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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s723014109
|
p03024
|
u492929439
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 104 |
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()
WinNum = S.count('o')
if(15-len(S) >= (8-WinNum)):
print("Yes")
else:
print("No")
|
s714619938
|
Accepted
| 17 | 2,940 | 105 |
S = input()
WinNum = S.count('o')
if(15-len(S) >= (8-WinNum)):
print("YES")
else:
print("NO")
|
s670621552
|
p03738
|
u558836062
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 113 |
You are given two positive integers A and B. Compare the magnitudes of these numbers.
|
A = int(input())
B = int(input())
if A > B:
print('GREATER')
if A == B:
print('EQUAL')
else:
print('LESS')
|
s141160838
|
Accepted
| 17 | 2,940 | 114 |
A = int(input())
B = int(input())
if A > B:
print("GREATER")
if A < B:
print("LESS")
if A == B:
print("EQUAL")
|
s168168327
|
p03998
|
u970308980
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 297 |
Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.
|
def abc():
players = dict()
players['a'] = list(input().rstrip())
players['b'] = list(input().rstrip())
players['c'] = list(input().rstrip())
p = 'a'
while True:
if len(players[p]) == 0:
print(p)
return
p = players[p].pop()
abc()
|
s686507078
|
Accepted
| 17 | 3,060 | 306 |
def abc():
players = dict()
players['a'] = list(input().rstrip())
players['b'] = list(input().rstrip())
players['c'] = list(input().rstrip())
p = 'a'
while True:
if len(players[p]) == 0:
print(p.upper())
return
p = players[p].pop(0)
abc()
|
s759998848
|
p02831
|
u464823755
| 2,000 | 1,048,576 |
Wrong Answer
| 24 | 9,096 | 72 |
Takahashi is organizing a party. At the party, each guest will receive one or more snack pieces. Takahashi predicts that the number of guests at this party will be A or B. Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted. We assume that a piece cannot be divided and distributed to multiple guests.
|
import math
a, b= map(int,input().split())
d= math.gcd(a,b)
print(a*b/d)
|
s406509707
|
Accepted
| 29 | 9,116 | 74 |
import math
a, b= map(int,input().split())
d= math.gcd(a,b)
print(a*b//d)
|
s851232789
|
p03712
|
u249895018
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 420 |
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.
|
import sys
if __name__ == '__main__':
count = 0
for line in sys.stdin:
D = []
if count == 0:
line = line.rstrip("\n").split(" ")
H = int(line[0])
W = int(line[1])
count+=1
else:
D.append(line.rstrip("\n"))
print("".join(["#"]*(W+1)))
for i in range(len(D)):
print("#"+D[i]+"#")
print("".join(["#"]*(W+1)))
|
s925602513
|
Accepted
| 17 | 3,064 | 416 |
import sys
if __name__ == '__main__':
count = 0
D = []
for line in sys.stdin:
if count == 0:
line = line.rstrip("\n").split(" ")
H = int(line[0])
W = int(line[1])
count+=1
else:
D.append(line.rstrip("\n"))
print("".join(["#"]*(W+2)))
for i in range(len(D)):
print("#"+D[i]+"#")
print("".join(["#"]*(W+2)))
|
s543622001
|
p02831
|
u094948011
| 2,000 | 1,048,576 |
Wrong Answer
| 67 | 3,064 | 399 |
Takahashi is organizing a party. At the party, each guest will receive one or more snack pieces. Takahashi predicts that the number of guests at this party will be A or B. Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted. We assume that a piece cannot be divided and distributed to multiple guests.
|
A,B = map(int,input().split())
p = 2
if A > B:
for y in range(100000):
t = (A * p) / B
o = (A * p)
if t % 2 == 0:
print(int(o))
break
else:
p += 1
else:
for r in range(100000):
t = (B * p) / A
o = (B * p)
if t % 2 == 0:
print(int(o))
break
else:
p += 1
|
s523693728
|
Accepted
| 54 | 3,064 | 398 |
A,B = map(int,input().split())
p = 1
if A > B:
for y in range(100000000):
e = (A * p) % B
o = (A * p)
if e == 0:
print(int(o))
break
else:
p += 1
else:
for r in range(100000000):
e = (B * p) % A
o = (B * p)
if e == 0:
print(int(o))
break
else:
p += 1
|
s884097258
|
p03999
|
u820351940
| 2,000 | 262,144 |
Wrong Answer
| 25 | 3,188 | 288 |
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.
|
(lambda nums: (lambda index, itertools=__import__("itertools"): print(sum([eval("".join((lambda t: [[t.insert(j, "+") for j in reversed(i)], t][1])(list(nums)))) for i in itertools.chain(*[itertools.combinations(index, x) for x in range(len(index) + 1)])])))(range(1, len(nums))))("1234")
|
s535813696
|
Accepted
| 31 | 3,064 | 289 |
(lambda nums: (lambda index, itertools=__import__("itertools"): print(sum([eval("".join((lambda t: [[t.insert(j, "+") for j in reversed(i)], t][1])(list(nums)))) for i in itertools.chain(*[itertools.combinations(index, x) for x in range(len(index) + 1)])])))(range(1, len(nums))))(input())
|
s680213219
|
p03471
|
u231218215
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 3,064 | 712 |
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.
|
#encoding: utf-8
import os, sys
import math
def read_items_in_line(cast=int):
return [cast(x) for x in sys.stdin.readline().strip(' \n').split(' ')]
def checK(remain_N, remain_Y, a, b, c):
if remain_N == 0 or remain_Y == 0:
print("{} {} {}".format(a, b, c))
exit()
N, Y = read_items_in_line()
for a_i in range(int(Y / 10000)):
remain_N = N - a_i
remain_Y = Y - a_i * 10000
for b_i in range(int(remain_Y / 5000)):
remain_N = N - b_i
remain_Y = Y - b_i * 5000
for c_i in range(int(remain_Y / 1000)):
remain_N = N - c_i
remain_Y = Y - c_i * 1000
checK(remain_N, remain_Y, a_i, b_i, c_i)
### unsat
print("-1 -1 -1")
|
s562071543
|
Accepted
| 1,746 | 3,064 | 1,005 |
#encoding: utf-8
import os, sys
import math
def read_items_in_line(cast=int):
return [cast(x) for x in sys.stdin.readline().strip(' \n').split(' ')]
def checK(remain_N, remain_Y, a, b, c):
if remain_N == 0 and remain_Y == 0:
print("{} {} {}".format(a, b, c))
exit(0)
N, Y = read_items_in_line()
for a_i in reversed(range(int(Y / 10000) + 1)):
remain_N = N - a_i
remain_Y = Y - a_i * 10000
if remain_N < 0 or remain_Y < 0:
continue
for b_i in reversed(range(int(remain_Y / 5000) + 1)):
remain_N = N - a_i - b_i
remain_Y = Y - a_i * 10000 - b_i * 5000
if remain_N < 0 or remain_Y < 0:
continue
if remain_N == remain_Y / 1000:
c_i = int(remain_Y / 1000)
remain_N = N - a_i - b_i - c_i
remain_Y = Y - a_i * 10000 - b_i * 5000 - c_i * 1000
checK(remain_N, remain_Y, a_i, b_i, c_i)
### unsat
print("-1 -1 -1")
|
s479425366
|
p04043
|
u912867658
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 120 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
s = list(input().split())
a = s.count("5")
b = s.count("7")
if a == 2 and b == 1:
print("Yes")
else:
print("No")
|
s296153071
|
Accepted
| 17 | 2,940 | 120 |
s = list(input().split())
a = s.count("5")
b = s.count("7")
if a == 2 and b == 1:
print("YES")
else:
print("NO")
|
s906586815
|
p03352
|
u681917640
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,060 | 240 |
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.
|
candidates = [ 1 ]
for b in range(2, int(1000**1/2)+1):
p = 0
while b**p <= 1000:
candidates.append(b**p)
p += 1
candidates.sort(reverse=True)
X = int(input())
for c in candidates:
if c <= X:
print(c)
exit()
|
s399567003
|
Accepted
| 18 | 2,940 | 227 |
candidates = set([1])
for b in range(2, int(1000**1/2)+1):
p = 2
while b**p <= 1000:
candidates.add(b**p)
p += 1
X = int(input())
for c in sorted(candidates, reverse=True):
if c <= X:
print(c)
exit()
|
s351354081
|
p03386
|
u276204978
| 2,000 | 262,144 |
Wrong Answer
| 2,107 | 18,256 | 110 |
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())
for i in range(A, B+1):
if i <= A + K or i <= B + K:
print(i)
|
s770899857
|
Accepted
| 17 | 3,060 | 280 |
A, B, K = map(int, input().split())
# if (A <= i and i <= A+K-1) or (B-K+1<= i and i <= B):
# print(i)
S=set()
for i in range(A, min(B+1, A+K)):
S.add(i)
for i in range(max(A, B-K+1), B+1):
S.add(i)
for k in sorted(S):
print(k)
|
s020438786
|
p04012
|
u007381711
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,096 | 190 |
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.
|
# -*- coding:utf-8 -*-
import sys
input = sys.stdin.readline
w = input()
w_set = set(w)
for i in w_set:
if w.count(i)%2==1:
print("No")
sys.exit()
else:
continue
print("Yes")
|
s732517226
|
Accepted
| 26 | 9,024 | 198 |
# -*- coding:utf-8 -*-
import sys
# input = sys.stdin.readline
w = list(input())
w_set = set(w)
for i in w_set:
if w.count(i)%2==1:
print("No")
sys.exit()
else:
continue
print("Yes")
|
s506206210
|
p00015
|
u917432951
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,588 | 233 |
A country has a budget of more than 81 trillion yen. We want to process such data, but conventional integer type which uses signed 32 bit can represent up to 2,147,483,647. Your task is to write a program which reads two integers (more than or equal to zero), and prints a sum of these integers. If given integers or the sum have more than 80 digits, print "overflow".
