ** 20
acc = -INF
left = []
for i in range(w):
a = alst[i]
if a == 0:
acc = INF
elif a > 0:
acc -= 1
elif a < 0:
acc = min(acc - 1, -a)
left.append(acc)
acc = -INF
right = []
for i in range(w):
a = alst[w - i - 1]
if a == 0:
acc = INF
elif a > 0:
acc -= 1
elif a < 0:
acc = min(acc - 1, -a)
right.append(acc)
right.reverse()
print(sum([max(0, min(alst[i], max(left[i], right[i]))) for i in range(w)]))
|
s890583725 | p03636 | u241234955 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 97 | The word `internationalization` is sometimes abbreviated to `i18n`. This comes from the fact that there are 18 letters between the first `i` and the last `n`. You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way. | s = input()
s_mid_len = len(s) - 2
print(s_mid_len)
T = s[0] + str(s_mid_len) + s[-1]
print(T)
| s281713006 | Accepted | 17 | 2,940 | 106 | s = input()
s_mid_len = len(s) - 2
T = s[0] + str(s_mid_len) + s[-1]
print(T)
|
s081148284 | p03151 | u825027400 | 2,000 | 1,048,576 | Wrong Answer | 124 | 18,292 | 554 | A university student, Takahashi, has to take N examinations and pass all of them. Currently, his _readiness_ for the i-th examination is A_{i}, and according to his investigation, it is known that he needs readiness of at least B_{i} in order to pass the i-th examination. Takahashi thinks that he may not be able to pass all the examinations, and he has decided to ask a magician, Aoki, to change the readiness for as few examinations as possible so that he can pass all of them, while not changing the total readiness. For Takahashi, find the minimum possible number of indices i such that A_i and C_i are different, for a sequence C_1, C_2, ..., C_{N} that satisfies the following conditions: * The sum of the sequence A_1, A_2, ..., A_{N} and the sum of the sequence C_1, C_2, ..., C_{N} are equal. * For every i, B_i \leq C_i holds. If such a sequence C_1, C_2, ..., C_{N} cannot be constructed, print -1. | N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
total = 0
diff_list = []
smaller_num = 0
for n in range(N):
if A[n] < B[n]:
total += B[n] - A[n]
smaller_num += 1
else:
diff_list.append(A[n] - B[n])
diff_list = sorted(diff_list, reverse = True)
diff_total = 0
diff_count = 0
for i, diff in enumerate(diff_list):
if diff_total >= total:
diff_count = i
break
diff_total += diff
if diff_total < total:
print(-1)
else:
print(smaller_num + diff_count)
| s934408461 | Accepted | 122 | 19,060 | 594 | N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
total = 0
diff_list = []
smaller_num = 0
for n in range(N):
if A[n] < B[n]:
total += B[n] - A[n]
smaller_num += 1
else:
diff_list.append(A[n] - B[n])
diff_list = sorted(diff_list, reverse = True)
diff_total = 0
diff_count = 0
for i, diff in enumerate(diff_list):
diff_total += diff
if diff_total >= total:
diff_count = i + 1
break
if smaller_num == 0:
print(0)
elif diff_total < total:
print(-1)
else:
print(smaller_num + diff_count)
|
s911620993 | p03548 | u217627525 | 2,000 | 262,144 | Wrong Answer | 33 | 2,940 | 94 | We have a long seat of width X centimeters. There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters. We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two people, and between the end of the seat and a person. At most how many people can sit on the seat? | x,y,z=map(int,input().split())
w=y+2*z
ans=1
while w<x-(y+z):
w+=y+z
ans+=1
print(ans) | s746654505 | Accepted | 17 | 2,940 | 58 | x,y,z=map(int,input().split())
ans=(x-z)//(y+z)
print(ans) |
s798662730 | p03478 | u022269787 | 2,000 | 262,144 | Wrong Answer | 21 | 3,064 | 273 | 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())
array = []
for i in range(0,1,1):
for j in range(0,9,1):
for k in range(0,9,1):
for l in range(0,9,1):
for m in range(0,9,1):
if A < i+j+k+l+m < B:
array.append(i+j+k+l+m)
sum(array)
print(array) | s488695731 | Accepted | 37 | 3,060 | 143 | N, A, B = map(int, input().split())
num = 0
for i in range(N+1):
if A <= sum(list(map(int, list(str(i))))) <= B:
num += i
print(num)
|
s659137492 | p03814 | u104922648 | 2,000 | 262,144 | Wrong Answer | 19 | 3,560 | 85 | 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`. | word = input()
start = word.index('A')
end = word.index('Z')
print(word[start:end+1]) | s019654215 | Accepted | 59 | 11,308 | 197 | word = input()
start = []
end = []
for idx, w in enumerate(word):
if w == 'A':
start.append(idx)
elif w == 'Z':
end.append(idx)
else:
continue
print(len(word[start[0]:end[-1]+1])) |
s030475482 | p03377 | u973013625 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 85 | There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals. | a, b, x = map(int, input().split())
print('Yes' if a <= x and (a + b) >= x else 'No') | s367322554 | Accepted | 17 | 2,940 | 77 | a, b, x = map(int, input().split())
print('YES' if a <= x <= a + b else 'NO') |
s816120197 | p03607 | u816428863 | 2,000 | 262,144 | Time Limit Exceeded | 2,117 | 193,816 | 248 | You are playing the following game with Joisino. * Initially, you have a blank sheet of paper. * Joisino announces a number. If that number is written on the sheet, erase the number from the sheet; if not, write the number on the sheet. This process is repeated N times. * Then, you are asked a question: How many numbers are written on the sheet now? The numbers announced by Joisino are given as A_1, ... ,A_N in the order she announces them. How many numbers will be written on the sheet at the end of the game? | # coding: utf-8
# Here your code !
list = []
for i in range(1000000000):
list.append(0)
N = int(input())
for i in range(N):
a = int(input())
if list[a - 1] == 0:
list[a-1] = 1
else:
list[a-1] = 0
print(list.count(1)) | s317923571 | Accepted | 211 | 11,884 | 158 | N = int(input())
card = set()
for i in range(N):
A = int(input())
if A in card:
card.remove(A)
else:
card.add(A)
print(len(card)) |
s714925361 | p00015 | u661290476 | 1,000 | 131,072 | Wrong Answer | 20 | 7,584 | 126 | A country has a budget of more than 81 trillion yen. We want to process such data, but conventional integer type which uses signed 32 bit can represent up to 2,147,483,647. Your task is to write a program which reads two integers (more than or equal to zero), and prints a sum of these integers. If given integers or the sum have more than 80 digits, print "overflow". | n=int(input())
for _ in range(n):
a=int(input())
b=int(input())
ab=a+b
print(ab if ab<=10**80 else "overflow") | s190220426 | Accepted | 30 | 7,512 | 125 | n=int(input())
for _ in range(n):
a=int(input())
b=int(input())
ab=a+b
print(ab if ab<10**80 else "overflow") |
s625574605 | p02842 | u274635633 | 2,000 | 1,048,576 | Wrong Answer | 24 | 5,016 | 95 | Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact. | n=int(input())
x=[(i*108)//100 for i in range(1,n+1)]
if n in x:
print(n)
else:
print(':(') | s325830999 | Accepted | 24 | 5,016 | 133 | import bisect
n=int(input())
x=[(i*108)//100 for i in range(1,n+1)]
if n in x:
print(bisect.bisect_left(x,n)+1)
else:
print(':(') |
s476848479 | p03943 | u172873334 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 87 | Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students. | a=[int(i) for i in input().split()]
a.sort()
print('YES' if a[0]+a[1]==a[2] else 'NO')
| s401994130 | Accepted | 20 | 2,940 | 87 | a=[int(i) for i in input().split()]
a.sort()
print('Yes' if a[0]+a[1]==a[2] else 'No')
|
s259492242 | p03478 | u615850856 | 2,000 | 262,144 | Wrong Answer | 34 | 3,060 | 163 | Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive). | N,A,B = (int(x) for x in input().split())
ans = 0
for n in range(1,N+1):
temp = [int(x) for x in str(n)]
if A <= sum(temp) <= B:
ans+= sum(temp)
print(ans) | s544411559 | Accepted | 33 | 3,060 | 155 | N,A,B = (int(x) for x in input().split())
ans = 0
for n in range(1,N+1):
temp = [int(x) for x in str(n)]
if A <= sum(temp) <= B:
ans+= n
print(ans) |
s086213670 | p03157 | u619819312 | 2,000 | 1,048,576 | Wrong Answer | 509 | 7,008 | 1,242 | There is a grid with H rows and W columns, where each square is painted black or white. You are given H strings S_1, S_2, ..., S_H, each of length W. If the square at the i-th row from the top and the j-th column from the left is painted black, the j-th character in the string S_i is `#`; if that square is painted white, the j-th character in the string S_i is `.`. Find the number of pairs of a black square c_1 and a white square c_2 that satisfy the following condition: * There is a path from the square c_1 to the square c_2 where we repeatedly move to a vertically or horizontally adjacent square through an alternating sequence of black and white squares: black, white, black, white... | h,w=map(int,input().split())
s=[[0]+list(input())+[0]for i in range(h)]
s.insert(0,[0]*(w+2))
s.append([0]*(w+2))
c=["#","."]
