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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s881332816
|
p03149
|
u371467115
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 88
|
You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974".
|
n=list(map(int,input().split()))
s=[1,9,4,7]
if n==s:
print("YES")
else:
print("NO")
|
s648632399
|
Accepted
| 17
| 2,940
| 98
|
n=list(map(int,input().split()))
s=[1,9,4,7]
if set(n)==set(s):
print("YES")
else:
print("NO")
|
s935552964
|
p03493
|
u571710867
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 30
|
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 = str(input())
s.count('1')
|
s774529980
|
Accepted
| 17
| 2,940
| 36
|
s = str(input())
print(s.count('1'))
|
s458171720
|
p03719
|
u079699418
| 2,000
| 262,144
|
Wrong Answer
| 29
| 9,088
| 95
|
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
a,b,c=map(int, input().split())
if a <= c <= b:
print('YES')
else:
print('NO')
|
s314659956
|
Accepted
| 26
| 9,076
| 85
|
a,b,c=map(int,input().split())
if a <= c <= b:
print('Yes')
else:
print('No')
|
s461752545
|
p02614
|
u595353654
| 1,000
| 1,048,576
|
Wrong Answer
| 123
| 27,192
| 3,292
|
We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is `.`, and black if c_{i,j} is `#`. Consider doing the following operation: * Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns. You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices.
|
##a = int(stdin.readline().rstrip())
##b, c = [int(x) for x in stdin.readline().rstrip().split()]
##s = stdin.readline().rstrip()
##a = list(map(int,[int(x) for x in stdin.readline().rstrip().split()])) a[0, 1, 2, ...]
##a = [[0] * 2 for _ in range(n)] a[0,0]
# -*- coding: utf-8 -*-
from sys import stdin
from operator import itemgetter
from collections import deque, Counter
import math
import pprint
from functools import reduce
import numpy as np
import random
import bisect
import copy
MOD = 1000000007
INF = float('inf')
alpha = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
def keta(kazu):
kazu_str = str(kazu)
kazu_list = [int(kazu_str[i]) for i in range(0, len(kazu_str))]
return kazu_list
def gcd(*numbers):
return reduce(math.gcd, numbers)
def combination(m,n): # mCn
if n > m:
return 'すまん'
return math.factorial(m) // (math.factorial(m-n) * math.factorial(n))
def pow_k(x,n):
if n == 0:
return 1
K = 1
while n > 1:
if n % 2 != 0:
K *= x
x *= x
n //= 2
return K * x
def fact(n):
arr = {}
temp = n
for i in range(2,int(n**0.5)+1):
if temp % i == 0:
cnt = 0
while temp % i == 0:
cnt += 1
temp //= i
arr[i] = cnt
if temp != 1:
arr[temp] = 1
if arr == {}:
arr[n] = 1
return arr
def main():
h, w, k = [int(x) for x in stdin.readline().rstrip().split()]
board = []
kuro = 0
pattern_h = [[0]*h for _ in range(2 ** h)]
pattern_w = [[0]*w for _ in range(2 ** w)]
ans = 0
nokori = 0
for _ in range(h):
s = stdin.readline().rstrip()
v = [s[i] for i in range(0, len(s))]
kuro += v.count("#")
board.append(v)
memo = copy.deepcopy(board)
for i in range(2 ** h):
for j in range(h):
if (i >> j) & 1:
pattern_h[i][j] = j + 1
for i in range(2 ** w):
for j in range(w):
if (i >> j) & 1:
pattern_w[i][j] = j + 1
for i in range(2 ** w):
for j in range(w):
if pattern_w[i][j] != 0:
for l in range(h):
memo[l][j] = "x"
print(memo)
for i2 in range(2 ** h):
for j2 in range(h):
if pattern_h[i2][j2] == 0:
nokori += memo[j2].count("#")
if k == nokori:
ans += 1
nokori = 0
memo = copy.deepcopy(board)
print(ans)
main()
|
s536159774
|
Accepted
| 122
| 27,288
| 3,272
|
##a = int(stdin.readline().rstrip())
##b, c = [int(x) for x in stdin.readline().rstrip().split()]
##s = stdin.readline().rstrip()
##a = list(map(int,[int(x) for x in stdin.readline().rstrip().split()])) a[0, 1, 2, ...]
##a = [[0] * 2 for _ in range(n)] a[0,0]
# -*- coding: utf-8 -*-
from sys import stdin
from operator import itemgetter
from collections import deque, Counter
import math
import pprint
from functools import reduce
import numpy as np
import random
import bisect
import copy
MOD = 1000000007
INF = float('inf')
alpha = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
def keta(kazu):
kazu_str = str(kazu)
kazu_list = [int(kazu_str[i]) for i in range(0, len(kazu_str))]
return kazu_list
def gcd(*numbers):
return reduce(math.gcd, numbers)
def combination(m,n): # mCn
if n > m:
return 'すまん'
return math.factorial(m) // (math.factorial(m-n) * math.factorial(n))
def pow_k(x,n):
if n == 0:
return 1
K = 1
while n > 1:
if n % 2 != 0:
K *= x
x *= x
n //= 2
return K * x
def fact(n):
arr = {}
temp = n
for i in range(2,int(n**0.5)+1):
if temp % i == 0:
cnt = 0
while temp % i == 0:
cnt += 1
temp //= i
arr[i] = cnt
if temp != 1:
arr[temp] = 1
if arr == {}:
arr[n] = 1
return arr
def main():
h, w, k = [int(x) for x in stdin.readline().rstrip().split()]
board = []
kuro = 0
pattern_h = [[0]*h for _ in range(2 ** h)]
pattern_w = [[0]*w for _ in range(2 ** w)]
ans = 0
nokori = 0
for _ in range(h):
s = stdin.readline().rstrip()
v = [s[i] for i in range(0, len(s))]
kuro += v.count("#")
board.append(v)
memo = copy.deepcopy(board)
for i in range(2 ** h):
for j in range(h):
if (i >> j) & 1:
pattern_h[i][j] = j + 1
for i in range(2 ** w):
for j in range(w):
if (i >> j) & 1:
pattern_w[i][j] = j + 1
for i in range(2 ** w):
for j in range(w):
if pattern_w[i][j] != 0:
for l in range(h):
memo[l][j] = "x"
for i2 in range(2 ** h):
for j2 in range(h):
if pattern_h[i2][j2] == 0:
nokori += memo[j2].count("#")
if k == nokori:
ans += 1
nokori = 0
memo = copy.deepcopy(board)
print(ans)
main()
|
s043944742
|
p03448
|
u186893542
| 2,000
| 262,144
|
Wrong Answer
| 56
| 3,064
| 204
|
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())
res = 0
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
if a*i + b*j + c*k == x:
res += 1
print(res)
|
s817983712
|
Accepted
| 49
| 3,064
| 210
|
a = int(input())
b = int(input())
c = int(input())
x = int(input())
res = 0
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
if 500*i + 100*j + 50*k == x:
res += 1
print(res)
|
s871643059
|
p02418
|
u914146430
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,404
| 70
|
Write a program which finds a pattern $p$ in a ring shaped text $s$.
|
s=input()
p=input()
if p in s:
print("Yes")
else:
print("No")
|
s055686084
|
Accepted
| 20
| 7,476
| 76
|
s=input()
p=input()
s+=s
if p in s:
print("Yes")
else:
print("No")
|
s909330694
|
p02690
|
u768496010
| 2,000
| 1,048,576
|
Wrong Answer
| 44
| 9,164
| 250
|
Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X.
|
x = int(input())
def fun(x):
res = []
for a in range(-100, 100):
for b in range(-100, 100):
mult = a**5 - b**5
if mult == x:
res = a, b
break
print(str(res)[1:-1])
fun(x)
|
s817400918
|
Accepted
| 1,988
| 9,116
| 254
|
x = int(input())
def fun(x):
res = []
for a in range(-1000, 1000):
for b in range(-1000, 1000):
mult = a**5 - b**5
if mult == x:
res = a, b
break
print(res[0], res[1])
fun(x)
|
s391738219
|
p03680
|
u021337285
| 2,000
| 262,144
|
Wrong Answer
| 175
| 13,392
| 218
|
Takahashi wants to gain muscle, and decides to work out at AtCoder Gym. The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it. Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up. Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.
|
N = int(input())
a = list(map(int, [input() for i in range(N)]))
current_index = 1
for i in range(N - 1):
current_index = a[current_index - 1]
if current_index == 2:
print(i)
exit()
print(-1)
|
s906436112
|
Accepted
| 173
| 13,356
| 221
|
N = int(input())
a = list(map(int, [input() for i in range(N)]))
current_index = 1
for i in range(N - 1):
current_index = a[current_index - 1]
if current_index == 2:
print(i + 1)
exit()
print(-1)
|
s379112734
|
p03407
|
u488127128
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 66
|
An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.
|
a,b,c = map(int, input().split())
print('Yes' if a+b<=c else 'No')
|
s543550640
|
Accepted
| 17
| 2,940
| 66
|
a,b,c = map(int, input().split())
print('Yes' if a+b>=c else 'No')
|
s756743664
|
p03493
|
u254091062
| 2,000
| 262,144
|
Wrong Answer
| 26
| 9,048
| 24
|
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.
|
a = input()
a.count('1')
|
s263759916
|
Accepted
| 26
| 9,076
| 31
|
a = input()
print(a.count('1'))
|
s605004580
|
p03416
|
u419686324
| 2,000
| 262,144
|
Wrong Answer
| 63
| 2,940
| 102
|
Find the number of _palindromic numbers_ among the integers between A and B (inclusive). Here, a palindromic number is a positive integer whose string representation in base 10 (without leading zeros) reads the same forward and backward.