|
if __name__ == '__main__':
n = (int)(input())
for _ in range(n):
sNumber = (int)(input())
tNumber = (int)(input())
uNumber = sNumber+tNumber
print("over flow" if uNumber > 10**80 else uNumber)
|
s663132432
|
Accepted
| 20 | 5,584 | 233 |
if __name__ == '__main__':
n = (int)(input())
for _ in range(n):
sNumber = (int)(input())
tNumber = (int)(input())
uNumber = sNumber+tNumber
print("overflow" if uNumber >= 10**80 else uNumber)
|
s285525334
|
p03608
|
u477320129
| 2,000 | 262,144 |
Wrong Answer
| 1,162 | 39,620 | 1,204 |
There are N towns in the State of Atcoder, connected by M bidirectional roads. The i-th road connects Town A_i and B_i and has a length of C_i. Joisino is visiting R towns in the state, r_1,r_2,..,r_R (not necessarily in this order). She will fly to the first town she visits, and fly back from the last town she visits, but for the rest of the trip she will have to travel by road. If she visits the towns in the order that minimizes the distance traveled by road, what will that distance be?
|
#!/usr/bin/env python3
import sys
def solve(N: int, M: int, R: int, r: "List[int]", A: "List[int]", B: "List[int]", C: "List[int]"):
from scipy.sparse import coo_matrix
from scipy.sparse.csgraph import floyd_warshall
import numpy as np
# coo_matrix((data, (i, j)), [shape=(M, N)])
mat = floyd_warshall(coo_matrix((C, (A, B)), shape=(N+1, N+1), dtype=np.int32).tocsr())
# Generated by 1.1.7.1 https://github.com/kyuridenamida/atcoder-tools
def main():
def iterate_tokens():
for line in sys.stdin:
for word in line.split():
yield word
tokens = iterate_tokens()
N = int(next(tokens)) # type: int
M = int(next(tokens)) # type: int
R = int(next(tokens)) # type: int
r = [int(next(tokens)) for _ in range(R)] # type: "List[int]"
A = [int()] * (M) # type: "List[int]"
B = [int()] * (M) # type: "List[int]"
C = [int()] * (M) # type: "List[int]"
for i in range(M):
A[i] = int(next(tokens))
B[i] = int(next(tokens))
C[i] = int(next(tokens))
print(solve(N, M, R, r, A, B, C))
def test():
import doctest
doctest.testmod()
if __name__ == '__main__':
#test()
main()
|
s827237071
|
Accepted
| 1,310 | 39,564 | 1,373 |
#!/usr/bin/env python3
import sys
def solve(N: int, M: int, R: int, r: "List[int]", A: "List[int]", B: "List[int]", C: "List[int]"):
from scipy.sparse import coo_matrix
from scipy.sparse.csgraph import floyd_warshall
from itertools import permutations
from functools import reduce
# coo_matrix((data, (i, j)), [shape=(M, N)])
mat = floyd_warshall(coo_matrix((C, (A, B)), shape=(N+1, N+1)).tocsr(), directed=False)
f = lambda a, b: (a[0]+mat[a[1]][b], b)
return int(min(reduce(f, q, (0, s))[0] for s, *q in permutations(r)))
# Generated by 1.1.7.1 https://github.com/kyuridenamida/atcoder-tools
def main():
def iterate_tokens():
for line in sys.stdin:
for word in line.split():
yield word
tokens = iterate_tokens()
N = int(next(tokens)) # type: int
M = int(next(tokens)) # type: int
R = int(next(tokens)) # type: int
r = [int(next(tokens)) for _ in range(R)] # type: "List[int]"
A = [int()] * (M) # type: "List[int]"
B = [int()] * (M) # type: "List[int]"
C = [int()] * (M) # type: "List[int]"
for i in range(M):
A[i] = int(next(tokens))
B[i] = int(next(tokens))
C[i] = int(next(tokens))
print(solve(N, M, R, r, A, B, C))
def test():
import doctest
doctest.testmod()
if __name__ == '__main__':
#test()
main()
|
s938175597
|
p03696
|
u259190728
| 2,000 | 262,144 |
Wrong Answer
| 25 | 9,180 | 190 |
You are given a string S of length N consisting of `(` and `)`. Your task is to insert some number of `(` and `)` into S to obtain a _correct bracket sequence_. Here, a correct bracket sequence is defined as follows: * `()` is a correct bracket sequence. * If X is a correct bracket sequence, the concatenation of `(`, X and `)` in this order is also a correct bracket sequence. * If X and Y are correct bracket sequences, the concatenation of X and Y in this order is also a correct bracket sequence. * Every correct bracket sequence can be derived from the rules above. Find the shortest correct bracket sequence that can be obtained. If there is more than one such sequence, find the lexicographically smallest one.
|
n=int(input())
s=input()
l=[]
for i in s:
if i==')':
if len(l)>0:
if l[-1]=='(':
l.pop()
else:
l.append(i)
else:
l.append(i)
print(len(l)*'(' + s)
|
s655311482
|
Accepted
| 29 | 9,156 | 239 |
n=int(input())
s=input()
l=[]
for i in s:
if i==')':
if len(l)>0:
if l[-1]=='(':
l.pop()
else:
l.append(i)
else:
l.append(i)
else:
l.append(i)
print(l.count(')')*'(' + s +l.count('(')*')')
|
s240611688
|
p03711
|
u021916304
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 351 |
Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group.
|
def ii():return int(input())
def iim():return map(int,input().split())
def iil():return list(map(int,input().split()))
def ism():return map(str,input().split())
def isl():return list(map(str,input().split()))
l1 = [1,3,5,7,8,10,12]
l2 = [4,6,9,11]
x,y = iim()
if (x in l1 and y in l2) or (x in l2 and y in l2):
print('Yes')
else:
print('No')
|
s237864550
|
Accepted
| 17 | 3,064 | 351 |
def ii():return int(input())
def iim():return map(int,input().split())
def iil():return list(map(int,input().split()))
def ism():return map(str,input().split())
def isl():return list(map(str,input().split()))
l1 = [1,3,5,7,8,10,12]
l2 = [4,6,9,11]
x,y = iim()
if (x in l1 and y in l1) or (x in l2 and y in l2):
print('Yes')
else:
print('No')
|
s463126680
|
p03611
|
u357949405
| 2,000 | 262,144 |
Wrong Answer
| 102 | 13,964 | 260 |
You are given an integer sequence of length N, a_1,a_2,...,a_N. For each 1≤i≤N, you have three choices: add 1 to a_i, subtract 1 from a_i or do nothing. After these operations, you select an integer X and count the number of i such that a_i=X. Maximize this count by making optimal choices.
|
N = int(input())
A = [int(i) for i in input().split()]
ave = sum(A) // len(A)
ans1 = len(list(filter(lambda x: x == ave or x+1 == ave or x-1 == ave, A)))
ans2 = len(list(filter(lambda x: x == ave+1 or x+1 == ave+1 or x-1 == ave+1, A)))
print(max(ans1, ans2))
|
s776440711
|
Accepted
| 297 | 31,060 | 440 |
from collections import Counter
N = int(input())
A = [int(i) for i in input().split()]
l = []
for i in A:
l.extend([i-1, i, i+1])
count = Counter(l)
cand = sorted(count.most_common(), key=lambda x: x[1], reverse=True)[:3]
cand = list(filter(lambda x: cand[2][1]<=x[1], count.most_common()))
ans = 0
for tpl in cand:
n = len(list(filter(lambda x: x == tpl[0] or x+1 == tpl[0] or x-1 == tpl[0], A)))
ans = max(ans, n)
print(ans)
|
s161121433
|
p03605
|
u071061942
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 82 |
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()
for i in N:
if i == 9:
print("Yes")
else:
print("No")
|
s774170701
|
Accepted
| 19 | 2,940 | 93 |
N = list(map(int,input()))
if N[0] == 9 or N[1] == 9:
print("Yes")
else:
print("No")
|
s820196400
|
p03251
|
u748562597
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,064 | 288 |
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
N, M, X, Y= map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
xx = max(x)
yy = min(y)
if (X > Y):
print('War')
elif (xx > yy):
print('War')
elif (yy < X):
print('War')
elif (Y < xx):
print('War')
else:
print('No war')
|
s342785135
|
Accepted
| 17 | 3,064 | 292 |
N, M, X, Y= map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
xx = max(x)
yy = min(y)
if (X >= Y):
print('War')
elif (xx >= yy):
print('War')
elif (yy <= X):
print('War')
elif (Y <= xx):
print('War')
else:
print('No War')
|
s616874130
|
p03485
|
u543954314
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 76 |
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 ==1:
a += 1
print((a+b)/2)
|
s419535127
|
Accepted
| 19 | 2,940 | 81 |
a, b = map(int, input().split())
if (a + b) % 2 ==1:
a += 1
print(int((a+b)/2))
|
s086933593
|
p02401
|
u731896389
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,604 | 290 |
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 1:
a,b,c = list(map(str,input().split()))
num_a=int(a)
num_c=int(c)
if b=="+":
print(num_a+num_c)
elif b=="-":
print(num_a-num_c)
elif b=="*":
print(num_a*num_c)
elif b=="/":
print(num_a/num_c)
elif b=="?":
break
|
s774214670
|
Accepted
| 20 | 7,692 | 291 |
while 1:
a,b,c = list(map(str,input().split()))
num_a=int(a)
num_c=int(c)
if b=="+":
print(num_a+num_c)
elif b=="-":
print(num_a-num_c)
elif b=="*":
print(num_a*num_c)
elif b=="/":
print(num_a//num_c)
elif b=="?":
break
|
s359857754
|
p03337
|
u575653048
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 55 |
You are given two integers A and B. Find the largest value among A+B, A-B and A \times B.