n=0
print(s)
for i in range(1,h+1):
for j in range(1,w+1):
g=[0,0]
if s[i][j]!=0:
p=c.index(s[i][j])
g[c.index(s[i][j])]+=1
k=[[i,j]]
s[i][j]=0
while len(k)>0:
b=[]
for f in range(len(k)):
if s[k[f][0]+1][k[f][1]]==c[(p+1)%2]:
b.append([k[f][0]+1,k[f][1]])
s[k[f][0]+1][k[f][1]]=0
if s[k[f][0]-1][k[f][1]]==c[(p+1)%2]:
b.append([k[f][0]-1,k[f][1]])
s[k[f][0]-1][k[f][1]]=0
if s[k[f][0]][k[f][1]+1]==c[(p+1)%2]:
b.append([k[f][0],k[f][1]+1])
s[k[f][0]][k[f][1]+1]=0
if s[k[f][0]][k[f][1]-1]==c[(p+1)%2]:
b.append([k[f][0],k[f][1]-1])
s[k[f][0]][k[f][1]-1]=0
else:
k=b
g[(p+1)%2]+=len(b)
p+=1
else:
n+=g[0]*g[1]
else:
print(n) | s366168237 | Accepted | 496 | 4,468 | 1,233 | h,w=map(int,input().split())
s=[[0]+list(input())+[0]for i in range(h)]
s.insert(0,[0]*(w+2))
s.append([0]*(w+2))
c=["#","."]
n=0
for i in range(1,h+1):
for j in range(1,w+1):
g=[0,0]
if s[i][j]!=0:
p=c.index(s[i][j])
g[c.index(s[i][j])]+=1
k=[[i,j]]
s[i][j]=0
while len(k)>0:
b=[]
for f in range(len(k)):
if s[k[f][0]+1][k[f][1]]==c[(p+1)%2]:
b.append([k[f][0]+1,k[f][1]])
s[k[f][0]+1][k[f][1]]=0
if s[k[f][0]-1][k[f][1]]==c[(p+1)%2]:
b.append([k[f][0]-1,k[f][1]])
s[k[f][0]-1][k[f][1]]=0
if s[k[f][0]][k[f][1]+1]==c[(p+1)%2]:
b.append([k[f][0],k[f][1]+1])
s[k[f][0]][k[f][1]+1]=0
if s[k[f][0]][k[f][1]-1]==c[(p+1)%2]:
b.append([k[f][0],k[f][1]-1])
s[k[f][0]][k[f][1]-1]=0
else:
k=b
g[(p+1)%2]+=len(b)
p+=1
else:
n+=g[0]*g[1]
else:
print(n) |
s128093951 | p03844 | u816631826 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 275 | Joisino wants to evaluate the formula "A op B". Here, A and B are integers, and the binary operator op is either `+` or `-`. Your task is to evaluate the formula instead of her. | # python
user_input = input("Input in format A op B")
input_sep = user_input.split()
input_a = int(input_sep[0])
operator = input_sep[1]
input_b = int(input_sep[2])
if operator is '+':
print(int(input_a + input_b))
elif operator is '-':
print(int(input_a - input_b)) | s816754038 | Accepted | 17 | 2,940 | 109 | a=list(input().split(" "))
if a[1]=='+':
print(int(a[0])+int(a[2]))
else:
print(int(a[0])-int(a[2]))
|
s907698811 | p03696 | u248670337 | 2,000 | 262,144 | Wrong Answer | 29 | 9,128 | 96 | 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. | input()
s=t=input()
l,r=o="()"
exec('s=s'+'.replace(o,"")'*50)
print(l*(c:=s.count)(l)+t+r*c(r)) | s038377839 | Accepted | 26 | 9,200 | 96 | input()
s=t=input()
l,r=o="()"
exec('s=s'+'.replace(o,"")'*50)
print(l*(c:=s.count)(r)+t+r*c(l)) |
s882378695 | p03160 | u816265237 | 2,000 | 1,048,576 | Wrong Answer | 141 | 14,672 | 317 | There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N. | n = int(input())
ln = list(map(int, input().split()))
min_ln = []
for i in range(n):
if i == 0:
min_ln.append(0)
elif i == 1:
min_ln.append(abs(ln[1] - ln[0]))
else:
min_ln.append(min(abs(ln[i] - ln[i-1]) + min_ln[i-1], abs(ln[i] - ln[i-2]) + min_ln[i-2]))
print(min_ln,min_ln[-1]) | s593131843 | Accepted | 132 | 13,980 | 310 | n = int(input())
ln = list(map(int, input().split()))
min_ln = []
for i in range(n):
if i == 0:
min_ln.append(0)
elif i == 1:
min_ln.append(abs(ln[1] - ln[0]))
else:
min_ln.append(min(abs(ln[i] - ln[i-1]) + min_ln[i-1], abs(ln[i] - ln[i-2]) + min_ln[i-2]))
print(min_ln[-1]) |
s017987547 | p03524 | u818349438 | 2,000 | 262,144 | Wrong Answer | 18 | 3,188 | 136 | Snuke has a string S consisting of three kinds of letters: `a`, `b` and `c`. He has a phobia for palindromes, and wants to permute the characters in S so that S will not contain a palindrome of length 2 or more as a substring. Determine whether this is possible. | s = input()
a = s.count('a')
b = s.count('b')
c = s.count('c')
p = [a,b,c]
if max(p) - min(p) <=1:
print('Yes')
else:print('No') | s189632435 | Accepted | 18 | 3,188 | 136 | s = input()
a = s.count('a')
b = s.count('b')
c = s.count('c')
p = [a,b,c]
if max(p) - min(p) <=1:
print('YES')
else:print('NO') |
s928785153 | p03759 | u569272329 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 94 | 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 a - b == b - c:
print("Yes")
else:
print("No")
| s565176627 | Accepted | 18 | 2,940 | 94 | a, b, c = map(int, input().split())
if b - a == c - b:
print("YES")
else:
print("NO")
|
s967932803 | p03140 | u762424840 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,060 | 224 | You are given three strings A, B and C. Each of these is a string of length N consisting of lowercase English letters. Our objective is to make all these three strings equal. For that, you can repeatedly perform the following operation: * Operation: Choose one of the strings A, B and C, and specify an integer i between 1 and N (inclusive). Change the i-th character from the beginning of the chosen string to some other lowercase English letter. What is the minimum number of operations required to achieve the objective? | # -*- coding: utf-8 -*-
N = int(input())
A = input()
B = input()
C = input()
cnt = 0
for i in range(N):
if A[i] != B[i]:
cnt+=1
for i in range(N):
if A[i] != C[i]:
cnt+=1
print("{}".format(cnt))
| s420603768 | Accepted | 17 | 3,060 | 338 | # -*- coding: utf-8 -*-
N = int(input())
A = input()
B = input()
C = input()
min = 0
for i in range(N):
if A[i] == B[i]:
if A[i] != C[i]:
min += 1
else:
if A[i] == C[i]:
min += 1
elif C[i] == B[i]:
min += 1
else:
min += 2
print("{}".format(min))
|
s679283840 | p03470 | u186893542 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 71 | An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have? | d = {}
for i in range(int(input())):
d[i] = i
print(len(d.keys())) | s365639813 | Accepted | 18 | 2,940 | 90 | d = {}
for i in range(int(input())):
k = int(input())
d[k] = 1
print(len(d.keys())) |
s613627670 | p02844 | u282657760 | 2,000 | 1,048,576 | Wrong Answer | 2,104 | 5,236 | 455 | AtCoder Inc. has decided to lock the door of its office with a 3-digit PIN code. The company has an N-digit lucky number, S. Takahashi, the president, will erase N-3 digits from S and concatenate the remaining 3 digits without changing the order to set the PIN code. How many different PIN codes can he set this way? Both the lucky number and the PIN code may begin with a 0. | import re
N = int(input())
S = input()
ans = 0
tmp = [[m.start() for m in re.finditer(str(i), S)] for i in range(10)]
print(tmp)
for i in range(10):
if len(tmp[i]) == 0:
continue
for j in range(10):
if len(tmp[j]) == 0:
continue
for k in range(10):
if len(tmp[k]) == 0:
continue
l = tmp[i][0]
r = tmp[k][-1]
for m in range(l+1, r):
if m in tmp[j]:
ans += 1
