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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s524453222 | p04029 | u961469795 | 2,000 | 262,144 | Wrong Answer | 34 | 9,012 | 69 | 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())
ans = 0
for i in range(N):
ans += i
print(ans) | s454585669 | Accepted | 23 | 9,064 | 73 | N = int(input())
ans = 0
for i in range(N):
ans += i + 1
print(ans) |
s965460554 | p03338 | u870297120 | 2,000 | 1,048,576 | Wrong Answer | 18 | 3,060 | 219 | 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 = input()
x = 0
for i in range(1, n):
print(i)
a,b = set(s[:i]), set(s[i:])
if max(len(a),len(b))-max(len(a-b),len(b-a))>x:
x = max(len(a),len(b))-max(len(a-b),len(b-a))
print(x) | s482934084 | Accepted | 17 | 2,940 | 139 | n = int(input())
s = input()
x = 0
for i in range(1, n):
a,b = set(s[:i]), set(s[i:])
if len(a&b)>x:
x = len(a&b)
print(x) |
s861558772 | p03361 | u785578220 | 2,000 | 262,144 | Wrong Answer | 19 | 3,064 | 786 | We have a canvas divided into a grid with H rows and W columns. The square at the i-th row from the top and the j-th column from the left is represented as (i, j). Initially, all the squares are white. square1001 wants to draw a picture with black paint. His specific objective is to make Square (i, j) black when s_{i, j}= `#`, and to make Square (i, j) white when s_{i, j}= `.`. However, since he is not a good painter, he can only choose two squares that are horizontally or vertically adjacent and paint those squares black, for some number of times (possibly zero). He may choose squares that are already painted black, in which case the color of those squares remain black. Determine if square1001 can achieve his objective. | N,M = map(int, input().split())
field =[]
field.append(list("."*(M+2)))
for i in range(N):
t = list("."+input()+".")
field.append(t)
field.append(list("."*(M+2)))
print(field)
res = 1
def dfs(x,y):
r = 0
#field[x][y] = "."
for dx in range(-1,2):
for dy in range(-1,2):
if abs(dx+dy) ==1:
nx = x + dx
ny = y + dy
if field[nx][ny]==".":
r+=1
if r==4: return 0
else: return 1
for i in range(0,N+2):
for j in range(0,M+2):
if field[i][j] == "W":
res = res *dfs(i,j)
if res:print("Yes")
else:print("No") | s867028911 | Accepted | 24 | 3,064 | 773 | N,M = map(int, input().split())
field =[]
field.append(list("."*(M+2)))
for i in range(N):
t = list("."+input()+".")
field.append(t)
field.append(list("."*(M+2)))
res = 1
def dfs(x,y):
r = 0
#field[x][y] = "."
for dx in range(-1,2):
for dy in range(-1,2):
if abs(dx+dy) ==1:
nx = x + dx
ny = y + dy
if field[nx][ny]==".":
r+=1
if r==4: return 0
else: return 1
for i in range(0,N+2):
for j in range(0,M+2):
if field[i][j] == "#":
res = res *dfs(i,j)
if res:print("Yes")
else:print("No") |
s446849752 | p04044 | u924308178 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 89 | Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m. | N,L = list(map(int,input().split(" ")))
S = [input() for i in range(N)]
S.sort()
print(S) | s781949191 | Accepted | 17 | 3,060 | 98 | N,L = list(map(int,input().split(" ")))
S = [input() for i in range(N)]
S.sort()
print("".join(S)) |
s934030769 | p03946 | u254871849 | 2,000 | 262,144 | Wrong Answer | 112 | 15,840 | 313 | There are N towns located in a line, conveniently numbered 1 through N. Takahashi the merchant is going on a travel from town 1 to town N, buying and selling apples. Takahashi will begin the travel at town 1, with no apple in his possession. The actions that can be performed during the travel are as follows: * _Move_ : When at town i (i < N), move to town i + 1. * _Merchandise_ : Buy or sell an arbitrary number of apples at the current town. Here, it is assumed that one apple can always be bought and sold for A_i yen (the currency of Japan) at town i (1 ≦ i ≦ N), where A_i are distinct integers. Also, you can assume that he has an infinite supply of money. For some reason, there is a constraint on merchandising apple during the travel: the sum of the number of apples bought and the number of apples sold during the whole travel, must be at most T. (Note that a single apple can be counted in both.) During the travel, Takahashi will perform actions so that the _profit_ of the travel is maximized. Here, the profit of the travel is the amount of money that is gained by selling apples, minus the amount of money that is spent on buying apples. Note that we are not interested in apples in his possession at the end of the travel. Aoki, a business rival of Takahashi, wants to trouble Takahashi by manipulating the market price of apples. Prior to the beginning of Takahashi's travel, Aoki can change A_i into another arbitrary non-negative integer A_i' for any town i, any number of times. The cost of performing this operation is |A_i - A_i'|. After performing this operation, different towns may have equal values of A_i. Aoki's objective is to decrease Takahashi's expected profit by at least 1 yen. Find the minimum total cost to achieve it. You may assume that Takahashi's expected profit is initially at least 1 yen. | import sys
from bisect import bisect_left as bi_l
n, t, *a = map(int, sys.stdin.read().split())
def main():
cand = []
mi = a[0]
for x in a[1:]:
cand.append(x - mi)
mi = min(mi, x)
cand.sort()
print(cand)
ans = n - bi_l(cand, cand[-1]) - 1
print(ans)
if __name__ == '__main__':
main() | s215836469 | Accepted | 102 | 14,052 | 299 | import sys
from bisect import bisect_left as bi_l
n, t, *a = map(int, sys.stdin.read().split())
def main():
cand = []
mi = a[0]
for x in a[1:]:
cand.append(x - mi)
mi = min(mi, x)
cand.sort()
ans = n - bi_l(cand, cand[-1]) - 1
print(ans)
if __name__ == '__main__':
main() |
s616182503 | p03999 | u670133596 | 2,000 | 262,144 | Wrong Answer | 27 | 3,064 | 533 | 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. Evaluate all possible formulas, and print the sum of the results. | S = input()
combination = 2 ** (len(S) - 1)
ans = 0
for i in range(combination):
bin_str = bin(i)[2:]
bin_str = '0' * (len(S) - len(bin_str) - 1) + bin_str
formula = ''
for i, s_char in enumerate(S):
plus = ''
if len(S) - 1 != i and bin_str[i] == '1':
plus = '+'
formula += s_char + plus
ans += eval(formula) | s137141202 | Accepted | 28 | 3,064 | 545 | S = input()
combination = 2 ** (len(S) - 1)
ans = 0
for i in range(combination):
bin_str = bin(i)[2:]
bin_str = '0' * (len(S) - len(bin_str) - 1) + bin_str
formula = ''
for i, s_char in enumerate(S):
plus = ''
if len(S) - 1 != i and bin_str[i] == '1':
plus = '+'
formula += s_char + plus
ans += eval(formula)
print(ans) |
s164530075 | p03433 | u146714526 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 225 | E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins. |
def main():
N = int(input())
a = int(input())
r = N%500
if a >= r:
print ('yes')
else:
print ('No')
if __name__ == "__main__":
# global stime
# stime = time.clock()
main()
| s985928036 | Accepted | 17 | 2,940 | 264 |
def main():
N = int(input())
a = int(input())
r = N%500
if r == 0:
print ('Yes')
elif a >= r:
print ('Yes')
else:
print ('No')
if __name__ == "__main__":
# global stime
# stime = time.clock()
main()
|
s415270895 | p02536 | u395672550 | 2,000 | 1,048,576 | Wrong Answer | 314 | 12,768 | 899 | There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M. Road i connects City A_i and City B_i. Snuke can perform the following operation zero or more times: * Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities. After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times). What is the minimum number of roads he must build to achieve the goal? | class union_find():
def __init__(self,n):
self.n=n
self.root=[-1]*(n+1)
self.rank=[0]*(n+1)
def find_root(self,x):
if self.root[x]<0:
return x
else:
self.root[x]=self.find_root(self.root[x])
return self.root[x]
def unite(self,x,y):
x=self.find_root(x)
y=self.find_root(y)
if x==y:
return
elif self.rank[x]>self.rank[y]:
self.root[x]+=self.root[y]
self.root[y]=x
else:
self.root[y]+=self.root[x]
self.root[x]=y
if self.rank[x]==self.rank[y]:
self.rank[y]+=1
N,M = map(int,input().split())
g = union_find(N)
group = 0
for i in range(M):
A,B= map(int,input().split())
g.unite(A,B)
print(g.root)
for j in g.root:
if j < 0:
group += 1
print(group - 2)
| s087619570 | Accepted | 333 | 11,576 | 884 | class union_find():
def __init__(self,n):
self.n=n
self.root=[-1]*(n+1)
self.rank=[0]*(n+1)
def find_root(self,x):
if self.root[x]<0:
return x
else:
self.root[x]=self.find_root(self.root[x])
return self.root[x]
def unite(self,x,y):
x=self.find_root(x)
y=self.find_root(y)
if x==y:
return
elif self.rank[x]>self.rank[y]:
self.root[x]+=self.root[y]
self.root[y]=x
else:
self.root[y]+=self.root[x]
self.root[x]=y
if self.rank[x]==self.rank[y]:
self.rank[y]+=1
N,M = map(int,input().split())
g = union_find(N)
group = 0
for i in range(M):
A,B= map(int,input().split())
g.unite(A,B)
for j in g.root:
if j < 0:
group += 1
print(group - 2)
|
