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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s920773708
|
p03005
|
u560988566
| 2,000 | 1,048,576 |
Wrong Answer
| 20 | 3,316 | 77 |
Takahashi is distributing N balls to K persons. If each person has to receive at least one ball, what is the maximum possible difference in the number of balls received between the person with the most balls and the person with the fewest balls?
|
n,k = map(int, input().split())
ans = n-k+1
if k == 1:
ans = 0
print(ans)
|
s266633828
|
Accepted
| 18 | 2,940 | 75 |
n,k = map(int, input().split())
ans = n-k
if k == 1:
ans = 0
print(ans)
|
s319779475
|
p03386
|
u833492079
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 285 |
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()) # 5 7 2
ans=[]
count=0
for i in range(a, b+1):
ans.append(i)
count += 1
if count > k: break
count=0
for i in range(b, a-1, -1):
ans.append(i)
count += 1
if count > k: break
ans = sorted(set(ans), key=ans.index)
for i in ans:
print(i)
|
s928313336
|
Accepted
| 17 | 3,060 | 272 |
a, b, k = map(int, input().split()) # 5 7 2
ans=[]
count=0
for i in range(a, b+1):
ans.append(i)
count += 1
if count >= k: break
count=0
for i in range(b, a-1, -1):
ans.append(i)
count += 1
if count >= k: break
ans = sorted(set(ans))
for i in ans:
print(i)
|
s084057759
|
p02393
|
u964416376
| 1,000 | 131,072 |
Wrong Answer
| 40 | 7,652 | 53 |
Write a program which reads three integers, and prints them in ascending order.
|
a = [int(i) for i in input().split()]
a.sort
print(a)
|
s641380080
|
Accepted
| 20 | 7,732 | 80 |
a = [int(i) for i in input().split()]
print(' '.join(list(map(str, sorted(a)))))
|
s676482019
|
p00275
|
u766477342
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,720 | 405 |
百人一首の札を使った遊戯の１つに、「坊主めくり」というものがあります。絵札だけを使う簡単な遊戯なので広く楽しまれています。きまりには様々な派生型がありますが、ここで考える坊主めくりはN人の参加者で、以下のようなルールで行います。 * ６４枚の「男」、１５枚の「坊主」、２１枚の「姫」、計１００枚の札を使う。 * 絵が見えないように札を裏がえしにしてよく混ぜ、「札山」をつくる。 * 参加者の一人目から順番に１枚ずつ札山の札を引く。N人目の次は、また一人目から繰り返す。 * 引いた札が男なら、引いた人はその札を手に入れる。 * 引いた札が坊主なら、引いた人はその札を含め、持っている札をすべて「場」に出す。 * 引いた札が姫なら、引いた人はその札を含め、場にある札をすべて手に入れる。 * 札山の札がなくなったら終了で、一番多くの札を持っている人の勝ち。 参加者数と札山に積まれた札の順番が与えられたとき、遊戯が終了した時点で各参加者が持っている札数を昇順で並べたものと、場に残っている札数を出力するプログラムを作成してください。
|
while 1:
ba = 0
p = [0 for i in range(int(input()))]
if len(p) == 0:break
for i,v in enumerate(list(input())):
p[i%len(p)] += 1
if v == 'L':
p[i%len(p)] += ba
ba = 0
elif v== 'S':
ba += p[i%len(p)]
p[i%len(p)] = 0
result = ''
for v in p:
result += '%d ' % v
result += str(ba)
print(result)
|
s242144991
|
Accepted
| 50 | 6,724 | 413 |
while 1:
ba = 0
p = [0 for i in range(int(input()))]
if len(p) == 0:break
for i,v in enumerate(list(input())):
p[i%len(p)] += 1
if v == 'L':
p[i%len(p)] += ba
ba = 0
elif v== 'S':
ba += p[i%len(p)]
p[i%len(p)] = 0
result = ''
for v in sorted(p):
result += '%d ' % v
result += str(ba)
print(result)
|
s492577172
|
p03855
|
u389910364
| 2,000 | 262,144 |
Wrong Answer
| 1,142 | 73,740 | 2,716 |
There are N cities. There are also K roads and L railways, extending between the cities. The i-th road bidirectionally connects the p_i-th and q_i-th cities, and the i-th railway bidirectionally connects the r_i-th and s_i-th cities. No two roads connect the same pair of cities. Similarly, no two railways connect the same pair of cities. We will say city A and B are _connected by roads_ if city B is reachable from city A by traversing some number of roads. Here, any city is considered to be connected to itself by roads. We will also define _connectivity by railways_ similarly. For each city, find the number of the cities connected to that city by both roads and railways.
|
import bisect
import heapq
import itertools
import math
import os
import re
import string
import sys
from collections import Counter, deque, defaultdict
from decimal import Decimal
from fractions import gcd
from functools import lru_cache, reduce
from operator import itemgetter
import numpy as np
if os.getenv("LOCAL"):
sys.stdin = open("_in.txt", "r")
sys.setrecursionlimit(2147483647)
INF = float("inf")
IINF = 10 ** 18
N, K, L = list(map(int, sys.stdin.readline().split()))
P, Q = list(zip(*[map(int, sys.stdin.readline().split()) for _ in range(K)]))
R, S = list(zip(*[map(int, sys.stdin.readline().split()) for _ in range(L)]))
class UnionFind:
def __init__(self, size=None, nodes=None):
assert size is not None or nodes is not None
if size is not None:
self._parents = [i for i in range(size)]
self._ranks = [0 for _ in range(size)]
self._sizes = [1 for _ in range(size)]
else:
self._parents = {k: k for k in nodes}
self._ranks = {k: 0 for k in nodes}
self._sizes = {k: 1 for k in nodes}
def unite(self, x, y):
x = self.find(x)
y = self.find(y)
if x == y:
return
if self._ranks[x] > self._ranks[y]:
self._parents[y] = x
self._sizes[x] += self._sizes[y]
else:
self._parents[x] = y
self._sizes[y] += self._sizes[x]
if self._ranks[x] == self._ranks[y]:
self._ranks[y] += 1
def find(self, x):
if self._parents[x] == x:
return x
self._parents[x] = self.find(self._parents[x])
return self._parents[x]
def size(self, x):
return self._sizes[self.find(x)]
uf1 = UnionFind(size=N + 1)
for p, q in zip(P, Q):
uf1.unite(p, q)
uf2 = UnionFind(size=N + 1)
for r, s in zip(R, S):
if uf1.find(r) == uf1.find(s):
uf2.unite(r, s)
ans = []
for i in range(1, N + 1):
ans.append(uf2.size(i))
print(' '.join(map(str, ans)))
|
s937461274
|
Accepted
| 1,343 | 84,908 | 2,555 |
import os
import sys
from collections import defaultdict
if os.getenv("LOCAL"):
sys.stdin = open("_in.txt", "r")
sys.setrecursionlimit(2147483647)
INF = float("inf")
IINF = 10 ** 18
N, K, L = list(map(int, sys.stdin.readline().split()))
P, Q = list(zip(*[map(int, sys.stdin.readline().split()) for _ in range(K)]))
R, S = list(zip(*[map(int, sys.stdin.readline().split()) for _ in range(L)]))
class UnionFind:
def __init__(self, size=None, nodes=None):
assert size is not None or nodes is not None
if size is not None:
self._parents = [i for i in range(size)]
self._ranks = [0 for _ in range(size)]
self._sizes = [1 for _ in range(size)]
else:
self._parents = {k: k for k in nodes}
self._ranks = {k: 0 for k in nodes}
self._sizes = {k: 1 for k in nodes}
def unite(self, x, y):
x = self.find(x)
y = self.find(y)
if x == y:
return
if self._ranks[x] > self._ranks[y]:
self._parents[y] = x
self._sizes[x] += self._sizes[y]
else:
self._parents[x] = y
self._sizes[y] += self._sizes[x]
if self._ranks[x] == self._ranks[y]:
self._ranks[y] += 1
def find(self, x):
if self._parents[x] == x:
return x
self._parents[x] = self.find(self._parents[x])
return self._parents[x]
def size(self, x):
return self._sizes[self.find(x)]
uf1 = UnionFind(size=N + 1)
for p, q in zip(P, Q):
uf1.unite(p, q)
uf2 = UnionFind(size=N + 1)
for r, s in zip(R, S):
uf2.unite(r, s)
counts = defaultdict(int)
for i in range(1, N + 1):
counts[(uf1.find(i), uf2.find(i))] += 1
ans = []
for i in range(1, N + 1):
ans.append(counts[(uf1.find(i), uf2.find(i))])
print(' '.join(map(str, ans)))
|
s477523380
|
p03502
|
u319818856
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 255 |
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 f(x: int) -> int:
s = 0
while x > 0:
s += x % 10
x //= 10
return s
def harshad_number(N: int) -> bool:
return N % f(N) == 0
if __name__ == "__main__":
N = int(input())
ans = harshad_number(N)
print(ans)
|
s439767701
|
Accepted
| 17 | 2,940 | 274 |
def f(x: int) -> int:
s = 0
while x > 0:
s += x % 10
x //= 10
return s
def harshad_number(N: int) -> bool:
return N % f(N) == 0
if __name__ == "__main__":
N = int(input())
yes = harshad_number(N)
print('Yes' if yes else 'No')
|
s497590684
|
p02412
|
u138628845
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,596 | 497 |
Write a program which identifies the number of combinations of three integers which satisfy the following conditions: * You should select three distinct integers from 1 to n. * A total sum of the three integers is x. For example, there are two combinations for n = 5 and x = 9. * 1 + 3 + 5 = 9 * 2 + 3 + 4 = 9
|
ele_and_tar = []
rs = []
flag1 = 1
flag2 = 0
while flag1:
data = [int(x) for x in input().split()]
if data == [0,0]:
flag1 = 0
else:
ele_and_tar.append(data)
for i in range(len(ele_and_tar)):
rs.append(0)
for math in ele_and_tar:
for i in range(1,math[0]+1):
for j in range(i+1,math[0]+1):
for k in range(j+1,math[0]+1):
if (i + j + k) == math[1]:
rs[flag2] = rs[flag2] + 1
flag2 = flag2 + 1
|
s966396303
|
Accepted
| 520 | 5,604 | 316 |
while True:
res = 0
data = [int(x) for x in input().split()]
if data == [0,0]:
break
for i in range(1,data[0]+1):
for j in range(i+1,data[0]+1):
for k in range(j+1,data[0]+1):
if (i + j + k) == data[1]:
res = res + 1
print(res)
|
s349200353
|
p03448
|
u282151316
| 2,000 | 262,144 |
Wrong Answer
| 48 | 3,060 | 227 |
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
A = int(input())
B = int(input())
C = int(input())
X = int(input())
count = 0
for i in range(A):
for j in range(B):
for k in range(C):
if (X== 500*i +100*j +50*k):
count += 1
print(count)
|
s112025710
|
Accepted
| 55 | 3,060 | 233 |
A = int(input())
B = int(input())
C = int(input())
X = int(input())
count = 0
for i in range(A+1):
for j in range(B+1):
for k in range(C+1):
if (X== 500*i +100*j +50*k):
count += 1
print(count)
|
s669472275
|
p03449
|
u006425112
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 250 |
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j). You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down. The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them. At most how many candies can you collect when you choose the best way to travel?