break
print(ans) | s191278427 | Accepted | 136 | 4,552 | 540 | import re
N = int(input())
S = input()
ans = 0
tmp = [[m.start() for m in re.finditer(str(i), S)] for i in range(10)]
for i in range(10):
if len(tmp[i]) == 0:
continue
for j in range(10):
if len(tmp[j]) == 0:
continue
for k in range(10):
if len(tmp[k]) == 0:
continue
l = tmp[i][0]
flag = True
for m in tmp[j]:
if m > l:
flag = False
break
if flag:
continue
for n in tmp[k]:
if n > m:
ans += 1
break
print(ans) |
s093362765 | p03636 | u106778233 | 2,000 | 262,144 | Wrong Answer | 24 | 9,036 | 71 | The word `internationalization` is sometimes abbreviated to `i18n`. This comes from the fact that there are 18 letters between the first `i` and the last `n`. You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way. | s =input()
length = len(s)
ans = s[0] + str(length) + s[-1]
print(ans) | s561031446 | Accepted | 29 | 8,996 | 74 | s =input()
length = len(s)- 2
ans = s[0] + str(length) + s[-1]
print(ans) |
s641500854 | p02603 | u695197008 | 2,000 | 1,048,576 | Wrong Answer | 30 | 8,940 | 348 | To become a millionaire, M-kun has decided to make money by trading in the next N days. Currently, he has 1000 yen and no stocks - only one kind of stock is issued in the country where he lives. He is famous across the country for his ability to foresee the future. He already knows that the price of one stock in the next N days will be as follows: * A_1 yen on the 1-st day, A_2 yen on the 2-nd day, ..., A_N yen on the N-th day. In the i-th day, M-kun can make the following trade **any number of times** (possibly zero), **within the amount of money and stocks that he has at the time**. * Buy stock: Pay A_i yen and receive one stock. * Sell stock: Sell one stock for A_i yen. What is the maximum possible amount of money that M-kun can have in the end by trading optimally? | N = int(input())
A = list(map(int, input().split()))
yen = 1000
stocks = 0
for i in range(N):
if i == N-1:
yen += stocks * A[i]
stocks = 0
break
if A[i] > A[i-1]:
# sell everything
yen += stocks * A[i]
stocks = 0
else:
# buy everything
num = yen // A[i]
stocks += num
yen -= A[i] * num
print(yen) | s648307161 | Accepted | 31 | 9,188 | 349 | N = int(input())
A = list(map(int, input().split()))
yen = 1000
stocks = 0
for i in range(N):
if i == N-1:
yen += stocks * A[i]
stocks = 0
break
if A[i] > A[i+1]:
# sell everything
yen += stocks * A[i]
stocks = 0
else:
# buy everything
num = yen // A[i]
stocks += num
yen -= A[i] * num
print(yen)
|
s559657181 | p02603 | u276686572 | 2,000 | 1,048,576 | Wrong Answer | 31 | 9,228 | 321 | To become a millionaire, M-kun has decided to make money by trading in the next N days. Currently, he has 1000 yen and no stocks - only one kind of stock is issued in the country where he lives. He is famous across the country for his ability to foresee the future. He already knows that the price of one stock in the next N days will be as follows: * A_1 yen on the 1-st day, A_2 yen on the 2-nd day, ..., A_N yen on the N-th day. In the i-th day, M-kun can make the following trade **any number of times** (possibly zero), **within the amount of money and stocks that he has at the time**. * Buy stock: Pay A_i yen and receive one stock. * Sell stock: Sell one stock for A_i yen. What is the maximum possible amount of money that M-kun can have in the end by trading optimally? | days = int(input())
yen = 1000
prices = list(map(int, input().split()))
stocks = 0
for i in range(0, days-1):
if prices[i] < prices[i+1]:
yen %= prices[i]
stocks = yen // prices[i]
elif prices[i] > prices[i+1]:
yen += stocks*prices[i]
stocks = 0
print(yen)
yen += stocks*prices[days-1]
print(yen) | s681930741 | Accepted | 30 | 9,032 | 311 | days = int(input())
yen = 1000
prices = list(map(int, input().split()))
stocks = 0
for i in range(0, days-1):
if prices[i] < prices[i+1]:
stocks += yen // prices[i]
yen %= prices[i]
elif prices[i] > prices[i+1]:
yen += stocks*prices[i]
stocks = 0
yen += stocks*prices[days-1]
print(yen)
|
s409726518 | p03129 | u223646582 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 79 | Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1. | N,K=map(int,input().split())
if K<=(N-1)//2:
print('YES')
else:
print('NO') | s937587531 | Accepted | 20 | 2,940 | 79 | N,K=map(int,input().split())
if K<=(N+1)//2:
print('YES')
else:
print('NO') |
s316426353 | p03455 | u511421299 | 2,000 | 262,144 | Wrong Answer | 18 | 3,064 | 72 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | a, b = map(int, input().split())
print('even' if a*b%2 == 0 else 'odd')
| s573768171 | Accepted | 17 | 2,940 | 71 | a, b = map(int, input().split())
print('Even' if a*b%2 == 0 else 'Odd') |
s708893690 | p03578 | u999449420 | 2,000 | 262,144 | Wrong Answer | 2,105 | 35,420 | 558 | 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, M, K = map(int, input().split())
# N = int(input())
N = int(input())
D = list(map(int, input().split()))
M = int(input())
T = list(map(int, input().split()))
flag = False
ans_flag = True
for i in T:
for n, j in enumerate(D):
if i == j:
D.pop(n)
# print(D)
flag = True
break
if flag == False:
print("No")
ans_flag = False
break
else:
flag = False
if ans_flag == True:
print("Yes")
| s608735393 | Accepted | 323 | 57,056 | 429 | import collections
N = int(input())
D = list(map(int, input().split()))
M = int(input())
T = list(map(int, input().split()))
flag = False
ans_flag = True
d = collections.Counter(D)
t = collections.Counter(T)
# print(d.most_common())
# print(t.most_common())
for i in t:
i_str = str(i)
d_count = d[i]
if d_count < t[i]:
print("NO")
ans_flag = False
break
if ans_flag == True:
print("YES") |
s244654200 | p02390 | u505411588 | 1,000 | 131,072 | Wrong Answer | 50 | 7,568 | 95 | Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively. | S = int(input())
h = str(S//360)
M = S%360
m = str(M//60)
s = str(M%60)
print(h+":"+m+":"+s) | s698313340 | Accepted | 30 | 7,648 | 97 | S = int(input())
h = str(S//3600)
M = S%3600
m = str(M//60)
s = str(M%60)
print(h+":"+m+":"+s) |
s108074663 | p03861 | u369212307 | 2,000 | 262,144 | Wrong Answer | 18 | 3,064 | 103 | You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x? | a, b, x = map(int, input().split())
ans = 0
if a == 0:
ans = 1
ans += int((b - a) / x )
print(ans) | s839258348 | Accepted | 18 | 2,940 | 254 | a, b, x = map(int, input().split())
ans = 0
def fanc(n, x):
if n == -1:
f = 0
else:
f = n // x + 1
return f
print(fanc(b, x) - fanc(a - 1, x)) |