s821806564 | p03777 | u464912173 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 56 | Two deer, AtCoDeer and TopCoDeer, are playing a game called _Honest or Dishonest_. In this game, an honest player always tells the truth, and an dishonest player always tell lies. You are given two characters a and b as the input. Each of them is either `H` or `D`, and carries the following information: If a=`H`, AtCoDeer is honest; if a=`D`, AtCoDeer is dishonest. If b=`H`, AtCoDeer is saying that TopCoDeer is honest; if b=`D`, AtCoDeer is saying that TopCoDeer is dishonest. Given this information, determine whether TopCoDeer is honest. | a,b = input().split()
print("H" if 'a' == 'b' else "D") | s914200893 | Accepted | 17 | 2,940 | 52 | a,b = input().split()
print("H" if a == b else "D") |
s510342938 | p03089 | u706929073 | 2,000 | 1,048,576 | Wrong Answer | 18 | 3,064 | 372 | Snuke has an empty sequence a. He will perform N operations on this sequence. In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1). You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it. | n = int(input())
b = list(map(int, input().split()))
c = []
while 0 < len(b):
candidate = []
for i, b_i in enumerate(b):
if (i + 1) == b_i:
candidate.append(i)
if 0 == len(candidate):
print(-1)
exit(0)
c.append(b[candidate[-1]])
b.pop(candidate[-1])
if 0 != len(b):
print(-1)
exit(0)
c.reverse()
print(c)
| s798873744 | Accepted | 19 | 3,060 | 310 | n = int(input())
b = list(map(int, input().split()))
result = []
for i in range(len(b)):
for j in reversed(list(range(len(b)))):
if b[j] == j + 1:
result.append(b.pop(j))
break
if len(b) != 0:
print(-1)
exit(0)
for result_i in reversed(result):
print(result_i) |
s900113896 | p03574 | u686390526 | 2,000 | 262,144 | Wrong Answer | 153 | 12,896 | 916 | You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process. | import numpy as np
def check_mass(i,j):
c=0
if i>0:
if j>0:
if mass[i-1][j-1]=="#":
c+=1
if mass[i-1][j]=="#":
c+=1
if j<W-1:
if mass[i-1][j+1]=="#":
c+=1
if True:
if mass[i][j-1]=="#":
c+=1
if mass[i][j]=="#":
c+=1
if j<W-1:
if mass[i][j+1]=="#":
c+=1
if i<H-1:
if mass[i+1][j-1]=="#":
c+=1
if mass[i+1][j]=="#":
c+=1
if j<W-1:
if mass[i+1][j+1]=="#":
c+=1
return c
H,W= list(map(int, input().split()))
mass= []
for i in range(H):
s = list(input())
mass.append(s)
print(mass)
for i in range(H):
for j in range(W):
if mass[i][j] == ".":
c=check_mass(i,j)
mass[i][j] = str(c)
for i in range(H):
for j in range(W):
print(mass[i][j], end="")
print()
| s197029699 | Accepted | 155 | 12,892 | 954 | import numpy as np
def check_mass(i,j):
c=0
if i>0:
if j>0:
if mass[i-1][j-1]=="#":
c+=1
if mass[i-1][j]=="#":
c+=1
if j<W-1:
if mass[i-1][j+1]=="#":
c+=1
if True:
if j>0:
if mass[i][j-1]=="#":
c+=1
if mass[i][j]=="#":
c+=1
if j<W-1:
if mass[i][j+1]=="#":
c+=1
if i<H-1:
if j>0:
if mass[i+1][j-1]=="#":
c+=1
if mass[i+1][j]=="#":
c+=1
if j<W-1:
if mass[i+1][j+1]=="#":
c+=1
return c
H,W= list(map(int, input().split()))
mass= []
for i in range(H):
s = list(input())
mass.append(s)
# print(mass)
for i in range(H):
for j in range(W):
if mass[i][j] == ".":
c=check_mass(i,j)
mass[i][j] = str(c)
for i in range(H):
for j in range(W):
print(mass[i][j], end="")
print()
|
s556797747 | p03170 | u386195929 | 2,000 | 1,048,576 | Wrong Answer | 2,206 | 9,732 | 267 | There is a set A = \\{ a_1, a_2, \ldots, a_N \\} consisting of N positive integers. Taro and Jiro will play the following game against each other. Initially, we have a pile consisting of K stones. The two players perform the following operation alternately, starting from Taro: * Choose an element x in A, and remove exactly x stones from the pile. A player loses when he becomes unable to play. Assuming that both players play optimally, determine the winner. | n, k = map(int, input().strip().split())
arr = list(map(int, input().strip().split()))
dp = [0 for i in range(k+1)]
for i in range(len(dp)):
for j in arr:
if j+i < k+1:
dp[j+i] = (dp[i]+1)%2
ans = "First" if dp[-1] else "Second"
print(ans)
| s295209454 | Accepted | 120 | 9,864 | 287 | n, k = map(int, input().strip().split())
arr = list(map(int, input().strip().split()))
dp = [0 for i in range(k+1)]
for i in range(len(dp)):
if dp[i] == 0:
for j in arr:
if j+i < k+1:
dp[j+i] = 1
ans = "First" if dp[-1] else "Second"
print(ans) |
s276417116 | p03485 | u597436499 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 190 | You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer. | import sys
stdin = sys.stdin
ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
ns = lambda: stdin.readline()
a,b = na()
import math
print(math.ceil((a+b / 2)))
| s332747620 | Accepted | 17 | 3,060 | 192 | import sys
stdin = sys.stdin
ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
ns = lambda: stdin.readline()
a,b = na()
import math
print(math.ceil(((a+b) / 2)))
|
s442497734 | p03080 | u249685005 | 2,000 | 1,048,576 | Wrong Answer | 24 | 2,940 | 94 | There are N people numbered 1 to N. Each person wears a red hat or a blue hat. You are given a string s representing the colors of the people. Person i wears a red hat if s_i is `R`, and a blue hat if s_i is `B`. Determine if there are more people wearing a red hat than people wearing a blue hat. | s=list(input())
x=s.count('R')
y=s.count('B')
if x>>y:
print('Yes')
else:
print('No')
| s972463412 | Accepted | 17 | 2,940 | 103 | a=input()
s=list(input())
x=s.count('R')
y=s.count('B')
if x>y:
print('Yes')
else:
print('No')
|
s947771645 | p03852 | u863841238 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 76 | Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: `a`, `e`, `i`, `o` and `u`. | s = input()
if s in "aiueo":
print("Vowel")
else:
print("Consonant") | s660560073 | Accepted | 17 | 2,940 | 77 | s = input()
if s in "aiueo":
print("vowel")
else:
print("consonant")
|
s188878321 | p03386 | u564368158 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 187 | Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers. | a, b, k = map(int, input().split())
ans = []
if b - a < k:
k = b - a
for i in range(k):
ans.append(a+i)
ans.append(b-i)
ans = list(set(sorted(ans)))
for i in ans:
print(i) | s289578618 | Accepted | 17 | 3,060 | 197 | a, b, k = map(int, input().split())
if 2*k >= b-a+1:
for i in range(a, b+1):
print(i)
else:
for i in range(a, a+k):
print(i)
for j in range(b-k+1, b+1):
print(j) |
s474085697 | p03796 | u477142306 | 2,000 | 262,144 | Wrong Answer | 2,205 | 9,420 | 64 | Snuke loves working out. He is now exercising N times. Before he starts exercising, his _power_ is 1. After he exercises for the i-th time, his power gets multiplied by i. Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7. | s=int(input())
ans=1
for i in range(1,s):
ans*=i
print(ans) | s289449611 | Accepted | 43 | 9,112 | 92 | s=int(input())
ans=1
for i in range(1,s+1):
ans=ans%(10**9+7)*i
print(ans%(10**9+7)) |
s830598848 | p03474 | u932465688 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 193 | 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()
L = [0,1,2,3,4,5,6,7,8,9]
for i in (range(A) and range(A+1,A+B+1)):
if (S[i] in L):
if (S[A] == '-'):
print('Yes')
else:
print('No') | s662978840 | Accepted | 22 | 3,316 | 281 | A,B = map(int,input().split())
S = list(input())
flag = True
L = ['1','2','3','4','5','6','7','8','9','0']
if (S[A] == '-'):
del S[A]
for i in range(len(S)):
if (S[i] not in L):
flag = False
break
else:
flag = False
if flag:
print('Yes')
else:
print('No') |
s341465029 | p03997 | u000123984 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 69 | 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 ) | s556193269 | Accepted | 17 | 2,940 | 78 | a = int(input())
b = int(input())
h = int(input())
s = int((a+b)*h/2)
print(s) |
s317598806 | p03729 | u603234915 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 73 | You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`. | a,b,c = input().split()
print(('No','Yes')[a[-1]==b[0] and b[-1]==c[0]])
| s005713832 | Accepted | 17 | 2,940 | 73 | a,b,c = input().split()
print(('NO','YES')[a[-1]==b[0] and b[-1]==c[0]])
|
s926073088 | p02407 | u369313788 | 1,000 | 131,072 | Wrong Answer | 20 | 7,596 | 86 | Write a program which reads a sequence and prints it in the reverse order. | input()
numbers = [int(i) for i in input().split()]
numbers.reverse()
print(numbers) | s461562436 | Accepted | 30 | 7,424 | 78 | input()
numbers = input().split()
numbers.reverse()
print(" ".join(numbers)) |
s478187885 | p03645 | u796424048 | 2,000 | 262,144 | Wrong Answer | 747 | 39,284 | 301 | 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 Snuke is on Island 1 now, and wants to go to Island N. However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services. Help him. | N,M = map(int,input().split())
a = [[0 for i in range(2)] for j in range(M)]
b = []
for l in range(M):
a[l][0] , a[l][1] = map(int,input().split())
for k in range(M):
if a[k][0] == a[0][1]:
b.append(k)
for m in b:
if a[m][1] == N:
print ('possible')
else:
print('impossible')
| s864993030 | Accepted | 814 | 48,584 | 448 | import sys
