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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s232151030 | p03401 | u272557899 | 2,000 | 262,144 | Wrong Answer | 226 | 14,048 | 430 | There are N sightseeing spots on the x-axis, numbered 1, 2, ..., N. Spot i is at the point with coordinate A_i. It costs |a - b| yen (the currency of Japan) to travel from a point with coordinate a to another point with coordinate b along the axis. You planned a trip along the axis. In this plan, you first depart from... | n = int(input())
a = [int(x) for x in input().split()]
b = []
for i in range(n):
if i == 0:
b.append(abs(a[i]))
if i == n - 1:
b.append(abs(a[i]))
else:
b.append(abs(a[i] - a[i - 1]))
sd = sum(b)
for j in range(n):
if j == 0:
print(sd - b[j] - b[j + 1] + abs(a[j + 1]))
if j == n - 1:
... | s463851647 | Accepted | 215 | 14,048 | 439 | n = int(input())
a = [int(x) for x in input().split()]
b = []
for i in range(n + 1):
if i == 0:
b.append(abs(a[i]))
elif i == n:
b.append(abs(a[i - 1]))
else:
b.append(abs(a[i] - a[i - 1]))
sd = sum(b)
for j in range(n):
if j == 0:
print(sd - b[j] - b[j + 1] + abs(a[j + 1]))
elif j == n ... |
s481158991 | p03079 | u144980750 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 78 | You are given three integers A, B and C. Determine if there exists an equilateral triangle whose sides have lengths A, B and C. | A,B,C=map(int,input().split())
if A-B==2*C:
print("Yes")
else:
print("No") | s236457095 | Accepted | 17 | 2,940 | 83 | A,B,C=map(int,input().split())
if A==B and A==C:
print("Yes")
else:
print("No") |
s098667577 | p03434 | u813450984 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 189 | 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 c... | n = int(input())
l = list(map(int, input().split()))
l.sort(reverse=True)
a, b = 0, 0
print(l)
for i in range(len(l)):
if i % 2 == 0:
a += l[i]
else:
b += l[i]
print(abs(a-b))
| s168143748 | Accepted | 17 | 2,940 | 180 | n = int(input())
l = list(map(int, input().split()))
l.sort(reverse=True)
a, b = 0, 0
for i in range(len(l)):
if i % 2 == 0:
a += l[i]
else:
b += l[i]
print(abs(a-b))
|
s176851211 | p02806 | u094191970 | 2,525 | 1,048,576 | Wrong Answer | 21 | 3,316 | 196 | Niwango created a playlist of N songs. The title and the duration of the i-th song are s_i and t_i seconds, respectively. It is guaranteed that s_1,\ldots,s_N are all distinct. Niwango was doing some work while playing this playlist. (That is, all the songs were played once, in the order they appear in the playlist, w... | n=int(input())
l=[input().split() for i in range(n)]
x=input()
ans=0
sum=0
for i,j in l:
print(i,j)
if i!=x:
ans+=int(j)
else:
ans+=int(j)
break
for i,j in l:
sum+=int(j)
print(sum-ans) | s439843186 | Accepted | 31 | 9,156 | 269 | from sys import stdin
nii=lambda:map(int,stdin.readline().split())
lnii=lambda:list(map(int,stdin.readline().split()))
n=int(input())
l=[input().split() for i in range(n)]
x=input()
ans=0
c=False
for s,t in l:
if c:
ans+=int(t)
if x==s:
c=True
print(ans) |
s443848828 | p03129 | u671239754 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 159 | Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1. | A, B = map(int, input().split())
line = []
for k in range(0,A):
line.append(k)
line = line[1::2]
if len(line) >= B:
print("YES")
else:
print("NO")
| s923194643 | Accepted | 17 | 2,940 | 164 |
A, B = map(int, input().split())
line = []
for k in range(0,A+1):
line.append(k)
line = line[1::2]
if len(line) >= B:
print("YES")
else:
print("NO")
|
s364751745 | p02275 | u022407960 | 1,000 | 131,072 | Wrong Answer | 30 | 7,864 | 695 | Counting sort can be used for sorting elements in an array which each of the n input elements is an integer in the range 0 to k. The idea of counting sort is to determine, for each input element x, the number of elements less than x as C[x]. This information can be used to place element x directly into its position in ... | #!/usr/bin/env python
# -*- coding: utf-8 -*-
import sys
def counting_sort(A, k):
B = [0] * k
C = [0] * k
for j in range(array_length):
C[A[j]] += 1
print('1', C[:array_length])
for i in range(1, k):
C[i] += C[i - 1]
print('2', C[:array_length])
for m in range(array_len... | s279301699 | Accepted | 2,380 | 220,968 | 789 | #!/usr/bin/env python
# -*- coding: utf-8 -*-
import sys
def counting_sort(A, max_occurance):
result = [0] * array_length
count_list = [0] * max_occurance
for j in range(array_length):
count_list[A[j]] += 1
# print('step-1', C[:array_length])
for i in range(1, max_occurance):
c... |
s093486317 | p03359 | u050708958 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 63 | 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 ... | a, b = map(int, input().split())
print(a - (0 if a < b else 1)) | s774680618 | Accepted | 17 | 2,940 | 64 | a, b = map(int, input().split())
print(a - (0 if a <= b else 1)) |
s094355466 | p02612 | u299645128 | 2,000 | 1,048,576 | Wrong Answer | 29 | 9,144 | 24 | 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. | print(int(input())%1000) | s047665739 | Accepted | 28 | 9,156 | 84 | X = int(input())
lack = X % 1000
if lack == 0:
print(0)
else:
print(1000 - lack) |
s518913832 | p03457 | u279229189 | 2,000 | 262,144 | Wrong Answer | 629 | 13,252 | 640 | 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... |
t = []
x = []
y = []
t.append(0)
x.append(0)
y.append(0)
for i in (range(0, 1000000, 1)):
try:
c = input()
if i == 0:
N = int(c)
else:
t.append(int(c.split(" ")[0]))
x.append(int(c.split(" ")[1]))
y.append(int(c.split(" ")[2]))
except:
break
for i in range(0, len(t) - 1, ... | s978324313 | Accepted | 436 | 11,844 | 603 |
t = []
x = []
y = []
t.append(0)
x.append(0)
y.append(0)
for i in (range(0, 1000000, 1)):
try:
c = input()
if i == 0:
N = int(c)
else:
t.append(int(c.split(" ")[0]))
x.append(int(c.split(" ")[1]))
y.append(int(c.split(" ")[2]))
except:
break
for i in range(0, len(t) - 1, ... |
s390763625 | p03697 | u594762426 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 71 | You are given two integers A and B as the input. Output the value of A + B. However, if A + B is 10 or greater, output `error` instead. | s = str(input())
if s == set(s):
print("yes")
else:
print("no") | s836843782 | Accepted | 17 | 2,940 | 92 | h , w = map(int, input().split())
if(h + w < 10):
print(h + w)
else:
print("error") |
s754895803 | p03377 | u271469978 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 81 | 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') | s209537698 | Accepted | 17 | 2,940 | 81 | a, b, x = map(int, input().split())
print('YES' if a <= x and a+b >= x else 'NO') |
s093687609 | p03129 | u470717435 | 2,000 | 1,048,576 | Wrong Answer | 18 | 3,060 | 116 | Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1. | import math
n, k = [int(i) for i in input().split()]
if math.ceil(n-1) < k:
print("No")
else:
print("Yes") | s474900228 | Accepted | 17 | 2,940 | 116 | import math
n, k = [int(i) for i in input().split()]
if math.ceil(n/2) < k:
print("NO")
else:
print("YES") |
s450735130 | p03457 | u201565171 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 297 | 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... | N=int(input())
pre_t=0
pre_x=0
pre_y=0
for i in range(N):
now_t,now_x,now_y=list(map(int,input().split()))
d=abs(pre_x-now_x)+abs(pre_y-now_y)
t=pre_t-now_t