s272963357 | p03609 | u023229441 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 46 | We have a sandglass that runs for X seconds. The sand drops from the upper bulb at a rate of 1 gram per second. That is, the upper bulb initially contains X grams of sand. How many grams of sand will the upper bulb contains after t seconds? | a,b=map(int,input().split())
print(max(0,b-a)) | s036911314 | Accepted | 18 | 2,940 | 46 | a,b=map(int,input().split())
print(max(0,a-b)) |
s687913332 | p04031 | u492447501 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 212 | Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal **integers** by transforming some of them. He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2). Find the minimum total cost to achieve his objective. | n = int(input())
*a, = map(int, input().split())
stack = []
for i in range(n):
stack.append(a[i])
stack.reverse()
st = ""
for i in range(len(stack)):
st = st + str(stack[i])
print(" ".join(st)) | s482762352 | Accepted | 25 | 3,060 | 235 | N = int(input())
*A, = map(int, input().split())
min = float("inf")
for y in range(-100, 101, 1):
sum = 0
for a in range(len(A)):
sum = sum + (A[a]-y)**2
if min > sum:
min = sum
ans = y
print(min) |
s932900851 | p02397 | u403901064 | 1,000 | 131,072 | Wrong Answer | 20 | 7,700 | 168 | Write a program which reads two integers x and y, and prints them in ascending order. | #!/usr/bin/env python3
# -*- coding: utf-8 -*-
def main():
nums = input().split(" ")
nums.sort(key = int)
print(" ".join(nums))
if __name__ == '__main__':
main() | s617648903 | Accepted | 40 | 8,312 | 249 | #!/usr/bin/env python3
# -*- coding: utf-8 -*-
def main():
nums = []
while True:
x = input().split(" ")
if x == ["0","0"]:
break
x.sort(key = int)
nums.append(x)
for i in nums:
print(" ".join(i))
if __name__ == '__main__':
main() |
s055251086 | p02831 | u183422236 | 2,000 | 1,048,576 | Wrong Answer | 2,103 | 2,940 | 151 | Takahashi is organizing a party. At the party, each guest will receive one or more snack pieces. Takahashi predicts that the number of guests at this party will be A or B. Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted. We assume that a piece cannot be divided and distributed to multiple guests. | a, b = map(int, input().split())
c = min(a, b)
print(c)
for i in range(c, a * b + 1):
if i % a == 0 and i % b == 0:
print(i)
exit() | s472862959 | Accepted | 38 | 2,940 | 204 | a, b = map(int, input().split())
c = a * b
d = min(a, b)
j = 1
for i in range(2, d + 1):
j += 1
if a % j == 0 and b % j == 0:
c //= j
a //= j
b //= j
j = 1
print(c) |
s656421900 | p02417 | u256678932 | 1,000 | 131,072 | Wrong Answer | 30 | 7,916 | 193 | Write a program which counts and reports the number of each alphabetical letter. Ignore the case of characters. | import string
chs = dict([(ch, 0) for ch in string.ascii_lowercase])
for ch in input():
if ch not in chs:
continue
chs[ch] += 1
for k, v in chs.items():
print(k, ':', v) | s310646296 | Accepted | 40 | 6,684 | 340 | import string
dic = { ch:0 for ch in string.ascii_lowercase }
while True:
try:
line = input()
for ch in line:
c = ch.lower()
if c not in dic:
continue
dic[c] += 1
except EOFError:
break
for ch in string.ascii_lowercase:
print(ch, ':', dic[ch])
|
s657234034 | p02747 | u572012241 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 220 | A Hitachi string is a concatenation of one or more copies of the string `hi`. For example, `hi` and `hihi` are Hitachi strings, while `ha` and `hii` are not. Given a string S, determine whether S is a Hitachi string. | s = input()
for i in range(len(s)-1):
if (s[i]== "h" and s[i+1] == "i"):
if i+1 == len(s)-1:
print("Yes")
elif s[i+2] != "i":
print("No")
else:
print("Yes")
else:
print("No")
| s220592223 | Accepted | 58 | 5,732 | 451 | from collections import Counter,defaultdict,deque
from heapq import heappop,heappush,heapify
import sys,bisect,math,itertools,fractions,pprint
sys.setrecursionlimit(10**8)
mod = 10**9+7
INF = float('inf')
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
s = input()
if s == 'hi' or s == 'hihi' or s == 'hihihi' or s == 'hihihihi' or s == 'hihihihihi':
print('Yes')
else:
print('No')
|
s808652396 | p03067 | u813450934 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 108 | 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 min(a,c)<= b and max(a,c) >= b:
print("Yes")
else :
print("No")
| s195464566 | Accepted | 17 | 2,940 | 108 | a,b,c = map(int,input().split())
if min(a,b)<= c and max(a,b) >= c:
print("Yes")
else :
print("No")
|
s647314565 | p03493 | u254871849 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 100 | 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. | ls = list(map(int, input().split()))
count = 0
for i in ls:
if i == 1:
count += 1
print(count) | s025728749 | Accepted | 21 | 3,316 | 174 | import sys
from collections import Counter
s = sys.stdin.readline().rstrip()
def main():
c = Counter(s)
print(c.get('1', 0))
if __name__ == '__main__':
main() |
s298986865 | p03434 | u197078193 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 132 | We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score. | N = int(input())
A = [int(a) for a in input().split()]
A.sort()
d = 0
for a in A[::2]:
d += a
for a in A[1::2]:
d -= a
print(d)
| s380020416 | Accepted | 17 | 2,940 | 144 | N = int(input())
A = [int(a) for a in input().split()]
A.sort()
A.reverse()
d = 0
for a in A[::2]:
d += a
for a in A[1::2]:
d -= a
print(d)
|
s625601563 | p04043 | u622045059 | 2,000 | 262,144 | Wrong Answer | 20 | 3,316 | 122 | Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order. | import collections
abc = collections.Counter(input().split())
print('Yes' if abc['5'] == 2 and abc['7'] == 1 else 'No')
| s561787483 | Accepted | 20 | 3,316 | 122 | import collections
abc = collections.Counter(input().split())
print('YES' if abc['5'] == 2 and abc['7'] == 1 else 'NO')
|
s207182963 | p03162 | u905582793 | 2,000 | 1,048,576 | Wrong Answer | 655 | 55,008 | 507 | Taro's summer vacation starts tomorrow, and he has decided to make plans for it now. The vacation consists of N days. For each i (1 \leq i \leq N), Taro will choose one of the following activities and do it on the i-th day: * A: Swim in the sea. Gain a_i points of happiness. * B: Catch bugs in the mountains. Gain b_i points of happiness. * C: Do homework at home. Gain c_i points of happiness. As Taro gets bored easily, he cannot do the same activities for two or more consecutive days. Find the maximum possible total points of happiness that Taro gains. | import sys
N = int(input())
summer = []
for i in range(N):