N,M = map(int,input().split())
a = [[0 for i in range(2)] for j in range(M)]
b = []
c = []
for i in range(M):
a[i][0] , a[i][1] = map(int,input().split())
first, second = a[i][0],a[i][1]
if first == 1:
b.append(second)
if second == N:
c.append(first)
b_set = set(b)
c_set = set(c)
if list(b_set & c_set) is None or len(list(b_set & c_set)):
res = ('POSSIBLE')
else:
res = ('IMPOSSIBLE')
print(res)
|
s334346399 | p03196 | u821432765 | 2,000 | 1,048,576 | Wrong Answer | 123 | 3,064 | 313 | There are N integers a_1, a_2, ..., a_N not less than 1. The values of a_1, a_2, ..., a_N are not known, but it is known that a_1 \times a_2 \times ... \times a_N = P. Find the maximum possible greatest common divisor of a_1, a_2, ..., a_N. | N, P = [int(i) for i in input().split()]
div = {}
x = P
y = 2
while y*y <= x:
while x % y == 0:
try:
div[y] += 1
except KeyError:
div[y] = 1
x //= y
y += 1
if x > 1:
div[x] = 1
res = 1
for k, v in div.items():
res *= max(1, (v//4)) * k
print(res) | s633136387 | Accepted | 125 | 3,064 | 304 | N, P = [int(i) for i in input().split()]
div = {}
x = P
y = 2
while y*y <= x:
while x % y == 0:
try:
div[y] += 1
except KeyError:
div[y] = 1
x //= y
y += 1
if x > 1:
div[x] = 1
res = 1
for k, v in div.items():
res *= k**(v//N)
print(res) |
s536582805 | p03797 | u595289165 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 107 | Snuke loves puzzles. Today, he is working on a puzzle using `S`\- and `c`-shaped pieces. In this puzzle, you can combine two `c`-shaped pieces into one `S`-shaped piece, as shown in the figure below: Snuke decided to create as many `Scc` groups as possible by putting together one `S`-shaped piece and two `c`-shaped pieces. Find the maximum number of `Scc` groups that can be created when Snuke has N `S`-shaped pieces and M `c`-shaped pieces. | n, m = map(int, input().split())
if 2*n >= m:
ans = m//2
else:
ans = n - (2*n - m)//4 -1
print(ans) | s851356626 | Accepted | 17 | 2,940 | 104 | n, m = map(int, input().split())
if 2*n >= m:
ans = m//2
else:
ans = n + (m - 2*n)//4
print(ans) |
s964261904 | p03860 | u941884460 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 47 | Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name. | a,b,c=map(str,input().split())
print('A'+b+'C') | s577604848 | Accepted | 17 | 2,940 | 59 | a,b,c=map(str,input().split())
print('A'+b[0].upper()+'C')
|
s632884719 | p02854 | u598229387 | 2,000 | 1,048,576 | Wrong Answer | 99 | 26,060 | 656 | Takahashi, who works at DISCO, is standing before an iron bar. The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters. Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length. However, this may not be possible as is, so he will do the following operations some number of times **before** he does the cut: * Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan). * Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen. Find the minimum amount of money needed before cutting the bar into two parts with the same length. | n=int(input())
a = [int(i) for i in input().split()]
length = sum(a)
if length %2 ==0:
check = 0
half = length//2
idx = 0
for i in range(len(a)):
check+=a[i]
if check > half:
idx = i
break
if check - half > half - (check-a[idx]):
check -=a[idx]
ans = (half-check)*2
else:
check = 0
half = length//2
idx = 0
for i in range(len(a)):
check+=a[i]
if check > half:
idx = i
break
if check - half > half - (check-a[idx]):
check -=a[idx]
ans = abs((half-check))*2
print(ans)
| s969274929 | Accepted | 104 | 26,220 | 737 | n=int(input())
a = [int(i) for i in input().split()]
length = sum(a)
if length %2 ==0:
check = 0
half = length//2
idx = 0
for i in range(len(a)):
check+=a[i]
if check > half:
idx = i
break
if check - half > half - (check-a[idx]):
check -=a[idx]
ans = abs((half-check)*2)
else:
check = 0
half = length//2
idx = 0
for i in range(len(a)):
check+=a[i]
if check > half:
idx = i
break
check -=a[idx]
afte = sum(a[idx:])
ans = half-check+afte-half
check +=a[idx]
afte = sum(a[idx+1:])
ans = min(ans,abs(check-half)+abs(afte-half))
print(ans)
|
s559921933 | p03719 | u576432509 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 137 | You are given three integers A, B and C. Determine whether C is not less than A and not greater than B. | icase=0
if icase==0:
a,b,c=map(int,input().split())
if a<=c and c<=b:
print("YES")
else:
print("NO")
| s853029441 | Accepted | 17 | 2,940 | 130 | icase=61
if icase==61:
a,b,c=map(int,input().split())
if a<=c and c<=b:
print("Yes")
else:
print("No") |
s121626477 | p03150 | u922901775 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,060 | 411 | A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string. | S = input()
def judge():
L = len(S)
keyence = 'keyence'
if S[0] != 'k':
if S[-7:] != 'keyence':
return False
else:
return True
else:
for i in range(8):
if S[0:i] == keyence[0:i] and S[L-(7-i):L] == keyence[i:7]:
return True
return False
def main():
print(judge())
if __name__ == '__main__':
main()
| s239606327 | Accepted | 17 | 3,060 | 459 | S = input()
def judge():
L = len(S)
keyence = 'keyence'
if S[0] != 'k':
if S[-7:] != 'keyence':
return False
else:
return True
else:
for i in range(8):
if S[0:i] == keyence[0:i] and S[L-(7-i):L] == keyence[i:7]:
return True
return False
def main():
if judge():
print('YES')
else:
print('NO')
if __name__ == '__main__':
main()
|
s964402848 | p03721 | u536034761 | 2,000 | 262,144 | Wrong Answer | 255 | 18,024 | 282 | There is an empty array. The following N operations will be performed to insert integers into the array. In the i-th operation (1≤i≤N), b_i copies of an integer a_i are inserted into the array. Find the K-th smallest integer in the array after the N operations. For example, the 4-th smallest integer in the array \\{1,2,2,3,3,3\\} is 3. | n, k = map(int, input().split())
S = set()
D = dict()
for _ in range(n):
a, b = map(int, input().split())
if a in S:
D[a] += b
else:
S.add(a)
D[a] = b
L = sorted(S)
cnt=0
for l in L:
if cnt >= k:
print(l)
break
cnt += D[l] | s271964029 | Accepted | 229 | 17,924 | 284 | n, k = map(int, input().split())
S = set()
D = dict()
for _ in range(n):
a, b = map(int, input().split())
if a in S:
D[a] += b
else:
S.add(a)
D[a] = b
L = sorted(S)
cnt = 0
for l in L:
cnt += D[l]
if cnt >= k:
print(l)
break |
s329298906 | p02612 | u044220565 | 2,000 | 1,048,576 | Wrong Answer | 27 | 9,152 | 523 | 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. | # coding: utf-8
import sys
# from operator import itemgetter
sysread = sys.stdin.readline
read = sys.stdin.read
sys.setrecursionlimit(10 ** 7)
#from heapq import heappop, heappush
#from collections import OrderedDict, defaultdict
#import math
#from itertools import product, accumulate, combinations, product
#import numpy as np
#from copy import deepcopy
#from collections import deque
#import numba
def run():
N = int(input())
print(N % 1000)
if __name__ == "__main__":
run() | s961389876 | Accepted | 29 | 9,160 | 553 | # coding: utf-8
import sys
# from operator import itemgetter
sysread = sys.stdin.readline
read = sys.stdin.read
sys.setrecursionlimit(10 ** 7)
#from heapq import heappop, heappush
#from collections import OrderedDict, defaultdict
#import math
#from itertools import product, accumulate, combinations, product
#import numpy as np
#from copy import deepcopy
#from collections import deque
#import numba
def run():
N = int(input())
a = (N + 999) // 1000
print(a * 1000 - N)
if __name__ == "__main__":
run() |
s447218526 | p03495 | u914529932 | 2,000 | 262,144 | Wrong Answer | 270 | 50,176 | 439 | Takahashi has N balls. Initially, an integer A_i is written on the i-th ball. He would like to rewrite the integer on some balls so that there are at most K different integers written on the N balls. Find the minimum number of balls that Takahashi needs to rewrite the integers on them. | dbflag=0
def debug(*args):
if dbflag: print(*args)
result = -1
N, K = map(int, input().split())
A = [int(i) for i in input().split(' ')]
bag = {}
for x in A:
if x in bag:
bag['x'+str(x)]+=1
else:
bag['x'+str(x)] = 1
debug(bag)
L = sorted(bag.items(),key= lambda x:x[1])
debug(L)
cnt = len(L) - K
if cnt > 0:
result = sum(y[1] for y in L[0:K])
debug([y[1] for y in L[0:K]])
else:
result = 0
print(result)
| s024304585 | Accepted | 332 | 52,408 | 460 | dbflag=0
def debug(*args):
if dbflag: print(*args)
result = 0
N, K = map(int, input().split())
A = [int(i) for i in input().split(' ')]
bag = {}
for x in A:
if 'x'+str(x) in bag:
bag['x'+str(x)]+=1
else:
bag['x'+str(x)] = 1
debug(bag)
L = sorted(bag.items(),key= lambda x:x[1])
debug(L,len(L),K)
cnt = len(L) - K
if cnt > 0:
result = sum(y[1] for y in L[0:cnt])
debug([y[1] for y in L[0:cnt]])
else:
result = 0
print(result)
|
s958152825 | p03712 | u760794812 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 216 | You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1. | H,W = map(int,input().split())
List = []
for _ in range(H):
S = input()
List.append(S)
print('*'*(W+2))
for i in range(H):
print('*'.strip(),end='')