|
a, b = map(int, input().split())
print([a+b, a-b, a*b])
|
s685539627
|
Accepted
| 17 | 2,940 | 60 |
a, b = map(int, input().split())
print(max([a+b, a-b, a*b]))
|
s708193218
|
p03591
|
u997023236
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 67 |
Ringo is giving a present to Snuke. Ringo has found out that Snuke loves _yakiniku_ (a Japanese term meaning grilled meat. _yaki_ : grilled, _niku_ : meat). He supposes that Snuke likes grilled things starting with `YAKI` in Japanese, and does not like other things. You are given a string S representing the Japanese name of Ringo's present to Snuke. Determine whether S starts with `YAKI`.
|
S=input()
if S[0:3]=='YAKI':
print('Yes')
else:
print('No')
|
s452145937
|
Accepted
| 17 | 2,940 | 65 |
if input().startswith("YAKI"):
print("Yes")
else:
print("No")
|
s596117891
|
p03160
|
u858929490
| 2,000 | 1,048,576 |
Wrong Answer
| 124 | 13,900 | 300 |
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
n=int(input())
a=list(map(int,input().split()))
ans=0
i=0
while(i<n-2):
k=min(abs(a[i+2]-a[i]),abs(a[i+1]-a[i])+abs(a[i+2]-a[i+1]))
ans+=k
if abs(a[i+2]-a[i])<=abs(a[i+1]-a[i])+abs(a[i+2]-a[i+1]):
i+=2
else:
i+=1
if i==n-2:
ans+=abs(a[n-1]-a[n-2])
print(ans)
|
s424963886
|
Accepted
| 138 | 13,980 | 225 |
n=int(input())
a=list(map(int,input().split()))
dp=[0]*n
for i in range(1,n):
if i==1:
dp[i]=abs(a[i]-a[i-1])
else:
dp[i]=min(dp[i-1]+abs(a[i]-a[i-1]),dp[i-2]+abs(a[i]-a[i-2]))
print(dp[n-1])
|
s135327933
|
p02796
|
u232652798
| 2,000 | 1,048,576 |
Wrong Answer
| 2,105 | 93,980 | 334 |
In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep.
|
N = int(input())
L = []
for i in range(N):
x = [int(x) for x in input().split()]
L.append(x)
for i in range(len(L)):
L[i][0] = L[i][0] - L[i][1]
L[i][1] = L[i][0] + L[i][1]
S =[]
Q = sorted(L)
for i in range(0,len(Q)-1):
if Q[i+1][0] >=Q[i][1] :
S.append([Q[i][0],Q[i][1]])
print(S)
print(len(S)+1)
|
s342032014
|
Accepted
| 516 | 44,496 | 410 |
N = int(input())
XL = [list(map(int, input().split())) for _ in range(N)]
RL = [[0, 0] for _ in range(N)]
for i in range(N):
left = XL[i][0] - XL[i][1]
right = XL[i][0] + XL[i][1]
RL[i][0] = left
RL[i][1] = right
RL.sort(key=lambda x: x[1])
now_right = RL[0][1]
cnt = 0
for i in range(1, N):
if now_right <= RL[i][0]:
now_right = RL[i][1]
else:
cnt += 1
print(N - cnt)
|
s880084226
|
p02744
|
u377370946
| 2,000 | 1,048,576 |
Wrong Answer
| 59 | 3,316 | 541 |
In this problem, we only consider strings consisting of lowercase English letters. Strings s and t are said to be **isomorphic** when the following conditions are satisfied: * |s| = |t| holds. * For every pair i, j, one of the following holds: * s_i = s_j and t_i = t_j. * s_i \neq s_j and t_i \neq t_j. For example, `abcac` and `zyxzx` are isomorphic, while `abcac` and `ppppp` are not. A string s is said to be in **normal form** when the following condition is satisfied: * For every string t that is isomorphic to s, s \leq t holds. Here \leq denotes lexicographic comparison. For example, `abcac` is in normal form, but `zyxzx` is not since it is isomorphic to `abcac`, which is lexicographically smaller than `zyxzx`. You are given an integer N. Print all strings of length N that are in normal form, in lexicographically ascending order.
|
n = int(input())
tmp_list = list('a' * n)
num_list = [97] * n
target = n - 1
tmp_target=n-1
while tmp_target>0:
print("".join(list(map(chr, num_list))))
if num_list[target-1] >= num_list[target]:
num_list[target] += 1
else:
num_list[target]=97
tmp_target=target-1
while tmp_target>0:
if num_list[tmp_target-1] >=num_list[tmp_target]:
num_list[tmp_target]+=1
break
else:
num_list[tmp_target]=97
tmp_target-=1
|
s935679157
|
Accepted
| 368 | 4,340 | 584 |
n = int(input())
num_list = [97] * n
target = n - 1
tmp_target = n - 1
max_num = 97
if n==1:
print('a')
while tmp_target > 0:
print("".join(list(map(chr, num_list))))
if max(num_list[0:target:]) >= num_list[target]:
num_list[target] += 1
else:
num_list[target] = 97
tmp_target = target - 1
while tmp_target > 0:
if max(num_list[0:tmp_target:]) >= num_list[tmp_target]:
num_list[tmp_target] += 1
break
else:
num_list[tmp_target] = 97
tmp_target -= 1
|
s554467375
|
p03574
|
u804085889
| 2,000 | 262,144 |
Wrong Answer
| 27 | 3,064 | 420 |
You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process.
|
h,w = map(int, input().split())
S=[["." for _ in range(w+2)]]
S+=[["."]+[i for i in input()]+["."]for _ in range(h)]
S+=[["." for _ in range(w+2)]]
print(S)
for i in range(1, h+1):
l=""
for j in range(1, w+1):
if S[i][j]=="#":
l+="#"
else:
# l+=str(int(S[i][j]=="."))
l+=str(sum(S[i+n][j+m]=="#" for n in range(-1, 2, 1) for m in range(-1, 2, 1)))
print(l)
|
s070477753
|
Accepted
| 27 | 3,064 | 371 |
h,w = map(int, input().split())
S=[["." for _ in range(w+2)]]
S+=[["."]+[i for i in input()]+["."]for _ in range(h)]
S+=[["." for _ in range(w+2)]]
for i in range(1, h+1):
l=""
for j in range(1, w+1):
if S[i][j]=="#":
l+="#"
else:
l+=str(sum(S[i+n][j+m]=="#" for n in range(-1, 2, 1) for m in range(-1, 2, 1)))
print(l)
|
s420345459
|
p03779
|
u667458133
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 84 |
There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0. During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right. That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i. The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible. Find the earliest possible time to reach coordinate X.
|
X = int(input())
r = 1
num = 0
while num >= X:
num += r
r += 1
print(r-1)
|
s687523136
|
Accepted
| 29 | 2,940 | 84 |
X = int(input())
r = 1
num = 0
while num < X:
num += r
r += 1
print(r-1)
|
s824896430
|
p02646
|
u188226132
| 2,000 | 1,048,576 |
Wrong Answer
| 2,205 | 9,180 | 232 |
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())
for i in range(t):
if b<a:
a-=v
b-=w
else:
a+=v
b+=w
if a==b:
print("Yes")
exit()
print("No")
|
s159513124
|
Accepted
| 24 | 9,168 | 200 |
a,v=map(int,input().split())
b,w=map(int,input().split())
t=int(input())
if w>=v:
print("NO")
else:#v>w
dif=v-w
if dif*t>=abs(a-b):
print("YES")
else:
print("NO")
|
s908223208
|
p03698
|
u498620941
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 203 |
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
s = list(input())
size = len(s)
count = 0
ans = {}
for i in s:
if i in ans.keys() :
count += 1
else :
ans[i] = 1
pass
if count == 0 : print("YES")
else: print("NO")
|
s359948910
|
Accepted
| 17 | 3,060 | 203 |
s = list(input())
size = len(s)
count = 0
ans = {}
for i in s:
if i in ans.keys() :
count += 1
else :
ans[i] = 1
pass
if count == 0 : print("yes")
else: print("no")
|
s127912211
|
p03434
|
u826263061
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 169 |
We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.
|
n = int(input())
a = list(map(int, input().split()))
a.sort(reverse=True)
alice = sum([i for i in a if i%2 == 0])
bob = sum([i for i in a if i%2 == 1])
print(alice-bob)
|
s605274227
|
Accepted
| 18 | 3,060 | 173 |
n = int(input())
a = list(map(int, input().split()))
a.sort(reverse=True)
alice = sum([a[i] for i in range(0,n,2)])
bob = sum([a[i] for i in range(1,n,2)])
print(alice-bob)
|
s667615634
|
p03353
|
u729133443
| 2,000 | 1,048,576 |
Wrong Answer
| 34 | 4,388 | 86 |
You are given a string s. Among the **different** substrings of s, print the K-th lexicographically smallest one. A substring of s is a string obtained by taking out a non-empty contiguous part in s. For example, if s = `ababc`, `a`, `bab` and `ababc` are substrings of s, while `ac`, `z` and an empty string are not. Also, we say that substrings are different when they are different as strings. Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.
|
s=input();print(sorted(s[i:i+j]for i in range(len(s))for j in range(6))[int(input())])
|
s520002226
|
Accepted
| 36 | 4,584 | 78 |
s=input();print(sorted({s[I//6:I//6+I%6]for I in range(30000)})[int(input())])
|
s885479296
|
p03779
|
u127499732
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,188 | 47 |
There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0. During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right. That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i. The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible. Find the earliest possible time to reach coordinate X.