|
a, b = map(int, input().split())
sum(map(lambda i: 1 if str(i) == str(i)[::-1] else 0, range(a, b+1)))
|
s087235690
|
Accepted
| 62
| 3,064
| 109
|
a, b = map(int, input().split())
print(sum(map(lambda i: 1 if str(i) == str(i)[::-1] else 0, range(a, b+1))))
|
s260091503
|
p02646
|
u457460736
| 2,000
| 1,048,576
|
Wrong Answer
| 31
| 9,612
| 343
|
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.
|
import math
import collections
import itertools
import copy
def YesNo(Bool):
if(Bool):
print("YES")
else:
print("NO")
return
def resolve():
A,V=map(int,input().split())
B,W=map(int,input().split())
T=int(input())
if(A+V*T >= B+W*T):
print("Yes")
else:
print("No")
resolve()
|
s975048352
|
Accepted
| 33
| 9,456
| 589
|
import math
import collections
import itertools
import copy
def YesNo(Bool):
if(Bool):
print("YES")
else:
print("NO")
return
def resolve():
A,V=map(int,input().split())
B,W=map(int,input().split())
T=int(input())
if(A<B and V<=W):
print("NO")
return
if(A>B and V<=W):
print("NO")
return
if(A<B):
if(A+V*T >= B+W*T):
print("YES")
else:
print("NO")
else:
if(A-V*T <= B-W*T):
print("YES")
else:
print("NO")
resolve()
|
s459116939
|
p03962
|
u393229280
| 2,000
| 262,144
|
Wrong Answer
| 25
| 3,444
| 79
|
AtCoDeer the deer recently bought three paint cans. The color of the one he bought two days ago is a, the color of the one he bought yesterday is b, and the color of the one he bought today is c. Here, the color of each paint can is represented by an integer between 1 and 100, inclusive. Since he is forgetful, he might have bought more than one paint can in the same color. Count the number of different kinds of colors of these paint cans and tell him.
|
import collections
a = input()
c = collections.Counter(list(a))
print(len(c))
|
s499888370
|
Accepted
| 20
| 3,316
| 82
|
import collections
a = input()
c = collections.Counter(a.split())
print(len(c))
|
s324874509
|
p02612
|
u724742135
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 9,152
| 70
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
from sys import stdin
N = int(stdin.readline().rstrip())
print(N%1000)
|
s172613928
|
Accepted
| 25
| 9,056
| 82
|
from sys import stdin
N = int(stdin.readline().rstrip())
print((1000-N%1000)%1000)
|
s392146572
|
p03379
|
u735008991
| 2,000
| 262,144
|
Wrong Answer
| 491
| 25,228
| 165
|
When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N.
|
import bisect
N = int(input())
X = list(map(int, input().split()))
Y = sorted(X.copy())
for x in X:
print(Y[N//2-1] if bisect.bisect(Y, x) >= N//2 else Y[N//2])
|
s727332240
|
Accepted
| 496
| 25,224
| 164
|
import bisect
N = int(input())
X = list(map(int, input().split()))
Y = sorted(X.copy())
for x in X:
print(Y[N//2-1] if bisect.bisect(Y, x) > N//2 else Y[N//2])
|
s708259337
|
p03623
|
u665038048
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 66
|
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x, a, b = map(int, input().split())
print(min(abs(x-a), abs(x-b)))
|
s667967166
|
Accepted
| 17
| 2,940
| 95
|
x, a, b = map(int, input().split())
if abs(x-a) > abs(x-b):
print('B')
else:
print('A')
|
s177474630
|
p03777
|
u789364190
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 98
|
Two deer, AtCoDeer and TopCoDeer, are playing a game called _Honest or Dishonest_. In this game, an honest player always tells the truth, and an dishonest player always tell lies. You are given two characters a and b as the input. Each of them is either `H` or `D`, and carries the following information: If a=`H`, AtCoDeer is honest; if a=`D`, AtCoDeer is dishonest. If b=`H`, AtCoDeer is saying that TopCoDeer is honest; if b=`D`, AtCoDeer is saying that TopCoDeer is dishonest. Given this information, determine whether TopCoDeer is honest.
|
str = input().split(' ')
a = str[0]
b = str[1]
res = 'H'
if a != b:
res = 'D'
else:
res = 'H'
|
s190605197
|
Accepted
| 17
| 2,940
| 111
|
str = input().split(' ')
a = str[0]
b = str[1]
res = 'H'
if a != b:
res = 'D'
else:
res = 'H'
print(res)
|
s638941644
|
p01125
|
u124909914
| 8,000
| 131,072
|
Wrong Answer
| 90
| 6,780
| 859
|
宇宙暦 1603〜1867 年，人々はその時代のことを EDO 時代と呼ぶ．EDO とは当時最先端の宇宙航行技術，Enhanced Driving Operation のことであり、1603 年に Dr.Izy によって開発された． あなたは宇宙冒険家であり，宇宙を飛び回って様々な惑星を冒険していた．その冒険の途中で，あなたはとても不思議な惑星を発見した．その惑星には，至るところに七色に輝く不思議な宝石が落ちていた．あなたはその惑星への降下を試みようと考えたが，重大な問題のためにそれは不可能であることがわかった．その惑星の空気には，人間が触れただけで即死してしまうような猛毒の成分が含まれていたのだ． そこで，あなたはロボットを使って宝石を回収することを考えついた．あなたはその惑星の周回軌道にて待機する．そして，降下させたロボットを遠隔操作することによって宝石を回収するのだ．あなたは，ロボットに「移動する方向」と「移動する距離」の組からなる命令の列によってロボットを遠隔操作する．ロボットは移動経路上（到達点を含む）に宝石を見つけると，それらを全て回収する． あなたの仕事は，ロボットが与えられた全ての命令の実行を終えたときに，全ての宝石を回収することができたかどうかを判定するプログラムを書くことである． なお，ロボットが不思議な宝石を回収する範囲はそれほど広くない．そのため，ロボットが移動する範囲は全て 2 次元の平面で表すことができる．そして，ロボットが移動する範囲は (0,0) および (20,20) をそれぞれ左下隅および右上隅とする正方形の内部（境界線を含む）である．ロボットは常に範囲の中央，すなわち (10,10) の座標に降下する．また，全ての宝石は中央以外の格子点上にあることが保証されている．
|
#!/usr/bin/python3
while True:
n = int(input())
if n == 0 : break
gems = []
for i in range(n):
gem = tuple(map(int, input().split()))
gems.append(gem)
m = int(input())
bot = [10, 10]
history = [(10, 10)]
for j in range(m):
move = list(input().split())
move[1] = int(move[1])
if move[0] == 'N':
index = 1 #y
direc = 1 # pos
elif move[0] == 'S':
index = 1
derec = -1
elif move[0] == 'E':
index = 0 #x
direc = 1
else: #W
index = 0
direc = -1
for j in range(abs(move[1])):
bot[index] += direc
history.append(tuple(bot))
for g in gems:
if g not in history:
print("No")
break
print("Yes")
|
s985437313
|
Accepted
| 90
| 6,784
| 885
|
#!/usr/bin/python3
while True:
n = int(input())
if n == 0 : break
gems = []
for i in range(n):
gem = tuple(map(int, input().split()))
gems.append(gem)
m = int(input())
bot = [10, 10]
history = [(10, 10)]
for j in range(m):
move = list(input().split())
move[1] = int(move[1])
if move[0] == 'N':
index = 1 #y
direc = 1 # pos
elif move[0] == 'S':
index = 1
direc = -1
elif move[0] == 'E':
index = 0 #x
direc = 1
else: #W
index = 0
direc = -1
for j in range(abs(move[1])):
bot[index] += direc
history.append(tuple(bot))
f = True
for g in gems:
if g not in history:
f = False
break
print("Yes" if f else "No")
|
s588012097
|
p03712
|
u370413678
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 195
|
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.
|
size=[int(i) for i in input().split()]
image=[[str(input())] for i in range(size[0])]
print(image)
print("#"*(size[1]+2))
for i in image:
print(''.join(["#"]+i+["#"]))
print("#"*(size[1]+2))
|
s330578164
|
Accepted
| 17
| 3,060
| 182
|
size=[int(i) for i in input().split()]
image=[[str(input())] for i in range(size[0])]
print("#"*(size[1]+2))
for i in image:
print(''.join(["#"]+i+["#"]))
print("#"*(size[1]+2))
|
s665558346
|
p02694
|
u746154235
| 2,000
| 1,048,576
|
Wrong Answer
| 25
| 9,248
| 150
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
X = int(input())
import math
price=100
year=0
while True:
price += round(price*0.01)
year+=1
print(price)
if price >= X:
break
print(year)
|
s813456506
|
Accepted
| 23
| 9,144
| 120
|
import math
X = int(input())
price=100
year=0
while price < X:
price += math.floor(price*0.01)
year+=1
print(year)
|
s929133122
|
p03796
|
u807021746
| 2,000
| 262,144
|
Wrong Answer
| 2,206
| 9,924
| 87
|
Snuke loves working out. He is now exercising N times. Before he starts exercising, his _power_ is 1. After he exercises for the i-th time, his power gets multiplied by i. Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7.