|
n = int(input())
A = []
A.append(list(map(int, input().split())))
A.append(list(map(int, input().split())))
print(A)
up = A[0][0]
bottom = A[1][0] + up
for i in range(n-1):
up += A[0][i+1]
bottom = max(up, bottom) + A[1][i+1]
print(bottom)
|
s656652866
|
Accepted
| 17 | 3,064 | 240 |
n = int(input())
A = []
A.append(list(map(int, input().split())))
A.append(list(map(int, input().split())))
up = A[0][0]
bottom = A[1][0] + up
for i in range(n-1):
up += A[0][i+1]
bottom = max(up, bottom) + A[1][i+1]
print(bottom)
|
s908054820
|
p03836
|
u810625173
| 2,000 | 262,144 |
Wrong Answer
| 29 | 9,096 | 256 |
Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him.
|
sx, sy, tx, ty = map(int, input().split())
ans = ""
ans += "U" * (ty - sy) + "R" * (tx - sx)
ans += "D" * (ty - sy) + "L" * (tx - sx)
ans += "L" + "U" * (ty - sy + 1) + "R" * (tx - sx + 1)
ans += "R" + "D" * (ty - sy + 1) + "L" * (tx - sx + 1)
print(ans)
|
s271349577
|
Accepted
| 28 | 9,148 | 268 |
sx, sy, tx, ty = map(int, input().split())
ans = ""
ans += "U" * (ty - sy) + "R" * (tx - sx)
ans += "D" * (ty - sy) + "L" * (tx - sx)
ans += "L" + "U" * (ty - sy + 1) + "R" * (tx - sx + 1) + "D"
ans += "R" + "D" * (ty - sy + 1) + "L" * (tx - sx + 1) + "U"
print(ans)
|
s087806453
|
p03447
|
u384679440
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 88 |
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())
ans = X - A
ans -= ans / B
print(ans)
|
s683155861
|
Accepted
| 17 | 2,940 | 96 |
X = int(input())
A = int(input())
B = int(input())
ans = X - A
ans -= (ans // B ) * B
print(ans)
|
s354698116
|
p04045
|
u371467115
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,064 | 188 |
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())
d=list(map(str,input().split()))
for i in range(n,10000):
for j in list(str(i).replace(""," ").split()):
if int(j) in d:
break
print(i)
break
|
s376192340
|
Accepted
| 111 | 2,940 | 194 |
n,k=map(int,input().split())
d=list(map(str,input().split()))
for i in range(n,88889):
for j in list(str(i).replace(""," ").split()):
if j in d:
break
else:
print(i)
break
|
s302801549
|
p03494
|
u350997995
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 374 |
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())
A = list(map(int,input().split()))
def gcd(x,y):
if x<y:
z = x
x = y
y = z
if y==0:
return x
return gcd(y, x%y)
for i in range(N):
if i==0:
all_gcd = gcd(A[i],A[i+1])
else:
all_gcd = gcd(all_gcd,A[i])
print(all_gcd)
for i in range(50):
if all_gcd<2**i:
print(i-1)
break
|
s620357471
|
Accepted
| 17 | 3,064 | 359 |
N = int(input())
A = list(map(int,input().split()))
def gcd(x,y):
if x<y:
z = x
x = y
y = z
if y==0:
return x
return gcd(y, x%y)
for i in range(N):
if i==0:
all_gcd = gcd(A[i],A[i+1])
else:
all_gcd = gcd(all_gcd,A[i])
for i in range(50):
if all_gcd<2**i:
print(i-1)
break
|
s783304930
|
p03485
|
u921773161
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 93 |
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())
x = (a+b)/2
if x%1 == 0 :
print(x)
else :
print(x//1 + 1)
|
s369625217
|
Accepted
| 17 | 2,940 | 112 |
a,b = map(int, input().split())
x = (a+b)/2
if x%1 == 0 :
print(round(x))
else :
print(round(x//1 + 1))
|
s393371071
|
p03844
|
u679817762
| 2,000 | 262,144 |
Wrong Answer
| 23 | 9,148 | 20 |
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(exec(input()))
|
s451947788
|
Accepted
| 29 | 9,004 | 30 |
exec('print(' + input() + ')')
|
s023483132
|
p03068
|
u814986259
| 2,000 | 1,048,576 |
Wrong Answer
| 29 | 9,128 | 171 |
You are given a string S of length N consisting of lowercase English letters, and an integer K. Print the string obtained by replacing every character in S that differs from the K-th character of S, with `*`.
|
N = int(input())
S = input()
K = int(input())
e = S[K-1]
ans = []
for i in range(N):
if S[i] != 'e':
ans.append("*")
else:
ans.append(S[i])
print(''.join(ans))
|
s998151464
|
Accepted
| 25 | 9,012 | 170 |
N = int(input())
S = input()
K = int(input())
e = S[K-1]
ans = []
for i in range(N):
if S[i] != e:
ans.append("*")
else:
ans.append(S[i])
print(''.join(ans))
|
s594481388
|
p04031
|
u113255362
| 2,000 | 262,144 |
Wrong Answer
| 31 | 9,136 | 268 |
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal **integers** by transforming some of them. He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2). Find the minimum total cost to achieve his objective.
|
N=int(input())
List = list(map(int, input().split()))
sumS = 0
for i in range(N):
sumS +=List[i]
trial = sumS // N
mid = 0
res = 10000000
for i in range(N):
trial += i
mid = 0
for j in range(N):
mid += (trial - List[i])**2
res = max(res, mid)
print(res)
|
s424781376
|
Accepted
| 33 | 9,132 | 268 |
N=int(input())
List = list(map(int, input().split()))
sumS = 0
for i in range(N):
sumS +=List[i]
trial = sumS // N
mid = 0
res = 10000000
for i in range(N):
trial += i
mid = 0
for j in range(N):
mid += (trial - List[j])**2
res = min(res, mid)
print(res)
|
s846514940
|
p03861
|
u609061751
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 163 |
You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x?
|
import sys
input = sys.stdin.readline
a, b, x = [int(x) for x in input().split()]
B = 1 + b // x
if a == 0:
print(B)
else:
A = 1 + a // x
print(B - A)
|
s444272977
|
Accepted
| 17 | 3,060 | 220 |
import sys
input = sys.stdin.readline
a, b, x = [int(x) for x in input().split()]
B = 1 + b // x
if a == 0:
print(B)
elif a % x == 0:
A = 1 + a // x
print(B - A + 1)
else:
A = 1 + a // x
print(B - A)
|
s957768707
|
p03659
|
u303019816
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 24,832 | 224 |
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()))
cum_sum = [sum(A[:i+1]) for i in range(len(A))]
b = cum_sum[-1]
print(cum_sum)
ans = float("inf")
for a in cum_sum[:-1]:
ans = min(ans, abs(2 * a - b))
print(ans)
|
s971545327
|
Accepted
| 169 | 24,832 | 184 |
n = int(input())
A = list(map(int, input().split()))
b = sum(A)
ans = float("inf")
cum_sum = 0
for a in A[:-1]:
cum_sum += a
ans = min(ans, abs(2 * cum_sum - b))
print(ans)
|
s092627368
|
p03485
|
u693211869
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 75 |
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 math
a, b = map(int, input().split())
x = math.ceil(a / b)
print(x)
|
s987810066
|
Accepted
| 17 | 2,940 | 87 |
import math
a, b = map(int, input().split())
i = ( a + b) / 2
x = math.ceil(i)
print(x)
|
s932904567
|
p03493
|
u266113953
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 32 |
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()
print(s.count("0"))
|
s064882624
|
Accepted
| 18 | 2,940 | 32 |
s = input()
print(s.count("1"))
|
s322914906
|
p03302
|
u399280934
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 105 |
You are given two integers a and b. Determine if a+b=15 or a\times b=15 or neither holds. Note that a+b=15 and a\times b=15 do not hold at the same time.