summer.append(list(map(int, sys.stdin.readline().rstrip().split(" "))))
print(summer)
INF = float('inf')
DP = [[-INF for i in range(3)] for i in range(N)]
print(DP)
for i in range(N):
if i == 0:
DP[i] = summer[i]
else:
DP[i][0] = max(DP[i-1][1], DP[i-1][2]) + summer[i][0]
DP[i][1] = max(DP[i-1][0], DP[i-1][2]) + summer[i][1]
DP[i][2] = max(DP[i-1][0], DP[i-1][1]) + summer[i][2]
print(max(DP[N-1])) | s254610747 | Accepted | 520 | 47,280 | 482 | import sys
N = int(input())
summer = []
for i in range(N):
summer.append(list(map(int, sys.stdin.readline().rstrip().split(" "))))
INF = float('inf')
DP = [[-INF for i in range(3)] for i in range(N)]
for i in range(N):
if i == 0:
DP[i] = summer[i]
else:
DP[i][0] = max(DP[i-1][1], DP[i-1][2]) + summer[i][0]
DP[i][1] = max(DP[i-1][0], DP[i-1][2]) + summer[i][1]
DP[i][2] = max(DP[i-1][0], DP[i-1][1]) + summer[i][2]
print(max(DP[N-1])) |
s839302832 | p02613 | u243312682 | 2,000 | 1,048,576 | Wrong Answer | 146 | 17,732 | 488 | Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format. | from collections import Counter
def main():
n = int(input())
s = [input() for i in range(n)]
print(s)
ac = 0
wa = 0
tle = 0
re = 0
for i in s:
if i == 'AC':
ac += 1
elif i == 'WA':
wa += 1
elif i == 'TLE':
tle += 1
elif i == 'RE':
re += 1
print('AC x', ac)
print('WA x', wa)
print('TLE x', tle)
print('RE x', re)
if __name__ == '__main__':
main() | s179711587 | Accepted | 144 | 16,204 | 115 | n = int(input())
s = [input() for i in range(n)]
for i in ['AC', 'WA', 'TLE', 'RE']:
print(i, 'x', s.count(i)) |
s677201892 | p03599 | u607074939 | 3,000 | 262,144 | Time Limit Exceeded | 3,184 | 433,900 | 626 | Snuke is making sugar water in a beaker. Initially, the beaker is empty. Snuke can perform the following four types of operations any number of times. He may choose not to perform some types of operations. * Operation 1: Pour 100A grams of water into the beaker. * Operation 2: Pour 100B grams of water into the beaker. * Operation 3: Put C grams of sugar into the beaker. * Operation 4: Put D grams of sugar into the beaker. In our experimental environment, E grams of sugar can dissolve into 100 grams of water. Snuke will make sugar water with the highest possible density. The beaker can contain at most F grams of substances (water and sugar combined), and there must not be any undissolved sugar in the beaker. Find the mass of the sugar water Snuke will make, and the mass of sugar dissolved in it. If there is more than one candidate, any of them will be accepted. We remind you that the sugar water that contains a grams of water and b grams of sugar is \frac{100b}{a + b} percent. Also, in this problem, pure water that does not contain any sugar is regarded as 0 percent density sugar water. | A,B,C,D,E,F = map(int,input().split())
W = []
S = []
for i in range(F+1):
for j in range(F+1):
x = 100*A*i + 100*B*j
W.append(x)
for i in range(F+1):
for j in range(F+1):
y = C*i + D*j
S.append(y)
maxcon = 0
limitcon = 100*E/(100+E)
ans1 = 0
ans2 = 0
for i in range(1,len(W)):
for j in range(1,len(S)):
sw = W[i] + S[j]
con = 100*S[j] / sw
if (sw <= F) and (con <= limitcon) and (maxcon < con):
maxcon = con
ans1 = sw
ans2 = S[j]
print("%s %s"%(ans1,ans2)) | s300537508 | Accepted | 753 | 3,064 | 742 | A,B,C,D,E,F=map(int,input().split())
best_sugarwater = 100*A
best_sugar = 0
best_con = 0
max_con = E/(100+E)
maxA = F//(100*A)
for i in range(maxA+1):
maxB = (F-100*A*i)//(100*B)
for j in range(maxB+1):
water = 100*A*i+100*B*j
maxC = (F-100*A*i-100*B*j)//C
for k in range(maxC+1):
maxD = (F-100*A*i-100*B*j-C*k)//D
for l in range (maxD+1):
sugar = C*k+D*l
if sugar + water == 0 :
pass
elif (best_con < sugar/(sugar + water) <= max_con ):
best_sugarwater = sugar + water
best_sugar = sugar
best_con = sugar/(sugar + water)
print(best_sugarwater,best_sugar) |
s359823904 | p03997 | u278143034 | 2,000 | 262,144 | Wrong Answer | 30 | 9,156 | 133 | 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())
area = (a + b) * h / 0.5
print(area) | s324984536 | Accepted | 26 | 9,168 | 137 |
a = int(input())
b = int(input())
h = int(input())
area = int((a + b) * h / 2)
print(area) |
s390685025 | p03672 | u407730443 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 142 | We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not. You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input. | s = input()
for i in range(len(s)):
s = s[:-1]
if len(s) % 2 == 0 and s[:len(s)//2] == s[-len(s)//2:]:
print(s)
break | s953912492 | Accepted | 17 | 2,940 | 147 | s = input()
for i in range(len(s)):
s = s[:-1]
if len(s) % 2 == 0 and s[:len(s)//2] == s[-len(s)//2:]:
print(len(s))
break |
s818630919 | p03149 | u225254556 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 209 | You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974". | # B_KEYENCE String
import sys
S = list(input())
key = ['k','e','y','e','n','c','e']
i = 0
for _ in range(len(S)) :
if S[_] == key[i] :
i += 1
if i > 6 :
print("YES")
else :
print("NO")
| s510478713 | Accepted | 18 | 3,064 | 513 | #A_Beginning
import sys
N = list(map(int, input().split()))
wk = [99 for _ in range(len(N))]
flg = True
for i in range(len(N)) :
if N[i] == 1 or N[i] == 9 or N[i] == 7 or N[i] == 4 :
for _ in range(len(wk)) :
if wk[_] == N[i] :
flg = False
break
if flg == False :
break
else :
wk[i] = N[i]
flg = True
else :
flg = False
break
if flg == True :
print("YES")
else :
print("NO") |
s823957903 | p03229 | u621100542 | 2,000 | 1,048,576 | Wrong Answer | 316 | 8,288 | 1,288 | You are given N integers; the i-th of them is A_i. Find the maximum possible sum of the absolute differences between the adjacent elements after arranging these integers in a row in any order you like. | # coding: utf-8
# Your code here!
S = int(input())
lis = []
for item in range(0,int(S)):
tmp = input()
lis.append(int(tmp))
ls = sorted(lis)
#print(ls)
#print(ls[int(S/2)])
i = 0
lo = 0
hg = int(S)-1
ans = ls[hg]- ls[int(S/2)]
#print("first",ans)
if S % 2 == 1 or S % 2 == 0:
for k in range(0,int(S/2)+1):
#print(i)
# TODO: write code...
# TODO: write code...
# TODO: write code...
#if(i != 0):
#hg-=1
if(i == int(S)-1):
break
ans = ans + ls[hg] - ls[lo]
i+=1
#print(ls[hg],ls[lo])
hg-=1
if(i == int(S)-1):
break
ans = ans + ls[hg] - ls[lo]
#print(ls[hg],ls[lo])
lo+=1
i+=1
else:
for i in range(0,int(S/2)+1):
# TODO: write code...
# TODO: write code...
# TODO: write code...
#if(i != 0):
#hg-=1
ans = ans + ls[hg] - ls[lo]
print(ls[hg],ls[lo])
if(hg == int(S/2)):
break
hg-=1
ans = ans + ls[hg] - ls[lo]
print(ls[hg],ls[lo])
lo+=1
print(ans)