|
print(((-1+(1+8*int(input()))**0.5)//2+1))
|
s952323795
|
Accepted
| 25 | 2,940 | 223 |
def main():
x = int(input())
k = int(x ** .5)
i = 0
while True:
p = i * (i + 1) // 2
if p >= x:
print(i)
break
i += 1
if __name__ == '__main__':
main()
|
s797846148
|
p03564
|
u445511055
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 424 |
Square1001 has seen an electric bulletin board displaying the integer 1. He can perform the following operations A and B to change this value: * Operation A: The displayed value is doubled. * Operation B: The displayed value increases by K. Square1001 needs to perform these operations N times in total. Find the minimum possible value displayed in the board after N operations.
|
# -*- coding: utf-8 -*-
def main():
"""Function."""
n = int(input())
k = int(input())
flag = 0
temp = 1
for _ in range(n):
print(_, temp)
if flag == 0:
if temp < k:
temp = temp * 2
else:
temp += k
flag = 1
elif flag == 1:
temp += k
print(temp)
if __name__ == "__main__":
main()
|
s014436264
|
Accepted
| 20 | 3,060 | 401 |
# -*- coding: utf-8 -*-
def main():
"""Function."""
n = int(input())
k = int(input())
flag = 0
temp = 1
for _ in range(n):
if flag == 0:
if temp < k:
temp = temp * 2
else:
temp += k
flag = 1
elif flag == 1:
temp += k
print(temp)
if __name__ == "__main__":
main()
|
s262329641
|
p03502
|
u504715104
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 277 |
An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10. Given an integer N, determine whether it is a Harshad number.
|
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Wed Oct 18 17:04:29 2017
@author: goto
"""
y=0
x=input()
print(type(x))
x_int=[int(x)for i in list(x)]
for i in range(len(x_int)):
y+=x_int[i]
if int(x)/y==0:
print("Yes")
else:
print("No")
|
s645837968
|
Accepted
| 17 | 3,060 | 269 |
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Wed Oct 18 17:04:29 2017
@author: goto
"""
y=0
x=input()
x_int=[int(i)for i in list(str(x))]
for i in range(len(x_int)):
y+=x_int[i]
if int(x)%y==0:
print("Yes")
else:
print("No")
|
s249872940
|
p03971
|
u627417051
| 2,000 | 262,144 |
Wrong Answer
| 111 | 4,712 | 290 |
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these. Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests. * A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B. * An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students. A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these. Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
|
N, A, B = list(map(int, input().split()))
S = list(input())
a = 0
b = 0
for i in range(N):
if S[i] == "a":
if a < A + B:
print("Yes")
else:
print("No")
a += 1
elif S[i] == "b":
if a + b < A + B and b < B:
print("Yes")
else:
print("No")
b += 1
else:
print("No")
|
s648227881
|
Accepted
| 116 | 4,712 | 322 |
N, A, B = list(map(int, input().split()))
S = list(input())
a = 0
b = 0
ps = 0
for i in range(N):
if S[i] == "a":
if ps < A + B:
print("Yes")
ps += 1
else:
print("No")
a += 1
elif S[i] == "b":
if ps < A + B and b < B:
print("Yes")
b += 1
ps += 1
else:
print("No")
else:
print("No")
|
s059574296
|
p02741
|
u649423802
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 124 |
Print the K-th element of the following sequence of length 32: 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51
|
K=int(input())
a=[1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51]
a[K-1]
|
s062026842
|
Accepted
| 17 | 3,060 | 131 |
K=int(input())
a=[1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51]
print(a[K-1])
|
s034458629
|
p02612
|
u508061226
| 2,000 | 1,048,576 |
Wrong Answer
| 26 | 9,052 | 33 |
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)
|
s405635097
|
Accepted
| 30 | 9,076 | 89 |
n = int(input())
amari = n % 1000
if amari != 0:
print(1000 - amari)
else:
print(0)
|
s656214174
|
p00773
|
u494573739
| 8,000 | 131,072 |
Wrong Answer
| 20 | 5,460 | 1 |
VAT (value-added tax) is a tax imposed at a certain rate proportional to the sale price. Our store uses the following rules to calculate the after-tax prices. * When the VAT rate is _x_ %, for an item with the before-tax price of _p_ yen, its after-tax price of the item is _p_ (100+ _x_ ) / 100 yen, fractions rounded off. * The total after-tax price of multiple items paid at once is the sum of after-tax prices of the items. The VAT rate is changed quite often. Our accountant has become aware that "different pairs of items that had the same total after-tax price may have different total after-tax prices after VAT rate changes." For example, when the VAT rate rises from 5% to 8%, a pair of items that had the total after-tax prices of 105 yen before can now have after-tax prices either of 107, 108, or 109 yen, as shown in the table below. Before-tax prices of two items| After-tax price with 5% VAT| After-tax price with 8% VAT ---|---|--- 20, 80| 21 + 84 = 105| 21 + 86 = 107 2, 99| 2 + 103 = 105| 2 + 106 = 108 13, 88| 13 + 92 = 105| 14 + 95 = 109 Our accountant is examining effects of VAT-rate changes on after-tax prices. You are asked to write a program that calculates the possible maximum total after-tax price of two items with the new VAT rate, knowing their total after- tax price before the VAT rate change.
|
s120049766
|
Accepted
| 310 | 5,668 | 427 |
from math import floor
while(True):
x,y,s=map(int,input().split())
if (x,y,s)==(0,0,0):
quit()
ans=0
for i in range(1,s):
j=s-floor(i*(100+x)/100)
if j<1:
continue
j=floor((j+1)*(100/(100+x)))
for k in [j-1,j]:
if floor(i*(100+x)/100)+floor(k*(100+x)/100)==s:
ans=max(ans,floor(i*(100+y)/100)+floor(k*(100+y)/100))
print(ans)
|
|
s290171928
|
p02260
|
u496045719
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,668 | 572 |
Write a program of the Selection Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: SelectionSort(A) 1 for i = 0 to A.length-1 2 mini = i 3 for j = i to A.length-1 4 if A[j] < A[mini] 5 mini = j 6 swap A[i] and A[mini] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 6 of the pseudocode in the case where i ≠ mini.
|
def main():
element_num = int(input())
elements = [int(s) for s in input().split()]
swap_count = 0
for i in range(0, element_num - 1):
max_val = elements[i + 1]
max_index = i + 1
for j in range(i + 1, element_num):
if max_val < elements[j]:
max_val = elements[j]
max_index = j
if elements[i] > elements[max_index]:
elements[i], elements[max_index] = elements[max_index], elements[i]
swap_count += 1
print(' '.join(list(map(str, elements))))
print(swap_count)
return 0
if __name__ == '__main__':
main()
|
s488871897
|
Accepted
| 40 | 7,740 | 572 |
def main():
element_num = int(input())
elements = [int(s) for s in input().split()]
swap_count = 0
for i in range(0, element_num - 1):
min_val = elements[i + 1]
min_index = i + 1
for j in range(i + 1, element_num):
if elements[j] < min_val:
min_val = elements[j]
min_index = j
if elements[i] > elements[min_index]:
elements[i], elements[min_index] = elements[min_index], elements[i]
swap_count += 1
print(' '.join(list(map(str, elements))))
print(swap_count)
return 0
if __name__ == '__main__':
main()
|
s988154175
|
p03635
|
u733738237
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 56 |
In _K-city_ , there are n streets running east-west, and m streets running north-south. Each street running east-west and each street running north-south cross each other. We will call the smallest area that is surrounded by four streets a block. How many blocks there are in K-city?
|
s = input()
l= len(s)-2
x=s[0]
y=s[-1]
print(x+str(l)+y)
|
s333853248
|
Accepted
| 17 | 2,940 | 47 |
n,m=map(int,input().split())
print((n-1)*(m-1))
|
s423501877
|
p03860
|
u813098295
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 41 |
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.
|
s = input().split()
print(s[0]+s[1]+s[2])
|
s111997500
|
Accepted
| 17 | 2,940 | 50 |
s = input().split()
print(s[0][0]+s[1][0]+s[2][0])
|
s296709088
|
p03854
|
u281610856
| 2,000 | 262,144 |
Wrong Answer
| 68 | 3,188 | 288 |
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
S = input()
while True:
if len(S) == 0:
print("Yes")
break
elif S[-5:] == 'dream' or S[-5:] == 'erase':
S = S[:-5]
elif S[-6:] == 'eraser':
S = S[:-6]
elif S[-7:] == 'dreamer':
S = S[:-7]
else:
print("No")
break
|
s642440362
|
Accepted
| 68 | 3,188 | 288 |
S = input()
while True:
if len(S) == 0:
print("YES")
break
elif S[-5:] == 'dream' or S[-5:] == 'erase':
S = S[:-5]
elif S[-6:] == 'eraser':
S = S[:-6]
elif S[-7:] == 'dreamer':
S = S[:-7]
else:
print("NO")
break
|
s327415551
|
p02608
|
u316603606
| 2,000 | 1,048,576 |
Wrong Answer
| 23 | 9,052 | 147 |
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).
|
N = int (input ())
z = 1
l = []
while True:
l.append ((2+z)**2-(1+z+z))
if N <= (2+z)**2-(1+z+z):
break
else:
print (z)
z += 1
|
s136335535
|
Accepted
| 540 | 9,164 | 207 |
N = int ( input())
ans = [0]*N
for x in range (1,101):
for y in range (1,101):
for z in range (1,101):
m = (x+y+z)**2-(x*y+y*z+z*x)
if m <= N:
ans[m-1] += 1
for i in ans:
print(i)
|
s553583654
|
p04011
|
u766407523
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 130 |
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.