|
a,b=map(int,input().split())
if a+b==15:
print('*')
elif a*b==15:
print('+')
else:
print('x')
|
s049859349
|
Accepted
| 18 | 2,940 | 105 |
a,b=map(int,input().split())
if a+b==15:
print('+')
elif a*b==15:
print('*')
else:
print('x')
|
s210190054
|
p03911
|
u052499405
| 2,000 | 262,144 |
Wrong Answer
| 418 | 22,036 | 1,247 |
On a planet far, far away, M languages are spoken. They are conveniently numbered 1 through M. For _CODE FESTIVAL 20XX_ held on this planet, N participants gathered from all over the planet. The i-th (1≦i≦N) participant can speak K_i languages numbered L_{i,1}, L_{i,2}, ..., L_{i,{}K_i}. Two participants A and B can _communicate_ with each other if and only if one of the following conditions is satisfied: * There exists a language that both A and B can speak. * There exists a participant X that both A and B can communicate with. Determine whether all N participants can communicate with all other participants.
|
n, m = [int(item) for item in input().split()]
query = []
appeared = set()
for i in range(n):
line = [int(item) for item in input().split()]
appeared.update(line[1:])
query.append(line[1:])
class UnionFind:
def __init__(self, n):
self.par = [i for i in range(n)]
self.size = [1] * n
self.rank = [0] * n
def find(self, x):
if self.par[x] == x:
return x
else:
self.par[x] = self.find(self.par[x])
return self.par[x]
def same_check(self, x, y):
return self.find(x) == self.find(y)
def get_size(self, x):
return self.size[self.find(x)]
def union(self, x, y):
x = self.find(x)
y = self.find(y)
if self.rank[x] < self.rank[y]:
self.par[x] = y
else:
self.par[y] = x
if self.rank[x] == self.rank[y]:
self.rank[x] += 1
uf = UnionFind(m)
for q in query:
if len(q) == 1:
continue
for item in q[1:]:
if not uf.same_check(q[0]-1, item-1):
uf.union(q[0]-1, item-1)
par = uf.find(appeared.pop() - 1)
for item in appeared:
if uf.find(item - 1) != par:
print("No")
exit()
print("Yes")
|
s046854250
|
Accepted
| 418 | 22,036 | 1,247 |
n, m = [int(item) for item in input().split()]
query = []
appeared = set()
for i in range(n):
line = [int(item) for item in input().split()]
appeared.update(line[1:])
query.append(line[1:])
class UnionFind:
def __init__(self, n):
self.par = [i for i in range(n)]
self.size = [1] * n
self.rank = [0] * n
def find(self, x):
if self.par[x] == x:
return x
else:
self.par[x] = self.find(self.par[x])
return self.par[x]
def same_check(self, x, y):
return self.find(x) == self.find(y)
def get_size(self, x):
return self.size[self.find(x)]
def union(self, x, y):
x = self.find(x)
y = self.find(y)
if self.rank[x] < self.rank[y]:
self.par[x] = y
else:
self.par[y] = x
if self.rank[x] == self.rank[y]:
self.rank[x] += 1
uf = UnionFind(m)
for q in query:
if len(q) == 1:
continue
for item in q[1:]:
if not uf.same_check(q[0]-1, item-1):
uf.union(q[0]-1, item-1)
par = uf.find(appeared.pop() - 1)
for item in appeared:
if uf.find(item - 1) != par:
print("NO")
exit()
print("YES")
|
s294547070
|
p02646
|
u757715307
| 2,000 | 1,048,576 |
Wrong Answer
| 22 | 9,192 | 271 |
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 = (int(i) for i in input().split(' '))
b, w = (int(i) for i in input().split(' '))
t = int(input())
l1 = a + v * t
l2 = b + w * t
flag = False
if v >= 0:
if l1 >= l2:
flag = True
else:
if l1 <= l2:
flag = True
print("Yes" if flag else "No")
|
s663375567
|
Accepted
| 22 | 9,128 | 329 |
a, v = (int(i) for i in input().split(' '))
b, w = (int(i) for i in input().split(' '))
t = int(input())
flag = False
if a < b:
l1 = a + v * t
l2 = b + w * t
if l1 >= l2:
flag = True
else:
l1 = a + (v * -1) * t
l2 = b + (w * -1) * t
if l1 <= l2:
flag = True
print("YES" if flag else "NO")
|
s190254218
|
p03760
|
u136090046
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 201 |
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions. You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
|
val = input()
val2 = input()
res = ""
for i in range(0, min(len(val), len(val2))):
res += val[i]
res += val2[i]
if len(val) > len(val2):
res += val[-1]
else:
res += val2[-1]
print(res)
|
s424607302
|
Accepted
| 17 | 3,064 | 239 |
val = input()
val2 = input()
res = ""
for i in range(0, min(len(val), len(val2))):
res += val[i]
res += val2[i]
if len(val) == len(val2):
pass
elif len(val) > len(val2):
res += val[-1]
else:
res += val2[-1]
print(res)
|
s232132404
|
p03998
|
u052499405
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,064 | 581 |
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.
|
sa = input() + "#"
sb = input() + "#"
sc = input() + "#"
ia = 0
ib = 0
ic = 0
who = 0
for i in range(300):
if who == 0:
ia += 1
if sa[ia] == "#":
print("a")
exit()
elif sa[ia] == "b":
who = 1
elif sa[ia] == "c":
who = 2
elif who == 1:
ib += 1
if sb[ib] == "#":
print("b")
exit()
elif sb[ib] == "a":
who = 0
elif sb[ib] == "c":
who = 2
elif who == 2:
ic += 1
if sc[ic] == "#":
print("c")
exit()
elif sc[ic] == "a":
who = 0
elif sc[ic] == "b":
who = 1
|
s629260013
|
Accepted
| 18 | 3,064 | 671 |
#!/usr/bin/env python3
import sys
sys.setrecursionlimit(10**8)
input = sys.stdin.readline
sa = input() + "#"
sb = input() + "#"
sc = input() + "#"
ia = 0
ib = 0
ic = 0
who = 0
for i in range(300):
if who == 0:
if sa[ia] == "#":
print("A")
exit()
elif sa[ia] == "b":
who = 1
elif sa[ia] == "c":
who = 2
ia += 1
elif who == 1:
if sb[ib] == "#":
print("B")
exit()
elif sb[ib] == "a":
who = 0
elif sb[ib] == "c":
who = 2
ib += 1
elif who == 2:
if sc[ic] == "#":
print("C")
exit()
elif sc[ic] == "a":
who = 0
elif sc[ic] == "b":
who = 1
ic += 1
|
s848995927
|
p03729
|
u629607744
| 2,000 | 262,144 |
Wrong Answer
| 29 | 9,000 | 118 |
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 = [str(i) for i in input().split()]
if a[0][-1] == a[1][0] and a[1][-1] == a[2][0]:
print("Yes")
else:
print("No")
|
s208751760
|
Accepted
| 27 | 9,032 | 119 |
a = [str(i) for i in input().split()]
if a[0][-1] == a[1][0] and a[1][-1] == a[2][0]:
print("YES")
else:
print("NO")
|
s509877314
|
p03644
|
u163501259
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 238 |
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())
counter = 0
cnt = [0]*(n+1)
for i in range(1,n+1):
j = i
while(j%2 == 0):
j = j/2
counter += 1
cnt[i] = counter
counter = 0
max_val = max(cnt)
ans = cnt.index(max_val)
print(cnt)
print(ans)
|
s800250124
|
Accepted
| 17 | 3,064 | 237 |
n = int(input())
counter = 0
cnt = [0]*(n+1)
for i in range(1,n+1):
j = i
while(j%2 == 0):
j = j/2
counter += 1
cnt[i] = counter
counter = 0
max_val = max(cnt[1:])
ans = cnt[1:].index(max_val)
print(ans+1)
|
s318311220
|
p03478
|
u806257533
| 2,000 | 262,144 |
Wrong Answer
| 33 | 3,060 | 216 |
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).
|
in_all = list(map(int, input().split()))
cnt = 0
for n in range(in_all[0]):
tmp = str(n)
all = 0
for t in tmp:
all += int(t)
if all>=in_all[1] and all<=in_all[2]:
cnt += 1
print(cnt)
|
s727797580
|
Accepted
| 33 | 3,060 | 229 |
in_all = list(map(int, input().split()))
total = 0
for n in range(1, in_all[0]+1):
tmp = str(n)
all = 0
for t in tmp:
all += int(t)
if all>=in_all[1] and all<=in_all[2]:
total += n
print(total)
|
s410628095
|
p03891
|
u297467990
| 2,000 | 262,144 |
Wrong Answer
| 23 | 3,064 | 341 |
A 3×3 grid with a integer written in each square, is called a magic square if and only if the integers in each row, the integers in each column, and the integers in each diagonal (from the top left corner to the bottom right corner, and from the top right corner to the bottom left corner), all add up to the same sum. You are given the integers written in the following three squares in a magic square: * The integer A at the upper row and left column * The integer B at the upper row and middle column * The integer C at the middle row and middle column Determine the integers written in the remaining squares in the magic square. It can be shown that there exists a unique magic square consistent with the given information.
|
#!/usr/bin/env python3
import itertools
a = int(input())
b = int(input())
c = int(input())
for k in itertools.count():
f = [
[ a, b, k-a-b ],
[ 2*k-2*a-b-2*c, c, 2*a+b+c-k ],
[ a+b+2*c-k, k-b-c, k-a-c ] ]
print(*f[0])
print(*f[1])
print(*f[2])
break
|
s702804732
|
Accepted
| 24 | 3,188 | 320 |
#!/usr/bin/env python3
import itertools
a = int(input())
b = int(input())
c = int(input())
k = 3*c
f = [
[ a, b, k-a-b ],
[ 2*k-2*a-b-2*c, c, 2*a+b+c-k ],
[ a+b+2*c-k, k-b-c, k-a-c ] ]
assert f[0][2] + f[1][1] + f[2][0] == k
for y in range(3):
print(*f[y])
|
s561551704
|
p03493
|
u027208253
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 104 |
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
input = input()
count = 0
for i in range(len(input)):
if input[i] == 1:
count += 1
print(count)
|
s605256914
|
Accepted
| 17 | 2,940 | 109 |
input = input()
count = 0
for i in range(len(input)):
if input[i] == "1":
count += 1
print(count)
|
s244790859
|
p02833
|
u475503988
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 19 |
For an integer n not less than 0, let us define f(n) as follows: * f(n) = 1 (if n < 2) * f(n) = n f(n-2) (if n \geq 2) Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N).