if d<=t and t%2==d%2:
flg=1
else:
flg=0
break
if flg==1:
print('Yes')
else:
print('No') | s898556346 | Accepted | 407 | 3,060 | 335 | N = int(input())
pre_t = 0
pre_x = 0
pre_y = 0
for i in range(N):
now_t,now_x,now_y = list(map(int,input().split()))
d = abs(now_x - pre_x) + abs(now_y - pre_y)
t = now_t - pre_t
if d <= t and t % 2 == d % 2:
flg = 1
else:
flg = 0
break
if flg == 1:
print('Yes')
else:
... |
s739390660 | p03637 | u107269063 | 2,000 | 262,144 | Wrong Answer | 64 | 14,252 | 296 | We have a sequence of length N, a = (a_1, a_2, ..., a_N). Each a_i is a positive integer. Snuke's objective is to permute the element in a so that the following condition is satisfied: * For each 1 ≤ i ≤ N - 1, the product of a_i and a_{i + 1} is a multiple of 4. Determine whether Snuke can achieve his objective. | N = int(input())
l = list(map(int,input().split()))
num_odd = 0
num_even = 0
num_four = 0
for i in l:
if i % 4 == 0:
num_four += 1
elif i % 2 == 0:
num_even += 1
else:
num_odd += 1
if num_four <= 2 * num_odd:
ans = "No"
else:
ans = "Yes"
print(ans)
| s807660638 | Accepted | 65 | 14,252 | 359 | N = int(input())
l = list(map(int,input().split()))
num_odd = 0
num_even = 0
num_four = 0
for i in l:
if i % 2 == 1:
num_odd += 1
elif i % 4 == 0:
num_four += 1
else:
num_even += 1
if num_four + 1 > num_odd:
ans = "Yes"
elif num_four + 1 == num_odd and num_even == 0:
ans =... |
s635793848 | p03435 | u466478199 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 343 | We have a 3 \times 3 grid. A number c_{i, j} is written in the square (i, j), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left. According to Takahashi, there are six integers a_1, a_2, a_3, b_1, b_2, b_3 whose values are fixed, and the number written in the square (i, j) ... | l=[]
for i in range(3):
l1=list(map(int,input().split()))
l.append(l1)
flag=0
for i in range(l[0][0]+1):
y0=l[0][0]-i
y1=l[1][0]-i
y2=l[2][0]-i
x1=l[0][1]-y1
x2=l[0][2]-y1
if l[1][1]==y1+x1 and l[1][2]==y1+x2 and l[2][1]==y2+x1 and l[2][2]==y2+x2:
flag=1
else:
continue
if flag==1:
print('Yes... | s289848634 | Accepted | 18 | 3,064 | 297 | l=[]
for i in range(3):
l1=list(map(int,input().split()))
l.append(l1)
flag=0
y0=l[0][0]-0
y1=l[1][0]-0
y2=l[2][0]-0
x1=l[0][1]-y0
x2=l[0][2]-y0
if l[1][1]==y1+x1 and l[1][2]==y1+x2 and l[2][1]==y2+x1 and l[2][2]==y2+x2:
flag=1
else:
pass
if flag==1:
print('Yes')
else:
print('No') |
s792193938 | p02865 | u366886346 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 31 | How many ways are there to choose two distinct positive integers totaling N, disregarding the order? | n=int(input())
print(-(n//-2))
| s637152220 | Accepted | 17 | 2,940 | 31 | n=int(input())
print((n-1)//2)
|
s617576533 | p02612 | u556610039 | 2,000 | 1,048,576 | Wrong Answer | 29 | 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. | num = int(input())
print(num % 1000) | s074228911 | Accepted | 31 | 9,152 | 81 | num = int(input())
val = num % 1000
if val == 0: print(0)
else: print(1000 - val) |
s871910720 | p03636 | u423624748 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 65 | 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()
slen =len(s)
ans = s[0:1]+str(slen)+s[-1:]
print(ans) | s028358773 | Accepted | 17 | 2,940 | 67 | s = input()
slen =len(s)
ans = s[0:1]+str(slen-2)+s[-1:]
print(ans) |
s249864500 | p03048 | u684120680 | 2,000 | 1,048,576 | Wrong Answer | 1,641 | 3,060 | 243 | Snuke has come to a store that sells boxes containing balls. The store sells the following three kinds of boxes: * Red boxes, each containing R red balls * Green boxes, each containing G green balls * Blue boxes, each containing B blue balls Snuke wants to get a total of exactly N balls by buying r red boxes, g... | r,g,b,n = [int(i) for i in input().split()]
cnt = 0
for i in range(n//r+1):
n_r = i*r
if n_r > n:
break
for j in range(n//g+1):
n_rg = j*g + n_r
if n_rg > n:
break
if n_rg%b == 0:
cnt += 1
print(cnt) | s602590805 | Accepted | 1,885 | 3,060 | 247 | r,g,b,n = [int(i) for i in input().split()]
cnt = 0
for i in range(n//r+1):
n_r = i*r
if n_r > n:
break
for j in range(n//g+1):
n_rg = j*g + n_r
if n_rg > n:
break
if (n-n_rg)%b == 0:
cnt += 1
print(cnt) |
s899520289 | p03456 | u553070631 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 177 | 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. | a,b=map(int,input().split())
ab=10*a+b
if ab==1 or ab==4 or ab ==9 or ab==16 or ab==25 or ab==36 or ab==49 or ab==64 or ab==81 or ab==100:
print('Yes')
else:
print('No') | s220109746 | Accepted | 17 | 2,940 | 149 | a,b=input().split()
ab=int(a+b)
a=0
for i in range (4,317):
if ab==i*i:
print('Yes')
a=1
break
if a==0:
print('No') |
s817169058 | p03151 | u489959379 | 2,000 | 1,048,576 | Wrong Answer | 163 | 20,332 | 442 | 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 pa... | n = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
diff_AB = sum(A) - sum(B)
if diff_AB < 0:
print(-1)
exit()
Diff = []
for i in range(n):
diff = A[i] - B[i]
Diff.append(diff)
Diff = sorted(Diff)
print(Diff)
res = 0
for i in range(n):
if Diff[i] < 0:
res += 1
else:
i... | s130492427 | Accepted | 125 | 19,112 | 733 | import sys
sys.setrecursionlimit(10 ** 7)
input = sys.stdin.readline
f_inf = float('inf')
mod = 10 ** 9 + 7
def resolve():
n = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
diff_AB = sum(A) - sum(B)
if diff_AB < 0:
print(-1)
exit()
Dif... |
s421963328 | p03693 | u016901717 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 71 | AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this i... | r,g,b = map(int,input().split())
print("YES" if (g+b)%4==0 else "NO" )
| s899093597 | Accepted | 18 | 2,940 | 74 | r,g,b = map(int,input().split())
print("YES" if (10*g+b)%4==0 else "NO" )
|
s039422281 | p00436 | u811733736 | 1,000 | 131,072 | Wrong Answer | 30 | 7,704 | 796 | 1 から 2n の数が書かれた 2n 枚のカードがあり，上から 1, 2, 3, ... , 2n の順に積み重なっている． このカードを，次の方法を何回か用いて並べ替える． **整数 k でカット** 上から k 枚のカードの山 A と 残りのカードの山 B に分けた後， 山 A の上に山 B をのせる． **リフルシャッフル** 上から n 枚の山 A と残りの山 B に分け， 上から A の1枚目， B の1枚目， A の2枚目， B の2枚目， …， A の n枚目， B の n枚目， となるようにして， 1 つの山にする． 入力の指示に従い，カードを並び替えたあとのカードの番号を，上から順番... | # -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0513
AC
"""
import sys
from sys import stdin
from itertools import chain
input = stdin.readline
def flatten(listOfLists):
"Flatten one level of nesting"
return chain.from_iterable(listOfLists)
def main(args):
n = int(in... | s600604529 | Accepted | 30 | 7,796 | 793 | # -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0513
AC
"""
import sys
from sys import stdin
from itertools import chain
input = stdin.readline
def flatten(listOfLists):
"Flatten one level of nesting"
return chain.from_iterable(listOfLists)
cards = []
def main(args):
... |
s657333910 | p02396 | u382692819 | 1,000 | 131,072 | Wrong Answer | 130 | 7,216 | 122 | 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 give... | s = input()
i=1
while True:
print("case {0}: {1}".format(i,s))
i += 1
s = input()
if s=="0":
break | s081656829 | Accepted | 130 | 7,784 | 137 | a = []
i = 0
while True:
n = input()
if n == "0":
break
a.append(n)
i += 1
print("Case {0}: {1}".format(i,n)) |