|
def inpn(): return int(input())
N, K, X, Y = inpn(), inpn(), inpn(), inpn()
if K <= N:
print(K*X)
else:
print(N*X+(N-K)*Y)
|
s968121351
|
Accepted
| 17 | 2,940 | 131 |
def inpn(): return int(input())
N, K, X, Y = inpn(), inpn(), inpn(), inpn()
if N <= K:
print(N*X)
else:
print(K*X+(N-K)*Y)
|
s594429565
|
p02659
|
u552145906
| 2,000 | 1,048,576 |
Wrong Answer
| 20 | 9,160 | 94 |
Compute A \times B, truncate its fractional part, and print the result as an integer.
|
A, B = input().split()
A = int(A)
B = float(B)
print(A)
print(B)
C = A * B
C = int(C)
print(C)
|
s140866757
|
Accepted
| 28 | 10,044 | 102 |
import decimal
A, B = input().split()
A = int(A)
B = decimal.Decimal(B)
C = A * B
C = int(C)
print(C)
|
s228478929
|
p03636
|
u737321654
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 86 |
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()
initial = s[0]
final = s[len(s) - 1]
print(initial + str(len(s)) + final)
|
s618259377
|
Accepted
| 17 | 2,940 | 89 |
s = input()
initial = s[0]
final = s[len(s) - 1]
print(initial + str(len(s)-2) + final)
|
s095005971
|
p03151
|
u050024609
| 2,000 | 1,048,576 |
Wrong Answer
| 121 | 20,912 | 551 |
A university student, Takahashi, has to take N examinations and pass all of them. Currently, his _readiness_ for the i-th examination is A_{i}, and according to his investigation, it is known that he needs readiness of at least B_{i} in order to pass the i-th examination. Takahashi thinks that he may not be able to pass all the examinations, and he has decided to ask a magician, Aoki, to change the readiness for as few examinations as possible so that he can pass all of them, while not changing the total readiness. For Takahashi, find the minimum possible number of indices i such that A_i and C_i are different, for a sequence C_1, C_2, ..., C_{N} that satisfies the following conditions: * The sum of the sequence A_1, A_2, ..., A_{N} and the sum of the sequence C_1, C_2, ..., C_{N} are equal. * For every i, B_i \leq C_i holds. If such a sequence C_1, C_2, ..., C_{N} cannot be constructed, print -1.
|
N=input()
A=list(map(int, input().split()))
B=list(map(int, input().split()))
if sum(A) < sum(B) :
print(-1)
else :
diff=[A[i] - B[i] for i in range(len(A))]
pos=sorted([x for x in diff if x > 0])
neg=[x for x in diff if x < 0]
nes=sum(neg)
if nes >= 0 :
print(0)
else:
print(diff)
items = 0
total_pos = 0
for p in reversed(pos):
items += 1
total_pos += p
if(total_pos + nes >= 0) :
print(items + len(neg))
break
|
s248118796
|
Accepted
| 111 | 18,356 | 531 |
N=input()
A=list(map(int, input().split()))
B=list(map(int, input().split()))
if sum(A) < sum(B) :
print(-1)
else :
diff=[A[i] - B[i] for i in range(len(A))]
pos=sorted([x for x in diff if x > 0])
neg=[x for x in diff if x < 0]
nes=sum(neg)
if nes >= 0 :
print(0)
else:
items = 0
total_pos = 0
for p in reversed(pos):
items += 1
total_pos += p
if(total_pos + nes >= 0) :
print(items + len(neg))
break
|
s552886780
|
p02420
|
u629874472
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,592 | 269 |
Your task is to shuffle a deck of n cards, each of which is marked by a alphabetical letter. A single shuffle action takes out h cards from the bottom of the deck and moves them to the top of the deck. The deck of cards is represented by a string as follows. abcdeefab The first character and the last character correspond to the card located at the bottom of the deck and the card on the top of the deck respectively. For example, a shuffle with h = 4 to the above deck, moves the first 4 characters "abcd" to the end of the remaining characters "eefab", and generates the following deck: eefababcd You can repeat such shuffle operations. Write a program which reads a deck (a string) and a sequence of h, and prints the final state (a string).
|
char = list(input())
m = int(input())
for i in range(m):
sl = input()
if sl == '-':
break
else:
sl = int(sl)
a = char[0:sl]
b = char.extend(a)
del char[0:sl] #del
print("".join(char))
|
s517042666
|
Accepted
| 20 | 5,592 | 332 |
while True:
char = input()
if char == '-':
break
else:
char = list(char)
m = int(input())
for i in range(m):
sl = int(input())
a = char[0:sl]
b = char.extend(a)
del char[0:sl] #del
print("".join(char))
|
s658233818
|
p03854
|
u086624329
| 2,000 | 262,144 |
Wrong Answer
| 69 | 3,188 | 453 |
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
s=input()
t1='maerd'
t2='remaerd'
t3='esare'
t4='resare'
t=[t1,t2,t3,t4]
s=s[::-1]
ans=0
while True:
if s.startswith(t1):
s=s[5::]
elif s.startswith(t2):
s=s[7::]
elif s.startswith(t3):
s=s[5::]
elif s.startswith(t4):
s=s[6::]
elif len(s)==0:
break
else:
print('NO')
ans=1
break
if ans==0:
print('Yes')
|
s041885009
|
Accepted
| 69 | 3,188 | 462 |
s=input()
t1='maerd'
t2='remaerd'
t3='esare'
t4='resare'
t=[t1,t2,t3,t4]
s=s[::-1]
ans=0
while True:
if s.startswith(t1):
s=s[5::]
elif s.startswith(t2):
s=s[7::]
elif s.startswith(t3):
s=s[5::]
elif s.startswith(t4):
s=s[6::]
elif len(s)==0:
break
else:
print('NO')
ans=1
break
if ans==0:
print('YES')
|
s477363618
|
p02417
|
u088337682
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,568 | 198 |
Write a program which counts and reports the number of each alphabetical letter. Ignore the case of characters.
|
ans = [0 for i in range(23)]
S = list(map(ord,input().upper()))
S = [e for e in S if 64<e<91]
print(S)
for i in S:
ans[i-65]+=1
for i in range(23):
print("{} : {}".format(chr(97+i),ans[i]))
|
s480363107
|
Accepted
| 20 | 5,560 | 334 |
import sys
input_str = sys.stdin.read()
table = [0]*26
letters = "abcdefghijklmnopqrstuvwxyz"
for A in input_str:
index = 0
for B in letters:
if A == B or A == B.upper():
table[index] += 1
break
index += 1
for i in range(len(letters)):
print("{} : {}".format(letters[i],table[i]))
|
s516065690
|
p02612
|
u704470185
| 2,000 | 1,048,576 |
Wrong Answer
| 31 | 9,132 | 34 |
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)
|
s570396870
|
Accepted
| 27 | 9,112 | 81 |
N = int(input())
a = N % 1000
if a == 0:
print(0)
else:
print(1000 - a)
|
s195117138
|
p03386
|
u216631280
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 241 |
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())
li = []
for i in range(a, a + k):
if b < i:
break
li.append(i)
for i in range(b - k, b + 1):
if i < a:
continue
li.append(i)
li = list(set(li))
for num in li:
print(num)
|
s802226938
|
Accepted
| 17 | 3,060 | 256 |
a, b, k = map(int, input().split())
li = []
for i in range(a, a + k):
if b < i:
break
li.append(i)
for i in range(b - k + 1, b + 1):
if i < a:
continue
li.append(i)
li = list(set(li))
li.sort()
for num in li:
print(num)
|
s070789225
|
p03583
|
u814986259
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 29 |
You are given an integer N. Find a triple of positive integers h, n and w such that 4/N = 1/h + 1/n + 1/w. If there are multiple solutions, any of them will be accepted.
|
N=int(input())
print(2*N,N,N)
|
s547322629
|
Accepted
| 1,589 | 3,060 | 347 |
N = int(input())
for h in range(1, 3501):
for n in range(1, 3501):
# 4/N - 1/h - 1/n = 1/w
# (4hn - Nn - Nh) / nhN = 1/w
# w = nhN / (4hn - Nn - Nh)
if (4*h*n - N*(n + h)) > 0 and (n*h*N) % (4*h*n - N*(n + h)) == 0:
w = (n*h*N) // (4*h*n - N*(n + h))
print(h, n, w)
exit(0)
|
s289387361
|
p03359
|
u453526259
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 90 |
In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year. For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year. In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For example, 5-5 is Takahashi. How many days from 2018-1-1 through 2018-a-b are Takahashi?
|
a, b = map(int, input().split())
if b <= a:
ans = a - 1
else:
ans = a
print(ans)
|
s174469027
|
Accepted
| 17 | 2,940 | 89 |
a, b = map(int, input().split())
if b < a:
ans = a - 1
else:
ans = a
print(ans)
|
s179808274
|
p03729
|
u599547273
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 87 |
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("Yes" if a[-1] == b[0] and b[-1] == c[0] else "No")
|
s695405748
|
Accepted
| 17 | 2,940 | 87 |
a, b, c = input().split(" ")
print("YES" if a[-1] == b[0] and b[-1] == c[0] else "NO")
|
s032561240
|
p02255
|
u351208175
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,720 | 280 |
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.