print(List[i].strip(),end='')
print('*')
print('*'*(W+2)) | s258566992 | Accepted | 18 | 3,060 | 216 | H,W = map(int,input().split())
List = []
for _ in range(H):
S = input()
List.append(S)
print('#'*(W+2))
for i in range(H):
print('#'.strip(),end='')
print(List[i].strip(),end='')
print('#')
print('#'*(W+2)) |
s805283294 | p03044 | u532966492 | 2,000 | 1,048,576 | Wrong Answer | 2,107 | 108,196 | 670 | We have a tree with N vertices numbered 1 to N. The i-th edge in the tree connects Vertex u_i and Vertex v_i, and its length is w_i. Your objective is to paint each vertex in the tree white or black (it is fine to paint all vertices the same color) so that the following condition is satisfied: * For any two vertices painted in the same color, the distance between them is an even number. Find a coloring of the vertices that satisfies the condition and print it. It can be proved that at least one such coloring exists under the constraints of this problem. | N=int(input())
uvw=[list(map(int,input().split())) for _ in range(N-1)]
q=[[] for _ in range(N)]
max=10**15
dist=[0]+[max for i in range(N-1)]
for i in range(N-1):
q[uvw[i][0]-1].append((uvw[i][1],uvw[i][2]))
q[uvw[i][1]-1].append((uvw[i][0],uvw[i][2]))
now=0
now_d=0
while(True):
print(q)
if q[q[now][-1][0]-1]==[]:
q[now].pop()
if q[now]==[]:
break
posi=q[now][-1]
if dist[posi[0]-1]==max:
dist[posi[0]-1]=now_d+posi[1]
now_d=dist[posi[0]-1]
q[now].pop()
now=posi[0]-1
posi=q[posi[0]-1][0]
print(dist)
for i in range(N):
if dist[i]%2==0:
print(0)
else:
print(1) | s253979333 | Accepted | 903 | 62,004 | 735 | N=int(input())
uvw=[list(map(int,input().split())) for _ in range(N-1)]
q=[[] for _ in range(N)]
max=10**15
dist=[0]+[max for i in range(N-1)]
for i in range(N-1):
q[uvw[i][0]-1].append((uvw[i][1]-1,uvw[i][2]))
q[uvw[i][1]-1].append((uvw[i][0]-1,uvw[i][2]))
now=0
now_d=0
while(True):
if q[q[now][-1][0]]==[]:
q[now].pop()
if q[now]==[]:
break
if dist[q[now][-1][0]]!=max:
if len(q[now])!=1:
q[now]=[q[now][-1]]+q[now][:-1]
posi=q[now][-1]
if dist[posi[0]]==max:
dist[posi[0]]=now_d+posi[1]
now_d=dist[posi[0]]
q[now].pop()
now=posi[0]
posi=q[posi[0]][0]
for i in range(N):
if dist[i]%2==0:
print(0)
else:
print(1) |
s700813115 | p02612 | u935511247 | 2,000 | 1,048,576 | Wrong Answer | 29 | 9,040 | 64 | 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. | t=int(input())
m=0
while 1000*m<t:
m=m+1
c=1000*m-t
print(t) | s153331437 | Accepted | 24 | 9,156 | 64 | t=int(input())
m=0
while 1000*m<t:
m=m+1
c=1000*m-t
print(c) |
s422910885 | p02277 | u247976584 | 1,000 | 131,072 | Wrong Answer | 20 | 7,732 | 1,157 | Let's arrange a deck of cards. Your task is to sort totally n cards. A card consists of a part of a suit (S, H, C or D) and an number. Write a program which sorts such cards based on the following pseudocode: Partition(A, p, r) 1 x = A[r] 2 i = p-1 3 for j = p to r-1 4 do if A[j] <= x 5 then i = i+1 6 exchange A[i] and A[j] 7 exchange A[i+1] and A[r] 8 return i+1 Quicksort(A, p, r) 1 if p < r 2 then q = Partition(A, p, r) 3 run Quicksort(A, p, q-1) 4 run Quicksort(A, q+1, r) Here, A is an array which represents a deck of cards and comparison operations are performed based on the numbers. Your program should also 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). | class Quicksort:
def quicksort(self, a, p, r):
if p < r:
q = self.partion(a, p, r)
self.quicksort(a, p, q - 1)
self.quicksort(a, q + 1, r)
return(a)
def partion(self, a, p, r):
x = a[r][1]
i = p - 1
for j in range(p, r):
if a[j][1] <= x:
i += 1
a[i], a[j] = a[j], a[i]
a[i + 1], a[r] = a[r], a[i + 1]
return(i+1)
def bubbleSort(a, n):
flag = 1
while flag == 1:
flag = 0
for j in range(n-1, 0, -1):
if a[j][1] < a[j - 1][1]:
a[j], a[j-1] = a[j-1], a[j]
flag = 1
return(a)
if __name__ == '__main__':
n = int(input().rstrip())
cards = []
for i in range(n):
tmp = input().rstrip().split(" ")
tmp[1] = int(tmp[1])
cards.append(tmp)
a = cards[:]
b = cards[:]
x = Quicksort()
a = x.quicksort(a, 0, n-1)
b = bubbleSort(b, n)
if a == b:
print("Stable")
else:
print("Not Stable")
for i in a:
i[1] = str(i[1])
print(" ".join(i)) | s969769806 | Accepted | 1,170 | 35,912 | 1,160 | class Quicksort:
def quicksort(self, a, p, r):
if p < r:
q = self.partion(a, p, r)
self.quicksort(a, p, q - 1)
self.quicksort(a, q + 1, r)
return(a)
def partion(self, a, p, r):
x = a[r][1]
i = p - 1
for j in range(p, r):
if a[j][1] <= x:
i += 1
a[i], a[j] = a[j], a[i]
a[i + 1], a[r] = a[r], a[i + 1]
return(i+1)
def bubbleSort(a, n):
flag = 1
while flag == 1:
flag = 0
for j in range(n-1, 0, -1):
if a[j][1] < a[j - 1][1]:
a[j], a[j-1] = a[j-1], a[j]
flag = 1
return(a)
if __name__ == '__main__':
n = int(input().rstrip())
cards = []
for i in range(n):
tmp = input().rstrip().split(" ")
tmp[1] = int(tmp[1])
cards.append(tmp)
a = cards[:]
b = cards[:]
x = Quicksort()
a = x.quicksort(a, 0, n-1)
b = sorted(b, key=lambda i: i[1])
if a == b:
print("Stable")
else:
print("Not stable")
for i in a:
i[1] = str(i[1])
print(" ".join(i)) |
s064279787 | p03610 | u979060189 | 2,000 | 262,144 | Wrong Answer | 27 | 3,572 | 107 | You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1. | s = str(input())
odd =[]
n = int(len(s)/2)
for i in range(n):
odd.append(s[2*i])
print(''.join(odd)) | s091188388 | Accepted | 18 | 3,188 | 36 | s = str(input())
s = s[::2]
print(s) |
s063311545 | p03721 | u102242691 | 2,000 | 262,144 | Wrong Answer | 330 | 5,744 | 228 | There is an empty array. The following N operations will be performed to insert integers into the array. In the i-th operation (1≤i≤N), b_i copies of an integer a_i are inserted into the array. Find the K-th smallest integer in the array after the N operations. For example, the 4-th smallest integer in the array \\{1,2,2,3,3,3\\} is 3. |
n,k = map(int,input().split())
x = [0]* (10**5)
for i in range(n):
a,b = map(int,input().split())
x[a-1] += b
#print(x)
count = 0
for i in range(n):
count += x[i]
if count >= k:
print(i)
break
| s758604098 | Accepted | 345 | 10,348 | 226 | N, K = map(int, input().split())
cnt = [0]*100001
S = set()
for i in range(N):
a, b = map(int, input().split())
cnt[a-1] += b
S.add(a)
acc = 0
for s in sorted(S):
acc += cnt[s-1]
if acc >= K:
print(s)
break
|
s162244794 | p03474 | u114954806 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 124 | 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()
print("True") if s[:a].isdigit() and s[a:b]=='-' and s[b:].isdigit() else print("No") | s488643774 | Accepted | 18 | 2,940 | 138 | a,b=map(int,input().split())
s=input()
if s[:a].isdigit() and s[a]=='-' and s[a+1:a+b+1].isdigit():
print("Yes")
else:
print("No") |
s562991520 | p03729 | u722189950 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 107 | 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`. | #ABC060A
A,B,C = input().split()
if A[-1] == B[0] and B[-1] == C[0]:
print("Yes")
else:
print("No") | s491486283 | Accepted | 17 | 2,940 | 107 | #ABC060A
A,B,C = input().split()
if A[-1] == B[0] and B[-1] == C[0]:
print("YES")
else:
print("NO") |
s997171650 | p03693 | u827765795 | 2,000 | 262,144 | Wrong Answer | 26 | 9,016 | 104 | 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())
x = 10 * g + b
if x % 4 == 0:
print('Yes')
else:
print('No') | s186228278 | Accepted | 28 | 9,156 | 104 | r, g, b = map(int, input().split())
x = 10 * g + b
if x % 4 == 0:
print('YES')
else:
print('NO') |
s486774694 | p03024 | u850266651 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 113 | Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility. | S = input()
r = 0
for s in S:
if s == "o":
r += 1
if r >= 8:
print("YES")
else:
print("NO")
| s451840516 | Accepted | 17 | 2,940 | 110 | S = input()
r = 0
for s in S:
if s == "x":
r += 1
if r < 8:
print("YES")
else:
print("NO") |
s002899941 | p03945 | u698919163 | 2,000 | 262,144 | Wrong Answer | 33 | 3,188 | 128 | Two foxes Jiro and Saburo are playing a game called _1D Reversi_. This game is played on a board, using black and white stones. On the board, stones are placed in a row, and each player places a new stone to either end of the row. Similarly to the original game of Reversi, when a white stone is placed, all black stones between the new white stone and another white stone, turn into white stones, and vice versa. In the middle of a game, something came up and Saburo has to leave the game. The state of the board at this point is described by a string S. There are |S| (the length of S) stones on the board, and each character in S represents the color of the i-th (1 ≦ i ≦ |S|) stone from the left. If the i-th character in S is `B`, it means that the color of the corresponding stone on the board is black. Similarly, if the i-th character in S is `W`, it means that the color of the corresponding stone is white. Jiro wants all stones on the board to be of the same color. For this purpose, he will place new stones on the board according to the rules. Find the minimum number of new stones that he needs to place. | S = input()
count = 0