s021312620 | p03853 | u648890761 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 107 | There is an image with a height of H pixels and a width of W pixels. Each of the pixels is represented by either `.` or `*`. The character representing the pixel at the i-th row from the top and the j-th column from the left, is denoted by C_{i,j}. Extend this image vertically so that its height is doubled. That is, p... |
a = input().split()
n = int(a[0])
m = a[1]
for i in range(1, n):
s = input()
print(s + "\n" + s)
| s820310349 | Accepted | 17 | 3,060 | 198 |
a = input().split()
n = int(a[0])
m = a[1]
s = "";
for i in range(0, n):
tmp = input()
if i != 0:
s = s + "\n" + tmp + "\n" + tmp
else:
s = s + tmp + "\n" + tmp
print(s) |
s122544739 | p02645 | u913294360 | 2,000 | 1,048,576 | Wrong Answer | 24 | 9,028 | 39 | When you asked some guy in your class his name, he called himself S, where S is a string of length between 3 and 20 (inclusive) consisting of lowercase English letters. You have decided to choose some three consecutive characters from S and make it his nickname. Print a string that is a valid nickname for him. | a=input()
str = list(a)
print(str[0:2]) | s799687535 | Accepted | 23 | 8,844 | 37 | a=input()
#str = list(a)
print(a[:3]) |
s837046772 | p03737 | u264265458 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 43 | You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words. | a,b,c=input().split()
print(a[0]+b[0]+c[0]) | s234963822 | Accepted | 17 | 2,940 | 67 | a,b,c=input().split()
print(a[0].upper()+b[0].upper()+c[0].upper()) |
s721319111 | p03447 | u128859393 | 2,000 | 262,144 | Wrong Answer | 24 | 3,060 | 233 | 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? | N = int(input())
cnt1 = list(map(int, input().split()))
cnt2 = list(map(int, input().split()))
max_candies = 0
for i in range(N):
temp = sum(cnt1[:i]) + sum(cnt2[i:])
max_candies = max(temp, max_candies)
print(max_candies) | s731954228 | Accepted | 17 | 2,940 | 61 | X, A, B = [int(input()) for _ in range(3)];print((X - A) % B) |
s974172348 | p03643 | u785066634 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 25 | This contest, _AtCoder Beginner Contest_ , is abbreviated as _ABC_. When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example, ABC680 is the 680th round of ABC. What is the abbreviation for the N-th round of ABC? Write a program to output the answer. | n=input()
print ('ABC',n) | s778399725 | Accepted | 17 | 2,940 | 30 | n=str(input())
print ('ABC'+n) |
s529669392 | p03409 | u947823593 | 2,000 | 262,144 | Wrong Answer | 24 | 3,064 | 683 | On a two-dimensional plane, there are N red points and N blue points. The coordinates of the i-th red point are (a_i, b_i), and the coordinates of the i-th blue point are (c_i, d_i). A red point and a blue point can form a _friendly pair_ when, the x-coordinate of the red point is smaller than that of the blue point, ... | #coding=utf-8
def solve(abs, cds):
abs = sorted(abs, key = lambda x: -1 * x[1]) #sorted by y
cds = sorted(cds, key = lambda x: x[0]) #sorted by x
count = 0
for r in cds:
print(r)
_abs = list(filter(lambda x: r[0] > x[0] and r[1] > x[1], abs))
if len(_abs) == 0:
con... | s516385186 | Accepted | 20 | 3,064 | 666 | #coding=utf-8
def solve(abs, cds):
abs = sorted(abs, key = lambda x: -1 * x[1]) #sorted by y
cds = sorted(cds, key = lambda x: x[0]) #sorted by x
count = 0
for r in cds:
_abs = list(filter(lambda x: r[0] > x[0] and r[1] > x[1], abs))
if len(_abs) == 0:
continue
del... |
s879834360 | p02645 | u481187938 | 2,000 | 1,048,576 | Wrong Answer | 24 | 9,364 | 712 | When you asked some guy in your class his name, he called himself S, where S is a string of length between 3 and 20 (inclusive) consisting of lowercase English letters. You have decided to choose some three consecutive characters from S and make it his nickname. Print a string that is a valid nickname for him. | from collections import defaultdict, deque
from heapq import heappush, heappop
from itertools import permutations, accumulate
import sys
import math
import bisect
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readlin... | s654651183 | Accepted | 26 | 9,384 | 737 | #!usr/bin/env pypy3
from collections import defaultdict, deque
from heapq import heappush, heappop
from itertools import permutations, accumulate
import sys
import math
import bisect
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x ... |
s771263054 | p04011 | u371132735 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 97 | There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee. | n = int(input())
k = int(input())
x = int(input())
y = int(input())
print(((n-k) * x) + (k * y))
| s493462608 | Accepted | 17 | 2,940 | 130 | n = int(input())
k = int(input())
x = int(input())
y = int(input())
if n-k>0:
print(((n-k) * y) + (k * x))
else:
print(n * x)
|
s785736866 | p03130 | u671239754 | 2,000 | 1,048,576 | Wrong Answer | 18 | 3,064 | 290 | 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 roa... | A = sorted([int(i) for i in input().split()])
B = sorted([int(i) for i in input().split()])
C = sorted([int(i) for i in input().split()])
k = 0
if A[1] - A[0] == 2:
k += 1
if B[1] - B[0] == 2:
k += 1
if C[1] - C[0] == 2:
k += 1
if k == 1:
print("No")
else:
print("Yes") | s821812770 | Accepted | 18 | 3,060 | 291 | A = sorted([int(i) for i in input().split()])
B = sorted([int(i) for i in input().split()])
C = sorted([int(i) for i in input().split()])
k = 0
if A[1] - A[0] == 2:
k += 1
if B[1] - B[0] == 2:
k += 1
if C[1] - C[0] == 2:
k += 1
if k == 1:
print("NO")
else:
print("YES")
|
s406752979 | p04011 | u777028980 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 109 | There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee. | a=int(input())
b=int(input())
c=int(input())
d=int(input())
if(a>b):
print(a*c+(b-a)*d)
else:
print(a*c) | s969825511 | Accepted | 18 | 2,940 | 111 | a=int(input())
b=int(input())
c=int(input())
d=int(input())
if(a>b):
print(b*c+(a-b)*d)
else:
print(a*c)
|
s374214293 | p03712 | u721425712 | 2,000 | 262,144 | Wrong Answer | 20 | 3,316 | 215 | You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1. | # B
h, w = map(int, input().split())
a = []
for _ in range(h):
a.append('#' + input() + '#')
from collections import deque
a = deque(a)
a.appendleft('#'*w)
a.append('#'*w)
for i in a:
print(''.join(i)) | s355422958 | Accepted | 31 | 3,444 | 223 | # B
h, w = map(int, input().split())
a = []
for _ in range(h):
a.append('#' + input() + '#')
from collections import deque
a = deque(a)
a.appendleft('#'*(w+2))
a.append('#'*(w+2))
for i in a:
print(''.join(i)) |
s092353407 | p02613 | u805011545 | 2,000 | 1,048,576 | Wrong Answer | 162 | 9,228 | 320 | 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`,... | N = int(input())
ac = 0
wa = 0
tle = 0
re = 0
for i in range(N):
st = str(input())
if(st == 'AC'):
ac += 1
elif(st == 'WA'):
wa += 1
elif(st == 'TLE'):
tle += 1
else:
re += 1
print('AC × {}'.format(ac))
print('WA × {}'.format(wa))
print('TLE × {}'.format(tle))
print('RE × {}'.format(re)) | s486686464 | Accepted | 164 | 9,212 | 316 | N = int(input())
ac = 0
wa = 0
tle = 0
re = 0
for i in range(N):
st = str(input())
if(st == 'AC'):
ac += 1
elif(st == 'WA'):
wa += 1
elif(st == 'TLE'):
tle += 1
else:
re += 1
print('AC x {}'.format(ac))
print('WA x {}'.format(wa))
print('TLE x {}'.format(tle))
print('RE x {}'.format(re)) |
s149866897 | p00015 | u408260374 | 1,000 | 131,072 | Wrong Answer | 30 | 6,724 | 159 | 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 t... | n = int(input())