# hg-=1
# print(ls[hg] , ls[lo])
# print(ans)
# lo+=1
# hg-=1
# print(ls[hg] , ls[lo])
# print(ans)
| s507518068 | Accepted | 1,825 | 8,928 | 682 | # coding: utf-8
# Your code here!
S = int(input())
lis = []
for item in range(0,int(S)):
tmp = input()
lis.append(int(tmp))
ls = sorted(lis)
lsb = sorted(lis)
center = ls[int(S/2-0.5)]
ans1 = ls[-1] - center
ls.remove(center)
for k in range(0,int(S/2)):
try:
ans1 = ans1 + ls.pop(-1) - ls[0]
ans1 = ans1 + ls[-1] -ls.pop(0)
except:
break
if S % 2 ==0:
print(ans1)
exit()
center = lsb[int(S/2)]
ans2 = center - lsb[0]
lsb.remove(center)
for k in range(0,int(S/2)):
try:
ans2 = ans2 + lsb[-1] - lsb.pop(0)
ans2 = ans2 + lsb.pop(-1) - lsb[0]
except:
break
print(max(ans1,ans2))
|
s872989577 | p03139 | u670180528 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 61 | We conducted a survey on newspaper subscriptions. More specifically, we asked each of the N respondents the following two questions: * Question 1: Are you subscribing to Newspaper X? * Question 2: Are you subscribing to Newspaper Y? As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2. What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y? Write a program to answer this question. | n,a,b=map(int,input().split());print(min(n,a+b),max(0,a+b-n)) | s823393574 | Accepted | 17 | 2,940 | 59 | n,a,b=map(int,input().split());print(min(a,b),max(0,a+b-n)) |
s335676683 | p03854 | u989074104 | 2,000 | 262,144 | Wrong Answer | 18 | 3,188 | 236 | You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`. | S=input()
str_list=["dream","dreamer","erase","eraser"]
S_tmp=""
while True:
S_tmp=S
for str in str_list:
S=S.replace(str,"")
if S_tmp==S:
break
if S=="":
print("Yes")
if S!="":
print("No") | s194836630 | Accepted | 100 | 3,340 | 260 | S=input()
str_list=["dreamer","eraser","dream","erase"]
while True:
S_tmp=S
for str in str_list:
if S[-len(str):]==str:
S=S[0:len(S)-len(str)]
if S_tmp==S:
break
if S=="":
print("YES")
if S!="":
print("NO") |
s597896463 | p02261 | u831244171 | 1,000 | 131,072 | Wrong Answer | 20 | 7,672 | 1,297 | Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance). | def selectionSort(a,b):
for i in range(len(a)):
mini = i
for j in range(i,len(a)):
if a[j] < a[mini]:
mini = j
if i != mini:
ret = a[i]
a[i] = a[mini]
a[mini] = ret
ret = b[i]
b[i] =b[mini]
b[mini] = ret
output = []
for i in range(len(a)):
output.append(str(b[i])+str(a[i]))
print(" ".join(output))
return output
def bubblesort(a,b):
swap = True
while swap:
swap = False
for j in range(1,len(a))[::-1]:
if a[j] < a[j-1]:
ret = a[j]
a[j] = a[j-1]
a[j-1] = ret
ret = b[j]
b[j] = b[j-1]
b[j-1] = ret
swap = True
output = []
for i in range(len(a)):
output.append(str(b[i])+str(a[i]))
print(" ".join(output))
n = int(input())
card = input().split()
number = []
mark = []
for i in range(n):
number.append(int(list(card[i])[1]))
mark.append(list(card[i])[0])
number2 = number
mark2 = mark
output1 = bubblesort(number,mark)
print("Stable")
output2 = selectionSort(number2, mark2)
print("Stable" if output1 == output2 else "Not table" ) | s252578315 | Accepted | 20 | 7,792 | 1,386 | def selectionSort(a,b):
for i in range(len(a)):
mini = i
for j in range(i,len(a)):
if a[j] < a[mini]:
mini = j
if i != mini:
ret = a[i]
a[i] = a[mini]
a[mini] = ret
ret = b[i]
b[i] =b[mini]
b[mini] = ret
output = []
for i in range(len(a)):
output.append(str(b[i])+str(a[i]))
print(" ".join(output))
return output
def bubblesort(a,b):
swap = True
while swap:
swap = False
for j in range(1,len(a))[::-1]:
if a[j] < a[j-1]:
ret = a[j]
a[j] = a[j-1]
a[j-1] = ret
ret = b[j]
b[j] = b[j-1]
b[j-1] = ret
swap = True
output = []
for i in range(len(a)):
output.append(str(b[i])+str(a[i]))
print(" ".join(output))
return output
n = int(input())
card = input().split()
number = []
mark = []
number2 = []
mark2 = []
for i in range(n):
number.append(int(list(card[i])[1]))
mark.append(list(card[i])[0])
number2.append(int(list(card[i])[1]))
mark2.append(list(card[i])[0])
output1 = bubblesort(number,mark)
print("Stable")
output2 = selectionSort(number2, mark2)
print("Stable" if output1 == output2 else "Not stable" ) |
s744702413 | p03385 | u710282484 | 2,000 | 262,144 | Wrong Answer | 30 | 8,772 | 85 | You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`. | s=input()
s=sorted(s)
l=['a','b','c']
if s==l:
print('YES')
else:
print('NO') | s953699815 | Accepted | 26 | 8,940 | 103 | s=input()
s=sorted(s)
if s[0]=='a' and s[1]=='b' and s[2]=='c':
print('Yes')
else:
print('No') |
s228711161 | p03455 | u220870679 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 85 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | n, m = map(int, input().split())
if (n*m)%2:
print("Even")
else:
print("Odd") | s249790589 | Accepted | 17 | 2,940 | 101 | a, b = map(int, input() .split())
c = (a * b) % 2
if c == 0:
print("Even")
else:
print("Odd") |
s215511232 | p03854 | u065578867 | 2,000 | 262,144 | Wrong Answer | 21 | 4,724 | 452 | You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`. | s_init = input()
s_list = list(reversed(s_init))
s = ''.join(s_list)
t = ""
words = ["maerd", "remaerd", 'esare', 'resare']
s_list = list(s)
t_list = list(t)
while s != t:
if len(s_list) < len(t_list):
break
elif t == 'gameover':
break
for word in words:
if s.startswith(t + word):
t = t + word
break
else:
t = 'gameover'
if s == t:
print('YES')
else:
print('NO')
| s270149890 | Accepted | 236 | 4,724 | 505 | s_init = input()
s_list = list(reversed(s_init))
s = ''.join(s_list)
t = ""
words = ["maerd", "remaerd", 'esare', 'resare']
s_list = list(s)
t_list = list(t)
while s != t:
t_fixed = t
if len(s_list) < len(t_list):
break
elif t == 'gameover':
break
else:
for word in words:
if s.startswith(t + word):
t = t + word
break
if t == t_fixed:
t = 'gameover'
if s == t:
print('YES')
else:
print('NO')
|
s069329418 | p03739 | u652057333 | 2,000 | 262,144 | Wrong Answer | 185 | 14,500 | 926 | 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. | n = int(input())
a = list(map(int, input().split()))
ans = 10**10
# 1 -1 1 -1...
cost1 = 0
s = 0
for i in range(n):
t = 1 if i % 2 == 0 else -1
c = 0
if i % 2 == 0:
if s + a[i] > 0:
s += a[i]
else:
c = abs(s - t)
s += c
cost1 += abs(c - a[i])
else:
if s + a[i] < 0:
s += a[i]
else:
c = abs(s - t)
s -= c
cost1 += abs(c - a[i])
cost2 = 0
s = 0
for i in range(n):
t = -1 if i % 2 == 0 else 1
c = 0
if i % 2 == 1:
if s + a[i] > 0:
s += a[i]
else:
c = abs(s - t)
s += c
cost2 += abs(c - a[i])
else:
if s + a[i] < 0:
s += a[i]
else:
c = abs(s - t)
s -= c
cost2 += abs(- c - a[i])
print(cost1, cost2)
print(min(cost1, cost2))
| s890568792 | Accepted | 186 | 14,332 | 908 | n = int(input())
a = list(map(int, input().split()))