|
1000000000000000000
|
s505388999
|
Accepted
| 17 | 2,940 | 147 |
N = int(input())
ans = N // 10
tmp = N // 10
while 5 <= tmp:
ans += tmp // 5
tmp = tmp // 5
if N % 2:
print(0)
else:
print(ans)
|
s383622263
|
p03478
|
u271469978
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 123 |
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
while n > 0:
if a <= n%10 <= b:
ans += n%10
n //= 10
print(ans)
|
s079789324
|
Accepted
| 28 | 3,064 | 354 |
n, a, b = map(int, input().split())
def check_a_b(n, a, b):
sum = 0
while n > 0:
if sum + n%10 <= b:
sum += n%10
n //= 10
else:
return False
if sum >= a:
return True
else:
return False
ans = 0
for i in range(n+1):
if check_a_b(i, a, b):
ans += i
print(ans)
|
s010946344
|
p03943
|
u117193815
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 125 |
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
a,b,c = map(int, input().split())
l=[a,b,c]
l.sort()
if l[0]+l[1]==l[2]:
print("YES")
else:
print("NO")
|
s656407631
|
Accepted
| 17 | 3,060 | 124 |
a,b,c = map(int, input().split())
l=[a,b,c]
l.sort()
if l[0]+l[1]==l[2]:
print("Yes")
else:
print("No")
|
s022203697
|
p03854
|
u579832365
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,316 | 234 |
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()
S = S[::-1]
print(S)
S = S.replace("resare","")
S = S.replace("esare","")
S = S.replace("remaerd","")
S = S.replace("maerd","")
if S == "":
print("YES")
else:
print("NO")
|
s326066559
|
Accepted
| 19 | 3,188 | 225 |
S = input()
S = S[::-1]
S = S.replace("resare","")
S = S.replace("esare","")
S = S.replace("remaerd","")
S = S.replace("maerd","")
if S == "":
print("YES")
else:
print("NO")
|
s611614729
|
p03370
|
u295416460
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 136 |
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts. Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition: * For each of the N kinds of doughnuts, make at least one doughnut of that kind. At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
|
N, X = map(int, input().split())
m_list = [int(input()) for _ in range(N)]
m_list.sort()
x = X - sum(m_list)
print(N + x // m_list[-1])
|
s471889455
|
Accepted
| 17 | 2,940 | 135 |
N, X = map(int, input().split())
m_list = [int(input()) for _ in range(N)]
m_list.sort()
x = X - sum(m_list)
print(N + x // m_list[0])
|
s915942336
|
p03502
|
u611090896
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 67 |
An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10. Given an integer N, determine whether it is a Harshad number.
|
N = input()
print('Yes' if int(N) // sum(map(int,N)) ==0 else 'No')
|
s628448025
|
Accepted
| 17 | 2,940 | 68 |
N = input()
print('Yes' if int(N) % sum(map(int,N)) == 0 else 'No')
|
s810379046
|
p03943
|
u583276018
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 89 |
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
s = map(int, input().split())
if(max(s) == sum(s)//2):
print("Yes")
else:
print("No")
|
s314974121
|
Accepted
| 18 | 2,940 | 143 |
s = list(map(int, input().split()))
if(sum(s)%2==1):
print("No")
else:
if(max(s) == (sum(s)//2)):
print("Yes")
else:
print("No")
|
s228134774
|
p03502
|
u341855122
| 2,000 | 262,144 |
Wrong Answer
| 155 | 12,388 | 136 |
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.
|
import numpy as np
a = input()
c = list(map(int,list(a)))
s = sum(np.array(c))
if int(a)%s == 0:
print("YES")
else:
print("NO")
|
s384663657
|
Accepted
| 152 | 12,416 | 136 |
import numpy as np
a = input()
c = list(map(int,list(a)))
s = sum(np.array(c))
if int(a)%s == 0:
print("Yes")
else:
print("No")
|
s386769088
|
p03556
|
u731665172
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 33 |
Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer.
|
N=int(input())
print(int(N**0.5))
|
s225298785
|
Accepted
| 17 | 3,060 | 38 |
N = int(input())
print(int(N**0.5)**2)
|
s753377389
|
p03697
|
u471684875
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 61 |
You are given two integers A and B as the input. Output the value of A + B. However, if A + B is 10 or greater, output `error` instead.
|
n,m=map(int,input().split())
print(n+m if n+m<9 else 'error')
|
s228464692
|
Accepted
| 17 | 2,940 | 62 |
n,m=map(int,input().split())
print(n+m if n+m<10 else 'error')
|
s303492210
|
p02578
|
u536034761
| 2,000 | 1,048,576 |
Wrong Answer
| 106 | 32,236 | 149 |
N persons are standing in a row. The height of the i-th person from the front is A_i. We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person: Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool. Find the minimum total height of the stools needed to meet this goal.
|
N = int(input())
A = list(map(int, input().split()))
pre = A[0]
ans = 0
for a in A[1:]:
if pre > a:
ans += pre - a
pre = a
print(ans)
|
s839867550
|
Accepted
| 112 | 32,364 | 217 |
N = int(input())
A = list(map(int, input().split()))
pre = A[0]
ans = 0
if N == 1:
print(0)
else:
for a in A[1:]:
if pre > a:
ans += pre - a
else:
pre = a
print(ans)
|
s887122830
|
p03486
|
u940743763
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 74 |
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.
|
s = input()
t = input()
if s < t:
print('Yes')
else:
print('No')
|
s724777654
|
Accepted
| 17 | 2,940 | 158 |
s = input()
t = input()
sol = ''.join((sorted(list(s))))
tol = ''.join(sorted(list(t), reverse=True))
if sol < tol:
print('Yes')
else:
print('No')
|
s177702158
|
p03635
|
u451017206
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 58 |
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?
|
n,m = list(map(int, input().split()))
print (n-1 * m - 1)
|
s149780504
|
Accepted
| 17 | 2,940 | 59 |
n,m = list(map(int, input().split()))
print ((n-1)*(m-1))
|
s259843175
|
p03545
|
u048034149
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 250 |
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
abcd = input()
op = ['+', '-']
for op1 in op:
for op2 in op:
for op3 in op:
tmp_ans = abcd[0] + op1 + abcd[1] + op2 + abcd[2] + op3 + abcd[3]
if eval(tmp_ans) == 7:
print(tmp_ans)
|
s957140813
|
Accepted
| 17 | 2,940 | 278 |
abcd = input()
op = ['+', '-']
for op1 in op:
for op2 in op:
for op3 in op:
tmp_ans = abcd[0] + op1 + abcd[1] + op2 + abcd[2] + op3 + abcd[3]
if eval(tmp_ans) == 7:
print(tmp_ans+'=7')
exit()
|
s050087409
|
p03434
|
u992519990
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 154 |
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.
|
# coding: utf-8
N = int(input())
a = list(map(int,input().split()))
a.sort(reverse=True)
print(a)
alice = sum(a[0::2])
bob = sum(a[1::2])
print(alice-bob)
|
s449964704
|
Accepted
| 17 | 2,940 | 145 |
# coding: utf-8
N = int(input())
a = list(map(int,input().split()))
a.sort(reverse=True)
alice = sum(a[0::2])
bob = sum(a[1::2])
print(alice-bob)
|
s483521401
|
p03854
|
u296150111
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,188 | 239 |
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()
i=1
while i<len(s)+1:
if s[-i:-i-5:-1]=="erase" \
or s[-i:-i-5:-1]=="dream":
print(s[-i:-i-5:-1])
i+=5
elif s[-i:-i-6:-1]=="eraser":
i+=6
elif s[-i:-i-7:-1]=="deaamer":
i+=7
else:
print("NO")
exit()
print("YES")
|
s547371473
|
Accepted
| 38 | 3,188 | 225 |
s=input()
i=1
while i<len(s):
if s[-i-4:-i]+s[-i]=="erase" \
or s[-i-4:-i]+s[-i]=="dream":
i+=5
elif s[-i-5:-i]+s[-i]=="eraser":
i+=6
elif s[-i-6:-i]+s[-i]=="dreamer":
i+=7
else:
print("NO")
exit()
print("YES")
|
s921363513
|
p03251
|
u339550873
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,060 | 243 |
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.
|
#! /usr/bin/env python3
# -*- coding: utf-8 -*-
N, M, X, Y = [int(c) for c in input().split()]
xls = [int(x) for x in input().split()]
yls = [int(c) for c in input().split()]
if max (xls) < min(yls):
print('War')
else:
print('No War')
|
s990384190
|
Accepted
| 17 | 3,060 | 277 |
#! /usr/bin/env python3
# -*- coding: utf-8 -*-
N, M, X, Y = [int(c) for c in input().split()]
xls = [int(x) for x in input().split()]
yls = [int(c) for c in input().split()]
if max (xls) < min(yls) and Y > max(xls) and X < min(yls):
print('No War')
else:
print('War')
|
s800989645
|
p03024
|
u630096262
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 82 |
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()
cnt = s.count('x')
if cnt <= 7:
print("Yes")
else:
print("No")
|
s832937986
|
Accepted
| 17 | 2,940 | 82 |
s = input()
cnt = s.count('x')
if cnt <= 7:
print("YES")
else:
print("NO")
|
s155779463
|
p03447
|
u786020649
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,168 | 1,202 |
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?