|
import math
N = int(input())
a = math.factorial(N)
print(a)
x = a % (10**9+7)
print(x)
|
s557619520
|
Accepted
| 157
| 9,892
| 77
|
import math
N = int(input())
a = math.factorial(N)
x = a % (10**9+7)
print(x)
|
s827181656
|
p03556
|
u223904637
| 2,000
| 262,144
|
Wrong Answer
| 23
| 2,940
| 64
|
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())
ans=0
while ans*ans<=n:
ans+=1
print(ans-1)
|
s954504624
|
Accepted
| 23
| 2,940
| 69
|
n=int(input())
ans=0
while ans*ans<=n:
ans+=1
print((ans-1)**2)
|
s864221065
|
p04046
|
u957084285
| 2,000
| 262,144
|
Wrong Answer
| 328
| 18,804
| 479
|
We have a large square grid with H rows and W columns. Iroha is now standing in the top-left cell. She will repeat going right or down to the adjacent cell, until she reaches the bottom-right cell. However, she cannot enter the cells in the intersection of the bottom A rows and the leftmost B columns. (That is, there are A×B forbidden cells.) There is no restriction on entering the other cells. Find the number of ways she can travel to the bottom-right cell. Since this number can be extremely large, print the number modulo 10^9+7.
|
MOD = 10**9+7
N = 200000
p = [1] * (N+1)
q = [1] * (N+1)
for i in range(1, N+1):
p[i] = (p[i-1]*(i)%MOD)
q[0] = pow(p[-1], MOD-2, MOD)
for i in range(1, N+1):
q[i] = (N-i+1)*q[i-1]%MOD
q.reverse()
def nCk(n,k):
if k > n or n == 0:
return 0
elif k == 0:
return 1
else:
return p[n]*q[k]%MOD*q[n-k]%MOD
h,w,a,b = map(int, input().split())
ans = 0
for i in range(b, w):
ans += nCk(h-a+i-1, i)*nCk(w-i-2+a, w-i-1)%MOD
print(ans%MOD)
|
s292088103
|
Accepted
| 335
| 18,804
| 492
|
MOD = 10**9+7
N = 200000
p = [1] * (N+1)
q = [1] * (N+1)
for i in range(1, N+1):
p[i] = (p[i-1]*(i)%MOD)
q[0] = pow(p[-1], MOD-2, MOD)
for i in range(1, N+1):
q[i] = (N-i+1)*q[i-1]%MOD
q.reverse()
def nCk(n,k):
if k > n or (k != 0 and n == 0):
return 0
elif k == 0:
return 1
else:
return p[n]*q[k]%MOD*q[n-k]%MOD
h,w,a,b = map(int, input().split())
ans = 0
for i in range(b, w):
ans += nCk(h-a+i-1, i)*nCk(w-i-2+a, w-i-1)%MOD
print(ans%MOD)
|
s112776093
|
p03759
|
u143051858
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 78
|
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful.
|
A,B,C=map(int,input().split())
if B-A==C-A:
print('YES')
else:
print('NO')
|
s881389677
|
Accepted
| 17
| 2,940
| 79
|
A,B,C=map(int,input().split())
if B-A==C-B:
print('YES')
else:
print('NO')
|
s758973735
|
p03836
|
u044459372
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 385
|
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.
|
def main():
sx, sy, tx, ty = map(int, input().split())
ans = solve(sx, sy, tx, ty)
print(ans)
def solve(sx, sy, tx, ty):
dx = tx - sx
dy = ty - sy
st = 'U' * dy + 'R' * dx
ts = 'D' * dy + 'L' * dy
rt1 = st + ts
rt2 = 'L' + 'U' + st + 'R' + 'D' + 'R' + 'D' + ts + 'L' + 'U'
ans = rt1 + rt2
return ans
if __name__ == '__main__':
main()
|
s394976088
|
Accepted
| 17
| 3,064
| 385
|
def main():
sx, sy, tx, ty = map(int, input().split())
ans = solve(sx, sy, tx, ty)
print(ans)
def solve(sx, sy, tx, ty):
dx = tx - sx
dy = ty - sy
st = 'U' * dy + 'R' * dx
ts = 'D' * dy + 'L' * dx
rt1 = st + ts
rt2 = 'L' + 'U' + st + 'R' + 'D' + 'R' + 'D' + ts + 'L' + 'U'
ans = rt1 + rt2
return ans
if __name__ == '__main__':
main()
|
s247548969
|
p03836
|
u830054172
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 235
|
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())
x = tx-sx
y = ty-sy
ans = ""
out1 = "R"*x+"U"*y
return1 = "L"*x+"D"*y
out2a = "LU"
out2b = "RD"
return2a = "RD"
return2b = "LU"
print(out1+return1+out2a+out1+out2b+return2a+return1+return2b)
|
s489536761
|
Accepted
| 17
| 3,060
| 172
|
sx, sy, tx, ty = map(int, input().split())
x = tx-sx
y = ty-sy
ans = ""
go = "U"*y+"R"*x
back = "D"*y+"L"*x
a1 = "LU"
a2 = "RDRD"
a3 = "LU"
print(go+back+a1+go+a2+back+a3)
|
s794576017
|
p03523
|
u601575292
| 2,000
| 262,144
|
Wrong Answer
| 2,103
| 3,064
| 424
|
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`?
|
S = list(input())
n = len(S)
def solve(S):
if n < 5 or n > 9:
return "No"
while len(S) < 9:
if S[0] != "A":
S.insert(0, "A")
elif S[4] != "A":
S.insert(4, "A")
elif S[6] != "A":
S.insert(6, "A")
elif S[-1] != "A":
S.append("A")
if "".join(S) == "AKIHABARA":
return "Yes"
else:
return "No"
print(solve(S))
|
s836532797
|
Accepted
| 19
| 3,188
| 112
|
import re
S = input()
if re.fullmatch(r"A?KIHA?BA?RA?",S):
print("YES")
else:
print("NO")
exit()
|
s638610427
|
p02831
|
u242196904
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 186
|
Takahashi is organizing a party. At the party, each guest will receive one or more snack pieces. Takahashi predicts that the number of guests at this party will be A or B. Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted. We assume that a piece cannot be divided and distributed to multiple guests.
|
def gcd(a, b):
if a < b :
a, b = b, a
while True:
a, b = b, a % b
if b == 0:
return a
a, b = map(int, input().split())
int(a * b / gcd(a, b))
|
s824102909
|
Accepted
| 17
| 3,060
| 193
|
def gcd(a, b):
if a < b :
a, b = b, a
while True:
a, b = b, a % b
if b == 0:
return a
a, b = map(int, input().split())
print(int(a * b / gcd(a, b)))
|
s119188371
|
p03567
|
u503111914
| 2,000
| 262,144
|
Wrong Answer
| 20
| 2,940
| 100
|
Snuke built an online judge to hold a programming contest. When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.) Determine whether the judge can return the string `AC` as the verdict to a program.
|
S = input()
for i in range(len(S)-1):
if S[i]+S[i] == "AC":
print("Yes")
break
print("No")
|
s979112607
|
Accepted
| 18
| 2,940
| 119
|
import sys
S = input()
for i in range(len(S)-1):
if S[i]+S[i+1] == "AC":
print("Yes")
sys.exit()
print("No")
|
s677176075
|
p03693
|
u993268357
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 28
|
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
a = input().strip()
print(a)
|
s853787901
|
Accepted
| 18
| 2,940
| 97
|
a = input().strip().split()
num = int(''.join(a))
if num%4==0:
print('YES')
else:
print('NO')
|
s808096725
|
p03486
|
u584520370
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 143
|
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 = list(input())
t = list(input())
s_a = ''.join(sorted(s))
t_a = ''.join(sorted(t))
if s_a < t_a :
print('Yes')
else :
print('No')
|
s217336597
|
Accepted
| 17
| 2,940
| 156
|
s = list(input())
t = list(input())
s_a = ''.join(sorted(s))
t_a = ''.join(sorted(t, reverse=True))
if s_a < t_a :
print('Yes')
else :
print('No')
|
s144138504
|
p02842
|
u759076129
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 9,100
| 194
|
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
n = int(input())
x = round(n / 1.08)
n_dash = int(x*1.08)
if n_dash == n:
print(n_dash)
elif n_dash-1 == n:
print(n_dash-1)
elif n_dash+1 == n:
print(n_dash+1)
else:
print(':(')
|
s368601915
|
Accepted
| 27
| 9,124
| 262
|
def n_cand(x):
return int(x*1.08)
n = int(input())
x = round(n / 1.08)
n_dash = n_cand(x)
n_dash_m = n_cand(x-1)
n_dash_p = n_cand(x+1)
if n_dash == n:
print(x)
elif n_dash_m == n:
print(x-1)
elif n_dash_p == n:
print(x+1)
else:
print(':(')
|
s894041042
|
p02396
|
u494048940
| 1,000
| 131,072
|
Wrong Answer
| 90
| 6,348
| 148
|
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.