check = S[0]
for i in S[1:]:
if check != S:
count+=1
check = i
print(count) | s658853591 | Accepted | 32 | 3,188 | 128 | S = input()
count = 0
check = S[0]
for i in S[1:]:
if check != i:
count+=1
check = i
print(count)
|
s982153608 | p04012 | u466105944 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 160 | 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()
def is_beutifull_str(char_str):
for s in w:
if w.count(s)%2 == 1:
return 'NO'
return 'YES'
print (is_beutifull_str(w)) | s031907105 | Accepted | 17 | 2,940 | 160 | w = input()
def is_beutifull_str(char_str):
for s in w:
if w.count(s)%2 == 1:
return 'No'
return 'Yes'
print (is_beutifull_str(w)) |
s416278174 | p03625 | u457683760 | 2,000 | 262,144 | Time Limit Exceeded | 2,144 | 513,924 | 117 | We have N sticks with negligible thickness. The length of the i-th stick is A_i. Snuke wants to select four different sticks from these sticks and form a rectangle (including a square), using the sticks as its sides. Find the maximum possible area of the rectangle. | N=int(input())
A=(int(i) for i in input().split())
C=[0 for i in range(1000000000)]
for i in A:
C[i]+=1
print(C) | s471663791 | Accepted | 1,540 | 14,244 | 394 | N=int(input())
A=[int(i) for i in input().split()]
A.sort(reverse=True)
x=0
y=0
for i in range(N-1):
if A[0]==A[1]:
x=A[0]
A.remove(x)
A.remove(x)
break
else:
A.remove(A[0])
if len(A)>=2:
for i in range(len(A)-1):
if A[0]==A[1]:
y=A[0]
break
else:
A.remove(A[0])
print(x*y) |
s486890674 | p02613 | u914043696 | 2,000 | 1,048,576 | Wrong Answer | 157 | 17,460 | 124 | 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. | L = []
for i in range(int(input())):
L.append(input())
print(L)
for j in ['AC','WA','TLE','RE']:
print(j,'x',L.count(j)) | s649429880 | Accepted | 145 | 16,252 | 115 | L = []
for i in range(int(input())):
L.append(input())
for j in ['AC','WA','TLE','RE']:
print(j,'x',L.count(j)) |
s317733012 | p04030 | u842689614 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 159 | Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now? | s=input()
out=[]
for c in list(s):
if c=='0':
out.append('0')
elif c=='1':
out.append('1')
else:
if len(out)>0:
del out[-1]
print(out) | s938095417 | Accepted | 17 | 3,060 | 168 | s=input()
out=[]
for c in list(s):
if c=='0':
out.append('0')
elif c=='1':
out.append('1')
else:
if len(out)>0:
del out[-1]
print(''.join(out)) |
s416788138 | p02613 | u984081384 | 2,000 | 1,048,576 | Wrong Answer | 149 | 9,156 | 292 | 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. | c = int(input())
j = [0]*4
for i in range(c):
i = input()
if i == "AC":
j[0] += 1
elif i == "WA":
j[1] += 1
elif i == "TLE":
j[2] += 1
else:
j[3] += 1
print("AC × "+ str(j[0]))
print("WA × "+ str(j[1]))
print("TLE × "+ str(j[2]))
print("RE × "+ str(j[3]))
| s251992026 | Accepted | 146 | 9,056 | 287 | c = int(input())
j = [0]*4
for i in range(c):
a = input()
if a == "AC":
j[0] += 1
elif a == "WA":
j[1] += 1
elif a == "TLE":
j[2] += 1
else:
j[3] += 1
print("AC x "+ str(j[0]))
print("WA x "+ str(j[1]))
print("TLE x "+ str(j[2]))
print("RE x "+ str(j[3])) |
s908086782 | p02608 | u276588887 | 2,000 | 1,048,576 | Wrong Answer | 504 | 10,008 | 192 | Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N). | import math
n = int(input())
v = [0]*(100000)
for x in range(101):
for y in range(101):
for z in range(101):
v[x*x+y*y+z*z+x*y+y*z+z*x] += 1
for i in range(n):
print(v[i+1])
| s370263917 | Accepted | 545 | 10,320 | 202 | import math
n = int(input())
v = [0]*(100000)
for x in range(1, 101):
for y in range(1, 101):
for z in range(1, 101):
v[x*x+y*y+z*z+x*y+y*z+z*x] += 1
print('\n'.join(map(str,v[1:n+1])))
|
s644347677 | p03543 | u258436671 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 126 | We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**? | s = input()
D = []
for i in range(10):
D.append(s.count('i'))
a = max(D)
if a >= 3:
print('Yes')
else:
print('No') | s469577631 | Accepted | 17 | 2,940 | 99 | s = input()
if s[0] == s[1] == s[2] or s[1] == s[2] == s[3]:
print('Yes')
else:
print('No') |
s191315050 | p02856 | u276588887 | 2,000 | 1,048,576 | Wrong Answer | 373 | 9,072 | 143 | N programmers are going to participate in the preliminary stage of DDCC 20XX. Due to the size of the venue, however, at most 9 contestants can participate in the finals. The preliminary stage consists of several rounds, which will take place as follows: * All the N contestants will participate in the first round. * When X contestants participate in some round, the number of contestants advancing to the next round will be decided as follows: * The organizer will choose two consecutive digits in the decimal notation of X, and replace them with the sum of these digits. The number resulted will be the number of contestants advancing to the next round. For example, when X = 2378, the number of contestants advancing to the next round will be 578 (if 2 and 3 are chosen), 2108 (if 3 and 7 are chosen), or 2315 (if 7 and 8 are chosen). When X = 100, the number of contestants advancing to the next round will be 10, no matter which two digits are chosen. * The preliminary stage ends when 9 or fewer contestants remain. Ringo, the chief organizer, wants to hold as many rounds as possible. Find the maximum possible number of rounds in the preliminary stage. Since the number of contestants, N, can be enormous, it is given to you as two integer sequences d_1, \ldots, d_M and c_1, \ldots, c_M, which means the following: the decimal notation of N consists of c_1 + c_2 + \ldots + c_M digits, whose first c_1 digits are all d_1, the following c_2 digits are all d_2, \ldots, and the last c_M digits are all d_M. | n=int(input())
dnum =0
dsum =0
for i in range(n):
d,c=map(int,input().split())
dnum += c
dsum += d*c
ans = c-1 + dsum//9
print(ans) | s922842504 | Accepted | 368 | 9,124 | 150 | n=int(input())
dnum =0
dsum =0
for i in range(n):
d,c=map(int,input().split())
dnum += c
dsum += d*c
ans = dnum-1 + (dsum-1)//9
print(ans) |
s640188361 | p03493 | u785220618 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 82 | 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()
ans=0
for i in range(3):
if s[i] == 1:
ans += 1
print(ans) | s684539719 | Accepted | 17 | 2,940 | 84 | s = input()
ans=0
for i in range(3):
if s[i] == '1':
ans += 1
print(ans) |
s558540891 | p03473 | u131405882 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 29 | How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December? | N = int(input())
print(24-N)
| s622844113 | Accepted | 17 | 2,940 | 29 | N = int(input())
print(48-N)
|
s255960672 | p03469 | u360617739 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 54 | On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it. | # 85 A
L = input()
L1 = L.replace("7","8")
print(L1) | s520922492 | Accepted | 19 | 3,060 | 66 |
s = input()
ss = s.replace("7","8",1)
print(ss)
|
s193076141 | p02408 | u678843586 | 1,000 | 131,072 | Wrong Answer | 20 | 5,604 | 234 | Taro is going to play a card game. However, now he has only n cards, even though there should be 52 cards (he has no Jokers). The 52 cards include 13 ranks of each of the four suits: spade, heart, club and diamond. | n = int(input())
cards = {}
for i in range(n):
card = input()
cards[card] = 1
for c in ['S', 'H', 'C', 'D']:
for n in range(1,14):
key = c + ''+ str(n)
if not key in cards:
print(key)
| s265906766 | Accepted | 20 | 5,600 | 246 | n = int(input())
cards = {}
for i in range(n):
card = input()
cards[card] = 1
for c in ['S', 'H', 'C', 'D']:
for n in range(1,14):
key = c + ' ' + str(n)
if not key in cards:
print(key)
|
s692827279 | p03730 | u711295009 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 255 | We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`. | a, b, c = map(int, input().split())
modL=[0]
while True:
amari = (modL[len(modL)-1]+a)%b
if amari == c:
print("Yes")
break
elif amari in modL:
print("No")
break
else:
modL.append(amari)
| s390292339 | Accepted | 17 | 3,060 | 242 | a, b, c = map(int, input().split())
modL=[0]
while True:
amari = (modL[len(modL)-1]+a)%b
if amari == c:
print("YES")
break
elif amari in modL:
print("NO")
break
else:
modL.append(amari) |
s717682151 | p02694 | u864276028 | 2,000 | 1,048,576 | Wrong Answer | 31 | 8,912 | 109 | Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time? | X = int(input())
M = 100
i = 1
while True:
I = 0.01*M
M += I
i += 1
if M >= X:
print(i)
break | s846373605 | Accepted | 31 | 9,064 | 75 | X = int(input())
A = 100
Y = 0
while A < X:
A += A//100
Y += 1
print(Y) |
s696685398 | p03997 | u213800869 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 76 | You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid. | a = int(input())
b = int(input())
h = int(input())
print((a + b) * h * 0.5) | s600688586 | Accepted | 18 | 3,188 | 91 | a = int(input())
b = int(input())
h = int(input())
s = (a + b) * h * 0.5
print(round(s))
|