|
import sys
sys.setrecursionlimit(10**9)
read=sys.stdin.read
class Unionfind():
def __init__(self,n):
self.parents=[-1]*n
self.dist=[0]*n
def find(self,x):
if self.parents[x]<0:
return self.dist[x],x
else:
tmp=self.find(self.parents[x])
self.dist[x]+=tmp[0]
self.parents[x]=tmp[1]
return self.dist[x], self.parents[x]
def union(self,x,y,d):
rx=self.find(x)[1]
ry=self.find(y)[1]
diff=self.dist[y]-self.dist[x]-d
if rx==ry:
if diff!=0:
return True
return False
if diff<0:
rx,ry=ry,rx
diff=-diff
self.parents[ry]=min(self.parents[ry],self.parents[rx]-diff)
self.parents[rx]=ry
self.dist[rx]=diff
return False
def main():
n,m,*lrd=map(int,read().split())
v=Unionfind(n)
for l,r,d in zip(*[iter(lrd)]*3):
if v.union(l-1,r-1,d):
print('No')
break
else:
if max(-d-1 for d in v.parents if d<0) >10**9:
print('No')
else:
print('Yes')
if __name__=='__main__':
main()
|
s094702218
|
Accepted
| 29 | 9,160 | 65 |
import sys
x,a,b=map(int,sys.stdin.read().split())
print((x-a)%b)
|
s297212628
|
p03997
|
u958210291
| 2,000 | 262,144 |
Wrong Answer
| 25 | 9,160 | 70 |
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)*2)/2)
|
s176089477
|
Accepted
| 30 | 9,104 | 76 |
a = int(input())
b =int(input())
h = int(input())
print(int(((a+b)*h)/2))
|
s768239308
|
p03386
|
u726439578
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,060 | 102 |
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,a+k):
print(i)
for i in range(b-k,b):
print(i)
|
s540519632
|
Accepted
| 17 | 3,060 | 189 |
a,b,k=map(int,input().split())
if a+k-1>=b-k+1:
for i in range(a,b+1):
print(i)
else:
for i in range(a,a+k):
print(i)
for i in range(b-k+1,b+1):
print(i)
|
s801036714
|
p03672
|
u676010704
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 276 |
We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not. You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
|
def main():
msg = input()
while len(msg) >= 2:
before = msg[:int(len(msg) / 2)]
after = msg[int(len(msg) / 2):]
if before == after:
print(len(before))
break
msg = msg[:-1]
if __name__ == '__main__':
main()
|
s127781838
|
Accepted
| 17 | 3,060 | 341 |
def main():
msg = input()
msg = msg[:-1]
if len(msg) % 2 != 0:
msg = msg[:-1]
while len(msg) >= 2:
before = msg[:int(len(msg) / 2)]
after = msg[int(len(msg) / 2):]
if before == after:
print(len(msg))
break
msg = msg[:-2]
if __name__ == '__main__':
main()
|
s923886336
|
p02413
|
u527848444
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,724 | 248 |
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) = [int(i) for i in input().split()]
table = []
for rc in range(r):
table.append([int(i) for i in input().split()])
table.append([0 for _ in range(c)])
for row in table:
for column in row:
print(column,end=' ')
print()
|
s887889421
|
Accepted
| 30 | 7,768 | 332 |
(r, c) = [int(i) for i in input().split()]
last_row = [0 for _ in range(c+1)]
for _ in range(r):
row = [int(i) for i in input().split()]
for cc in range(c):
last_row[cc] += int(row[cc])
print(row[cc], end=' ')
print(sum(row))
last_row[-1] = sum(last_row)
print(' '.join([str(a) for a in last_row]))
|
s170749623
|
p03089
|
u070163338
| 2,000 | 1,048,576 |
Wrong Answer
| 62 | 3,064 | 427 |
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 = [int(i) for i in input().split()]
pool = []
def search_num(num):
try:
nn = b.index(num)
b.remove(num)
return num
except:
if num-1 > 0:
return search_num(num-1)
else: return -1
for i in range(n):
num = search_num(i+1)
if num == -1:
print('-1')
break
else:
pool.append(str(num))
if i+1 == n:
text = '\n'.join(pool)
print(text)
break
|
s219490592
|
Accepted
| 23 | 3,188 | 551 |
n = int(input())
b = [int(i) for i in input().split()]
pool = []
def search_num(num):
try:
nn = b.index(num)
if nn + 1 == num:
b.remove(num)
return num
else: return search_num(num-1)
except:
if num-1 > 0:
return search_num(num-1)
else: return -1
if n != len(b):
print('-1')
else:
for i in range(n, 0, -1):
num = search_num(i+1)
if num == -1:
print('-1')
break
else:
pool.append(str(num))
if i == 1:
for i in (reversed(pool)):
print(i)
break
|
s906160200
|
p04031
|
u230717961
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 384 |
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal **integers** by transforming some of them. He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2). Find the minimum total cost to achieve his objective.
|
def solve(s):
sum1 = 0
sum2 = 0
length = 0
for i in s.split():
x = int(i)
sum1 += x**2
sum2 += x
length += 1
score = 10**8
for y in range(100,-101, -1):
tmp = sum1 - sum2*2*y + length*(y**2)
score = min(tmp, score)
return score
if __name__ == "__main__":
n = int(input())
s = input()
solve(s)
|
s792832302
|
Accepted
| 17 | 3,064 | 391 |
def solve(s):
sum1 = 0
sum2 = 0
length = 0
for i in s.split():
x = int(i)
sum1 += x**2
sum2 += x
length += 1
score = 10**8
for y in range(100,-101, -1):
tmp = sum1 - sum2*2*y + length*(y**2)
score = min(tmp, score)
return score
if __name__ == "__main__":
n = int(input())
s = input()
print(solve(s))
|
s770632383
|
p03597
|
u185037583
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 36 |
We have an N \times N square grid. We will paint each square in the grid either black or white. If we paint exactly A squares white, how many squares will be painted black?
|
print(int(input())*2 - int(input()))
|
s028598898
|
Accepted
| 17 | 2,940 | 37 |
print(int(input())**2 - int(input()))
|
s705132194
|
p03469
|
u761087127
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 55 |
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.
|
print(c if i!=3 else '8' for i,c in enumerate(input()))
|
s370800386
|
Accepted
| 18 | 2,940 | 66 |
print("".join([c if i!=3 else '8' for i,c in enumerate(input())]))
|
s282693995
|
p02415
|
u805716376
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,532 | 34 |
Write a program which converts uppercase/lowercase letters to lowercase/uppercase for a given string.
|
a = input()
a.swapcase()
print(a)
|
s014068254
|
Accepted
| 20 | 5,540 | 28 |
print(input().swapcase())
|
s145491241
|
p03712
|
u041075929
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,064 | 468 |
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, os
f = lambda:list(map(int,input().split()))
if 'local' in os.environ :
sys.stdin = open('./input.txt', 'r')
def solve():
n, m = f()
s = [ input() for i in range(n)]
print(s)
for i in range(n+2):
if i == 0 or i == n+1:
for j in range(m+2):
print('#', end = '')
print()
else:
print('#',end = '')
print(s[i-1], end = '')
print('#')
solve()
|
s268885373
|
Accepted
| 18 | 3,060 | 455 |
import sys, os
f = lambda:list(map(int,input().split()))
if 'local' in os.environ :
sys.stdin = open('./input.txt', 'r')
def solve():
n, m = f()
s = [ input() for i in range(n)]
for i in range(n+2):
if i == 0 or i == n+1:
for j in range(m+2):
print('#', end = '')
print()
else:
print('#',end = '')
print(s[i-1], end = '')
print('#')
solve()
|
s596670317
|
p03456
|
u636311816
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 69 |
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;['No','Yes'][math.sqrt(int(input().replace(' ','')))%1>0]
|
s882625044
|
Accepted
| 17 | 2,940 | 76 |
import math;print(['Yes','No'][math.sqrt(int(input().replace(' ','')))%1>0])
|
s610173268
|
p04043
|
u190195462
| 2,000 | 262,144 |
Wrong Answer
| 30 | 8,956 | 246 |
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, B, C = map(str, input().split())
D = 0
E = 0
if A == 5:
D += 1
if A == 7:
E += 1
if B == 5:
D += 1
if B == 7:
E += 1
if C == 5:
D += 1
if C == 7:
E += 1
if D == 2 and E==1:
print("YES")
else:
print("NO")
|
s125313788
|
Accepted
| 24 | 8,936 | 272 |
A, B, C = map(str, input().split())
D = 0
E = 0
if A == "5":
D =D + 1
if A == "7":
E =E + 1
if B == "5":
D =D + 1
if B == "7":
E =E + 1
if C == "5":
D =D + 1
if C == "7":
E =E + 1
if D == 2 and E == 1:
print("YES")
else:
print("NO")
|
s775450118
|
p02277
|
u126478680
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,340 | 1,037 |
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).