|
i=0
a=list()
while True :
a.append(int(input()))
i+=1
if a[i-1]==0:
break
for j in range(i) :
print("Case" , j+1 ,":",a[j])
|
s836998032
|
Accepted
| 140
| 5,616
| 117
|
i=0
a=list()
while True :
a=int(input())
if a == 0:
break
i+=1
print('Case %d: %d'%(i,a))
|
s481812899
|
p03455
|
u609814378
| 2,000
| 262,144
|
Wrong Answer
| 132
| 3,064
| 227
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a,b = map(str, input().split())
abint = int(a+b)
ans = 0
for i in range(1,1001):
for j in range(1,1001):
if i*j == abint and i == j:
ans = ans + 1
if ans == 0:
print("No")
else:
print("Yes")
|
s644333971
|
Accepted
| 17
| 2,940
| 103
|
a,b = map(int, input().split())
a_b = ((a*b)%2)
if a_b == 0:
print("Even")
else:
print("Odd")
|
s429059584
|
p03377
|
u379142263
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,316
| 289
|
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
import sys
import itertools
sys.setrecursionlimit(1000000000)
from heapq import heapify,heappop,heappush,heappushpop
import collections
a,b,x = map(int,input().split())
ab = a+b
least = x - a
if least<0:
print("No")
sys.exit()
if b>=least:
print("Yes")
else:
print("No")
|
s232673525
|
Accepted
| 21
| 3,316
| 237
|
import sys
import itertools
sys.setrecursionlimit(1000000000)
from heapq import heapify,heappop,heappush,heappushpop
import collections
a,b,x = map(int,input().split())
ab = a+b
if x>=a and x<=ab:
print("YES")
else:
print("NO")
|
s662993721
|
p02972
|
u224346910
| 2,000
| 1,048,576
|
Wrong Answer
| 413
| 22,716
| 276
|
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
def main():
N = int(input())
a = [None] + list(map(int, input().split()))
b = [0]*(N+1)
for i in range(N, 0, -1):
ball = [b[j] for j in range(i, N+1, i)]
if sum(ball) % 2 != a[i]:
b[i] = 1
print(' '.join(map(str, b[1:])))
main()
|
s084302712
|
Accepted
| 427
| 19,248
| 351
|
def main():
N = int(input())
a = [None] + list(map(int, input().split()))
b = [None] + [0]*N
for i in range(N, 0, -1):
ball = [b[j] for j in range(i, N+1, i)]
if sum(ball) % 2 != a[i]:
b[i] = 1
ans = [str(i) for i, n in enumerate(b) if n==1]
print(len(ans))
print(' '.join(map(str, ans)))
main()
|
s479107176
|
p03229
|
u455696302
| 2,000
| 1,048,576
|
Wrong Answer
| 1,025
| 11,072
| 874
|
You are given N integers; the i-th of them is A_i. Find the maximum possible sum of the absolute differences between the adjacent elements after arranging these integers in a row in any order you like.
|
N = int(input())
A = [int(input()) for _ in range(N)]
A = sorted(A)
print(A)
res = 0
if len(A) % 2 == 1:
if len(A) > 3:
mid = int(len(A)/2)
now_num = A.pop(mid)
for i in range(len(A)):
if len(A) == 0:
break
if i % 2 == 0:
res += A[-1] - now_num
now_num = A.pop(-1)
else:
res += now_num - A[0]
now_num = A.pop(0)
print(res)
else:
res += A[-1] - A[0]
res += A[-2] - A[0]
print(res)
else:
mid = int(len(A)/2) - 1
now_num = A.pop(mid)
for i in range(len(A)):
if len(A) == 0:
break
if i % 2 == 0:
res += A[-1] - now_num
now_num = A.pop(-1)
else:
res += now_num - A[0]
now_num = A.pop(0)
print(res)
|
s277470786
|
Accepted
| 1,884
| 8,928
| 818
|
N = int(input())
A_in = [int(input()) for _ in range(N)]
A = sorted(A_in)
res1 = 0
A2 = []
for i in range(len(A)):
if len(A) == 1:
if abs(A[0] - A2[0]) < abs(A[0] - A2[-1]):
A2.append(A[0])
else:
A2.insert(0,A[0])
break
if i % 2 == 0:
A2.append(A.pop(-1))
else:
A2.append(A.pop(0))
for i in range(len(A2)-1):
res1 += abs(A2[i] - A2[i+1])
A = sorted(A_in,reverse=True)
res2 = 0
A2 = []
for i in range(len(A)):
if len(A) == 1:
if abs(A[0] - A2[0]) < abs(A[0] - A2[-1]):
A2.append(A[0])
else:
A2.insert(0,A[0])
break
if i % 2 == 0:
A2.append(A.pop(-1))
else:
A2.append(A.pop(0))
for i in range(len(A2)-1):
res2 += abs(A2[i] - A2[i+1])
print(max(res1,res2))
|
s197378441
|
p03997
|
u980503157
| 2,000
| 262,144
|
Wrong Answer
| 26
| 9,100
| 68
|
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a = int(input())
b = int(input())
h = int(input())
print((a+b)/2*h)
|
s772938432
|
Accepted
| 21
| 9,148
| 74
|
a = int(input())
b = int(input())
h = int(input())
print(int((a+b)/2*h))
|
s041765243
|
p03400
|
u667024514
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 132
|
Some number of chocolate pieces were prepared for a training camp. The camp had N participants and lasted for D days. The i-th participant (1 \leq i \leq N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on. As a result, there were X chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces. Find the number of chocolate pieces prepared at the beginning of the camp.
|
import math
n = int(input())
d,x = map(int,input().split())
cou = x
for i in range(n):
cou += math.ceil(int(input())/d)
print(cou)
|
s520006661
|
Accepted
| 18
| 3,060
| 162
|
import math
n = int(input())
d,x = map(int,input().split())
lis = [int(input()) for i in range(n)]
cou = x
for i in range(n):cou += math.ceil(d/lis[i])
print(cou)
|
s134099473
|
p03671
|
u243159381
| 2,000
| 262,144
|
Wrong Answer
| 28
| 9,056
| 115
|
Snuke is buying a bicycle. The bicycle of his choice does not come with a bell, so he has to buy one separately. He has very high awareness of safety, and decides to buy two bells, one for each hand. The store sells three kinds of bells for the price of a, b and c yen (the currency of Japan), respectively. Find the minimum total price of two different bells.
|
a,b,c=map(int,input().split())
if a>b and a>c:
print(a+b)
elif b>a and b>c:
print(a+c)
else:
print(b+c)
|
s087745270
|
Accepted
| 27
| 8,916
| 62
|
a=list(map(int,input().split()))
b=sorted(a)
print(b[0]+b[1])
|
s222426929
|
p02972
|
u536325690
| 2,000
| 1,048,576
|
Wrong Answer
| 2,104
| 7,716
| 402
|
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
N = int(input())
a = [-1]
a.extend([int(i) for i in input().split()])
ans = [-1]
ans.extend([0 for i in range(N)])
for i in reversed(range(1, N+1)):
tmp = 0
for j in reversed(range(i,N+1)):
if j == i:
if tmp % 2 == 0:
ans[j] = 1
else:
ans[j] = 0
if j % i == 0:
tmp += ans[j]
print(len(ans)-1)
print(ans[1:])
|
s261058482
|
Accepted
| 615
| 14,136
| 350
|
N = int(input())
a = [int(i) for i in input().split()]
ans = [0 for i in range(N)]
for i in reversed(range(1, N+1)):
tmp = 0
for j in range(i, N+1, i):
tmp += ans[j-1]
if tmp % 2 != a[i-1]:
ans[i-1] = 1
anslist = []
for i in range(N):
if ans[i] == 1:
anslist.append(i+1)
print(len(anslist))
print(*anslist)
|
s931737730
|
p03472
|
u552176911
| 2,000
| 262,144
|
Wrong Answer
| 288
| 29,856
| 396
|
You are going out for a walk, when you suddenly encounter a monster. Fortunately, you have N katana (swords), Katana 1, Katana 2, …, Katana N, and can perform the following two kinds of attacks in any order: * Wield one of the katana you have. When you wield Katana i (1 ≤ i ≤ N), the monster receives a_i points of damage. The same katana can be wielded any number of times. * Throw one of the katana you have. When you throw Katana i (1 ≤ i ≤ N) at the monster, it receives b_i points of damage, and you lose the katana. That is, you can no longer wield or throw that katana. The monster will vanish when the total damage it has received is H points or more. At least how many attacks do you need in order to vanish it in total?
|
import math
n, h = map(int, input().split(" "))
abL = [list(map(int, input().split(" "))) for _ in range(n)]
aL = []
bL = []
for a, b in abL:
aL.append(a)
bL.append(b)
aL = sorted(aL)
bL = sorted(bL)
aMax = max(aL)
ans = 0
while h != 0:
if h <= 0:
break
if len(bL) == 0 or bL[-1] < aMax:
ans += h // aMax
break
ans += 1
h -= bL.pop()
print(ans)
|
s835162110
|
Accepted
| 286
| 29,732
| 406
|
import math
n, h = map(int, input().split(" "))
abL = [list(map(int, input().split(" "))) for _ in range(n)]
aL = []
bL = []
for a, b in abL:
aL.append(a)
bL.append(b)
aL = sorted(aL)
bL = sorted(bL)
aMax = max(aL)
ans = 0
while h != 0:
if h <= 0:
break
if len(bL) == 0 or bL[-1] < aMax:
ans += math.ceil(h / aMax)
break
ans += 1
h -= bL.pop()
print(ans)
|
s468760567
|
p03447
|
u626228246
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 82
|
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())
buf = ((X-A)//B)*B
print(X-A-B)
|
s638756614
|
Accepted
| 17
| 2,940
| 85
|
X = int(input())
A = int(input())
B = int(input())
buf = ((X-A)//B)*B
print(X-A-buf)
|
s377632261
|
p02396
|
u482227082
| 1,000
| 131,072
|
Wrong Answer
| 130
| 5,568
| 122
|
In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem.
|
for i in range(10000):
line = input()
if line == "0":
break
else:
print("Case ", "i: ", line)
|
s454904321
|
Accepted
| 130
| 5,628
| 197
|
#
# 3b
#
def main():
i = 1
while True:
x = int(input())
if x == 0:
break
print(f"Case {i}: {x}")
i += 1
if __name__ == '__main__':
main()
|
s142918294
|
p03816
|
u665038048
| 2,000
| 262,144
|
Wrong Answer
| 46
| 14,564
| 78
|
Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written. He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept. Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.