s599880538 | p03737 | u873849550 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 93 | 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. | s1, s2, s3 = input().split()
print(s1[0].upper() + ' ' + s2[0].upper() + ' ' + s3[0].upper()) | s724871514 | Accepted | 17 | 2,940 | 81 | s1, s2, s3 = input().split()
print(s1[0].upper() + s2[0].upper() + s3[0].upper()) |
s906718248 | p03605 | u803647747 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 108 | 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? | str_list = list(input())
if (str_list[0]!="9") and (str_list[1]!="9"):
print("No")
else:
print("No") | s375785563 | Accepted | 16 | 2,940 | 109 | str_list = list(input())
if (str_list[0]!="9") and (str_list[1]!="9"):
print("No")
else:
print("Yes") |
s415807529 | p02646 | u377989038 | 2,000 | 1,048,576 | Wrong Answer | 20 | 9,120 | 191 | Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally. | a, v = map(int, input().split())
b, w = map(int, input().split())
t = int(input())
if v <= w:
print("NO")
exit()
if (v - w) * t <= abs(a - b):
print("YES")
else:
print("NO") | s456583166 | Accepted | 23 | 9,140 | 152 | a, v = map(int, input().split())
b, w = map(int, input().split())
t = int(input())
if (v - w) * t >= abs(a - b):
print("YES")
else:
print("NO") |
s255732181 | p03447 | u350049649 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 61 | 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? | X=int(input())
A=int(input())
B=int(input())
print((X-A)//B) | s072350447 | Accepted | 18 | 2,940 | 61 | X=int(input())
A=int(input())
B=int(input())
print((X-A)%B)
|
s322899309 | p03139 | u712259362 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 131 | 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 = [int(i) for i in input().split()]
X = A if A > B else B
Y = 0 if N > (A+B) else N - (A+B)
print("{0} {1}".format(X, Y)) | s476881438 | Accepted | 17 | 2,940 | 125 | N, A, B = [int(i) for i in input().split()]
X = A if A < B else B
Y = 0 if N > A+B else A+B-N
print("{0} {1}".format(X, Y)) |
s598025381 | p03975 | u328755070 | 1,000 | 262,144 | Wrong Answer | 17 | 2,940 | 195 | Summer vacation ended at last and the second semester has begun. You, a Kyoto University student, came to university and heard a rumor that somebody will barricade the entrance of your classroom. The barricade will be built just before the start of the A-th class and removed by Kyoto University students just before the start of the B-th class. All the classes conducted when the barricade is blocking the entrance will be cancelled and you will not be able to attend them. Today you take N classes and class i is conducted in the t_i- th period. You take at most one class in each period. Find the number of classes you can attend. | N, A, B = list(map(int, input().split()))
t = [int(input()) for x in range(N)]
ans = 0
for i in range(N):
if t[i] < A or t[i] > B:
continue
else:
ans += 1
print(ans)
| s182936378 | Accepted | 17 | 2,940 | 201 | N, A, B = list(map(int, input().split()))
t = [int(input()) for x in range(N)]
ans = 0
for i in range(N):
if t[i] < A or t[i] >= B:
continue
else:
ans += 1
print(N - ans)
|
s024977792 | p03352 | u920299620 | 2,000 | 1,048,576 | Time Limit Exceeded | 2,105 | 29,316 | 127 | You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2. | n=int(input())
i=2
ans=[1]
while(i**2 <= n ):
j=2
while( i**(j+1) <=n ):
j+=1
ans.append( i**j)
print( max(ans) )
| s828755997 | Accepted | 17 | 2,940 | 134 | n=int(input())
i=2
ans=[1]
while(i**2 <= n ):
j=2
while( i**(j+1) <=n ):
j+=1
ans.append( i**j)
i+=1
print( max(ans) )
|
s383085066 | p03997 | u957198490 | 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) | s326391081 | Accepted | 17 | 2,940 | 68 | a = int(input())
b = int(input())
h = int(input())
print((a+b)*h//2) |
s255775439 | p02612 | u290886932 | 2,000 | 1,048,576 | Wrong Answer | 28 | 9,132 | 55 | 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())
while N >= 1000:
N -= 1000
print(N)
| s847256041 | Accepted | 29 | 9,140 | 75 | N = int(input())
pay = 1000
while pay < N:
pay += 1000
print(pay - N)
|
s321610676 | p03456 | u868600519 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 111 | AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number. | import math
ab = int(input().replace(' ', ''))
r = math.sqrt(ab)
print(round(r) if r.is_integer() else 'No')
| s636507639 | Accepted | 17 | 2,940 | 102 | import math
ab = int(input().replace(' ', ''))
print('Yes' if math.sqrt(ab).is_integer() else 'No')
|
s374932783 | p03721 | u686036872 | 2,000 | 262,144 | Wrong Answer | 378 | 21,248 | 243 | There is an empty array. The following N operations will be performed to insert integers into the array. In the i-th operation (1≤i≤N), b_i copies of an integer a_i are inserted into the array. Find the K-th smallest integer in the array after the N operations. For example, the 4-th smallest integer in the array \\{1,2,2,3,3,3\\} is 3. | N, K = map(int, input().split())
A=[]
for i in range(N):
a, b = map(int, input().split())
A.append([a, b])
A.sort(key=lambda x:x[0])
cnt = 0
for i in range(N):
cnt += A[i][1]
if cnt >= N:
print(A[i][0])
break | s289075263 | Accepted | 407 | 21,248 | 243 | N, K = map(int, input().split())
A=[]
for i in range(N):
a, b = map(int, input().split())
A.append([a, b])
A.sort(key=lambda x:x[0])
cnt = 0
for i in range(N):
cnt += A[i][1]
if cnt >= K:
print(A[i][0])
break |
s311098689 | p03605 | u611090896 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 59 | 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 = input()
if N in "9":
print("Yes")
else:
print("No") | s431550388 | Accepted | 18 | 2,940 | 59 | N = input()
if "9" in N:
print("Yes")
else:
print("No") |
s273386065 | p03351 | u396890425 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 95 | Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate. | a,b,c,d=map(int, input().split())
print('Yes' if abs(c-a)<=d or abs(a-b)+abs(b-c)<=d else 'No') | s509965178 | Accepted | 17 | 2,940 | 102 | a,b,c,d=map(int, input().split())
print('Yes' if abs(c-a)<=d or abs(a-b)<=d and abs(b-c)<=d else 'No') |
s453535798 | p03455 | u840438868 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 138 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | import math
b, c = map(str, input().split())
d = int(b + c)
e = math.sqrt(d)
f = int(e)
if(e-f==0):
print('Yes')
else:
print('No') | s308992810 | Accepted | 17 | 2,940 | 96 | b, c = map(int, input().split())
if ((b%2)+(c%2) == 2):
print('Odd')
else:
print('Even') |
s523156435 | p03860 | u786020649 | 2,000 | 262,144 | Wrong Answer | 30 | 9,016 | 49 | Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name. | a,s,c=input().split()
print('A{}X'.format(s[0])) | s356504951 | Accepted | 26 | 8,996 | 50 | a,s,c=input().split()
print('A{}C'.format(s[0])) |
s583683803 | p02413 | u429841998 | 1,000 | 131,072 | Wrong Answer | 20 | 5,592 | 442 | Your task is to perform a simple table calculation. Write a program which reads the number of rows r, columns c and a table of r × c elements, and prints a new table, which includes the total sum for each row and column. | r, c = map(int, input().split())
d = [0] * (c + 1)
print(d)
for i in range(r):
d[i] = list(map(int, input().split()))
d[i].append(0)
for j in range(c):
d[i][c] += d[i][j]
d[r] = [0] * (c + 1)
for i in range(c + 1):
for j in range(r):
d[r][i] += d[j][i]
for i in range(r + 1):
row = ''
for j in range(c + 1):
row += str(d[i][j])
if j <= c - 1:
row += ' '
print(row)
| s580213062 | Accepted | 30 | 5,696 | 433 | r, c = map(int, input().split())
d = [0] * (r + 1)
for i in range(r):
d[i] = list(map(int, input().split()))
d[i].append(0)
for j in range(c):
d[i][c] += d[i][j]
d[r] = [0] * (c + 1)
for i in range(c + 1):
for j in range(r):
d[r][i] += d[j][i]
for i in range(r + 1):
row = ''
for j in range(c + 1):
row += str(d[i][j])
if j <= c - 1:
row += ' '
print(row)
|
s208264347 | p03548 | u408791346 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 66 | 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())
ans = (x-2)//(y+z)
print(ans) | s016737000 | Accepted | 20 | 3,316 | 56 | x, y, z = map(int, input().split())
print((x-z)//(y+z)) |
s188978639 | p03478 | u735069283 | 2,000 | 262,144 | Wrong Answer | 19 | 2,940 | 109 | 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())
result=0
for i in range(n):
if a<=i//10+i%10<=b:
result+=i
print(result) | s499353771 | Accepted | 46 | 3,060 | 164 | N,A,B=map(int,input().split())
r=[0]*37
for i in range(N+1):
count=0
for j in range(len(str(i))):
count +=int(str(i)[j])
r[count] +=i
print(sum(r[A:B+1])) |
s727414505 | p03577 | u987164499 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 67 | Rng is going to a festival. The name of the festival is given to you as a string S, which ends with `FESTIVAL`, from input. Answer the question: "Rng is going to a festival of what?" Output the answer. Here, assume that the name of "a festival of s" is a string obtained by appending `FESTIVAL` to the end of s. For example, `CODEFESTIVAL` is a festival of `CODE`. | from sys import stdin