|
import copy
def num_from_card(card):
return int(card[2:])
def partition(A, p, r):
x = num_from_card(A[r])
i = p-1
for j in range(p, r):
if num_from_card(A[j]) <= 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 quick_sort(A_i, p, r):
A = copy.copy(A_i)
if p < r:
q = partition(A, p, r)
quick_sort(A, p, q-1)
quick_sort(A, q+1, r)
return A
def is_stable(A, sorted_A, N):
A_nums = [int(x[1:]) for x in A]
same_num = [x for x in set(A_nums) if A_nums.count(x) > 1]
A_symbols, sorted_A_symbols = [], []
flag = 'Stable'
for num in same_num:
if [card[0] for card in A if int(card[1:]) == num] != [card[0] for card in sorted_A if int(card[1:]) == num]:
flag = 'Not stable'
break
return flag
n = int(input())
cards = [input() for i in range(n)]
sorted_cards = quick_sort(cards, 0, n-1)
print(is_stable(cards, sorted_cards, n))
for c in sorted_cards:
print(c)
|
s458138949
|
Accepted
| 4,210 | 14,296 | 1,229 |
import copy
def num_from_card(card):
return int(card[2:])
def partition(A, p, r):
x = num_from_card(A[r])
i = p-1
for j in range(p, r):
if num_from_card(A[j]) <= 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 quick_sort(A, p, r):
if p < r:
q = partition(A, p, r)
quick_sort(A, p, q-1)
quick_sort(A, q+1, r)
def merge(A, left, mid, right):
L = A[left: mid]
R = A[mid: right]
L.append(' ' + str(int(10e9 + 1)))
R.append(' ' + str(int(10e9 + 1)))
i, j = 0, 0
for k in range(left, right):
if num_from_card(L[i]) <= num_from_card(R[j]):
A[k] = L[i]
i += 1
else:
A[k] = R[j]
j += 1
def merge_sort(A, left, right):
if left + 1 < right:
mid = int((left + right)/2)
merge_sort(A, left, mid)
merge_sort(A, mid, right)
merge(A, left, mid, right)
n = int(input())
cards = [input() for i in range(n)]
merge_cards = copy.copy(cards)
quick_sort(cards, 0, n-1)
merge_sort(merge_cards, 0, n)
if merge_cards == cards:
print('Stable')
else:
print('Not stable')
for c in cards:
print(c)
|
s810266246
|
p03386
|
u693007703
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 228 |
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 = [int(i) for i in input().split()]
numbers = set()
for i in range(A, A+K):
if i <= B:
numbers.add(i)
for j in range(B-K, B):
if j >= A:
numbers.add(j+1)
for n in numbers:
print(n)
|
s433522579
|
Accepted
| 17 | 3,060 | 259 |
A, B, K = [int(i) for i in input().split()]
numbers = set()
for i in range(A, A+K):
if i <= B:
numbers.add(i)
for j in range(B-K, B):
if j >= A:
numbers.add(j+1)
numbers = sorted(numbers)
for n in numbers:
print(n)
|
s221793222
|
p03555
|
u394731058
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 133 |
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.
|
i = list(input())
t = list(input())
for _ in range(3):
a = i.pop(-1)
i.insert(0, a)
if i == t:
print("YES")
else:
print("NO")
|
s460484041
|
Accepted
| 17 | 3,060 | 134 |
i = list(input())
t = list(input())
e = i.pop(-1)
s = i.pop(0)
i.insert(0,e)
i.append(s)
if i == t:
print("YES")
else:
print("NO")
|
s088685550
|
p03997
|
u872056745
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,060 | 74 |
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)
|
s222753346
|
Accepted
| 17 | 2,940 | 80 |
a = int(input())
b = int(input())
h = int(input())
print(int((a + b) * h / 2))
|
s395324234
|
p04025
|
u146575240
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 271 |
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal **integers** by transforming some of them. He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2). Find the minimum total cost to achieve his objective.
|
N = int(input())
A = list(map(int, input().split()))
A.sort()
A_s = set(A)
A_s_a = int(sum(A_s)//len(A_s))
if A[N//2+1-1] <= A_s_a:
pass
else:
A_s_a = A_s_a + 1
ans = 0
for i in range(N):
ans += (A[i]-A_s_a)**2
print(ans)
|
s395050797
|
Accepted
| 26 | 3,060 | 258 |
N = int(input())
A = list(map(int, input().split()))
ans = 10**8
for X in range(-100,101):
tmp = 0
for i in range(N):
tmp += (A[i]-X)**2
if ans >= tmp:
ans = tmp
else:
pass
print(ans)
|
s699525045
|
p03478
|
u812867074
| 2,000 | 262,144 |
Wrong Answer
| 30 | 2,940 | 122 |
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):
if a <= sum(map(int,str(i))) <= b:
ans+=1
print(ans)
|
s307708802
|
Accepted
| 30 | 2,940 | 123 |
n,a,b=map(int,input().split())
ans=0
for i in range(1,n+1):
if a <= sum(map(int,str(i))) <= b:
ans+=i
print(ans)
|
s893676923
|
p02396
|
u088372268
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,716 | 165 |
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.
|
x = list(input().split("\n"))
i = 1
for j in x:
int_j = int(j)
if int_j != 0:
print("Case", str(i)+":", int_j)
i += 1
else:
break
|
s354972007
|
Accepted
| 90 | 8,120 | 233 |
x = []
for k in range(10000):
n = input()
if n == "0":
break
x.append(int(n))
i = 1
for j in x:
int_j = int(j)
if int_j != 0:
print("Case", str(i)+":", int_j)
i += 1
else:
break
|
s813448030
|
p03795
|
u883232818
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 62 |
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())
x = 800 * N
y = (N / 15) * 200
print(x - y)
|
s275348563
|
Accepted
| 18 | 2,940 | 65 |
N = int(input())
x = 800 * N
y = int(N / 15) * 200
print(x - y)
|
s266125775
|
p03860
|
u268792407
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 37 |
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()[1]
print("A"+s+"C")
|
s412437186
|
Accepted
| 17 | 2,940 | 40 |
s=input().split()[1]
print("A"+s[0]+"C")
|
s517171083
|
p02267
|
u569960318
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,404 | 310 |
You are given a sequence of _n_ integers S and a sequence of different _q_ integers T. Write a program which outputs C, the number of integers in T which are also in the set S.
|
def linearSearch(S,t):
L = S + [t]
i = 0
while L[i] != t: i += 1
if i == len(L)-1: return 0
else : return 1
if __name__=='__main__':
n=input()
S=input().split()
q=input()
T=int,input().split()
cnt = 0
for t in T: cnt += linearSearch(S,t)
print(cnt)
|
s289446140
|
Accepted
| 60 | 8,092 | 306 |
def linearSearch(S,t):
L = S + [t]
i = 0
while L[i] != t: i += 1
if i == len(L)-1: return 0
else : return 1
if __name__=='__main__':
n=input()
S=input().split()
q=input()
T=input().split()
cnt = 0
for t in T: cnt += linearSearch(S,t)
print(cnt)
|
s323933744
|
p02393
|
u494314211
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,444 | 50 |
Write a program which reads three integers, and prints them in ascending order.
|
a=list(map(int,input().split()))
a.sort()
print(a)
|
s311938559
|
Accepted
| 20 | 7,644 | 63 |
a=list(map(int,input().split()))
a.sort()
print(a[0],a[1],a[2])
|
s055417074
|
p03679
|
u657512990
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 147 |
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
|
#list(map(int, input().split()))
x,a,b=list(map(int, input().split()))
if a>=b:print('delicious')
elif b-a>=x:print('safe')
else:print('dangerous')
|
s886145469
|
Accepted
| 17 | 3,060 | 147 |
#list(map(int, input().split()))
x,a,b=list(map(int, input().split()))
if a>=b:print('delicious')
elif b-a<=x:print('safe')
else:print('dangerous')
|
s186041159
|
p03795
|
u393512980
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 37 |
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
n=int(input())
print(800*n-200*n//15)
|
s641825121
|
Accepted
| 17 | 2,940 | 40 |
n=int(input())
print(800*n-200*(n//15))
|
s203458091
|
p03386
|
u626337957
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 111 |
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 num in (A, B+1):
if num <= A+K-1 or num >= A-K+1:
print(num)
|
s721903538
|
Accepted
| 18 | 3,060 | 212 |
A, B, K = map(int, input().split())
nums = []
for num in range(A, min(A+K, B+1)):
nums.append(num)
for num in range(max(B-K+1, A), B+1):
nums.append(num)
ans = sorted(set(nums))
for num in ans:
print(num)
|
s858679519
|
p02401
|
u682357930
| 1,000 | 131,072 |
Wrong Answer
| 40 | 7,676 | 351 |
Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.
|
while True:
l = input().split()
a = int(l[0])
b = int(l[2])
op = l[1]
if op == '?':
break
if op == '+':
ans = a + b
print(ans)
elif op == '-':
ans = a - b
print(ans)
elif op == '*':
ans = a * b
print(ans)
elif op == '/':
ans = a / b
print(ans)
|
s671109004
|
Accepted
| 30 | 7,668 | 356 |
while True:
l = input().split()
a = int(l[0])
b = int(l[2])
op = l[1]
if op == '?':
break
if op == '+':
ans = a + b
print(ans)
elif op == '-':
ans = a - b
print(ans)
elif op == '*':
ans = a * b
print(ans)
elif op == '/':
ans = int(a / b)
print(ans)
|
s812299821
|
p04012
|
u714533789
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 79 |
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()
odd=[i for i in set(w) if w.count(i)%2]
print('NO' if odd else 'YES')
|
s606447192
|
Accepted
| 17 | 2,940 | 79 |
w=input()
odd=[i for i in set(w) if w.count(i)%2]
print('No' if odd else 'Yes')
|
s565454161
|
p03449
|
u790877102
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 167 |
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j). You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down. The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them. At most how many candies can you collect when you choose the best way to travel?
|
n = int(input())
a1 = list(map(int,input().split()))
a2 = list(map(int,input().split()))
m = 0
for i in range(n):
m = max(m, sum(a1[0:i+1])+sum(a2[i:0]))
print(m)
|
s989360304
|
Accepted
| 17 | 2,940 | 163 |
n = int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
m = 0
for i in range(n):
m = max(m, sum(a[0:i+1])+sum(b[i:n]))
print(m)
|
s207342881
|
p03795
|
u543954314
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 42 |
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 - (n%15)*200)
|
s893925391
|
Accepted
| 17 | 2,940 | 44 |
n = int(input())
print(n*800 - (n//15)*200)
|
s101855145
|
p03548
|
u667024514
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 58 |
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?