|
n = int(input())
a = list(map(int, input().split()))
print(len(list(set(a))))
|
s363604985
|
Accepted
| 51
| 14,388
| 152
|
n = int(input())
a = list(map(int, input().split()))
if len(list(set(a))) % 2 == 1:
print(len(list(set(a))))
else:
print(len(list(set(a))) - 1)
|
s793571711
|
p03599
|
u145889196
| 3,000
| 262,144
|
Wrong Answer
| 79
| 3,188
| 832
|
Snuke is making sugar water in a beaker. Initially, the beaker is empty. Snuke can perform the following four types of operations any number of times. He may choose not to perform some types of operations. * Operation 1: Pour 100A grams of water into the beaker. * Operation 2: Pour 100B grams of water into the beaker. * Operation 3: Put C grams of sugar into the beaker. * Operation 4: Put D grams of sugar into the beaker. In our experimental environment, E grams of sugar can dissolve into 100 grams of water. Snuke will make sugar water with the highest possible density. The beaker can contain at most F grams of substances (water and sugar combined), and there must not be any undissolved sugar in the beaker. Find the mass of the sugar water Snuke will make, and the mass of sugar dissolved in it. If there is more than one candidate, any of them will be accepted. We remind you that the sugar water that contains a grams of water and b grams of sugar is \frac{100b}{a + b} percent. Also, in this problem, pure water that does not contain any sugar is regarded as 0 percent density sugar water.
|
A,B,C,D,E,F=map(int,input().split())
density=0
Ans1=0
Ans2=0
water=set()
sugar=set()
bm=F//(100*B)
for i in range(bm+1):
rw=F-100*i*B
am=rw//(100*A)
for j in range(am+1):
water.add(100*i*B+100*j*A)
dm=F//D
for i in range(dm+1):
rs=F-i*D
cm=rs//C
for j in range(cm+1):
sugar.add(i*D+j*C)
sugar=sorted(sugar)
sugar.pop(0)
water=sorted(water)
water.pop(0)
for s in sugar:
for w in water:
if w/100*E<s:
break
elif s+w>F:
break
else:
den=s/w
if den>density:
density=den
Ans1=s+w
Ans2=s
else:
break
print(Ans1)
print(Ans2)
|
s688037081
|
Accepted
| 76
| 3,188
| 639
|
#C
A,B,C,D,E,F=map(int,input().split())
density=0
Ans1=0
Ans2=0
water=set()
sugar=set()
bm=F//(100*B)
for i in range(bm+1):
rw=F-100*i*B
am=rw//(100*A)
for j in range(am+1):
water.add(100*i*B+100*j*A)
dm=F//D
for i in range(dm+1):
rs=F-i*D
cm=rs//C
for j in range(cm+1):
sugar.add(i*D+j*C)
sugar=sorted(sugar)
water=sorted(water)
for s in sugar:
for w in water:
if w/100*E>=s and s+w<=F and s+w!=0:
den=s/(s+w)
if den>=density:
density=den
Ans1=s+w
Ans2=s
else:
break
print(Ans1,Ans2)
|
s659408777
|
p03623
|
u922487073
| 2,000
| 262,144
|
Wrong Answer
| 16
| 2,940
| 64
|
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x,a,b = map(int, input().split())
print(min(abs(x-a), abs(x-b)))
|
s988057523
|
Accepted
| 17
| 2,940
| 76
|
x,a,b = map(int, input().split())
print("B" if abs(x-a) > abs(x-b) else "A")
|
s204991007
|
p03477
|
u320098990
| 2,000
| 262,144
|
Wrong Answer
| 26
| 9,016
| 230
|
A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R. Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan. Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right.
|
in_list = input().split(' ')
A = in_list[0]
B = in_list[1]
C = in_list[2]
D = in_list[3]
L = int(A) + int(B)
R = int(C) + int(D)
print(L, R)
if L < R:
print('Right')
elif L > R:
print('Left')
else:
print('Balanced')
|
s850650347
|
Accepted
| 25
| 9,056
| 217
|
in_list = input().split(' ')
A = in_list[0]
B = in_list[1]
C = in_list[2]
D = in_list[3]
L = int(A) + int(B)
R = int(C) + int(D)
if L < R:
print('Right')
elif L > R:
print('Left')
else:
print('Balanced')
|
s838874266
|
p03372
|
u794173881
| 2,000
| 262,144
|
Wrong Answer
| 794
| 47,040
| 1,016
|
"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter. Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is x_i meters. Also, the i-th sushi has a nutritive value of v_i kilocalories. Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter. Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.
|
n,c = map(int,input().split())
info = [[0,0]]+[list(map(int,input().split())) for i in range(n)]+[[c,0]]
left_info = info[::-1]
right_ruiseki = [0]*(n+1)
left_ruiseki = [0]*(n+1)
ou_right_ruiseki = [0]*(n+1)
ou_left_ruiseki = [0]*(n+1)
for i in range(n):
right_ruiseki[i+1] = right_ruiseki[i] + info[i+1][1] - info[i+1][0] + info[i][0]
left_ruiseki[i+1] = left_ruiseki[i] + left_info[i+1][1] + left_info[i+1][0] - left_info[i][0]
ou_right_ruiseki[i+1] = ou_right_ruiseki[i] + info[i+1][1] - 2*info[i+1][0] + 2*info[i][0]
ou_left_ruiseki[i+1] = ou_left_ruiseki[i] + left_info[i+1][1] + 2*left_info[i+1][0] - 2*left_info[i][0]
for i in range(n):
right_ruiseki[i+1]=max(right_ruiseki[i],right_ruiseki[i+1])
left_ruiseki[i+1]=max(left_ruiseki[i],left_ruiseki[i+1])
ans = 0
for i in range(n):
tmp1 = ou_right_ruiseki[i+1]+left_ruiseki[n-(i+1)]
tmp2 = ou_left_ruiseki[i+1]+right_ruiseki[n-(i+1)]
ans = max(tmp1,tmp2,ans)
print(ans)
|
s741712898
|
Accepted
| 802
| 47,040
| 1,075
|
n,c = map(int,input().split())
info = [[0,0]]+[list(map(int,input().split())) for i in range(n)]+[[c,0]]
left_info = info[::-1]
right_ruiseki = [0]*(n+1)
left_ruiseki = [0]*(n+1)
ou_right_ruiseki = [0]*(n+1)
ou_left_ruiseki = [0]*(n+1)
for i in range(n):
right_ruiseki[i+1] = right_ruiseki[i] + info[i+1][1] - info[i+1][0] + info[i][0]
left_ruiseki[i+1] = left_ruiseki[i] + left_info[i+1][1] + left_info[i+1][0] - left_info[i][0]
ou_right_ruiseki[i+1] = ou_right_ruiseki[i] + info[i+1][1] - 2*info[i+1][0] + 2*info[i][0]
ou_left_ruiseki[i+1] = ou_left_ruiseki[i] + left_info[i+1][1] + 2*left_info[i+1][0] - 2*left_info[i][0]
for i in range(n):
right_ruiseki[i+1]=max(right_ruiseki[i],right_ruiseki[i+1])
left_ruiseki[i+1]=max(left_ruiseki[i],left_ruiseki[i+1])
ans = 0
for i in range(n+1):
tmp1 = ou_right_ruiseki[i]+left_ruiseki[n-i]
tmp2 = ou_left_ruiseki[i]+right_ruiseki[n-i]
ans = max(tmp1,tmp2,ans)
print(ans)
|
s188772834
|
p03605
|
u153419200
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 67
|
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
a=input()
if a[0]==9 or a[1]==9:
print('Yes')
else:
print('No')
|
s826759821
|
Accepted
| 17
| 2,940
| 70
|
a=int(input())
if a>=90 or a%10==9:
print('Yes')
else:
print('No')
|
s295826595
|
p02238
|
u923668099
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,720
| 807
|
Depth-first search (DFS) follows the strategy to search ”deeper” in the graph whenever possible. In DFS, edges are recursively explored out of the most recently discovered vertex $v$ that still has unexplored edges leaving it. When all of $v$'s edges have been explored, the search ”backtracks” to explore edges leaving the vertex from which $v$ was discovered. This process continues until all the vertices that are reachable from the original source vertex have been discovered. If any undiscovered vertices remain, then one of them is selected as a new source and the search is repeated from that source. DFS timestamps each vertex as follows: * $d[v]$ records when $v$ is first discovered. * $f[v]$ records when the search finishes examining $v$’s adjacency list. Write a program which reads a directed graph $G = (V, E)$ and demonstrates DFS on the graph based on the following rules: * $G$ is given in an adjacency-list. Vertices are identified by IDs $1, 2,... n$ respectively. * IDs in the adjacency list are arranged in ascending order. * The program should report the discover time and the finish time for each vertex. * When there are several candidates to visit during DFS, the algorithm should select the vertex with the smallest ID. * The timestamp starts with 1.