ans = 10**10
# 1 -1 1 -1...
cost1 = 0
s = 0
for i in range(n):
t = 1 if i % 2 == 0 else -1
c = 0
if i % 2 == 0:
if s + a[i] > 0:
s += a[i]
else:
c = abs(s - t)
s += c
cost1 += abs(c - a[i])
else:
if s + a[i] < 0:
s += a[i]
else:
c = abs(s - t)
s -= c
cost1 += abs(- c - a[i])
cost2 = 0
s = 0
for i in range(n):
t = -1 if i % 2 == 0 else 1
c = 0
if i % 2 == 1:
if s + a[i] > 0:
s += a[i]
else:
c = abs(s - t)
s += c
cost2 += abs(c - a[i])
else:
if s + a[i] < 0:
s += a[i]
else:
c = abs(s - t)
s -= c
cost2 += abs(- c - a[i])
print(min(cost1, cost2))
|
s628898090 | p03359 | u969848070 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 93 | In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year. For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year. In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For example, 5-5 is Takahashi. How many days from 2018-1-1 through 2018-a-b are Takahashi? | a, b = map(int, input().split())
c = 0
for i in range(1,a+1):
if i == b:
c +=1
print(c) | s464721803 | Accepted | 17 | 2,940 | 79 | a, b = map(int, input().split())
if a < b:
print(a)
else:
print(max(a-1,b)) |
s235618486 | p03679 | u788856752 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 135 | Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache. | X, A, B = map(int, input().split())
if A <= B:
print("delicious")
elif (A - B) > X:
print("dagerous")
else:
print("safe")
| s966728885 | Accepted | 17 | 2,940 | 137 | X, A, B = map(int, input().split())
if A >= B:
print("delicious")
elif (B - A) <= X:
print("safe")
else:
print("dangerous")
|
s221424591 | p03351 | u302292660 | 2,000 | 1,048,576 | Wrong Answer | 19 | 2,940 | 89 | Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate. | a,b,c,d = map(int,input().split())
if b-a<d and c-b<d:
print("Yes")
else:
print("No") | s730738663 | Accepted | 17 | 3,060 | 135 | a,b,c,d = map(int,input().split())
if abs(b-a)<=d and abs(c-b)<=d:
print("Yes")
elif abs(c-a) <=d:
print("Yes")
else:
print("No") |
s629009006 | p03759 | u835575472 | 2,000 | 262,144 | Wrong Answer | 24 | 9,148 | 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 a - b == c - b:
print("YES")
else:
print("NO") | s944421492 | Accepted | 32 | 9,132 | 90 | a, b, c = map(int, input().split())
if b - a == c - b:
print("YES")
else:
print("NO")
|
s189448543 | p03545 | u576432509 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 517 | 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. | import itertools
a=list(map(int,list(input())))
#print(a)
for i in itertools.product([0,1], repeat=3):
r=list(i)
sump=a[0]
for ii in range(3):
if r[ii]==0:
sump=sump-a[ii+1]
else:
sump=sump+a[ii+1]
# print(sump,r)
if sump==7:
stra=str(a[0])
for ii in range(3):
if r[ii]==0:
stra=stra+"-"+str(a[ii+1])
else :
stra=stra+"+"+str(a[ii+1])
print(stra)
break
| s950810280 | Accepted | 17 | 3,064 | 389 | import itertools
a=list(map(int,list(input())))
ans=0
for i in itertools.product([0,1], repeat=3):
asum=a[0]
astr=str(a[0])
for ii in range(len(i)):
if i[ii]==0:
asum+=a[ii+1]
astr=astr +"+" + str(a[ii+1])
else:
asum-=a[ii+1]
astr=astr +"-" + str(a[ii+1])
if asum==7:
break
print(astr+"=7") |
s257614908 | p03609 | u540631540 | 2,000 | 262,144 | Wrong Answer | 26 | 9,068 | 53 | We have a sandglass that runs for X seconds. The sand drops from the upper bulb at a rate of 1 gram per second. That is, the upper bulb initially contains X grams of sand. How many grams of sand will the upper bulb contains after t seconds? | x, t = map(int, input().split())
print(min(0, x - t)) | s087675514 | Accepted | 28 | 8,992 | 53 | x, t = map(int, input().split())
print(max(0, x - t)) |
s827117269 | p03433 | u879761680 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 102 | E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins. | a = int(input())
c = int(input())
b = 0
b = c % 500
if c >= b:
print('YES')
else:
print("NO") | s246374901 | Accepted | 17 | 2,940 | 102 | a = int(input())
c = int(input())
b = 0
b = a% 500
if c >= b:
print('Yes')
else:
print("No") |
s790064649 | p03693 | u675073679 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 115 | AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4? | r, g, b = map(int, input().split())
RGB = int(str(r+g+b))
if RGB % 4 == 0:
print("YES")
else:
print("NO") | s038002567 | Accepted | 17 | 2,940 | 121 | r, g, b = map(int, input().split())
RGB = int(r*100 + g*10 + b)
if RGB % 4 == 0:
print("YES")
else:
print("NO") |
s412539038 | p03493 | u164261323 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 37 | 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. | n = input().split()
print(n.count(1)) | s005839784 | Accepted | 17 | 2,940 | 31 | n = input()
print(n.count("1")) |
s293688848 | p03005 | u271384833 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 100 | Takahashi is distributing N balls to K persons. If each person has to receive at least one ball, what is the maximum possible difference in the number of balls received between the person with the most balls and the person with the fewest balls? | if __name__ == "__main__":
n,k = input().split()
n = int(n)
k = int(k)
print(n-k+1)
| s070168909 | Accepted | 17 | 2,940 | 143 | if __name__ == "__main__":
n,k = input().split()
n = int(n)
k = int(k)
if k == 1:
print(0)
else:
print(n-k) |
s921743680 | p03605 | u163874353 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 116 | 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())
cnt = 0
for i in n:
if n == "9":
cnt += 1
if cnt >= 1:
print("Yes")
else:
print("No") | s872961247 | Accepted | 17 | 2,940 | 121 | n = str(input())
cnt = 0
for i in n:
if i == "9":
cnt += 1
if cnt >= 1:
print("Yes")
else:
print("No") |
s469467880 | p02608 | u688649934 | 2,000 | 1,048,576 | Time Limit Exceeded | 2,206 | 9,160 | 315 | Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N). | max_number = int(input())
list_numbers = [i+1 for i in range(max_number)]
count = 0
for i in list_numbers:
count = 0
for x in range(1,100):
for y in range(1,100):
for z in range(1,100):
tot = x**2 + y**2 + z**2 + x * y + y * z + z * x
if (tot == i):
count += 1
print(count) | s472808412 | Accepted | 832 | 9,432 | 241 | N=int(input())
a=[0]*10**4
for j in range(1,101):
for k in range(1,101):
for h in range(1,101):
tmp=h**2+j**2+k**2+h*j+j*k+k*h
if tmp<=10000:
a[tmp-1]+=1
for i in range(N):
print(a[i])
|
s476998716 | p03598 | u340781749 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 125 | There are N balls in the xy-plane. The coordinates of the i-th of them is (x_i, i). Thus, we have one ball on each of the N lines y = 1, y = 2, ..., y = N. In order to collect these balls, Snuke prepared 2N robots, N of type A and N of type B. Then, he placed the i-th type-A robot at coordinates (0, i), and the i-th type-B robot at coordinates (K, i). Thus, now we have one type-A robot and one type-B robot on each of the N lines y = 1, y = 2, ..., y = N. When activated, each type of robot will operate as follows. * When a type-A robot is activated at coordinates (0, a), it will move to the position of the ball on the line y = a, collect the ball, move back to its original position (0, a) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. * When a type-B robot is activated at coordinates (K, b), it will move to the position of the ball on the line y = b, collect the ball, move back to its original position (K, b) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. Snuke will activate some of the 2N robots to collect all of the balls. Find the minimum possible total distance covered by robots. | n = int(input())
k = int(input())
xxx = list(map(int, input().split()))
ans = sum(min(x, abs(k - x)) for x in xxx)
print(ans) | s305911883 | Accepted | 18 | 2,940 | 130 | n = int(input())
k = int(input())
xxx = list(map(int, input().split()))
ans = sum(min(x, abs(k - x)) for x in xxx) * 2
print(ans)
|
s840540027 | p03588 | u225388820 | 2,000 | 262,144 | Wrong Answer | 475 | 28,652 | 92 | A group of people played a game. All players had distinct scores, which are positive integers. Takahashi knows N facts on the players' scores. The i-th fact is as follows: the A_i-th highest score among the players is B_i. Find the maximum possible number of players in the game. | n=int(input())
a=sorted([list(map(int,input().split())) for i in range(n)])
print(sum(a[0])) | s728137987 | Accepted | 482 | 28,652 | 93 | n=int(input())