s = stdin.readline().rstrip()
print(s[:-9]) | s133648258 | Accepted | 17 | 2,940 | 67 | from sys import stdin
s = stdin.readline().rstrip()
print(s[:-8]) |
s722502184 | p02612 | u189326411 | 2,000 | 1,048,576 | Wrong Answer | 23 | 8,984 | 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)
| s851996561 | Accepted | 25 | 9,144 | 73 | n = int(input())
if n%1000==0:
print(0)
else:
print(1000-n%1000)
|
s796749373 | p03693 | u305965165 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 104 | 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 = (int(i) for i in input().split())
if (r*100+g*10+b) % 4:
print("YES")
else:
print("NO") | s973770135 | Accepted | 17 | 2,940 | 109 | r,g,b = (int(i) for i in input().split())
if (r*100+g*10+b) % 4 == 0:
print("YES")
else:
print("NO") |
s887776435 | p04011 | u640603056 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 135 | 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:
cost = n*k + (n-k)*y
else:
cost = n*x
print(cost) | s160634389 | Accepted | 17 | 2,940 | 135 | n = int(input())
k = int(input())
x = int(input())
y = int(input())
if n > k:
cost = k*x + (n-k)*y
else:
cost = n*x
print(cost) |
s268783588 | p03909 | u539517139 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 186 | There is a grid with H rows and W columns. The square at the i-th row and j-th column contains a string S_{i,j} of length 5. The rows are labeled with the numbers from 1 through H, and the columns are labeled with the uppercase English letters from `A` through the W-th letter of the alphabet. Exactly one of the squares in the grid contains the string `snuke`. Find this square and report its location. For example, the square at the 6-th row and 8-th column should be reported as `H6`. | h,w=map(int,input().split())
s=[list(input().split()) for _ in range(h)]
for i in range(h):
for j in range(w):
if s[i][j]=='snuke':
print(chr(ord('A')+j) +str(i))
break | s186931404 | Accepted | 17 | 3,060 | 187 | h,w=map(int,input().split())
s=[list(input().split()) for _ in range(h)]
for i in range(h):
for j in range(w):
if s[i][j]=='snuke':
print(chr(ord('A')+j)+str(i+1))
break |
s299146056 | p03251 | u255001744 | 2,000 | 1,048,576 | Wrong Answer | 18 | 3,060 | 189 | Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out. | n,m,x,y = map(int, input().split())
x_list = list(map(int,input().split()))
y_list = list(map(int,input().split()))
if max(x_list) - min(y_list) > 1:
print("No War")
else:
print("War") | s957543923 | Accepted | 18 | 3,064 | 368 | n,m,x,y = map(int, input().split())
x_list = list(map(int,input().split()))
y_list = list(map(int,input().split()))
x_max = max(x_list)
y_min = min(y_list)
def is_war(z):
if x_max < z and y_min >=z and x<z and z<=y:
return True
else:
return False
flag = False
for i in range(-100,100,1):
if is_war(i):
flag = True
print("No War" if flag else "War") |
s229684511 | p03155 | u263830634 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 85 | It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board. The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns. How many ways are there to choose where to put the notice so that it completely covers exactly HW squares? | N = int(input())
H = int(input())
W = int(input())
print ((N - H + 1) * (N - W - 1)) | s544606993 | Accepted | 17 | 2,940 | 85 | N = int(input())
H = int(input())
W = int(input())
print ((N - H + 1) * (N - W + 1)) |
s049359351 | p02606 | u634046173 | 2,000 | 1,048,576 | Wrong Answer | 30 | 9,012 | 89 | How many multiples of d are there among the integers between L and R (inclusive)? | L,R,d = map(int,input().split())
c =0
for i in (L,R+1):
if i % d ==0:
c+=1
print(c) | s276532047 | Accepted | 32 | 9,028 | 96 | L,R,d = map(int,input().split())
c = 0
for i in range(L,R+1):
if i % d ==0:
c+=1
print(c)
|
s764346379 | p03563 | u331036636 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 54 | 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. | nr = float(input())
p = float(input())
print((p-nr)+p) | s255910946 | Accepted | 17 | 2,940 | 37 | print(-int(input()) + 2*int(input())) |
s177132884 | p03593 | u792670114 | 2,000 | 262,144 | Wrong Answer | 19 | 3,064 | 717 | We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter. Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition: * Every row and column in A' can be read as a palindrome. Determine whether he can create a matrix satisfying the condition. | H, W = map(int, input().split())
ns = {}
for i in range(H):
As = input()
for a in As:
if a not in ns: ns[a] = 0
ns[a] += 1
ns4 = [0, 0, 0, 0]
for a in ns:
n = ns[a]
ns4[n%4] += 1
print(ns4)
if H%2 == 0 and W%2 == 0:
if ns4[1] > 0 or ns4[2] > 0 or ns4[3] > 0:
print("No")
else:
print("Yes")
if H%2 == 1 and W%2 == 0:
if ns4[1]+ns4[3] > 0:
print("No")
elif ns4[2] > W//2:
print("No")
else:
print("Yes")
if H%2 == 0 and W%2 == 1:
if ns4[1]+ns4[3] > 0:
print("No")
elif ns4[2] > H//2:
print("No")
else:
print("Yes")
if H%2 == 1 and W%2 == 1:
if ns4[1]+ns4[3] > 1:
print("No")
elif ns4[2]+ns4[3] > H//2+W//2:
print("No")
else:
print("Yes")
| s020907799 | Accepted | 19 | 3,064 | 718 | H, W = map(int, input().split())
ns = {}
for i in range(H):
As = input()
for a in As:
if a not in ns: ns[a] = 0
ns[a] += 1
ns4 = [0, 0, 0, 0]
for a in ns:
n = ns[a]
ns4[n%4] += 1
#print(ns4)
if H%2 == 0 and W%2 == 0:
if ns4[1] > 0 or ns4[2] > 0 or ns4[3] > 0:
print("No")
else:
print("Yes")
if H%2 == 1 and W%2 == 0:
if ns4[1]+ns4[3] > 0:
print("No")
elif ns4[2] > W//2:
print("No")
else:
print("Yes")
if H%2 == 0 and W%2 == 1:
if ns4[1]+ns4[3] > 0:
print("No")
elif ns4[2] > H//2:
print("No")
else:
print("Yes")
if H%2 == 1 and W%2 == 1:
if ns4[1]+ns4[3] > 1:
print("No")
elif ns4[2]+ns4[3] > H//2+W//2:
print("No")
else:
print("Yes")
|
s178574265 | p02413 | u144068724 | 1,000 | 131,072 | Wrong Answer | 30 | 7,624 | 164 | Your task is to perform a simple table calculation. Write a program which reads the number of rows r, columns c and a table of r × c elements, and prints a new table, which includes the total sum for each row and column. | r,c = map(int,input().split())
ss = [[int(i) for i in input().split()]for j in range(r)]
for row in ss:
row.append(sum(row))
print(' '.join(map(str, row))) | s207612603 | Accepted | 30 | 7,732 | 282 | r,c = map(int,input().split())
ss = [[int(i) for i in input().split()]for j in range(r)]
sum_ss = [0 for i in range(c + 1)]
for row in ss:
row.append(sum(row))
sum_ss = [x+y for (x, y) in zip(sum_ss, row)]
print(' '.join(map(str, row)))
print(' '.join(map(str, sum_ss))) |
s592388460 | p03371 | u870575557 | 2,000 | 262,144 | Wrong Answer | 18 | 3,064 | 207 | "Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively. Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas. | a,b,c,x,y = list(map(int, input().split()))
s1 = a*x + b*y
s2 = c*(max(x,y)*2)
if x>y:
s3 = c*(y*2) + a*(x-y)
elif y>x:
s3 = c*(x*2) + b*(y-x)
else: s3 = 10**10
print(s1,s2,s3)
print(min(s1,s2,s3)) | s240439532 | Accepted | 19 | 3,064 | 191 | a,b,c,x,y = list(map(int, input().split()))
s1 = a*x + b*y
s2 = c*(max(x,y)*2)
if x>y:
s3 = c*(y*2) + a*(x-y)
elif y>x:
s3 = c*(x*2) + b*(y-x)
else: s3 = 10**10
print(min(s1,s2,s3)) |
s331151720 | p03455 | u619850971 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 90 | 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("even")
else:
print("odd") | s661254216 | Accepted | 17 | 2,940 | 90 | a, b = map(int, input().split())
if (a*b)%2 == 0:
print("Even")
else:
print("Odd") |
s107447364 | p03739 | u731467249 | 2,000 | 262,144 | Wrong Answer | 102 | 19,892 | 612 | 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()))
cntA, sumA = 0, 0
for i in range(N):
sumA += A[i]
if i % 2 == 0:
if sumA < 0:
cntA += abs(sumA) + 1
sumA += abs(sumA) + 1
else:
if sumA > 0:
cntA += abs(sumA) + 1
sumA -= abs(sumA) + 1
cntB, sumB = 0, 0
for i in range(N):
sumB += A[i]
if i % 2 != 0:
if sumB > 0:
cntB += abs(sumB) + 1
sumB += abs(sumB) + 1
else:
if sumB > 0:
cntB += abs(sumB) + 1
sumB -= abs(sumB) + 1
print(min(cntA, cntB)) | s314947078 | Accepted | 117 | 20,144 | 608 | N = int(input())
A = list(map(int, input().split()))
cntA, sumA = 0, 0
for i in range(N):
sumA += A[i]
if i % 2 == 0:
if sumA <= 0:
cntA += abs(sumA) + 1
sumA += abs(sumA) + 1
else:
if sumA >= 0:
cntA += abs(sumA) + 1
sumA -= abs(sumA) + 1
cntB, sumB = 0, 0
for i in range(N):
sumB += A[i]
if i % 2 != 0:
if sumB <= 0:
cntB += abs(sumB) + 1
sumB += abs(sumB) + 1
else:
if sumB >= 0:
cntB += abs(sumB) + 1
sumB -= abs(sumB) + 1
print(min(cntA, cntB)) |
s004498338 | p02600 | u634046173 | 2,000 | 1,048,576 | Wrong Answer | 33 | 9,176 | 294 | M-kun is a competitor in AtCoder, whose highest rating is X. In this site, a competitor is given a _kyu_ (class) according to his/her highest rating. For ratings from 400 through 1999, the following kyus are given: * From 400 through 599: 8-kyu * From 600 through 799: 7-kyu * From 800 through 999: 6-kyu * From 1000 through 1199: 5-kyu * From 1200 through 1399: 4-kyu * From 1400 through 1599: 3-kyu * From 1600 through 1799: 2-kyu * From 1800 through 1999: 1-kyu What kyu does M-kun have? | N = int(input())