|
import sys
def debug(x, table):
for name, val in table.items():
if x is val:
print('DEBUG:{} -> {}'.format(name, val), file=sys.stderr)
return None
time = 0
def dfs(u, adj, d, f, sumi):
global time
time += 1
d[u] = time
for v in adj[u]:
if v not in sumi:
dfs(v, adj, d, f, sumi)
sumi.add(v)
time += 1
f[u] = time
def solve():
n = int(input())
adj = [None] * n
for i in range(n):
u, k, *vs = [int(j) - 1 for j in input().split()]
adj[u] = vs
d = [0] * n
f = [0] * n
sumi = {0}
dfs(0, adj, d, f, sumi)
for i in range(n):
print(i, d[i], f[i])
pass
if __name__ == '__main__':
solve()
|
s574399946
|
Accepted
| 20
| 7,824
| 1,037
|
import sys
def debug(x, table):
for name, val in table.items():
if x is val:
print('DEBUG:{} -> {}'.format(name, val), file=sys.stderr)
return None
time = 0
def dfs(u, adj, d, f, sumi):
global time
sumi.add(u)
time += 1
d[u] = time
for v in adj[u]:
if v not in sumi:
dfs(v, adj, d, f, sumi)
time += 1
f[u] = time
def solve():
n = int(input())
adj = [None] * n
for i in range(n):
u, k, *vs = [int(j) - 1 for j in input().split()]
adj[u] = vs
d = [0] * n
f = [0] * n
sumi = {0}
mitan = set(range(1, n))
dfs(0, adj, d, f, sumi)
mitan -= sumi
while mitan:
u = min(mitan)
dfs(u, adj, d, f, sumi)
mitan -= sumi
for i in range(n):
print(i + 1, d[i], f[i])
pass
if __name__ == '__main__':
solve()
|
s917661406
|
p03151
|
u369402805
| 2,000
| 1,048,576
|
Wrong Answer
| 125
| 18,356
| 582
|
A university student, Takahashi, has to take N examinations and pass all of them. Currently, his _readiness_ for the i-th examination is A_{i}, and according to his investigation, it is known that he needs readiness of at least B_{i} in order to pass the i-th examination. Takahashi thinks that he may not be able to pass all the examinations, and he has decided to ask a magician, Aoki, to change the readiness for as few examinations as possible so that he can pass all of them, while not changing the total readiness. For Takahashi, find the minimum possible number of indices i such that A_i and C_i are different, for a sequence C_1, C_2, ..., C_{N} that satisfies the following conditions: * The sum of the sequence A_1, A_2, ..., A_{N} and the sum of the sequence C_1, C_2, ..., C_{N} are equal. * For every i, B_i \leq C_i holds. If such a sequence C_1, C_2, ..., C_{N} cannot be constructed, print -1.
|
N = int(input())
a_list = [int(a) for a in input().split()]
b_list = [int(a) for a in input().split()]
# N = 3
# a_list = [17, 7, 1]
# b_list = [25, 6, 14]
if sum(a_list) < sum(b_list):
print("-1")
else:
d_list = [a - b for a, b in zip(a_list, b_list)]
d_list.sort()
minus = [d for d in d_list if d < 0]
deficit = sum(minus)
ans = len(minus)
if deficit == 0:
print("0")
else:
for i in range(N):
deficit += d_list[-i]
if deficit >= 0:
ans += i + 1
break
print(str(ans))
|
s308027651
|
Accepted
| 133
| 19,068
| 533
|
N = int(input())
a_list = [int(a) for a in input().split()]
b_list = [int(a) for a in input().split()]
if sum(a_list) < sum(b_list):
print("-1")
else:
d_list = [a - b for a, b in zip(a_list, b_list)]
d_list.sort()
minus = [d for d in d_list if d < 0]
deficit = sum(minus)
ans = len(minus)
if deficit == 0:
print("0")
else:
for i in range(N):
deficit += d_list[-i - 1]
if deficit >= 0:
ans += i + 1
break
print(str(ans))
|
s069504883
|
p02612
|
u118665579
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 9,132
| 30
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n = int(input())
print(n%1000)
|
s130002546
|
Accepted
| 28
| 9,148
| 87
|
S = int(input())
if S % 1000 == 0:
W = 0
else:
W = 1000-(S % 1000)
print(W)
190
|
s318210278
|
p03795
|
u779073299
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 52
|
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.
|
a = int(input())
ans = a*800 - (a%15)*200
print(ans)
|
s130428997
|
Accepted
| 17
| 2,940
| 53
|
a = int(input())
ans = a*800 - (a//15)*200
print(ans)
|
s291561457
|
p02264
|
u684241248
| 1,000
| 131,072
|
Wrong Answer
| 30
| 6,000
| 378
|
_n_ _q_ _name 1 time1_ _name 2 time2_ ... _name n timen_ In the first line the number of processes _n_ and the quantum _q_ are given separated by a single space. In the following _n_ lines, names and times for the _n_ processes are given. _name i_ and _time i_ are separated by a single space.
|
from collections import deque
n, q = [int(_) for _ in input().split()]
processes = deque([tuple(input().split()) for i in range(n)])
time = 0
while processes:
process = processes.popleft()
if int(process[1]) < q:
time += int(process[1])
print(process[0], time)
else:
time += q
processes.append((process[0], int(process[1]) - q))
|
s038718871
|
Accepted
| 340
| 12,644
| 379
|
from collections import deque
n, q = [int(_) for _ in input().split()]
processes = deque([tuple(input().split()) for i in range(n)])
time = 0
while processes:
process = processes.popleft()
if int(process[1]) <= q:
time += int(process[1])
print(process[0], time)
else:
time += q
processes.append((process[0], int(process[1]) - q))
|
s034053733
|
p02401
|
u757827098
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,648
| 263
|
Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.
|
while True:
a,op,b = input().split()
a =int(a)
b =int(b)
if op == '?':
break
if op == '+':
print(a+b)
if op == '-':
print(a-b)
if op == '*':
print(a*b)
if op == '/':
print(a/b)
|
s016587724
|
Accepted
| 30
| 7,656
| 264
|
while True:
a,op,b = input().split()
a =int(a)
b =int(b)
if op == '?':
break
if op == '+':
print(a+b)
if op == '-':
print(a-b)
if op == '*':
print(a*b)
if op == '/':
print(a//b)
|
s601799101
|
p03139
|
u371467115
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 79
|
We conducted a survey on newspaper subscriptions. More specifically, we asked each of the N respondents the following two questions: * Question 1: Are you subscribing to Newspaper X? * Question 2: Are you subscribing to Newspaper Y? As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2. What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y? Write a program to answer this question.
|
n,a,b=map(int,input().split())
s=min(a,b)
l=abs(a+b)-n
print(str(s)+" "+str(l))
|
s405710121
|
Accepted
| 17
| 2,940
| 102
|
n,a,b=map(int,input().split())
s=min(a,b)
l=0
if n<a+b:
l=abs(n-abs(b+a))
print("{} {}".format(s,l))
|
s081854195
|
p02422
|
u239721772
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,552
| 827
|
Write a program which performs a sequence of commands to a given string $str$. The command is one of: * print a b: print from the a-th character to the b-th character of $str$ * reverse a b: reverse from the a-th character to the b-th character of $str$ * replace a b p: replace from the a-th character to the b-th character of $str$ with p Note that the indices of $str$ start with 0.
|
string = list(input())
times = int(input())
for i in range(times):
order = input().split()
if order[0] == "replace":
replace_string = order[3]
count = 0
for j in range(int(order[1]), int(order[2]) + 1):
string[j] = replace_string[count]
count += 1
#print(count)
elif order[0] == "reverse":
reverse_string = string[int(order[1]):int(order[2]) + 1]
reverse_string.reverse()
for j in range(int(order[1]), int(order[2]) + 1):
string[j] = reverse_string[j]
#string[int(order[1]):int(order[2]) + 1].reverse()
print(string)
#string[int(order[1]):int(order[2])] = string[int(order[1]):int(order[2])].reverse()
elif order[0] == "print":
print("".join(string[int(order[1]):int(order[2]) + 1]))
|
s889440438
|
Accepted
| 20
| 7,700
| 611
|
string = list(input())
times = int(input())
for i in range(times):
order = input().split()
if order[0] == "replace":
replace_string = order[3]
count = 0
for j in range(int(order[1]), int(order[2]) + 1):
string[j] = replace_string[count]
count += 1
elif order[0] == "reverse":
reverse_string = string[int(order[1]):int(order[2]) + 1]
reverse_string.reverse()
string = string[:int(order[1])] + reverse_string + string[int(order[2])+1:]
elif order[0] == "print":
print("".join(string[int(order[1]):int(order[2]) + 1]))
|
s256496299
|
p02742
|
u545644875
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 176
|
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
h,w = map(int,input().split())
if h%2 == 0:
result = w*h/2
elif w%2 == 0:
result = w*(h - 1)/2 + w/2
else:
result = w*(h - 1)/2 + (w + 1)/2
print(result)
|
s090760819
|
Accepted
| 17
| 2,940
| 156
|
h, w = map(int, input().split())
x = (h * w) % 2
if h == 1 or w == 1:
print(1)
elif x == 0:
print((h * w) // 2)
else:
print(((h * w) // 2) + 1)
|
s372952012
|
p03160
|
u566264434
| 2,000
| 1,048,576
|
Wrong Answer
| 102
| 13,980
| 249
|
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
def dp1(n,h):
dp=[10^5]*n
dp[0]=0
dp[1]=abs(h[1]-h[0])
for i in range(2,n-1):
dp[i]=min(dp[i-1]+abs(h[i]-h[i-1]),dp[i-2]+abs(h[i]-h[i-2]))
return dp[n-1]
n=int(input())
h = list(map(int, input().split()))
print(dp1(n,h))
|
s334480225
|
Accepted
| 104
| 13,980
| 247
|
def dp1(n,h):
dp=[10^5]*n
dp[0]=0
dp[1]=abs(h[1]-h[0])
for i in range(2,n):
dp[i]=min(dp[i-1]+abs(h[i]-h[i-1]),dp[i-2]+abs(h[i]-h[i-2]))
return dp[n-1]
n=int(input())
h = list(map(int, input().split()))
print(dp1(n,h))
|
s856545293
|
p03697
|
u145600939
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 82
|
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.