|
def insertionSort(a,n):
for i in range(n):
v = a[i]
j = i-1
while j>=0 and a[j]>v:
a[j+1] = a[j]
j -= 1
a[j+1] = v
print(a)
return a
n = int(input())
a = [int(i) for i in input().split()]
insertionSort(a,n)
|
s203910636
|
Accepted
| 30 | 7,660 | 299 |
def insertionSort(a,n):
for i in range(n):
v = a[i]
j = i-1
while j>=0 and a[j]>v:
a[j+1] = a[j]
j -= 1
a[j+1] = v
print(" ".join(map(str,a)))
return a
n = int(input())
a = [int(i) for i in input().split()]
insertionSort(a,n)
|
s330058656
|
p02646
|
u325132311
| 2,000 | 1,048,576 |
Wrong Answer
| 24 | 9,208 | 304 |
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.
|
num_input = [input() for i in range(3)]
a = int(num_input[0].split()[0])
v = int(num_input[0].split()[1])
b = int(num_input[1].split()[0])
w = int(num_input[1].split()[1])
t = int(num_input[2].split()[0])
if v-w<=0:
ans="no"
elif int((b-a)/(v-w))<=t:
ans="yes"
else:
ans="no"
print(ans)
|
s710642399
|
Accepted
| 23 | 9,152 | 336 |
num_input = [input() for i in range(3)]
a = int(num_input[0].split()[0])
v = int(num_input[0].split()[1])
b = int(num_input[1].split()[0])
w = int(num_input[1].split()[1])
t = int(num_input[2].split()[0])
hoge=b-a
if hoge<0:
hoge=hoge*-1
if v-w<=0:
ans="NO"
elif t*(v-w)>=hoge:
ans="YES"
else:
ans="NO"
print(ans)
|
s117234672
|
p03729
|
u391731808
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 66 |
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("YNeos"[A[-1]!=B[0]or B[-1]!=C[0]::2])
|
s692236622
|
Accepted
| 17 | 2,940 | 66 |
A,B,C=input().split()
print("YNEOS"[A[-1]!=B[0]or B[-1]!=C[0]::2])
|
s484583953
|
p03377
|
u434329006
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 86 |
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())
if (x - b) >= a:
print("YES")
else:
print("NO")
|
s305356649
|
Accepted
| 17 | 2,940 | 118 |
a, b, x = map(int,input().split())
if a > x:
print("NO")
elif (a + b) < x:
print("NO")
else:
print("YES")
|
s019574670
|
p03457
|
u710921979
| 2,000 | 262,144 |
Wrong Answer
| 318 | 3,060 | 144 |
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())
for i in range(N):
t,x,y=map(int,input().split())
if x+y>t or (x+y+t)%2:
print("NO")
exit()
print("YES")
|
s590639026
|
Accepted
| 325 | 3,060 | 146 |
N=int(input())
for i in range(N):
t,x,y=map(int,input().split())
if x+y>t or (t+x+y)%2!=0:
print('No')
quit()
print('Yes')
|
s655852623
|
p02645
|
u023632682
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 9,136 | 20 |
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.
|
S=str(input())
S[:3]
|
s757097033
|
Accepted
| 23 | 9,028 | 27 |
S=str(input())
print(S[:3])
|
s902575118
|
p03338
|
u919905831
| 2,000 | 1,048,576 |
Wrong Answer
| 21 | 3,316 | 196 |
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.
|
num = int(input())
li = list(input())
i = int(num/2)
while(i>num):
fi = set(li[:i])
se = set(li[i:])
ans = len(list(fi & se))
if i < ans:
i = ans
else:
i = i+1
print(i)
|
s921487409
|
Accepted
| 17 | 3,060 | 177 |
num = int(input())
li = list(input())
ans = 0
for i in range(num):
fi = set(li[:i])
se = set(li[i:])
hoge = len(list(fi & se))
if hoge > ans:
ans = hoge
print(ans)
|
s052826176
|
p02396
|
u874395007
| 1,000 | 131,072 |
Wrong Answer
| 120 | 5,608 | 135 |
In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem.
|
repeat = True
i = 0
while repeat:
i += 1
num = int(input())
if num == 0:
repeat = False
print('Case {i}: num')
|
s703268845
|
Accepted
| 130 | 5,608 | 113 |
i = 0
while True:
i += 1
num = int(input())
if num == 0:
break
print(f'Case {i}: {num}')
|
s170616297
|
p03486
|
u837286475
| 2,000 | 262,144 |
Wrong Answer
| 23 | 3,700 | 302 |
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
|
from functools import reduce
ss = input()
st = input()
ls = [c for c in ss]
lt = [c for c in st]
#print(ls)
ls.sort()
lt.sort(reverse = True)
ans_s = reduce( lambda x,y: x+y, ls,'' )
ans_t = reduce(lambda x, y: x+y, lt)
print((ans_s, ans_t))
#ans_s.
print( 'Yes' if ans_s < ans_t else 'No' )
|
s668299516
|
Accepted
| 22 | 3,572 | 303 |
from functools import reduce
ss = input()
st = input()
ls = [c for c in ss]
lt = [c for c in st]
#print(ls)
ls.sort()
lt.sort(reverse = True)
ans_s = reduce( lambda x,y: x+y, ls,'' )
ans_t = reduce(lambda x, y: x+y, lt)
#print((ans_s, ans_t))
#ans_s.
print( 'Yes' if ans_s < ans_t else 'No' )
|
s631846659
|
p03644
|
u777283665
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 172 |
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
n = int(input())
alist = [2 ** i for i in range(1, 10) if 2 ** i <= 100]
for i in range(n, 0, -1):
if i == 1:
print(1)
elif i in alist:
print(i)
|
s389120549
|
Accepted
| 17 | 2,940 | 60 |
n = int(input())
x = 1
while x * 2 <= n:
x *= 2
print(x)
|
s137352597
|
p03149
|
u319805917
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 104 |
You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974".
|
A=list(input().split())
if (1 in A)and(9 in A)and(7 in A)and(4 in A):
print("YES")
else:
print("NO")
|
s967249750
|
Accepted
| 20 | 3,060 | 108 |
A=list(input().split())
A=sorted(A)
A=[int(s)for s in A]
if A==[1,4,7,9]:
print("YES")
else:
print("NO")
|
s349429555
|
p04043
|
u395816772
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 114 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
A = list(input().split())
a = A.count(5)
b = A.count(7)
if a == 2 and b == 1:
print('YES')
else:
print('NO')
|
s000228713
|
Accepted
| 17 | 2,940 | 119 |
A = list(input().split())
a = A.count('5')
b = A.count('7')
if a == 2 and b == 1:
print('YES')
else:
print('NO')
|
s285232418
|
p02255
|
u481175672
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,596 | 228 |
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.
|
n=int(input())
a=input().rstrip().split()
list(map(int,a))
for i in range(1,n):
v=a[i]
j=i-1
while(j>=0 and a[j]>v):
a[j+1]=a[j]
j-=1
a[j+1]=v
for k in range(n):
print(a[k],end=" ")
print()
|
s155674857
|
Accepted
| 20 | 5,984 | 193 |
n = int(input())
*A, = map(int, input().split())
for i in range(n):
v = A[i]
j = i - 1
while j >= 0 and A[j] > v:
A[j+1] = A[j]
j -= 1
A[j+1] = v
print(*A)
|
s335373313
|
p03720
|
u273326224
| 2,000 | 262,144 |
Wrong Answer
| 151 | 12,632 | 220 |
There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
|
import numpy as np
n, m = map(int, input().split(' '))
array = np.zeros(n)
for i in range(m):
a, b = map(int, input().split(' '))
array[a-1] += 1
array[b-1] += 1
for number in array:
print(number)
|
s569201773
|
Accepted
| 153 | 12,396 | 225 |
import numpy as np
n, m = map(int, input().split(' '))
array = np.zeros(n)
for i in range(m):
a, b = map(int, input().split(' '))
array[a-1] += 1
array[b-1] += 1
for number in array:
print(int(number))
|
s223646574
|
p03478
|
u350093546
| 2,000 | 262,144 |
Wrong Answer
| 41 | 9,132 | 154 |
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())
ans=0
for i in range(1,n+1):
i=str(i)
cnt=0
for j in i:
j=int(j)
cnt+=j
if a<=cnt<=b:
ans+=1
print(ans)
|
s246843377
|
Accepted
| 39 | 9,184 | 155 |
n,a,b=map(int,input().split())
ans=0
for i in range(1,n+1):
k=str(i)
cnt=0
for j in k:
j=int(j)
cnt+=j
if a<=cnt<=b:
ans+=i
print(ans)
|
s208122290
|
p03024
|
u602715823
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 85 |
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()
no = s.count('o')
r = 15 - len(s)
print('Yes' if r + no >= 8 else 'No')
|
s265663854
|
Accepted
| 18 | 2,940 | 85 |
s = input()
no = s.count('o')
r = 15 - len(s)
print('YES' if r + no >= 8 else 'NO')
|
s688271730
|
p03251
|
u741397536
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,060 | 223 |
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_arr = list(map(int, input().split()))
y_arr = list(map(int, input().split()))
x_arr.append(x)
y_arr.append(y)
if max(x_arr) < min(y_arr):
print("No war")
else:
print("War")
|
s451922641
|
Accepted
| 18 | 3,060 | 223 |
n, m, x, y = map(int, input().split())
x_arr = list(map(int, input().split()))
y_arr = list(map(int, input().split()))
x_arr.append(x)
y_arr.append(y)
if max(x_arr) < min(y_arr):
print("No War")
else:
print("War")
|
s632242849
|
p02744
|
u760642788
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 318 |
In this problem, we only consider strings consisting of lowercase English letters. Strings s and t are said to be **isomorphic** when the following conditions are satisfied: * |s| = |t| holds. * For every pair i, j, one of the following holds: * s_i = s_j and t_i = t_j. * s_i \neq s_j and t_i \neq t_j. For example, `abcac` and `zyxzx` are isomorphic, while `abcac` and `ppppp` are not. A string s is said to be in **normal form** when the following condition is satisfied: * For every string t that is isomorphic to s, s \leq t holds. Here \leq denotes lexicographic comparison. For example, `abcac` is in normal form, but `zyxzx` is not since it is isomorphic to `abcac`, which is lexicographically smaller than `zyxzx`. You are given an integer N. Print all strings of length N that are in normal form, in lexicographically ascending order.