|
a, b, c=map(int,input().split())
a=a-c
A=int(a/b)
print(A)
|
s084520857
|
Accepted
| 17 | 2,940 | 95 |
a = input().split()
b = int(a[0])
c = int(a[1])
d = int(a[2])
b = b-d
F = int(b/(c+d))
print(F)
|
s660147262
|
p03351
|
u461833298
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 131 |
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a, b, c, d = map(int, input().split())
if (abs(a-b) <= d & abs(c-b) <= d) | (abs(a-c) <= d):
print('Yes')
else:
print('No')
|
s389919057
|
Accepted
| 17 | 2,940 | 134 |
a, b, c, d = map(int, input().split())
if (abs(a-b) <= d and abs(c-b) <= d) or (abs(a-c) <= d):
print('Yes')
else:
print('No')
|
s110009806
|
p03545
|
u866833003
| 2,000 | 262,144 |
Wrong Answer
| 29 | 8,960 | 306 |
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
s = input()
for bit in range(1 << 3):
f = s[0]
ans = int(s[0])
for i in range(3):
if bit & (1 << i):
ans += int(s[i+1])
f += "+"
else:
ans -= int(s[i+1])
f += "-"
f += s[i+1]
if ans == 7:
print(f)
break
|
s541195311
|
Accepted
| 28 | 8,960 | 324 |
s = input()
for bit in range(1 << 3):
f = s[0]
ans = int(s[0])
for i in range(3):
if bit & (1 << i):
ans += int(s[i+1])
f += "+"
else:
ans -= int(s[i+1])
f += "-"
f += s[i+1]
if ans == 7:
f += "=7"
print(f)
break
|
s584820921
|
p02392
|
u464080148
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,588 | 87 |
Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No".
|
a,b,c=map(int,input().split())
if a<b and b<c:
print("YES")
else:
print("NO")
|
s843896394
|
Accepted
| 20 | 5,588 | 87 |
a,b,c=map(int,input().split())
if a<b and b<c:
print("Yes")
else:
print("No")
|
s963752354
|
p03394
|
u382423941
| 2,000 | 262,144 |
Wrong Answer
| 102 | 7,752 | 465 |
Nagase is a top student in high school. One day, she's analyzing some properties of special sets of positive integers. She thinks that a set S = \\{a_{1}, a_{2}, ..., a_{N}\\} of **distinct** positive integers is called **special** if for all 1 \leq i \leq N, the gcd (greatest common divisor) of a_{i} and the sum of the remaining elements of S is **not** 1. Nagase wants to find a **special** set of size N. However, this task is too easy, so she decided to ramp up the difficulty. Nagase challenges you to find a **special** set of size N such that the gcd of all elements are 1 and the elements of the set does not exceed 30000.
|
import fractions
def verify(seq):
sm = sum(seq)
print(all([fractions.gcd(2, sm-2) != 1 for s in seq]))
n = int(input())
cnt = 0
ans = []
for i in range(2, 30001):
if cnt == n - 1:
break
if i % 2 == 0 or i % 3 == 0:
ans.append(i)
cnt += 1
sm = sum(ans)
for j in range(i, 30001):
if ((sm + j) % (2*3)) == 0:
if j % 2 == 0 or j % 3 == 0:
ans.append(j)
break
print(' '.join(map(str, ans)))
|
s605582339
|
Accepted
| 38 | 5,180 | 661 |
n = int(input())
cnt = 0
ans = []
if n == 3:
print('2 5 63')
else:
for i in range(2, 30001):
if cnt == n - 2:
break
if i % 2 == 0 or i % 3 == 0 or i % 5 == 0:
ans.append(i)
cnt += 1
sm = sum(ans)
for j in range(i+1, 30001):
for k in range(j+1, 30001):
if (sm + j + k) % (2*3*5) == 0 and\
(j % 2 == 0 or j % 3 == 0 or j % 5 == 0) and\
(k % 2 == 0 or k % 3 == 0 or k % 5 == 0):
ans.append(j)
ans.append(k)
break
else:
continue
break
print(' '.join(map(str, ans)))
|
s292016952
|
p03971
|
u711238850
| 2,000 | 262,144 |
Wrong Answer
| 104 | 4,584 | 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 = tuple(map(int,input().split()))
s = list(input())
count = 0
foreign = 0
for s_i in s:
if s_i=='a':
if count<=a+b:
count+=1
print("Yes")
elif s_i=='b':
if count<=a+b and foreign <= b:
count+=1
foreign+=1
print("Yes")
else:
print("No")
|
s408310597
|
Accepted
| 106 | 4,712 | 370 |
n,a,b = tuple(map(int,input().split()))
s = list(input())
rank = 0
brank = 0
for s_i in s:
if s_i=='a':
if rank==a+b:
print("No")
continue
rank+=1
print("Yes")
elif s_i=='b':
if rank==a+b:
print("No")
continue
if brank==b:
print("No")
continue
rank+=1
brank+=1
print("Yes")
else:
print("No")
|
s234287629
|
p02936
|
u321035578
| 2,000 | 1,048,576 |
Wrong Answer
| 2,033 | 94,216 | 615 |
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations.
|
n,q = map(int,input().split())
bond = []
for i in range(n-1):
a,b = map(int,input().split())
a -= 1
b -= 1
if a > b:
tmp = a
a = b
b = tmp
bond.append((a,b))
bond.sort()
qry = []
for i in range(q):
p,x = map(int,input().split())
p -= 1
qry.append((p,x))
parent = [[] for i in range(n)]
for i, ab in enumerate(bond):
a,b = ab
parent[a].append(b)
ans = [0] * n
for i, qx in enumerate(qry):
q,x = qx
ans[q] += x
a = ''
for i, son in enumerate(parent):
a += str(ans[i]) + ' '
for j, s in enumerate(son):
ans[s] += ans[i]
print(a)
|
s163142080
|
Accepted
| 1,753 | 67,204 | 625 |
from collections import deque
n,q = map(int,input().split())
edge = [[] for i in range(n)]
for i in range(n-1):
a,b = map(int,input().split())
a -= 1
b -= 1
edge[a].append(b)
edge[b].append(a)
ans = [0] * n
for i in range(q):
p,x = map(int,input().split())
p -= 1
ans[p] += x
isVisited = [False] * n
dq = deque([0])
isVisited[0] = True
while len(dq) > 0:
parent = dq.popleft()
for j, son in enumerate(edge[parent]):
if isVisited[son]:
continue
ans[son] += ans[parent]
isVisited[son] = True
dq.append(son)
print(' '.join(map(str, ans)))
|
s271592913
|
p03129
|
u041196979
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 98 |
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.
|
N, K = map(int, input().split())
if (N + K - 1) // 2 >= K:
print("Yes")
else:
print("No")
|
s489758365
|
Accepted
| 17 | 2,940 | 94 |
N, K = map(int, input().split())
if (N + 1) // 2 >= K:
print("YES")
else:
print("NO")
|
s147741668
|
p03494
|
u724154852
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,104 | 2,940 | 227 |
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.
|
import sys
input = sys.stdin.readline
n = int(input())
a = (input().split())
b = 1
while True:
for i in range(n-1):
if int(a[i]) % 2 == 1:
break
a[i] = int(a[i]) / 2
b +=1
print(b)
|
s051674949
|
Accepted
| 17 | 3,060 | 270 |
import sys
input = sys.stdin.readline
n = int(input())
a = list(map(int,input().split()))
b = 0
c = 2
f = 0
while True:
for i in a:
if i % c > 0:
f = f +1
break
if f == 1:
break
b = b + 1
c = c * 2
print(b)
|
s273709004
|
p04045
|
u252160635
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,224 | 631 |
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 = list(map(int, input().split()))
D = list(map(int, input().split()))
availbale_numbers = [0,1,2,3,4,5,6,7,8,9]
for i in D:
availbale_numbers.remove(i)
print(availbale_numbers)
flg = False
for item1 in availbale_numbers:
for item2 in availbale_numbers:
for item3 in availbale_numbers:
for item4 in availbale_numbers:
if(N <= int(('').join([str(item1),str(item2),str(item3),str(item4)]))):
print(('').join([str(item1),str(item2),str(item3),str(item4)]))
exit()
Nmin = min(availbale_numbers)
print(f'{Nmin}{Nmin}{Nmin}{Nmin}{Nmin}')
|
s879855431
|
Accepted
| 100 | 9,076 | 292 |
N,K = list(map(int, input().split()))
D = list(map(int, input().split()))
availbale_numbers = [0,1,2,3,4,5,6,7,8,9]
for i in D:
availbale_numbers.remove(i)
for i in range(N,99999):
if(set(str(i)) <= set([str(a) for a in availbale_numbers])):
print(i)
break
|
s429020718
|
p03994
|
u436484848
| 2,000 | 262,144 |
Wrong Answer
| 72 | 3,700 | 309 |
Mr. Takahashi has a string s consisting of lowercase English letters. He repeats the following operation on s exactly K times. * Choose an arbitrary letter on s and change that letter to the next alphabet. Note that the next letter of `z` is `a`. For example, if you perform an operation for the second letter on `aaz`, `aaz` becomes `abz`. If you then perform an operation for the third letter on `abz`, `abz` becomes `aba`. Mr. Takahashi wants to have the lexicographically smallest string after performing exactly K operations on s. Find the such string.
|
str = input()
K = int(input())
tempK = K
result = ""
for i in range(len(str)):
changeNum = 122-ord(str[i])+1
if (changeNum <= tempK):
result += "a"
tempK -= changeNum
else:
result += str[i]
print(result)
if (tempK != 0):
result = result[:-1] + chr(ord(str[-1]) + tempK)
print(result)
|
s208917887
|
Accepted
| 83 | 3,572 | 451 |
str = input()
K = int(input())
tempK = K
result = ""
for i in range(len(str)-1):
changeNum = 122-ord(str[i])+1
if (changeNum <= tempK and str[i] != "a"):
result += "a"
tempK -= changeNum
else:
result += str[i]
tempK %= 26
if (tempK > 0):
changeNum = ord(str[-1]) + tempK
if (changeNum > 122):
changeNum = changeNum -122 + 97 -1
else:
changeNum = changeNum
result += chr(changeNum)
else:
result += str[-1]
print(result)
|
s090139485
|
p03657
|
u729294108
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 138 |
Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.