for _ in range(n):
a = sum([int(input()) for _ in range(2)])
if len(str(a)) < 80:
print(a)
else:
print('overflow') | s003076343 | Accepted | 30 | 6,720 | 104 | for _ in range(int(input())):
a=int(input())+int(input())
print(a if a < 10**80 else 'overflow') |
s222539102 | p02742 | u957098479 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,060 | 182 | We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the to... | H, W = map(int,input().split())
if H % 2 == 0:
sum = W * H / 2
elif W % 2 == 0:
sum = (W * (H-1) / 2) + (W / 2)
else:
sum = (W * (H-1) / 2) + ((W+1) / 2)
print(sum) | s292685250 | Accepted | 17 | 3,060 | 226 | H,W = map(int,input().split())
if (H == 1)or(W == 1):
sum = 1
elif H % 2 == 0:
sum = W * H // 2
elif W % 2 == 0:
sum = (W * (H-1) // 2) + (W // 2)
else:
sum = (W * (H-1) // 2) + ((W+1) // 2)
print(int(sum)) |
s740025450 | p03943 | u037901699 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 118 | 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. Not... | a, b, c = map(int, input().split())
if a == b + c or b == a + c or c == a + b:
print("YES")
else:
print("NO") | s959262272 | Accepted | 17 | 2,940 | 118 | a, b, c = map(int, input().split())
if a == b + c or b == a + c or c == a + b:
print("Yes")
else:
print("No") |
s258612145 | p04029 | u367130284 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 29 | There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total? | n=int(input());print(n*-~n/2) | s113091885 | Accepted | 17 | 2,940 | 33 | print(sum(range(int(input())+1))) |
s284039638 | p03555 | u131411061 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 109 | 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. | C = [input() for _ in range(2)]
tmp = C[0] + C[1]
if tmp == tmp[::-1]:
print('Yes')
else:
print('No') | s222507443 | Accepted | 17 | 2,940 | 109 | C = [input() for _ in range(2)]
tmp = C[0] + C[1]
if tmp == tmp[::-1]:
print('YES')
else:
print('NO') |
s804990844 | p02975 | u441575327 | 2,000 | 1,048,576 | Wrong Answer | 63 | 14,212 | 165 | Snuke has N hats. The i-th hat has an integer a_i written on it. There are N camels standing in a circle. Snuke will put one of his hats on each of these camels. If there exists a way to distribute the hats to the camels such that the following condition is satisfied for every camel, print `Yes`; otherwise, print `No... | N = int(input())
A = list(map(int, input().split()))
xor = A[0]
for i in range(N-1):
xor = xor ^ A[i+1]
if xor == 0:
print("yes")
else:
print("no") | s502007680 | Accepted | 65 | 14,116 | 165 | N = int(input())
A = list(map(int, input().split()))
xor = A[0]
for i in range(N-1):
xor = xor ^ A[i+1]
if xor == 0:
print("Yes")
else:
print("No") |
s882156649 | p03141 | u794173881 | 2,000 | 1,048,576 | Wrong Answer | 2,104 | 47,304 | 598 | There are N dishes of cuisine placed in front of Takahashi and Aoki. For convenience, we call these dishes Dish 1, Dish 2, ..., Dish N. When Takahashi eats Dish i, he earns A_i points of _happiness_ ; when Aoki eats Dish i, she earns B_i points of happiness. Starting from Takahashi, they alternately choose one dish a... | n = int(input())
info =[list(map(int,input().split())) for i in range(n)]
diff = [[0] for i in range(n)]
sorted_diff = [[0] for i in range(n)]
for i in range(n):
diff[i][0] = info[i][0]+info[i][1]
for i in range(n):
info[i].extend(diff[i])
#print(info)
sorted_diff = sorted(info, key=lambda x: x[2],reverse=Tru... | s244958536 | Accepted | 555 | 47,368 | 591 | n = int(input())
info =[list(map(int,input().split())) for i in range(n)]
diff = [[0] for i in range(n)]
sorted_diff = [[0] for i in range(n)]
for i in range(n):
diff[i][0] = info[i][0]+info[i][1]
for i in range(n):
info[i].extend(diff[i])
#print(info)
sorted_diff = sorted(info, key=lambda x: x[2])
takahash... |
s209525907 | p03612 | u735008991 | 2,000 | 262,144 | Wrong Answer | 60 | 14,008 | 135 | You are given a permutation p_1,p_2,...,p_N consisting of 1,2,..,N. You can perform the following operation any number of times (possibly zero): Operation: Swap two **adjacent** elements in the permutation. You want to have p_i ≠ i for all 1≤i≤N. Find the minimum required number of operations to achieve this. | N = int(input())
P = list(map(int, input().split()))
c = 0
for i, p in enumerate(P):
c += (i+1 == p)
print(c - 1 if c >= 2 else c)
| s483052115 | Accepted | 72 | 14,008 | 230 | N = int(input())
P = list(map(int, input().split()))
c = 0
prev = False
for i, p in enumerate(P):
if p == i+1:
c += 1
if prev:
c -= 1
prev = not prev
else:
prev = False
print(c)
|
s223553972 | p02612 | u453683890 | 2,000 | 1,048,576 | Wrong Answer | 33 | 9,076 | 30 | 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. | l = input()
print(int(l[-3:])) | s438994981 | Accepted | 26 | 9,152 | 76 | l = input()
if int(l[-3:]) == 0:
print(0)
else:
print(1000-int(l[-3:]))
|
s071095810 | p03433 | u901447859 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 79 | E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins. | if (int(input()) - int(input())) % 500 == 0:
print('Yes')
else:
print('No') | s297308079 | Accepted | 17 | 2,940 | 60 | print('Yes' if int(input()) % 500 <= int(input()) else 'No') |
s160360100 | p03455 | u315714363 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 110 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | # -*- coding: utf-8 -*-
a, b = map(int, input().split())
if a*b%2 == 0:
print("0dd")
else:
print("Even") | s300339101 | Accepted | 17 | 2,940 | 85 | a, b = map(int, input().split())
if a*b%2 == 0:
print('Even')
else:
print('Odd') |
s418173684 | p03469 | u973663980 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 46 | On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date col... | s=str(input())
print(s.replace("2018","2017")) | s355997331 | Accepted | 18 | 2,940 | 46 | s=str(input())
print(s.replace("2017","2018")) |
s647825355 | p02841 | u706786134 | 2,000 | 1,048,576 | Wrong Answer | 153 | 12,504 | 261 | In this problem, a date is written as Y-M-D. For example, 2019-11-30 means November 30, 2019. Integers M_1, D_1, M_2, and D_2 will be given as input. It is known that the date 2019-M_2-D_2 follows 2019-M_1-D_1. Determine whether the date 2019-M_1-D_1 is the last day of a month. | import numpy as np
m1, d1 = map(int, input().split())
m2, d2 = map(int, input().split())
x = [30] * 13
x[1] = x[3] = x[5] = x[7] = x[8] = x[10] = x[12] = 31
x[2] = 28
x[0] = 0
x = np.add.accumulate(x)
print('1' if x[m1 - 1] + d1 + 1 == x[m2 - 1] + d2 else '0')
| s474218335 | Accepted | 149 | 12,396 | 211 | import numpy as np
m1, d1 = map(int, input().split())
m2, d2 = map(int, input().split())
x = [30] * 13
x[1] = x[3] = x[5] = x[7] = x[8] = x[10] = x[12] = 31
x[2] = 28
x[0] = 0
print('1' if x[m1] == d1 else '0')
|
s769506378 | p04011 | u449555432 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 95 | There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee. | n=int(input())
k=int(input())
x=int(input())
y=int(input())
print(n*x if n<=k else n*x+(n-k)*y) | s175508335 | Accepted | 17 | 3,060 | 106 | n = int(input())
k = int(input())
x = int(input())
y = int(input())
print(n*x if n <= k else k*x+(n-k)*y)
|
s109582463 | p02401 | u316584871 | 1,000 | 131,072 | Wrong Answer | 20 | 5,600 | 292 | Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part. | while True:
a, op, b = input().split(' ')
if (op == '+'):
print(int(a) + int(b))
elif (op == '-'):
print(int(a) - int(b))
elif (op == '*'):
print(int(a) * int(b))
elif (op == '/'):
print(int(a) / int(b))
elif (op == '?'):
break
| s780580024 | Accepted | 20 | 5,648 | 333 | import math
while True:
a, op, b = input().split()
if (op == '+'):
print(int(a) + int(b))
elif (op == '-'):
print(int(a) - int(b))
elif (op == '*'):
print(int(a) * int(b))
elif (op == '/' and int(b) != 0):
print(math.trunc((int(a) / int(b))))
elif (op == '?' ):
... |
s934749083 | p02694 | u526094365 | 2,000 | 1,048,576 | Wrong Answer | 23 | 9,160 | 135 | Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or abo... | X = int(input())
ans = 100
cnt = 0
risok = 0
while ans <= X:
cnt += 1
risok = int(ans * 0.01)
ans = ans + risok
print(cnt)
| s592875039 | Accepted | 24 | 9,156 | 102 | X = int(input())
ans = 100
year = 0
while ans < X:
year += 1
ans = int(ans*1.01)
print(year)
|
s239589721 | p03828 | u677440371 | 2,000 | 262,144 | Wrong Answer | 32 | 3,552 | 2,351 | You are given an integer N. Find the number of the positive divisors of N!, modulo 10^9+7. | N = int(input())
def make_prime_list(num):
if num < 2:
return []
prime_list = []
for prime in range(2, num + 1):
if is_prime(prime):
prime_list.append(prime)
return prime_list
def is_prime(num):
if num < 2:
return False
if num == 2 or num == 3 or num == 5:... | s383917440 | Accepted | 40 | 3,064 | 1,071 | N = int(input())
import math
def is_prime(num):
if num < 2:
return False
if num == 2 or num == 3 or num == 5:
return True
if num % 2 == 0 or num % 3 == 0 or num % 5 == 0:
return False
prime = 7
step = 4
num_sqrt = math.sqrt(num)
while prime <= num_sqrt:
... |
s891857985 | p03214 | u782654209 | 2,525 | 1,048,576 | Wrong Answer | 17 | 2,940 | 30 | Niwango-kun is an employee of Dwango Co., Ltd. One day, he is asked to generate a thumbnail from a video a user submitted. To generate a thumbnail, he needs to select a frame of the video according to the following procedure: * Get an integer N and N integers a_0, a_1, ..., a_{N-1} as inputs. N denotes the numbe... | print(int((int(input())+1)/2)) | s866857551 | Accepted | 17 | 2,940 | 108 | N=int(input())
A=list(map(int,input().split(' ')))
m=sum(A)/N
B=[abs(a-m) for a in A]
print(B.index(min(B))) |
s319015254 | p04012 | u114233208 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 255 | Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful. | S = input()
ec = 0
c = {}
for each in S:
try:
c[each] += 1
if c[each] % 2 == 0:
ec -= 1
else:
ec += 1
except:
c[each] = 1
ec += 1
if (ec == 0):
print("YES")
else:
print("NO")
| s919793159 | Accepted | 17 | 2,940 | 255 | S = input()
ec = 0
c = {}
for each in S:
try:
c[each] += 1
if c[each] % 2 == 0:
ec -= 1
else:
ec += 1
except:
c[each] = 1
ec += 1
if (ec == 0):
print("Yes")
else:
print("No")
|
s098285367 | p03162 | u867826040 | 2,000 | 1,048,576 | Wrong Answer | 425 | 34,732 | 379 | 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... | n = int(input())
abc = [list(map(int, input().split())) for i in range(n)]
dp = [-1] * n
m = max(abc[0])
dp[0] = m
x = abc[0].index(m)
for i in range(1,n):
m = -1
t = -1
for j in range(3):
if x == j:
continue
print(j,x)
if m < abc[i][j]:
m = abc[i][j]
... | s866166841 | Accepted | 643 | 45,152 | 309 | n = int(input())
abc = [tuple(map(int,input().split())) for i in range(n)]
dp = [[0,0,0] for i in range(n+1)]
for i in range(n):
for j in range(3):
for k in range(3):
if j == k:
continue
dp[i+1][k] = max(dp[i+1][k],dp[i][j]+abc[i][k])
print(max(dp[-1])) |
s263605816 | p03598 | u063896676 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 217 | 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 t... | # -*- coding: utf-8 -*-
n = int(input())
k = int(input())
x = list(map(int, input().split())) #x[0->(n-1)]
sum = 0
for i in x:
if i > k - i:
sum = sum + i
else:
sum = sum + k - i
print(sum) | s193880275 | Accepted | 17 | 2,940 | 219 | # -*- coding: utf-8 -*-
n = int(input())
k = int(input())
x = list(map(int, input().split())) #x[0->(n-1)]
sum = 0
for i in x:
if i < k - i:
sum = sum + i
else:
sum = sum + k - i
print(sum*2) |
s694668422 | p03999 | u894623942 | 2,000 | 262,144 | Wrong Answer | 30 | 9,028 | 324 | You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion. All strings that can be obtained in this way can be evaluated as formulas. ... | def abc045_c():
S = list(input())
for i in range(len(S)-1):
i = i * 2 + 1
S.insert(i,'+')
abc045_c2(S)
def abc045_c2(S):
cnt = 0
for i in range(S.count('+')):
cnt += eval(''.join(S))
i = i + 1
del S[i]
print(cnt)
if __name__ == '__main__':
abc045... | s501776793 | Accepted | 27 | 9,120 | 199 | def dfs(i, f):
if i == n - 1:
return sum(list(map(int, f.split('+'))))
return dfs(i + 1, f + a[i + 1]) + dfs(i + 1, f + '+' + a[i + 1])
a = input()
n = len(a)
print(dfs(0, a[0]))
|
s598238941 | p00113 | u811733736 | 1,000 | 131,072 | Wrong Answer | 20 | 7,624 | 908 | 2 つの正の整数 p, q を入力し、 p / q を小数として正確に表現することを考えます。(ただし、0 < p < q < 106とします。) このとき、結果は * 有限の桁で正確に表現できる。 * ある桁の範囲を繰り返す循環小数となる。 のいずれかとなります。筆算と同じ手順で1桁ずつ小数部を求めていくと、 * 割り切れた(余りが 0 になった)なら、そこまでの桁で正確に表現できた。 * 1度出てきた余りが、再び現れたなら、循環した。 と区別できます。 2 つの整数 p, q を入力すると、 p / q を小数で表した時の、小数部を出力するプログラムを作成してください。 ただし、 * 結果が有限... | # -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0113
"""
import sys
def solve(p, q):
quotients = []
mods = []
cyclic = False
cycle_start = 0
cycle_length = 0
while True:
mods.append(p)
p *= 10
quotients.append(p // q)
p = p... | s841950956 | Accepted | 30 | 7,688 | 987 | # -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0113
"""
import sys
def solve(p, q):
quotients = []
mods = []
cyclic = False
cycle_start = 0
cycle_length = 0
while True:
mods.append(p)
p *= 10
quotients.append(p // q)
p = p... |
s721701318 | p02606 | u701638736 | 2,000 | 1,048,576 | Wrong Answer | 29 | 9,108 | 120 | How many multiples of d are there among the integers between L and R (inclusive)? | a,b,c = input().split()
count = 0
for i in range(int(b)):
i+=int(a)
if i % int(c) == 0:
count+=1
print(count) | s049997241 | Accepted | 32 | 9,156 | 133 | a,b,c = input().split()
count = 0
num = int(a)
while num <= int(b):
if num % int(c) == 0:
count+=1
num += 1
print(count) |
s696489772 | p03555 | u498401785 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 115 | 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[0] == b[2] and a[1] == b[1] and a[2] == b[0]):
print("Yes")
else:
print("No")
| s527783636 | Accepted | 17 | 2,940 | 114 | a = input()
b = input()
if(a[0] == b[2] and a[1] == b[1] and a[2] == b[0]):
print("YES")
else:
print("NO") |
s711034751 | p02678 | u346395915 | 2,000 | 1,048,576 | Wrong Answer | 20 | 9,088 | 11 | 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 th... | print('No') | s363539084 | Accepted | 1,380 | 191,460 | 548 | from collections import deque
n, m = map(int, input().split())
li = [list(map(int, input().split())) for _ in range(m)]
ans = [deque([]) for _ in range(m+10)]
ans2 = [[] for _ in range(m+10)]
for i in range(m):
s = li[i][0]
t = li[i][1]
ans[s].append(t)
ans[t].append(s)
stack = deque([1])
while stack... |