a=sorted([list(map(int,input().split())) for i in range(n)])
print(sum(a[-1])) |
s577567845 | p03563 | u788856752 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 56 | 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. | R = float(input())
G = float(input())
print(2 * G - R)
| s188094420 | Accepted | 18 | 2,940 | 52 | R = int(input())
G = int(input())
print(2 * G - R)
|
s218353352 | p03493 | u248192265 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 75 | 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. | s = input()
count = 0
for i in s:
if i == 0:
count += 0
print(count) | s332886922 | Accepted | 18 | 2,940 | 77 | s = input()
count = 0
for i in s:
if i == "1":
count += 1
print(count) |
s927775816 | p03844 | u353919145 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 103 | Joisino wants to evaluate the formula "A op B". Here, A and B are integers, and the binary operator op is either `+` or `-`. Your task is to evaluate the formula instead of her. | s = input()
if s[2]== "-":
print(int(s[0])-int(s[4]))
if s[1]== "+":
print(int(s[0])+int(s[4])) | s126924158 | Accepted | 30 | 9,000 | 128 | inputs = list(map(str,input().split()))
a=int(inputs[0])
b=int(inputs[2])
if inputs[1]=="-":
print(a-b)
else:
print(a+b) |
s226829572 | p03814 | u148471096 | 2,000 | 262,144 | Wrong Answer | 37 | 4,840 | 266 | 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`. | N=list(input())
N_num=len(N)
for i in range(N_num):
if N[i] == "A":
min_num=i
break
N.reverse()
for i in range(N_num):
if N[i] == "Z":
max_num=i
break
print(N_num-max_num-min_num+1) | s376727915 | Accepted | 39 | 4,840 | 284 | N=list(input())
N_num=len(N)
max_num=0
min_num=0
for i in range(N_num):
if N[i] == "A":
min_num=i
break
N.reverse()
for i in range(N_num):
if N[i] == "Z":
max_num=i
break
print(N_num-max_num-min_num)
|
s510872631 | p03067 | u918845030 | 2,000 | 1,048,576 | Wrong Answer | 19 | 2,940 | 136 | 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())
p_min = min(a, b)
p_max = min(a, b)
if p_min <= c <= p_max:
print('Yes')
else:
print('No')
| s127417676 | Accepted | 17 | 2,940 | 157 | a, b, c = map(int, input().split())
p_min = min(a, b)
p_max = max(a, b)
# print(p_min, p_max)
if p_min <= c <= p_max:
print('Yes')
else:
print('No') |
s850568732 | p00001 | u587151698 | 1,000 | 131,072 | Wrong Answer | 20 | 7,572 | 124 | There is a data which provides heights (in meter) of mountains. The data is only for ten mountains. Write a program which prints heights of the top three mountains in descending order. | hight = []
for i in range(10):
hight.append(int(input()))
hight.sort()
print(hight[0])
print(hight[1])
print(hight[2]) | s836858025 | Accepted | 30 | 7,664 | 136 | hight = []
for i in range(10):
hight.append(int(input()))
hight.sort(reverse=True)
print(hight[0])
print(hight[1])
print(hight[2]) |
s422890488 | p02396 | u234508244 | 1,000 | 131,072 | Wrong Answer | 50 | 6,296 | 145 | In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem. | import sys
for i, x in enumerate(sys.stdin.readlines(), 1):
if x == "0":
break
print("Case {0}: {1}".format(i, x.rstrip()))
| s648921199 | Accepted | 140 | 5,560 | 121 | i = 1
while True:
x = input()
if x == "0":
break
print("Case {0}: {1}".format(i, x))
i += 1
|
s207767363 | p02240 | u112247126 | 1,000 | 131,072 | Wrong Answer | 30 | 7,944 | 968 | Write a program which reads relations in a SNS (Social Network Service), and judges that given pairs of users are reachable each other through the network. | from collections import deque
def bfs(root):
global time
Q = deque()
distance[root] = 0
Q.append(root)
i = root
while len(Q) > 0:
label[i] = time
i = Q.popleft()
for v in range(n):
if v in adjDict[i] and color[v] == 'white':
Q.append(v)
color[v] = 'gray'
distance[v] = distance[i] + 1
color[i] = 'black'
firstline = list(map(int, input().split()))
n = firstline[0]
adjDict = {i: [] for i in range(n)}
color = ['white'] * n
distance = [-1] * n
label = [-1] * n
m = firstline[1]
for _ in range(m):
adj = list(map(int, input().split()))
adjDict[adj[0]].append(adj[1])
time = 0
for i in range(n):
if color[i] == 'white':
bfs(i)
time += 1
q = int(input())
for i in range(q):
question = list(map(int, input().split()))
if label[question[0]] == label[question[1]]:
print('yes')
else:
print('no') | s402786180 | Accepted | 760 | 46,412 | 985 | from collections import deque
def bfs(root):
global time
Q = deque()
distance[root] = 0
Q.append(root)
i = root
while len(Q) > 0:
i = Q.popleft()
for v in adjDict[i]:
if color[v] == 'white':
Q.append(v)
color[v] = 'gray'
distance[v] = distance[i] + 1
color[i] = 'black'
label[i] = time
firstline = list(map(int, input().split()))
n = firstline[0]
adjDict = {i: [] for i in range(n)}
color = ['white'] * n
distance = [-1] * n
label = [-1] * n
m = firstline[1]
for _ in range(m):
adj = list(map(int, input().split()))
adjDict[adj[0]].append(adj[1])
adjDict[adj[1]].append(adj[0])
time = 0
for i in range(n):
if color[i] == 'white':
bfs(i)
time += 1
q = int(input())
for _ in range(q):
question = list(map(int, input().split()))
if label[question[0]] == label[question[1]]:
print('yes')
else:
print('no') |
s768407074 | p00004 | u299798926 | 1,000 | 131,072 | Wrong Answer | 20 | 7,592 | 230 | Write a program which solve a simultaneous equation: ax + by = c dx + ey = f The program should print x and y for given a, b, c, d, e and f (-1,000 ≤ a, b, c, d, e, f ≤ 1,000). You can suppose that given equation has a unique solution. | while 1:
try:
a,b,c,d,e,f=[int(i) for i in input().split()]
y=float((c*d-f*a)/(b*d-e*a))
x=float((c-b*y)/a)
print("{0:.4f}".format(x)," ","{0:.4f}".format(y))
except EOFError:
break | s056732007 | Accepted | 30 | 7,604 | 226 | while 1:
try:
a,b,c,d,e,f=[int(i) for i in input().split()]
y=float((c*d-f*a)/(b*d-e*a))
x=float((c-b*y)/a)
print("{0:.3f}".format(x),"{0:.3f}".format(y))
except EOFError:
break |
s754756479 | p00780 | u078042885 | 8,000 | 131,072 | Wrong Answer | 2,130 | 8,924 | 327 | **Goldbach's Conjecture:** For any even number _n_ greater than or equal to 4, there exists at least one pair of prime numbers _p_ 1 and _p_ 2 such that _n_ = _p_ 1 \+ _p_ 2. This conjecture has not been proved nor refused yet. No one is sure whether this conjecture actually holds. However, one can find such a pair of prime numbers, if any, for a given even number. The problem here is to write a program that reports the number of all the pairs of prime numbers satisfying the condition in the conjecture for a given even number. A sequence of even numbers is given as input. Corresponding to each number, the program should output the number of pairs mentioned above. Notice that we are intereseted in the number of essentially different pairs and therefore you should not count ( _p_ 1, _p_ 2) and ( _p_ 2, _p_ 1) separately as two different pairs. | N=32768
p=[i for i in range(N)]
for i in range(2,int(N**0.5)+1):
if p[i]:
for j in range(i*i,N,i):p[j]=0
p=sorted(set(p))[2:]
print(p)
while 1:
n=int(input())
if n==0:break
c=0
if n&1:c=int(n-2 in p)
else:
for x in p:
if x>n//2:break
if n-x in p:c+=1
print(c) | s823809850 | Accepted | 2,120 | 8,900 | 318 | N=32768
p=[i for i in range(N)]
for i in range(2,int(N**0.5)+1):
if p[i]:
for j in range(i*i,N,i):p[j]=0
p=sorted(set(p))[2:]
while 1:
n=int(input())
if n==0:break
c=0
if n&1:c=int(n-2 in p)
else:
for x in p:
if x>n//2:break
if n-x in p:c+=1
print(c) |
s766838428 | p00330 | u350064373 | 1,000 | 262,144 | Wrong Answer | 30 | 7,284 | 23 | コンピュータで扱われるデータの最小単位をビット(bit)と呼び、複数のビットをまとめて表した情報量をワード(word)と呼びます。現在、多くのコンピュータでは１ワードを３２ビットとして処理しています。 １ワードを３２ビットで表すコンピュータについて、ワード単位で与えられたデータ量 W をビット単位で出力するプログラムを作成せよ。 | W = input()
print(W*32) | s936921987 | Accepted | 20 | 7,640 | 26 | print( int( input()) * 32) |
s924421242 | p02255 | u870370075 | 1,000 | 131,072 | Wrong Answer | 30 | 5,588 | 191 | 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())
arr=list(map(int,input().split()))
for i in range(1,n):
v=arr[i]
j=i-1
while j>=0 and arr[j]>v:
arr[j+1]=arr[j]
j-=1
arr[j+1]=v
print(' '.join(map(str,arr)))
| s562840371 | Accepted | 20 | 5,596 | 221 | n=int(input())
arr=list(map(int,input().split()))
print(' '.join(map(str,arr)))
for i in range(1,n):
v=arr[i]
j=i-1
while j>=0 and arr[j]>v:
arr[j+1]=arr[j]
j-=1
arr[j+1]=v
print(' '.join(map(str,arr)))
|
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