|
n = int(input())
lst = []
abc = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j"]
for i in range(n):
lst_ = lst
lst = []
for s in lst_:
for j in range(10):
if s[-1] == abc[j - 1]:
break
else:
lst.append(s + abc[j])
for s in lst:
print(s)
|
s984086860
|
Accepted
| 125 | 14,172 | 254 |
n = int(input())
lst = ["a"]
abc = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j"]
for i in range(n - 1):
lst_ = lst
lst = []
for s in lst_:
for t in abc[: len(set(s)) + 1]:
lst.append(s + t)
for s in lst:
print(s)
|
s850685931
|
p00169
|
u553058997
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,564 | 533 |
ブラックジャックはカジノで行われるカードゲームの一種で、1 〜 13 までの数が書かれたカードを使ってゲームが行われます。各カードの点数は次のように決まっています。 * 1 は 1 点あるいは 11 点 * 2 から 9 までは、書かれている数の通りの点数 * 10 から 13 までは、10 点 このゲームには親を含めた何人かの参加者がおり、それぞれが何枚かのカードの組を持っています。このカードの組のことを手と呼びます。手の点数はカードの点数の合計です。その計算は次のように行うものとします。 * カードの点数の合計が 21 より大きくなるときは、手の点数を 0 点とする * カードの点数として、1 は 1 点と計算しても 11 点と計算してもよいが、手の点数が最大となる方を選ぶこととする 配られたカードの情報を入力とし、手の点数を出力するプログラムを作成してください。
|
if input() == '1':
pass
while True:
inp = input()
if inp == '0': break
inp = inp.replace('11', '10')
inp = inp.replace('12', '10')
inp = inp.replace('13', '10')
cards = tuple(map(int, inp.split()))
ans = sum(cards)
for i in range(cards.count(1)):
if sum(cards) + 9 * (i+1) > 21:
print('test', ans)
break
elif ans < sum(cards) + 9 * (i+1):
print('test2', ans)
ans = sum(cards) + 9 * (i+1)
if ans > 21: ans = 0
print(ans)
|
s804192673
|
Accepted
| 20 | 7,532 | 444 |
while True:
inp = input()
if inp == '0': break
inp = inp.replace('11', '10')
inp = inp.replace('12', '10')
inp = inp.replace('13', '10')
cards = tuple(map(int, inp.split()))
ans = sum(cards)
for i in range(cards.count(1)):
if sum(cards) + 10 * (i+1) > 21:
break
elif ans < sum(cards) + 10 * (i+1):
ans = sum(cards) + 10 * (i+1)
if ans > 21: ans = 0
print(ans)
|
s817367025
|
p03773
|
u532087742
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 154 |
Dolphin loves programming contests. Today, he will take part in a contest in AtCoder. In this country, 24-hour clock is used. For example, 9:00 p.m. is referred to as "21 o'clock". The current time is A o'clock, and a contest will begin in exactly B hours. When will the contest begin? Answer in 24-hour time.
|
# coding: utf-8
import sys
for line in sys.stdin.readlines():
a = line.strip().split(" ")
a1 = int(a[0])
a2 = int(a[1])
print(a1+a2 % 24)
|
s337619875
|
Accepted
| 17 | 2,940 | 156 |
# coding: utf-8
import sys
for line in sys.stdin.readlines():
a = line.strip().split(" ")
a1 = int(a[0])
a2 = int(a[1])
print((a1+a2) % 24)
|
s936526167
|
p02612
|
u054825571
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 9,116 | 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.
|
print(1000-int(input())//1000)
|
s730623979
|
Accepted
| 28 | 9,116 | 48 |
n=int(input())%1000
print(1000-n if n!=0 else 0)
|
s250026878
|
p04012
|
u874741582
| 2,000 | 262,144 |
Wrong Answer
| 19 | 2,940 | 123 |
Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful.
|
w = input()
f=0
for i in w:
if w.count(i)%2 != 0:
f=0
elif i == w[-1]: f=1
print("YES" if f ==1 else "NO")
|
s082634066
|
Accepted
| 18 | 2,940 | 141 |
w = input()
f=0
def beautiful(w):
for i in w:
if w.count(i)%2 != 0:
return "No"
return "Yes"
print(beautiful(w))
|
s007636163
|
p03970
|
u300637346
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 125 |
CODE FESTIVAL 2016 is going to be held. For the occasion, Mr. Takahashi decided to make a signboard. He intended to write `CODEFESTIVAL2016` on it, but he mistakenly wrote a different string S. Fortunately, the string he wrote was the correct length. So Mr. Takahashi decided to perform an operation that replaces a certain character with another in the minimum number of iterations, changing the string to `CODEFESTIVAL2016`. Find the minimum number of iterations for the rewrite operation.
|
s=list(input())
t=list('CODEFESTIVAL2016')
count=0
for k in range(16):
if s[k] == t[k]:
count += 1
print(count)
|
s358911312
|
Accepted
| 17 | 2,940 | 124 |
s=input()
t='CODEFESTIVAL2016'
count=0
for num in range(len(s)):
if s[num] != t[num]:
count += 1
print(count)
|
s043831658
|
p03493
|
u501409901
| 2,000 | 262,144 |
Wrong Answer
| 32 | 9,076 | 218 |
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.
|
# Press the green button in the gutter to run the script.
if __name__ == '__main__':
s = input()
count = 0
for n in range(3):
if s[n] == 1:
count = count + 1
print(count)
|
s496625001
|
Accepted
| 27 | 8,944 | 213 |
# Press the green button in the gutter to run the script.
if __name__ == '__main__':
s = input()
count = 0
for n in range(3):
if s[n] == '1':
count = count + 1
print(count)
|
s551634148
|
p02613
|
u281152316
| 2,000 | 1,048,576 |
Wrong Answer
| 174 | 16,320 | 433 |
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
N = int(input())
x = ""
S = []
for i in range(N):
x = input()
S.append(x)
ans = [0]*4
for i in range(N):
if S[i] == "AC":
ans[0] += 1
elif S[i] == "WA":
ans[1] += 1
elif S[i] == "TLE":
ans[2] += 1
elif S[i] == "RE":
ans[3] += 1
print('AC × ',end='')
print(ans[0])
print('WA × ',end='')
print(ans[1])
print('TLE × ',end='')
print(ans[2])
print('RE × ',end='')
print(ans[3])
|
s255118506
|
Accepted
| 176 | 16,240 | 429 |
N = int(input())
x = ""
S = []
for i in range(N):
x = input()
S.append(x)
ans = [0]*4
for i in range(N):
if S[i] == "AC":
ans[0] += 1
elif S[i] == "WA":
ans[1] += 1
elif S[i] == "TLE":
ans[2] += 1
elif S[i] == "RE":
ans[3] += 1
print('AC x ',end='')
print(ans[0])
print('WA x ',end='')
print(ans[1])
print('TLE x ',end='')
print(ans[2])
print('RE x ',end='')
print(ans[3])
|
s476477791
|
p02261
|
u709538965
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,600 | 459 |
Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
n = int(input())
bLst = input().split()
iLst = bLst
for i in range(n):
for j in range(n-1,i,-1):
if bLst[j][1] < bLst[j-1][1]:
bLst[j],bLst[j-1] = bLst[j-1],bLst[j]
print(*bLst)
print("Stable")
for i in range(n):
min = i
for j in range(i,n):
if iLst[j][1] < iLst[min][1]:
min = j
iLst[i],iLst[min] = iLst[min],iLst[i]
print(*iLst)
if bLst == iLst:
print("Stable");
else:
print("Not stable")
|
s118440142
|
Accepted
| 20 | 5,608 | 463 |
n = int(input())
*bLst, = input().split()
iLst = bLst[:]
for i in range(n):
for j in range(n-1,i,-1):
if bLst[j][1] < bLst[j-1][1]:
bLst[j],bLst[j-1] = bLst[j-1],bLst[j]
print(*bLst)
print("Stable")
for i in range(n):
min = i
for j in range(i,n):
if iLst[j][1] < iLst[min][1]:
min = j
iLst[i],iLst[min] = iLst[min],iLst[i]
print(*iLst)
if bLst == iLst:
print("Stable")
else:
print("Not stable")
|
s994888592
|
p03844
|
u181215519
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 16 |
Joisino wants to evaluate the formula "A op B". Here, A and B are integers, and the binary operator op is either `+` or `-`. Your task is to evaluate the formula instead of her.
|
print( input() )
|
s322414673
|
Accepted
| 17 | 2,940 | 24 |
print( eval( input() ) )
|
s016762625
|
p03659
|
u663101675
| 2,000 | 262,144 |
Wrong Answer
| 172 | 23,800 | 231 |
Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer a_i written on it. They will share these cards. First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards. Here, both Snuke and Raccoon have to take at least one card. Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively. They would like to minimize |x-y|. Find the minimum possible value of |x-y|.
|
N = int(input())
A = list(map(int, input().split()))
S = sum(A)
count = 0
Near = 10 ** 15
for i in range(N-1):
count = count + A[i]
if abs(2 * Near - S) > abs(2 * count - S):
Near = count
print(abs(2 * count - S))
|
s403847835
|
Accepted
| 165 | 23,800 | 230 |
N = int(input())
A = list(map(int, input().split()))
S = sum(A)
count = 0
Near = 10 ** 15
for i in range(N-1):
count = count + A[i]
if abs(2 * Near - S) > abs(2 * count - S):
Near = count
print(abs(2 * Near - S))
|
s737897934
|
p03434
|
u671680336
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 247 |
We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.