s908233789 | p04046 | u786020649 | 2,000 | 262,144 | Time Limit Exceeded | 2,206 | 24,428 | 656 | We have a large square grid with H rows and W columns. Iroha is now standing in the top-left cell. She will repeat going right or down to the adjacent cell, until she reaches the bottom-right cell. However, she cannot enter the cells in the intersection of the bottom A rows and the leftmost B columns. (That is, there ... | h,w,a,b=map(int,input().split())
p=10**9+7
#p=127
def modp_factorial(n):
s=1
for x in range(1,h+1):
s=(s*x) % p
return s
def modp_prod(lst):
s=1
for x in lst:
s=(s*x)%p
return s
def inv(n):
s=1
q=p-2
while q>0:
if q&1:
s=s*n % p
n=n*n
... | s563161959 | Accepted | 245 | 24,744 | 670 | h,w,a,b=map(int,input().split())
p=10**9+7
#p=127
def modp_factorial(n):
s=1
for x in range(1,h+1):
s=(s*x) % p
return s
def modp_prod(lst):
s=1
for x in lst:
s=(s*x)%p
return s
def inv(n):
s=1
q=p-2
while q>0:
if q&1:
s=(s*n) % p
n=(n*n)... |
s539217672 | p03645 | u273010357 | 2,000 | 262,144 | Wrong Answer | 2,105 | 41,288 | 402 | In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands. For convenience, we will call them Island 1, Island 2, ..., Island N. There are M kinds of regular boat services between these islands. Each service connects two islands. The i-th service connects Island a_i and Island b_i. Cat Snuk... | from operator import itemgetter
N, M = map(int, input().split())
ab = []
for i in range(M):
a,b = map(int, input().split())
ab.append((a,b))
ab.sort(key=itemgetter(0))
print(ab)
fl = False
for i,j in ab:
if i!=1:
continue
for k,l in ab:
if i==1 and (j!=k or l!=N):
continue
... | s967619344 | Accepted | 750 | 48,144 | 259 | from collections import defaultdict
N, M = map(int, input().split())
d = defaultdict(list)
for i in range(M):
a,b = map(int, input().split())
d[a].append(b)
for j in d[1]:
if N in d[j]:
print('POSSIBLE')
exit()
print('IMPOSSIBLE') |
s017110861 | p03997 | u396495667 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 68 | 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/2) | s015899400 | Accepted | 17 | 2,940 | 87 | A = int(input())
B = int(input())
H = int(input())
answer = (H*(A+B)//2)
print(answer) |
s378601941 | p02612 | u916358282 | 2,000 | 1,048,576 | Wrong Answer | 31 | 9,072 | 28 | We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required. | def pay(N):
return N % 1000 | s266589897 | Accepted | 25 | 9,124 | 54 | N = int(input())
N = (1000 - N % 1000) % 1000
print(N) |
s408154657 | p02235 | u831244171 | 1,000 | 131,072 | Wrong Answer | 20 | 7,620 | 644 | For given two sequences $X$ and $Y$, a sequence $Z$ is a common subsequence of $X$ and $Y$ if $Z$ is a subsequence of both $X$ and $Y$. For example, if $X = \\{a,b,c,b,d,a,b\\}$ and $Y = \\{b,d,c,a,b,a\\}$, the sequence $\\{b,c,a\\}$ is a common subsequence of both $X$ and $Y$. On the other hand, the sequence $\\{b,c,a... | q = int(input())
for i in range(q):
x = list(input())
y = list(input())
max_count = 0
for j in range(len(x)):
for k in range(len(y)):
count = 0
if x[j] == y[k]:
count += 1
ret = j
ret_2 = k
while j < len(x)... | s227202220 | Accepted | 4,870 | 7,704 | 397 | q = int(input())
def lcss(x,y):
lcs = [0]*(len(x)+1)
for i in range(len(y)):
w1 = y[i]
lcs_2 = lcs[:]
for j in range(len(x)):
if x[j] == w1:
lcs[j+1] = lcs_2[j] + 1
elif lcs[j+1] < lcs[j]:
lcs[j+1] = lcs[j]
return lcs[... |
s899195852 | p02390 | u899891332 | 1,000 | 131,072 | Wrong Answer | 20 | 7,700 | 89 | 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. | a=int(input())
print(a/(60*60), end=':')
print(a%(60*60)/60, end=':')
print(a%(60*60*60)) | s517638230 | Accepted | 30 | 7,692 | 94 | a=int(input())
print(int(a/60/60), end=':')
print(int(a%(60*60)/60), end=':')
print(int(a%60)) |
s981782410 | p02401 | u592815095 | 1,000 | 131,072 | Wrong Answer | 20 | 5,560 | 61 | Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part. | a=input()
while a!="0 ? 0":
print(eval(a))
a=input()
| s325084715 | Accepted | 20 | 5,556 | 73 | while 1:
s=input()
if s.split()[1]=="?":exit()
print(int(eval(s)))
|
s844605595 | p02390 | u628732336 | 1,000 | 131,072 | Wrong Answer | 40 | 7,380 | 17 | 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. | print(38400/3600) | s451844176 | Accepted | 20 | 7,640 | 78 | S = int(input())
print(S // 3600, ":", S % 3600 // 60, ":", S % 60, sep = "") |
s274671295 | p02842 | u512895485 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 86 | 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).... | n = int(input())
x = n//1.08 + 1
m = x*1.08//1
if n==m:
print(x)
else:
print(':(') | s619334098 | Accepted | 42 | 2,940 | 122 | n = int(input())
for x in range(46297):
if (x+1)*1.08//1 == n:
print(x+1)
break
if x == 46296:
print(':(') |
s742639186 | p03606 | u993461026 | 2,000 | 262,144 | Wrong Answer | 20 | 3,060 | 114 | Joisino is working as a receptionist at a theater. The theater has 100000 seats, numbered from 1 to 100000. According to her memo, N groups of audiences have come so far, and the i-th group occupies the consecutive seats from Seat l_i to Seat r_i (inclusive). How many people are sitting at the theater now? | N = int(input())
total = 0
for i in range(N):
l, r = map(int, input().split())
total += l - r + 1
print(total) | s536407873 | Accepted | 20 | 3,060 | 114 | N = int(input())
total = 0
for i in range(N):
l, r = map(int, input().split())
total += r - l + 1
print(total) |
s628966173 | p03680 | u219180252 | 2,000 | 262,144 | Wrong Answer | 2,140 | 485,300 | 667 | Takahashi wants to gain muscle, and decides to work out at AtCoder Gym. The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It ... | n = int(input())
edges = [[0 for _ in range(n)] for _ in range(n)]
mem = [[0 for _ in range(n)] for _ in range(n)]
dist = []
for i in range(n):
edges[i][int(input())-1] = 1
for i in range(n):
print(''.join([str(e) for e in edges[i]]))
def has_edge(edges, v, t, d):
targets = [i for i, e in enumerate(edge... | s316620156 | Accepted | 202 | 17,892 | 221 | n = int(input())
edges = {i: int(input())-1 for i in range(n)}
def push_buttom(v, goal):
for i in range(n):
v = edges[v]
if v == goal:
return i+1
return -1
print(push_buttom(0, 1))
|
s475927568 | p03227 | u416223629 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,064 | 73 | You are given a string S of length 2 or 3 consisting of lowercase English letters. If the length of the string is 2, print it as is; if the length is 3, print the string after reversing it. | S=input()
if len(S)==2:
print(S)
else:
print(list(reversed(S)))
| s898546549 | Accepted | 17 | 2,940 | 82 | S=input()
if len(S)==2:
print(S)
else:
print(''.join(list(reversed(S))))
|
s306778940 | p03474 | u614181788 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 241 | The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom. | a,b = map(int,input().split())
s = input()
sw = 0
for i in range(len(s)):
if i == a and s[i] != '-':
sw = 1
elif i != a and s[i] == '-':
sw = 1
else:
pass
if sw == 1:
print('No')
else:
print('Yse') | s257098616 | Accepted | 17 | 3,064 | 241 | a,b = map(int,input().split())
s = input()
sw = 0
for i in range(len(s)):
if i == a and s[i] != '-':
sw = 1
elif i != a and s[i] == '-':
sw = 1
else:
pass
if sw == 1:
print('No')
else:
print('Yes') |
s307052975 | p02742 | u052221988 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 104 | We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the to... | h, w = map(int, input().split())
ans = h * w
if ans % 2 == 1:
print(ans//2+1)
else:
print(ans/2) | s831832053 | Accepted | 17 | 3,060 | 159 | h, w = map(int, input().split())
if h == 1 or w == 1:
ans = 1
elif h % 2 == 0 or w % 2 == 0:
ans = h * w // 2
else:
ans = h * w // 2 + 1
print(ans) |
s225123077 | p03469 | u238940874 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 45 | On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date col... | s = input()
s.replace("2017","2018")
print(s) | s304409340 | Accepted | 17 | 2,940 | 49 | s = input()
s = s.replace("2017","2018")
print(s) |
s285541087 | p02612 | u464950331 | 2,000 | 1,048,576 | Wrong Answer | 28 | 8,968 | 31 | 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(N%1000) | s663819677 | Accepted | 27 | 9,152 | 48 | N = int(input())
print((1000 - (N%1000))%1000) |
s598768407 | p03623 | u047197186 | 2,000 | 262,144 | Wrong Answer | 24 | 9,060 | 98 | 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... | x, a, b = map(int, input().split())
if abs(a - x) > abs(b - x):
print('A')
else:
print('B') | s229924855 | Accepted | 28 | 9,048 | 98 | x, a, b = map(int, input().split())
if abs(a - x) > abs(b - x):
print('B')
else:
print('A') |
s591571855 | p03457 | u642012866 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 284 | 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... | N = int(input())
nx = 0
ny = 0
nt = 0
for _ in range(N):
t, x, y = map(int, input().split())
k = abs(nx-x)+abs(ny-y)
if k > t:
print("No")
break
if (k-t)%2 != 0:
print("No")
break
nx = x
ny = y
nt = t
else:
print("Yes") | s986663088 | Accepted | 380 | 3,064 | 292 | N = int(input())
nx = 0
ny = 0
nt = 0
for _ in range(N):
t, x, y = map(int, input().split())
k = abs(nx-x)+abs(ny-y)
if k > t-nt:
print("No")
break
if (k-(t-nt))%2 != 0:
print("No")
break
nx = x
ny = y
nt = t
else:
print("Yes") |
s740095111 | p03693 | u642418876 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 89 | AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this i... | r,g,b=map(int,input().split())
if (100*r+10*g+b)%4==0:
print("Yes")
else:
print("No") | s069386878 | Accepted | 17 | 2,940 | 92 | r,g,b=map(int,input().split())
if (100*r+10*g+b)%4==0:
print("YES")
else:
print("NO")
|
s069301710 | p03079 | u629540524 | 2,000 | 1,048,576 | Wrong Answer | 23 | 8,912 | 70 | You are given three integers A, B and C. Determine if there exists an equilateral triangle whose sides have lengths A, B and C. | a=input().split()
if set(a)==a:
print('Yes')
else:
print('No') | s558277164 | Accepted | 29 | 9,028 | 75 | a=input().split()
if len(set(a))==1:
print('Yes')
else:
print('No') |
s523308401 | p00005 | u811773570 | 1,000 | 131,072 | Wrong Answer | 40 | 7,664 | 305 | Write a program which computes the greatest common divisor (GCD) and the least common multiple (LCM) of given a and b. | #GCD and LCM
while True:
try:
a, b = (int(i) for i in input(). split())
x = a * b
while True:
c = a % b
a = b
b = c
if a % b == 0:
break
x = x / a
print("%d %d" % (b, x))
except:
break | s812636317 | Accepted | 30 | 7,656 | 292 | def main():
gcd = lambda n, m : gcd(m, n % m) if m != 0 else n
while True:
try:
n, m = map(int, input().split())
g = gcd(n, m)
l = n * m // g
print(g, l)
except:
break
if __name__ == '__main__':
main() |
s849950812 | p03110 | u201382544 | 2,000 | 1,048,576 | Wrong Answer | 20 | 3,060 | 214 | Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000`... | n = int(input())
x = []
ans = 0
for i in range(n):
array = list(map(str,input().split()))
x.append(array)
if x[i][1] == "JPY":
ans += int(x[i][0])
else:
ans += int(float(x[i][0]) * 380000)
print(ans)
| s818133356 | Accepted | 17 | 2,940 | 209 | n = int(input())
x = []
ans = 0
for i in range(n):
array = list(map(str,input().split()))
x.append(array)
if x[i][1] == "JPY":
ans += int(x[i][0])
else:
ans += float(x[i][0]) * 380000
print(ans)
|
s155477142 | p03568 | u422104747 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 87 | We will say that two integer sequences of length N, x_1, x_2, ..., x_N and y_1, y_2, ..., y_N, are _similar_ when |x_i - y_i| \leq 1 holds for all i (1 \leq i \leq N). In particular, any integer sequence is similar to itself. You are given an integer N and an integer sequence of length N, A_1, A_2, ..., A_N. How man... | n=int(input())
s=input().split()
x=1
for i in s:
if(int(i)%2==0):
x*=2
print(2**n-x) | s696654678 | Accepted | 18 | 2,940 | 87 | n=int(input())
s=input().split()
x=1
for i in s:
if(int(i)%2==0):
x*=2
print(3**n-x) |
s961493739 | p04012 | u732844340 | 2,000 | 262,144 | Wrong Answer | 41 | 3,064 | 239 | Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful. | w = input()
flag = 0
for j in range(0,len(w)):
sum = 0
for i in range(0,len(w)):
if w[j] == w[i]:
sum += 1
if sum % 2 != 0:
print("NO")
flag = 1
break
if flag == 0:
print("YES") | s575317584 | Accepted | 171 | 3,572 | 208 | from collections import Counter
w = input()
flag = 0
counter = Counter(w)
for word, cnt in counter.most_common():
if cnt % 2 != 0:
flag = 1
if flag == 0:
print("Yes")
else:
print("No") |
s720289065 | p03854 | u089376182 | 2,000 | 262,144 | Wrong Answer | 23 | 6,516 | 95 | 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`. | import re
s = input()
print('Yes' if re.match('^(dream|dreamer|erase|eraser)+$', s) else 'No') | s823282616 | Accepted | 23 | 6,516 | 95 | import re
s = input()
print('YES' if re.match('^(dream|dreamer|erase|eraser)+$', s) else 'NO') |
s870128584 | p03455 | u078214750 | 2,000 | 262,144 | Wrong Answer | 29 | 9,016 | 84 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | a, b = map(int, input().split())
if (a*b)%2==0:
print('Odd')
else:
print('Even') | s530833610 | Accepted | 27 | 9,152 | 71 | a, b = map(int, input().split())
print('Even' if a*b%2 == 0 else 'Odd') |
s864331181 | p04011 | u934740772 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 100 | There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee. | N=int(input())
K=int(input())
X=int(input())
Y=int(input())
if N<=K:
print(X)
else:
print(Y) | s824702107 | Accepted | 17 | 2,940 | 114 | N=int(input())
K=int(input())
X=int(input())
Y=int(input())
if N<=K:
print(X*N)
else:
print(X*(K)+Y*(N-K)) |
s294597312 | p03644 | u514118270 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 123 | Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can... | N = int(input())
ans = 0
for i in range(100):
if N%2 == 0:
ans += 1
else:
print(ans)
exit() | s182466608 | Accepted | 17 | 2,940 | 131 | N = int(input())
if N == 1:
print(1)
exit()
for i in range(8):
if 2**i > N:
print(2**(i-1))
exit()
|
s047363245 | p03659 | u437215432 | 2,000 | 262,144 | Wrong Answer | 314 | 33,064 | 185 | 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 th... | import numpy as np
n = int(input())
a = list(map(int, input().split()))
total = sum(a)
Min = np.inf
s = 0
for i in range(len(a)-1):
s += a[i]
Min = min(Min, abs(2 * s- total))
| s978392472 | Accepted | 308 | 33,044 | 196 | import numpy as np
n = int(input())
a = list(map(int, input().split()))
total = sum(a)
Min = np.inf
s = 0
for i in range(len(a)-1):
s += a[i]
Min = min(Min, abs(2 * s- total))
print(Min)
|
s936553097 | p03574 | u395287676 | 2,000 | 262,144 | Wrong Answer | 37 | 3,952 | 3,329 | You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square con... | import sys
def main():
# Get Args
args = _input_args() # get arguments as an array from console/script parameters.