if 400 <= N <= 599:
print(8)
elif 600 <= N <= 799:
print(7)
elif 800 <= N <= 999:
print(6)
elif 1000 <= N <= 1199:
print(5)
if 1200 <= N <= 1399:
print(4)
elif 1400 <= N <= 1599:
print(3)
elif 1600 <= N <= 1799:
print(2)
else:
print(1)
| s637130391 | Accepted | 33 | 9,176 | 296 | N = int(input())
if 400 <= N <= 599:
print(8)
elif 600 <= N <= 799:
print(7)
elif 800 <= N <= 999:
print(6)
elif 1000 <= N <= 1199:
print(5)
elif 1200 <= N <= 1399:
print(4)
elif 1400 <= N <= 1599:
print(3)
elif 1600 <= N <= 1799:
print(2)
else:
print(1)
|
s656785987 | p03457 | u832381404 | 2,000 | 262,144 | Wrong Answer | 373 | 11,636 | 353 | AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan. | n = int(input())
t = [0] * (n+1)
x = [0] * (n+1)
y = [0] * (n+1)
for i in range(1, n+1):
t[i], x[i], y[i] = map(int, input().split())
can = 1
for i in range(1, n+1):
tm = t[i] - t[i-1]
xm = abs(x[i] - x[i-1])
ym = abs(y[i] - y[i-1])
if((tm < xm+ym) or ((tm-xm+ym)%2)):
can = 0
break;
print("YES" if can else "NO")
| s506432719 | Accepted | 370 | 11,636 | 353 | n = int(input())
t = [0] * (n+1)
x = [0] * (n+1)
y = [0] * (n+1)
for i in range(1, n+1):
t[i], x[i], y[i] = map(int, input().split())
can = 1
for i in range(1, n+1):
tm = t[i] - t[i-1]
xm = abs(x[i] - x[i-1])
ym = abs(y[i] - y[i-1])
if((tm < xm+ym) or ((tm-xm+ym)%2)):
can = 0
break;
print("Yes" if can else "No")
|
s586195595 | p02678 | u490553751 | 2,000 | 1,048,576 | Wrong Answer | 22 | 9,008 | 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 the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists. | print("No") | s728634313 | Accepted | 783 | 35,676 | 666 | #template
def inputlist(): return [int(k) for k in input().split()]
#template
N,M = inputlist()
graph = [[] for _ in range(N+1)]
for i in range(M):
A,B = inputlist()
graph[A].append(B)
graph[B].append(A)
ans = [0]*(N+1)
seen = [False]*(N+1)
from collections import deque
que = deque()
que.append(1)
while len(que) != 0:
k = que.popleft()
seen[k] = True
for i in graph[k]:
if seen[i]:
continue
ans[i] = k
que.append(i)
seen[i] = True
ansa = ans[2:]
n = len(ansa)
say = "Yes"
for i in range(n):
if ansa[i] == 0:
print("No")
exit()
print(say)
for i in range(n):
print(ansa[i]) |
s735746205 | p02534 | u958820283 | 2,000 | 1,048,576 | Wrong Answer | 26 | 9,152 | 72 | You are given an integer K. Print the string obtained by repeating the string `ACL` K times and concatenating them. For example, if K = 3, print `ACLACLACL`. | k = int(input())
ans=""
for i in range(k):
ans=ans+str(k)
print(ans) | s544229225 | Accepted | 23 | 9,148 | 73 | k = int(input())
ans=""
for i in range(k):
ans=ans+"ACL"
print(ans) |
s984651527 | p02422 | u896065593 | 1,000 | 131,072 | Wrong Answer | 20 | 7,680 | 1,075 | Write a program which performs a sequence of commands to a given string $str$. The command is one of: * print a b: print from the a-th character to the b-th character of $str$ * reverse a b: reverse from the a-th character to the b-th character of $str$ * replace a b p: replace from the a-th character to the b-th character of $str$ with p Note that the indices of $str$ start with 0. |
str = list(input())
str = "".join(str)
p = int(input())
# p??????????????????????????????????????????
orderList = [0 for i in range(p)]
for i in range(0, p):
orderList[i] = list(input())
orderList[i] = "".join(orderList[i]).split()
if orderList[i][0] == "print":
print("{0}".format(str[int(orderList[i][1]) - 1:int(orderList[i][2])]))
elif orderList[i][0] == "reverse":
str = str[0:int(orderList[i][1]) - 1] + str[-len(str) + int(orderList[i][2]) - 1:-len(str) + int(orderList[i][1]) - 2:-1] + str[int(orderList[i][2]):len(str)]
elif orderList[i][0] == "replace":
replace_str = str[:int(orderList[i][1]) - 1]
for j in range(0, int(orderList[i][2])):
replace_str += orderList[i][3]
replace_str += str[int(orderList[i][2]):]
str = replace_str | s308209150 | Accepted | 20 | 7,748 | 885 |
str = list(input())
str = "".join(str)
p = int(input())
# p??????????????????????????????????????????
orderList = [0 for i in range(p)]
for i in range(0, p):
orderList[i] = list(input())
orderList[i] = "".join(orderList[i]).split()
if orderList[i][0] == "print":
print("{0}".format(str[int(orderList[i][1]):int(orderList[i][2]) + 1]))
elif orderList[i][0] == "reverse":
str = str[0:int(orderList[i][1])] + str[-len(str) + int(orderList[i][2]):-len(str) + int(orderList[i][1]) - 1:-1] + str[int(orderList[i][2]) + 1:]
elif orderList[i][0] == "replace":
str = str[:int(orderList[i][1])] + orderList[i][3] + str[int(orderList[i][2]) + 1:] |
s368181673 | p03090 | u879870653 | 2,000 | 1,048,576 | Wrong Answer | 23 | 3,612 | 171 | You are given an integer N. Build an undirected graph with N vertices with indices 1 to N that satisfies the following two conditions: * The graph is simple and connected. * There exists an integer S such that, for every vertex, the sum of the indices of the vertices adjacent to that vertex is S. It can be proved that at least one such graph exists under the constraints of this problem. | N = int(input())
for u in range(1, N) :
for v in range(u+1, N+1) :
if u + v == N + (N % 2 != 1) :
continue
else :
print(u, v)
| s509175631 | Accepted | 26 | 3,956 | 263 | N = int(input())
cnt = 0
A = []
for u in range(1, N) :
for v in range(u+1, N+1) :
if u + v == N + (N % 2 != 1) :
continue
else :
cnt += 1
A.append((u, v))
print(cnt)
for i in range(cnt) :
print(*A[i])
|
s981863908 | p03485 | u256868077 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 85 | You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer. | a,b=map(int,input().split())
if (a+b)%2==0:
print((a+b)/2)
else:
print((a+b+1)/2) | s942849923 | Accepted | 17 | 2,940 | 96 | a,b=map(int,input().split())
if (a+b)%2==0:
print(int((a+b)/2))
else:
print(int((a+b+1)/2))
|
s279189698 | p03493 | u722117999 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 21 | 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. | print(int(input())%2) | s270741115 | Accepted | 17 | 2,940 | 25 | print(input().count('1')) |
s825398876 | p03352 | u197078193 | 2,000 | 1,048,576 | Wrong Answer | 19 | 3,316 | 282 | You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2. | def beki(n,X):
# the maximum of the form n**k under X
x = 1
while x*n <= X:
x *= n
if x == n:
return 1
else:
return(x)
X = int(input())
a = 1
for n in range(X)[2:]:
b = beki(n,X)
print(n,b)
if b > a:
a = b
print(a)
| s836153525 | Accepted | 17 | 2,940 | 266 | def beki(n,X):
# the maximum of the form n**k under X
x = 1
while x*n <= X:
x *= n
if x == n:
return 1
else:
return(x)
X = int(input())
a = 1
for n in range(X)[2:]:
b = beki(n,X)
if b > a:
a = b
print(a) |
s311294287 | p03044 | u163320134 | 2,000 | 1,048,576 | Wrong Answer | 423 | 12,652 | 444 | We have a tree with N vertices numbered 1 to N. The i-th edge in the tree connects Vertex u_i and Vertex v_i, and its length is w_i. Your objective is to paint each vertex in the tree white or black (it is fine to paint all vertices the same color) so that the following condition is satisfied: * For any two vertices painted in the same color, the distance between them is an even number. Find a coloring of the vertices that satisfies the condition and print it. It can be proved that at least one such coloring exists under the constraints of this problem. | n=int(input())
ans=[0]*(n+1)
s=set()
u,v,w=map(int,input().split())
s.add(u)
s.add(v)
if w%2==0:
ans[u]=1
ans[v]=1
elif w%2==1:
ans[u]=1
ans[v]=0
for i in range(n-2):
u,v,w=map(int,input().split())
if u in s:
s.add(v)
if w%2==0:
ans[v]=ans[u]
else:
ans[v]=abs(ans[u]-1)
elif v in s:
s.add(u)
if w%2==0:
ans[u]=ans[v]
else:
ans[u]=abs(ans[v]-1)
for i in range(1,n+1):
print(ans[i]) | s343333496 | Accepted | 660 | 39,956 | 454 | import collections
n=int(input())
g=[[] for _ in range(n+1)]
for _ in range(n-1):
a,b,w=map(int,input().split())
g[a].append((b,w))
g[b].append((a,w))
q=collections.deque()
q.append(1)
checked=[0]*(n+1)
ans=[0]*(n+1)
ans[1]=0
while len(q)!=0:
v=q.popleft()
checked[v]=1
for u,w in g[v]:
if checked[u]==0:
if w%2==0:
ans[u]=ans[v]
else:
ans[u]=ans[v]^1
q.append(u)
for i in range(1,n+1):
print(ans[i]) |
s273401767 | p03778 | u143492911 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 92 | AtCoDeer the deer found two rectangles lying on the table, each with height 1 and width W. If we consider the surface of the desk as a two-dimensional plane, the first rectangle covers the vertical range of AtCoDeer will move the second rectangle horizontally so that it connects with the first rectangle. Find the minimum distance it needs to be moved. | w,a,b=map(int,input().split())
if a<=b+w or b<=a+w:
print(0)
else:
print(abs(a+w-b)) | s057734832 | Accepted | 18 | 2,940 | 127 | w,a,b=map(int,input().split())
if a<=b<=a+w:
print(0)
elif a<=b+w<=a+w:
print(0)
else:
print(max(b-(a+w),a-(b+w)))
|
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