|
a,b = map(int,input().split())
if a+b >= 10:
print('error')
else:
print('a+b')
|
s950629509
|
Accepted
| 17
| 2,940
| 81
|
a,b = map(int,input().split())
if a+b >= 10:
print('error')
else:
print(a+b)
|
s919006603
|
p03379
|
u314089899
| 2,000
| 262,144
|
Wrong Answer
| 289
| 25,744
| 901
|
When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N.
|
#094c
N = int(input()) #N<200000=2 * 10**5
X_list = [int(e) for e in input().split()]
X_list.sort(reverse=False)
Bi_smaller = X_list[int(len(X_list)/2-1)]
Bi_bigger = X_list[int(len(X_list)/2)]
for i in range(int(N/2)):
#for j in range(N):
# if j!=i:
# print(X_list[j],end="")
print(Bi_bigger)
for i in range(int(N/2),N):
#for j in range(N):
# if j!=i:
# print(X_list[j],end="")
print(Bi_smaller)
|
s784722427
|
Accepted
| 330
| 25,732
| 363
|
#094c
import copy
N = int(input()) #N<200000=2 * 10**5
X_list = [int(e) for e in input().split()]
sort_X_list = copy.copy(X_list)
sort_X_list.sort(reverse=False)
Bi_smaller = sort_X_list[int(len(X_list)/2-1)]
Bi_bigger = sort_X_list[int(len(X_list)/2)]
for i in range(N):
if X_list[i]>=Bi_bigger:
print(Bi_smaller)
else:
print(Bi_bigger)
|
s052873271
|
p03478
|
u925950392
| 2,000
| 262,144
|
Wrong Answer
| 32
| 2,940
| 239
|
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(' '))
total = 0
for number in range(N):
total_number = 0
for number_str in str(number):
total_number += int(number_str)
if total_number >= A and total_number <= B:
total += number
print(total)
|
s180572099
|
Accepted
| 32
| 2,940
| 247
|
N, A, B = map(int, input().split(' '))
total = 0
for number in range(1, N + 1):
total_number = 0
for number_str in str(number):
total_number += int(number_str)
if total_number >= A and total_number <= B:
total += number
print(total)
|
s502752586
|
p03636
|
u007263493
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 51
|
The word `internationalization` is sometimes abbreviated to `i18n`. This comes from the fact that there are 18 letters between the first `i` and the last `n`. You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way.
|
s = str(input())
s = s[1:]
s = s[:-1]
print(len(s))
|
s217363704
|
Accepted
| 17
| 2,940
| 79
|
s = str(input())
a = s[0]
b = s[-1]
s = s[1:]
s = s[:-1]
print(a+str(len(s))+b)
|
s199727458
|
p03470
|
u379716238
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 175
|
An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?
|
N = int(input())
a = []
for i in range(N):
a.append(int(input()))
res = 0
data = 0
for i in range(N):
if a[i] > data:
res += 1
data = a[i]
print(res)
|
s256339812
|
Accepted
| 153
| 12,504
| 135
|
N = int(input())
d_list = []
for i in range(N):
d_list.append(int(input()))
import numpy as np
n = np.unique(d_list)
print(len(n))
|
s057038582
|
p02615
|
u039860745
| 2,000
| 1,048,576
|
Wrong Answer
| 218
| 31,444
| 408
|
Quickly after finishing the tutorial of the online game _ATChat_ , you have decided to visit a particular place with N-1 players who happen to be there. These N players, including you, are numbered 1 through N, and the **friendliness** of Player i is A_i. The N players will arrive at the place one by one in some order. To make sure nobody gets lost, you have set the following rule: players who have already arrived there should form a circle, and a player who has just arrived there should cut into the circle somewhere. When each player, except the first one to arrive, arrives at the place, the player gets **comfort** equal to the smaller of the friendliness of the clockwise adjacent player and that of the counter-clockwise adjacent player. The first player to arrive there gets the comfort of 0. What is the maximum total comfort the N players can get by optimally choosing the order of arrivals and the positions in the circle to cut into?
|
N = int(input())
A = list(map(int, input().split()))
A.sort(reverse=True)
# print(A)
limit = 0
ans = 0
count = 0
for i,a in enumerate(A):
limit = 2
if i == 0:
limit = 1
for j in range(limit, 1, -1):
# print(j)
if j > 0:
ans += a
count += 1
# print(count)
if count > N - 1:
break
# print(N - 1)
print(ans)
|
s164203470
|
Accepted
| 205
| 31,540
| 290
|
N = int(input())
A = list(map(int, input().split()))
A.sort(reverse=True)
limit = 0
ans = 0
count = 0
lim = 0
t = N - 1
for i in range(N):
lim = 2
if i == 0:
lim = 1
for j in range(lim):
if t > 0:
ans += A[i]
t -= 1
print(ans)
|
s863295345
|
p03214
|
u215115622
| 2,525
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 108
|
Niwango-kun is an employee of Dwango Co., Ltd. One day, he is asked to generate a thumbnail from a video a user submitted. To generate a thumbnail, he needs to select a frame of the video according to the following procedure: * Get an integer N and N integers a_0, a_1, ..., a_{N-1} as inputs. N denotes the number of the frames of the video, and each a_i denotes the representation of the i-th frame of the video. * Select t-th frame whose representation a_t is nearest to the average of all frame representations. * If there are multiple such frames, select the frame with the smallest index. Find the index t of the frame he should select to generate a thumbnail.
|
a=input().split()
a=[int(i) for i in a]
avg =sum(a)/len(a)
a=[abs(i -avg) for i in a]
print(a.index(min(a)))
|
s354782370
|
Accepted
| 17
| 2,940
| 118
|
a=input()
a=input().split()
a=[int(i) for i in a]
avg =sum(a)/len(a)
a=[abs(i -avg) for i in a]
print(a.index(min(a)))
|
s563375370
|
p03624
|
u697658632
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,188
| 131
|
You are given a string S consisting of lowercase English letters. Find the lexicographically (alphabetically) smallest lowercase English letter that does not occur in S. If every lowercase English letter occurs in S, print `None` instead.
|
s = input()
for i in range(ord('a'), ord('z') + 1):
c = chr(i)
if s.count(c) > 0:
print(c)
break
else:
print('None')
|
s270910350
|
Accepted
| 20
| 3,188
| 132
|
s = input()
for i in range(ord('a'), ord('z') + 1):
c = chr(i)
if s.count(c) == 0:
print(c)
break
else:
print('None')
|
s914950484
|
p03795
|
u275212209
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 51
|
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(round((800*n) - (200*(n/15))))
|
s956708140
|
Accepted
| 17
| 2,940
| 47
|
n=int(input())
x=800*n
y=(n//15)*200
print(x-y)
|
s622569548
|
p03457
|
u084069244
| 2,000
| 262,144
|
Wrong Answer
| 324
| 3,060
| 159
|
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
N = int(input())
for i in range(N):
t,x,y = map(int, input().split())
if x + y < t or (x + y + t) % 2:
print("No")
exit()
print("Yes")
|
s981831385
|
Accepted
| 329
| 3,060
| 163
|
N = int(input())
for _ in range(N):
t, x, y = map(int, input().split())
if x + y > t or (t + x + y)%2 != 0:
print("No")
quit()
print("Yes")
|
s468386248
|
p03160
|
u194894739
| 2,000
| 1,048,576
|
Wrong Answer
| 115
| 13,980
| 220
|
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
N = int(input())
h = list(map(int,input().split()))
dp = [0] * N
for i in range(1, N):
if i == 1:
dp[i] = abs(h[1] - h[0])
else:
dp[i] = min(abs(h[i] - h[i-1]), abs(h[i] - h[i-2]))
print(dp[N-1])
|
s376976071
|
Accepted
| 210
| 13,924
| 261
|
N = int(input())
H = list(map(int, input().split()))
dp = [10**9]*N
dp[0] = 0
for i in range(N):
if i+1 <= N-1:
dp[i+1] = min(dp[i+1], dp[i]+abs(H[i+1]-H[i]))
if i+2 <= N-1:
dp[i+2] = min(dp[i+2], dp[i]+abs(H[i+2]-H[i]))
print(dp[N-1])
|
s469790225
|
p02402
|
u106285852
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,584
| 864
|
Write a program which reads a sequence of $n$ integers $a_i (i = 1, 2, ... n)$, and prints the minimum value, maximum value and sum of the sequence.
|
# import
import sys
int(input())
inputData = map(int, input().split())
maxData = 0
minData = 0
totalData = 0
for inputVal in inputData:
if inputVal > maxData:
maxData = inputVal
if inputVal < minData:
minData = inputVal
totalData += inputVal
# ????????????
print(minData, maxData, totalData)
|
s785932453
|
Accepted
| 20
| 8,336
| 912
|
# import
import sys
n = int(input())
inputData = map(int, input().split())
maxData = -1000000
minData = 1000000
totalData = 0
for inputVal in inputData:
if inputVal > maxData:
maxData = inputVal
# minimum value check
if inputVal < minData:
minData = inputVal
totalData += inputVal
# ????????????
print(minData, maxData, totalData)
|
s768768181
|
p02612
|
u072717685
| 2,000
| 1,048,576
|
Wrong Answer
| 30
| 9,160
| 267
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
import sys
read = sys.stdin.read
readline = sys.stdin.readline
readlines = sys.stdin.readlines
#import numpy as np
def main():
n = int(input())
if n % 1000:
print(n // 1000 + 1)
else:
print(n // 1000)
if __name__ == '__main__':
main()
|
s932331581
|
Accepted
| 29
| 9,196
| 256
|
import sys
read = sys.stdin.read
readline = sys.stdin.readline
readlines = sys.stdin.readlines
from math import ceil
#import numpy as np
def main():
n = int(input())
r = ceil(n / 1000) * 1000 - n
print(r)
if __name__ == '__main__':
main()
|
s141739221
|
p03636
|
u825528847
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 56
|
The word `internationalization` is sometimes abbreviated to `i18n`. This comes from the fact that there are 18 letters between the first `i` and the last `n`. You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way.