# Call main Logic
result = _main(args)
# Output a result in a correct way.
_output_result(result)
def _main(args):
"""Write Main Logic here for the contest.
:param ... | s568474102 | Accepted | 26 | 3,316 | 3,293 | import sys
def main():
# Get Args
args = _input_args() # get arguments as an array from console/script parameters.
# Call main Logic
result = _main(args)
# Output a result in a correct way.
_output_result(result)
def _main(args):
"""Write Main Logic here for the contest.
:param ... |
s352835199 | p03379 | u474925961 | 2,000 | 262,144 | Wrong Answer | 1,277 | 34,156 | 981 | When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..... | import numpy as np
n=int(input())
l=list(map(int,input().split()))
arr=np.sort(np.array(l))
med=np.median(arr)
med_max=np.median(np.delete(arr,0))
med_min=np.median(np.delete(arr,n-1))
if n%2==1:
p=(n-1)//2
med_midmin=np.median(np.delete(arr,p+1))
med_mid=np.median(np.delete(arr,p))
med_midmax=np.med... | s378371826 | Accepted | 646 | 34,176 | 307 | import numpy as np
n=int(input())
l=list(map(int,input().split()))
arr=np.sort(np.array(l))
med=np.median(arr)
med_max=int(np.median(np.delete(arr,0)))
med_min=int(np.median(np.delete(arr,n-1)))
p=(n+1)//2
for i in range(n):
if l[i]<=arr[p-1]:
print(med_max)
else:
print(med_min) |
s155288221 | p03671 | u449555432 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 54 | Snuke is buying a bicycle. The bicycle of his choice does not come with a bell, so he has to buy one separately. He has very high awareness of safety, and decides to buy two bells, one for each hand. The store sells three kinds of bells for the price of a, b and c yen (the currency of Japan), respectively. Find the m... | a,b,c=map(int,input().split());print(max(a+b,b+c,a+c)) | s356065671 | Accepted | 18 | 2,940 | 54 | a,b,c=map(int,input().split());print(min(a+b,b+c,a+c)) |
s530197307 | p02678 | u633355062 | 2,000 | 1,048,576 | Wrong Answer | 22 | 9,140 | 11 | 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 th... | print('No') | s940257512 | Accepted | 687 | 34,748 | 528 | from collections import deque
n, m = map(int, input().split())
dir = [[] for _ in range(n+1)]
for i in range(m):
ta, tb = map(int, input().split())
dir[ta].append(tb)
dir[tb].append(ta)
ans = [0]*(n+1)
flag = [False]*(n+1)
flag[1] = True
que = deque([1])
while que:
now = que.popleft()
for to ... |
s050167814 | p03546 | u152061705 | 2,000 | 262,144 | Wrong Answer | 34 | 3,316 | 627 | Joisino the magical girl has decided to turn every single digit that exists on this world into 1. Rewriting a digit i with j (0≤i,j≤9) costs c_{i,j} MP (Magic Points). She is now standing before a wall. The wall is divided into HW squares in H rows and W columns, and at least one square contains a digit between 0 and... | # python3
# utf-8
import itertools
rows_nr, cols_nr = (int(x) for x in input().split())
d1___d2___cost = []
for i in range(10):
d1___d2___cost_i___j = [int(x) for x in input().split()]
d1___d2___cost.append(d1___d2___cost_i___j)
r_c___value = []
for row in range(rows_nr):
c___value = [int(x) for x in inpu... | s947865569 | Accepted | 32 | 3,316 | 625 | # python3
# utf-8
import itertools
rows_nr, cols_nr = (int(x) for x in input().split())
d1___d2___cost = []
for i in range(10):
d1___d2___cost_i___j = [int(x) for x in input().split()]
d1___d2___cost.append(d1___d2___cost_i___j)
r_c___value = []
for row in range(rows_nr):
c___value = [int(x) for x in inpu... |
s663334124 | p03400 | u403331159 | 2,000 | 262,144 | Wrong Answer | 19 | 3,316 | 287 | Some number of chocolate pieces were prepared for a training camp. The camp had N participants and lasted for D days. The i-th participant (1 \leq i \leq N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on. As a result, there were X ... | N=int(input())
D,X=map(int,input().split())
A=[int(input()) for i in range(N)]
cnt=0
for i in range(N):
cnt+=1
day=1
for j in range(1,101):
if D>=j*A[i]+1:
day=j*A[i]+1
cnt+=1
print(day)
else:
break
print(cnt+X) | s084920278 | Accepted | 19 | 3,060 | 264 | N=int(input())
D,X=map(int,input().split())
A=[int(input()) for i in range(N)]
cnt=0
for i in range(N):
cnt+=1
day=1
for j in range(1,101):
if D>=j*A[i]+1:
day=j*A[i]+1
cnt+=1
else:
break
print(cnt+X) |
s164385838 | p03545 | u773686010 | 2,000 | 262,144 | Wrong Answer | 29 | 9,220 | 355 | 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 in... | import itertools
moji = str(input())
A,B,C,D = int(moji[0]),int(moji[1]),int(moji[2]),int(moji[3])
Bn = [B,B*-1]
Cn = [C,C*-1]
Dn = [D,D*-1]
flg = 0
for i,j,k in itertools.product(Bn, Cn,Dn):
X = A + i + j + k
if X == 7:
break
print(str(A) + (str(i),"+"+str(i))[i >= 0]+ (str(j),"+"+str(j))[j >= 0]... | s879125583 | Accepted | 29 | 9,216 | 360 | import itertools
moji = str(input())
A,B,C,D = int(moji[0]),int(moji[1]),int(moji[2]),int(moji[3])
Bn = [B,B*-1]
Cn = [C,C*-1]
Dn = [D,D*-1]
flg = 0
for i,j,k in itertools.product(Bn, Cn,Dn):
X = A + i + j + k
if X == 7:
break
print(str(A) + (str(i),"+"+str(i))[i >= 0]+ (str(j),"+"+str(j))[j >= 0]... |
s317004688 | p03565 | u354915818 | 2,000 | 262,144 | Wrong Answer | 1,451 | 16,812 | 1,128 | E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One mo... | # coding: utf-8
# Here your code !
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import numpy as np
import copy as cp
S = input()
T = input()
s = len(S)
t = len(T)
rst = []
for i in range(t) :
flag = False
for j in reversed(range(i , s + i - t + 1)):
if (T[i] != S[j]) & (S[j] != "?") : co... | s429040596 | Accepted | 155 | 12,516 | 1,131 | # coding: utf-8
# Here your code !
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import numpy as np
import copy as cp
S = input()
T = input()
s = len(S)
t = len(T)
rst = []
for i in range(t) :
flag = False
for j in reversed(range(i , s + i - t + 1)):
if (T[i] != S[j]) & (S[j] != "?") : co... |
s636544483 | p03729 | u959421160 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 130 | 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 `Y... | Nstr = input().split()
result = True
for i in range(2):
if Nstr[i][-1] != Nstr[i + 1][0]:
result = False
print(result) | s972190421 | Accepted | 17 | 2,940 | 130 | Nstr = input().split()
result = 'YES'
for i in range(2):
if Nstr[i][-1] != Nstr[i + 1][0]:
result = 'NO'
print(result) |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.