|
a, b = map(int, input().strip().split())
if all([s % 3 == 0 for s in [a, b, a + b]]):
print('Possible')
else:
print('Impossible')
|
s739335406
|
Accepted
| 17 | 2,940 | 138 |
a, b = map(int, input().strip().split())
if any([s % 3 == 0 for s in [a, b, a + b]]):
print('Possible')
else:
print('Impossible')
|
s124686301
|
p03455
|
u794910686
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 100 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a,b=(int(x) for x in input().split())
if a*b%2 == 0:
print("even")
else:
print("odd")
|
s093168850
|
Accepted
| 17 | 2,940 | 95 |
a,b=(int(x) for x in input().split())
if a*b%2 == 0:
print("Even")
else:
print("Odd")
|
s226345284
|
p03485
|
u647767910
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,188 | 66 |
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 math
a, b = map(int, input().split())
print(math.ceil(a+b))
|
s506509071
|
Accepted
| 17 | 2,940 | 74 |
import math
a, b = map(int, input().split())
print(math.ceil((a + b) / 2))
|
s158740078
|
p03359
|
u892091780
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 168 |
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())
cnt = 0
if a <= b :
while(cnt <= a):
print(cnt+1)
cnt = cnt+1
else :
while(cnt < a):
print(cnt+1)
cnt = cnt+1
|
s730341281
|
Accepted
| 17 | 2,940 | 77 |
a,b = map(int,input().split())
if a <= b :
print(a)
else :
print(a-1)
|
s790365518
|
p03433
|
u767664985
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 109 |
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
N = int(input())
a = list(map(int, input().split()))
b = sorted(a)[:: -1]
print(sum(b[:: 2]) - sum(b[1:: 2]))
|
s049905369
|
Accepted
| 17 | 2,940 | 83 |
N, A = eval("int(input())," * 2)
if N % 500 <= A:
print("Yes")
else:
print("No")
|
s286850930
|
p03131
|
u859897687
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 207 |
Snuke has one biscuit and zero Japanese yen (the currency) in his pocket. He will perform the following operations exactly K times in total, in the order he likes: * Hit his pocket, which magically increases the number of biscuits by one. * Exchange A biscuits to 1 yen. * Exchange 1 yen to B biscuits. Find the maximum possible number of biscuits in Snuke's pocket after K operations.
|
k,a,b=map(int,input().split())
n=1
if b-a<=2:
print(k+1)
else:
if n < a:
n = a
k = k-a
m = k // 2
print(n,m,b-a)
n += (b-a)*m
if k % 2 ==1:
n+=1
print(n)
|
s523373720
|
Accepted
| 17 | 3,060 | 258 |
k,a,b=map(int,input().split())
n=1
if b-a<=2:
n += k
print(n)
elif k<a-1:
n += k
print(n)
else: #b-a>2 k-(a-1)>=0
n = a
k -= a-1 #>0
m = k // 2
n += (b-a)*m
if k - 2*m ==1:
n+=1
print(n)
|
s770532980
|
p03417
|
u902151549
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 25 |
There is a grid with infinitely many rows and columns. In this grid, there is a rectangular region with consecutive N rows and M columns, and a card is placed in each square in this region. The front and back sides of these cards can be distinguished, and initially every card faces up. We will perform the following operation once for each square contains a card: * For each of the following nine squares, flip the card in it if it exists: the target square itself and the eight squares that shares a corner or a side with the target square. It can be proved that, whether each card faces up or down after all the operations does not depend on the order the operations are performed. Find the number of cards that face down after all the operations.
|
print(2)
exit()
print(3)
|
s594189332
|
Accepted
| 42 | 5,724 | 3,571 |
# coding: utf-8
import re
import math
from collections import defaultdict
from collections import deque
from fractions import Fraction
import itertools
from copy import deepcopy
import random
import time
import os
import queue
import sys
import datetime
from functools import lru_cache
#@lru_cache(maxsize=None)
readline=sys.stdin.readline
sys.setrecursionlimit(2000000)
#import numpy as np
alphabet="abcdefghijklmnopqrstuvwxyz"
mod=int(10**9+7)
inf=int(10**20)
def yn(b):
if b:
print("yes")
else:
print("no")
def Yn(b):
if b:
print("Yes")
else:
print("No")
def YN(b):
if b:
print("YES")
else:
print("NO")
class union_find():
def __init__(self,n):
self.n=n
self.P=[a for a in range(n)]
self.rank=[0]*n
def find(self,x):
if(x!=self.P[x]):self.P[x]=self.find(self.P[x])
return self.P[x]
def same(self,x,y):
return self.find(x)==self.find(y)
def link(self,x,y):
if self.rank[x]<self.rank[y]:
self.P[x]=y
elif self.rank[y]<self.rank[x]:
self.P[y]=x
else:
self.P[x]=y
self.rank[y]+=1
def unite(self,x,y):
self.link(self.find(x),self.find(y))
def size(self):
S=set()
for a in range(self.n):
S.add(self.find(a))
return len(S)
def is_pow(a,b):
now=b
while now<a:
now*=b
if now==a:return True
else:return False
def getbin(num,size):
A=[0]*size
for a in range(size):
if (num>>(size-a-1))&1==1:
A[a]=1
else:
A[a]=0
return A
def get_facs(n,mod_=0):
A=[1]*(n+1)
for a in range(2,len(A)):
A[a]=A[a-1]*a
if(mod_>0):A[a]%=mod_
return A
def comb(n,r,mod,fac):
if(n-r<0):return 0
return (fac[n]*pow(fac[n-r],mod-2,mod)*pow(fac[r],mod-2,mod))%mod
def next_comb(num,size):
x=num&(-num)
y=num+x
z=num&(~y)
z//=x
z=z>>1
num=(y|z)
if(num>=(1<<size)):return False
else:return num
def get_primes(n,type="int"):
if n==0:
if type=="int":return []
else:return [False]
A=[True]*(n+1)
A[0]=False
A[1]=False
for a in range(2,n+1):
if A[a]:
for b in range(a*2,n+1,a):
A[b]=False
if(type=="bool"):return A
B=[]
for a in range(n+1):
if(A[a]):B.append(a)
return B
def is_prime(num):
if(num<=1):return False
i=2
while i*i<=num:
if(num%i==0):return False
i+=1
return True
def ifelse(a,b,c):
if a:return b
else:return c
def join(A,c=""):
n=len(A)
A=list(map(str,A))
s=""
for a in range(n):
s+=A[a]
if(a<n-1):s+=c
return s
def factorize(n,type_="dict"):
b = 2
list_ = []
while b * b <= n:
while n % b == 0:
n //= b
list_.append(b)
b+=1
if n > 1:list_.append(n)
if type_=="dict":
dic={}
for a in list_:
if a in dic:
dic[a]+=1
else:
dic[a]=1
return dic
elif type_=="list":
return list_
else:
return None
def pm(x):
return x//abs(x)
def inputintlist():
return list(map(int,input().split()))
######################################################################################################
H,W=map(int,input().split())
if H==1 and W==1:
ans=1
elif H==1 or W==1:
ans=max(H,W)-2
else:
ans=(H-2)*(W-2)
print(ans)
|
s695871062
|
p03523
|
u984529214
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,188 | 133 |
You are given a string S. Takahashi can insert the character `A` at any position in this string any number of times. Can he change S into `AKIHABARA`?
|
import re
akb_re = re.compile(r'A?KIHA?BA?RA?$')
s = input()
mo = akb_re.match(s)
if mo == None:
print('No')
else:
print('Yes')
|
s867687911
|
Accepted
| 19 | 3,188 | 133 |
import re
akb_re = re.compile(r'A?KIHA?BA?RA?$')
s = input()
mo = akb_re.match(s)
if mo == None:
print('NO')
else:
print('YES')
|
s262843712
|
p03141
|
u689562091
| 2,000 | 1,048,576 |
Wrong Answer
| 2,109 | 43,820 | 411 |
There are N dishes of cuisine placed in front of Takahashi and Aoki. For convenience, we call these dishes Dish 1, Dish 2, ..., Dish N. When Takahashi eats Dish i, he earns A_i points of _happiness_ ; when Aoki eats Dish i, she earns B_i points of happiness. Starting from Takahashi, they alternately choose one dish and eat it, until there is no more dish to eat. Here, both of them choose dishes so that the following value is maximized: "the sum of the happiness he/she will earn in the end" minus "the sum of the happiness the other person will earn in the end". Find the value: "the sum of the happiness Takahashi earns in the end" minus "the sum of the happiness Aoki earns in the end".
|
N = int(input())
l = []
for i in range(N):
l.append(list(map(int, input().split())))
l[i] += [i]
lA = sorted(l, key=lambda x: (x[0], -x[1]))
lB = sorted(l, key=lambda x: (-x[1]))
A = B = 0
while lA or lB:
#A
if i % 2 == 0:
tmp = lA.pop(0)
A += tmp[0]
lB.remove(tmp)
#B
else:
tmp = lB.pop(0)
B += tmp[1]
lA.remove(tmp)
i += 1
print(A-B)
|
s928555396
|
Accepted
| 369 | 7,388 | 208 |
N = int(input())
l = []
sumB = 0
for i in range(N):
a, b = map(int, input().split())
l.append(a + b)
sumB += b
l.sort(reverse=True)
A = 0
for i in range(0, N, 2):
A += l[i]
print(A - sumB)
|
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