|
N = int(input())
a_list = list(map(int, input().split()))
a_list.sort(reverse=True)
Alice = 0
Bob = 0
print(a_list)
for i in range(len(a_list)):
if i % 2 == 0:
Alice += a_list[i]
else:
Bob += a_list[i]
print(Alice-Bob)
|
s296122786
|
Accepted
| 17 | 3,060 | 232 |
N = int(input())
a_list = list(map(int, input().split()))
a_list.sort(reverse=True)
Alice = 0
Bob = 0
for i in range(len(a_list)):
if i % 2 == 0:
Alice += a_list[i]
else:
Bob += a_list[i]
print(Alice-Bob)
|
s638407630
|
p03962
|
u635252313
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 114 |
AtCoDeer the deer recently bought three paint cans. The color of the one he bought two days ago is a, the color of the one he bought yesterday is b, and the color of the one he bought today is c. Here, the color of each paint can is represented by an integer between 1 and 100, inclusive. Since he is forgetful, he might have bought more than one paint can in the same color. Count the number of different kinds of colors of these paint cans and tell him.
|
a,b,c=map(int,input().split())
if a==b==c:
print(3)
elif a==b or a==c or b==c:
print(2)
else:
print(1)
|
s375655786
|
Accepted
| 17 | 2,940 | 52 |
a=set(list(map(int,input().split())))
print(len(a))
|
s666335703
|
p03501
|
u506587641
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 50 |
You are parking at a parking lot. You can choose from the following two fee plans: * Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours. * Plan 2: The fee will be B yen, regardless of the duration. Find the minimum fee when you park for N hours.
|
n, a, b = map(int, input().split())
print(n*a, b)
|
s736268359
|
Accepted
| 17 | 2,940 | 55 |
n, a, b = map(int, input().split())
print(min(n*a, b))
|
s906285722
|
p04043
|
u037221289
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 190 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
X = [map(int,input().split(' '))]
five = 0
seven = 0
for i in X:
if i == 7:
seven += 1
elif i == 5:
five += 5
if five == 2 and seven == 1:
print('YES')
else:
print('NO')
|
s415152855
|
Accepted
| 17 | 2,940 | 84 |
X = sorted(list(map(int,input().split(' '))))
print('YES' if X == [5,5,7] else 'NO')
|
s740831634
|
p03575
|
u256464928
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 265 |
You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges. The i-th edge (1 \leq i \leq M) connects Vertex a_i and Vertex b_i. An edge whose removal disconnects the graph is called a _bridge_. Find the number of the edges that are bridges among the M edges.
|
V,E=map(int,input().split())
edges=[set() for i in range(V)]
for i in range(E):
a,b=map(int,input().split())
edges[a-1].add(b-1)
edges[b-1].add(a-1)
for i in range(V):
print(len({n for v in edges[i] for n in edges[v] if not n in edges[i] and n!=i}))
|
s207292594
|
Accepted
| 41 | 3,064 | 265 |
N,M=map(int,input().split())
edges=[list(map(int,input().split())) for i in range(M)]
ans=0
for x in edges:
l=list(range(N))
for y in edges:
if y!=x:l=[l[y[0]-1] if l[i]==l[y[1]-1] else l[i] for i in range(N)]
if len(set(l))!=1:ans+=1
print(ans)
|
s632994702
|
p04045
|
u757424147
| 2,000 | 262,144 |
Wrong Answer
| 26 | 9,160 | 303 |
Iroha is very particular about numbers. There are K digits that she dislikes: D_1, D_2, ..., D_K. She is shopping, and now paying at the cashier. Her total is N yen (the currency of Japan), thus she has to hand at least N yen to the cashier (and possibly receive the change). However, as mentioned before, she is very particular about numbers. When she hands money to the cashier, the decimal notation of the amount must not contain any digits that she dislikes. Under this condition, she will hand the minimum amount of money. Find the amount of money that she will hand to the cashier.
|
n,k = map(int,input().split())
if k == 0:
print(n)
exit()
dislike = list(map(int,input().split()))
usable = []
for i in range(10):
if i not in dislike:
usable.append(i)
print(usable)
ans = ''
for i in str(n):
while int(i) not in usable:
i = str((int(i)+1)%10)
ans+=i
print(ans)
|
s216416166
|
Accepted
| 61 | 9,132 | 232 |
n,k = map(int,input().split())
if k == 0:
print(n)
exit()
dislike = list(input().split())
ans = n-1
flag = 1
while flag == 1:
ans+=1
flag = 0
for i in str(ans):
if i in dislike:
flag = 1
break
print(ans)
|
s178922856
|
p03555
|
u487288850
| 2,000 | 262,144 |
Wrong Answer
| 29 | 8,888 | 64 |
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.
|
x =input()
y=input()
print('Yes' if x[2]+x[1]+x[0]==y else 'No')
|
s751285809
|
Accepted
| 26 | 8,960 | 64 |
x =input()
y=input()
print('YES' if x[2]+x[1]+x[0]==y else 'NO')
|
s770176120
|
p04011
|
u920438243
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 93 |
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.
|
num = [int(input()) for i in range(4)]
sum = num[0]*num[2]+(num[1]-num[0])*num[3]
print(sum)
|
s626371691
|
Accepted
| 17 | 2,940 | 152 |
num = [int(input()) for i in range(4)]
if num[0] <= num[1]:
sum = num[0]*num[2]
else:
sum = num[1]*num[2]+(num[0]-num[1])*num[3]
print(sum)
|
s969754065
|
p03487
|
u859897687
| 2,000 | 262,144 |
Wrong Answer
| 100 | 23,284 | 191 |
You are given a sequence of positive integers of length N, a = (a_1, a_2, ..., a_N). Your objective is to remove some of the elements in a so that a will be a **good sequence**. Here, an sequence b is a **good sequence** when the following condition holds true: * For each element x in b, the value x occurs exactly x times in b. For example, (3, 3, 3), (4, 2, 4, 1, 4, 2, 4) and () (an empty sequence) are good sequences, while (3, 3, 3, 3) and (2, 4, 1, 4, 2) are not. Find the minimum number of elements that needs to be removed so that a will be a good sequence.
|
n=int(input())
d=dict()
for i in map(int,input().split()):
if i not in d:
d[i]=1
else:
d[i]+=1
ans=0
for i in d.keys():
if i>=d[i]:
ans+=i-d[i]
else:
ans+=i
print(ans)
|
s161288591
|
Accepted
| 97 | 23,284 | 194 |
n=int(input())
d=dict()
for i in map(int,input().split()):
if i not in d:
d[i]=1
else:
d[i]+=1
ans=0
for i in d.keys():
if i<=d[i]:
ans+=d[i]-i
else:
ans+=d[i]
print(ans)
|
s205646938
|
p04011
|
u970198631
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,168 | 121 |
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(N*x)
else:
print(K*x + (K-N)*y)
|
s616467910
|
Accepted
| 24 | 9,032 | 121 |
N = int(input())
K = int(input())
x = int(input())
y = int(input())
if N <= K:
print(N*x)
else:
print(K*x + (N-K)*y)
|
s519860982
|
p03502
|
u163320134
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 109 |
An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10. Given an integer N, determine whether it is a Harshad number.
|
def n(num):
ret=0
for i in range(10):
ret+=int((num%(10**(i+1)))/(10**i))
return ret
print(n(1000))
|
s361396065
|
Accepted
| 17 | 2,940 | 162 |
def n(num):
ret=0
for i in range(10):
ret+=int((num%(10**(i+1)))/(10**i))
return ret
a=int(input())
if a%n(a)==0:
print('Yes')
else:
print('No')
|
s087369495
|
p04029
|
u731368968
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 44 |
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?
|
print(sum([i for i in range(int(input()))]))
|
s197121546
|
Accepted
| 17 | 2,940 | 46 |
print(sum([i+1 for i in range(int(input()))]))
|
s926318931
|
p03494
|
u439312138
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 290 |
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
N = int(input())
numList = list(map(int,input().split()))
def numEvener(N,numList):
for i in range(0,N-1):
if numList[i] // 2 == 0:
numList[i] = int(numList[i] / 2)
else:
numList[i] = 0
return numList
while not 0 in numList:
numList = numEvener(N,numList)
|
s371261732
|
Accepted
| 19 | 3,060 | 182 |
ans = 10**9
N = int(input())
A = list(map(int,input().split()))
for i in A:
p = 0
while i % 2 == 0:
i //= 2
p += 1
if p < ans:
ans = p
print(ans)
|
s097141200
|
p03795
|
u409064224
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 39 |
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
n = int(input())
print(n*800-200*n//15)
|
s320067579
|
Accepted
| 17 | 2,940 | 41 |
n = int(input())
print(n*800-200*(n//15))
|
s301204496
|
p03447
|
u031852574
| 2,000 | 262,144 |
Wrong Answer
| 25 | 9,136 | 12 |
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?
|
1234
150
100
|
s966132810
|
Accepted
| 27 | 9,084 | 62 |
x,a,b = [int(input()) for i in range(3)]
print((x - a) % b)
|
s651750948
|
p04029
|
u287431190
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 62 |
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())
s = 0
for i in range(n):
s = s + i
print(s)
|
s518533788
|
Accepted
| 17 | 2,940 | 69 |
n = int(input())
s = 0
for i in range(n+1):
s = s + i
print(int(s))
|
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