|
S = input()
print(S[0] + str(len(S[1:len(S)])) + S[-1])
|
s341790136
|
Accepted
| 17
| 2,940
| 58
|
S = input()
print(S[0] + str(len(S[1:len(S)-1])) + S[-1])
|
s390893750
|
p03997
|
u559313689
| 2,000
| 262,144
|
Wrong Answer
| 23
| 9,096
| 68
|
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a = int(input())
b = int(input())
h = int(input())
print((a+b)*h/2)
|
s367632735
|
Accepted
| 24
| 9,084
| 74
|
a = int(input())
b = int(input())
h = int(input())
print(int((a+b)*h/2))
|
s330380945
|
p03479
|
u039360403
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 2,940
| 132
|
As a token of his gratitude, Takahashi has decided to give his mother an integer sequence. The sequence A needs to satisfy the conditions below: * A consists of integers between X and Y (inclusive). * For each 1\leq i \leq |A|-1, A_{i+1} is a multiple of A_i and strictly greater than A_i. Find the maximum possible length of the sequence.
|
X,Y=map(int,input().split())
ans=0
for i in range(1,10**18):
if X*i<Y:
ans+=1
else:
print(ans)
break
|
s185636306
|
Accepted
| 17
| 2,940
| 146
|
X,Y=map(int,input().split())
ans=1
for i in range(1,10**18):
if X*2<=Y:
X*=2
ans+=1
else:
print(ans)
break
|
s860764679
|
p00005
|
u011621222
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,540
| 191
|
Write a program which computes the greatest common divisor (GCD) and the least common multiple (LCM) of given a and b.
|
a,b=map(int,input().split())
for i in range(1,a+1):
f=(b*i)%a
lcm=(b*i)
if f==0:
break
for j in range(1,a+1):
if a%j==0 and b%j==0 and j*lcm==a*b:
print(j,lcm)
|
s489812382
|
Accepted
| 20
| 5,664
| 147
|
import math
while True:
try:
a,b = map(int,input().split())
print(math.gcd(a,b),a*b//math.gcd(a,b))
except:
break
|
s967368512
|
p03005
|
u598229387
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 60
|
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())
if k==1:
ans=0
else:
ans=n-k
|
s311179619
|
Accepted
| 17
| 2,940
| 71
|
n,k=map(int,input().split())
if k==1:
ans=0
else:
ans=n-k
print(ans)
|
s895642537
|
p03478
|
u153729035
| 2,000
| 262,144
|
Wrong Answer
| 31
| 3,296
| 98
|
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())
sum([i for i in range(n+1) if a<=sum([int(c) for c in str(i)])<=b])
|
s146058599
|
Accepted
| 30
| 3,296
| 107
|
n,a,b=map(int,input().split())
print(sum([i for i in range(1,n+1) if a<=sum([int(c) for c in str(i)])<=b]))
|
s759359151
|
p02412
|
u340901659
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,732
| 457
|
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
|
def solve(n,x):
count = 0
for i in range(1,n+1):
for j in range(1,n+1):
if i == j :
continue
for k in range(1,n+1):
if j == k or i == k:
continue
if i+j+k == x:
print(i,j,k)
count += 1
return count//6
while True:
n,x = map(int,input().split())
if n == x == 0:
break;
print(solve(n,x))
|
s450055829
|
Accepted
| 40
| 7,644
| 297
|
def solve(n,x):
count = 0
for i in range(1,n+1):
for j in range(i+1,n+1):
k = x - i - j
if j < k and k <= n:
count += 1
return count
while True:
n,x = map(int,input().split())
if n == x == 0:
break;
print(solve(n,x))
|
s826900337
|
p04030
|
u894623942
| 2,000
| 262,144
|
Wrong Answer
| 26
| 8,948
| 256
|
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?
|
S = list(input())
print(S)
S_after = []
for i in S:
if i == '1':
S_after.append(1)
elif i == '0':
S_after.append(0)
elif i == 'B':
if S_after == []:
pass
else:
S_after.pop()
print(S_after)
|
s176697821
|
Accepted
| 27
| 8,736
| 260
|
S = list(input())
S_after = []
for i in S:
if i == '1':
S_after.append('1')
elif i == '0':
S_after.append('0')
elif i == 'B':
if S_after == []:
pass
else:
S_after.pop()
print(''.join(S_after))
|
s512112470
|
p02694
|
u131925051
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 9,164
| 140
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
X = int(input())
yen = 100
i = 0
def rishi(yen):
return int(yen * 1.01)
while 1:
yen = rishi(yen)
i += 1
if X < yen:
break
print(i)
|
s425347105
|
Accepted
| 20
| 9,156
| 141
|
X = int(input())
yen = 100
i = 0
def rishi(yen):
return int(yen * 1.01)
while 1:
i += 1
yen = rishi(yen)
if X <= yen:
break
print(i)
|
s819415091
|
p02613
|
u058259032
| 2,000
| 1,048,576
|
Wrong Answer
| 167
| 16,316
| 370
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
N = int(input())
list = []
for i in range(N):
list.append(input())
C0 = 0
C1 = 0
C2 = 0
C3 = 0
for i in range(N):
if list[i] == "AC":
C0 += 1
elif list[i] == "WA":
C1 += 1
elif list[i] == "TLE":
C2 += 1
elif list[i] == "RE":
C3 += 1
print("AC * ", C0)
print("WA * ", C1)
print("TLE * ", C2)
print("RE * ", C3)
|
s234590807
|
Accepted
| 167
| 16,240
| 365
|
N = int(input())
list = []
for i in range(N):
list.append(input())
C0 = 0
C1 = 0
C2 = 0
C3 = 0
for i in range(N):
if list[i] == "AC":
C0 += 1
elif list[i] == "WA":
C1 += 1
elif list[i] == "TLE":
C2 += 1
elif list[i] == "RE":
C3 += 1
print("AC x",C0)
print("WA x", C1)
print("TLE x", C2)
print("RE x", C3)
|
s505450612
|
p03854
|
u787059958
| 2,000
| 262,144
|
Wrong Answer
| 28
| 9,164
| 851
|
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()
idx = 0
while True:
if idx >= len(S) - 1:
break
if S[idx:idx +
7] == 'dreamer' and (S[idx +
5:idx +
10] != 'erase' or S[idx +
5:idx +
11] != 'eraser'):
idx += 7
elif S[idx:idx + 6] == 'eraser' and (S[idx +
4:idx +
9] != 'erase' or S[idx +
4:idx +
10] != 'eraser'):
idx += 6
elif S[idx:idx + 5] == 'dream' or S[idx:idx + 5] == 'erase':
idx += 5
else:
print('No')
exit()
print('Yes')
|
s067605989
|
Accepted
| 65
| 9,088
| 229
|
S = input()
L = ['maerd', 'remaerd', 'esare', 'resare']
T = ''
for i in range(len(S) - 1, -1, -1):
T += S[i]
if T not in L and len(T) > 7:
print('NO')
exit()
elif T in L:
T = ''
print('YES')
|
s187901479
|
p03730
|
u940743763
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 293
|
We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
a, b, c = list(map(int, input().split(' ')))
checker = {}
d = a
while(True):
m = (d - c) % b
checker.setdefault(m, False)
if m == 0:
print('Yes')
break
elif checker[m]:
print('No')
break
else:
checker[m] = True
d += a
|
s630083667
|
Accepted
| 17
| 3,060
| 293
|
a, b, c = list(map(int, input().split(' ')))
checker = {}
d = a
while(True):
m = (d - c) % b
checker.setdefault(m, False)
if m == 0:
print('YES')
break
elif checker[m]:
print('NO')
break
else:
checker[m] = True
d += a
|
s290047549
|
p04043
|
u448655578
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 152
|
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.
|
haiku = [5,7,5]
a = input().split(" ")
for i in a:
if i in haiku:
haiku.pop(haiku.index(i))
if len(haiku) == 0:
print("YES")
else:
print("NO")
|
s655539927
|
Accepted
| 17
| 2,940
| 119
|
haiku = [5,5,7]
a = list(map(int, input().split(" ")))
a = sorted(a)
if haiku == a:
print("YES")
else:
print("NO")
|
s428680437
|
p02612
|
u727787724
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 9,096
| 28
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n=int(input())
print(n%1000)
|
s747733270
|
Accepted
| 28
| 9,008
| 67
|
n=int(input())
if n%1000==0:
print(0)
else:
print(1000-n%1000)
|
s508058173
|
p03361
|
u796942881
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 623
|
We have a canvas divided into a grid with H rows and W columns. The square at the i-th row from the top and the j-th column from the left is represented as (i, j). Initially, all the squares are white. square1001 wants to draw a picture with black paint. His specific objective is to make Square (i, j) black when s_{i, j}= `#`, and to make Square (i, j) white when s_{i, j}= `.`. However, since he is not a good painter, he can only choose two squares that are horizontally or vertically adjacent and paint those squares black, for some number of times (possibly zero). He may choose squares that are already painted black, in which case the color of those squares remain black. Determine if square1001 can achieve his objective.
|
def main():
dy = [1, 0, -1, 0]
dx = [0, 1, 0, -1]
H, W = map(int, input().split())
s = ["." * (W + 2)]
s += ["." + input() + "." for i in range(H)]
s.append("." * (W + 2))
print(s)
for y in range(1, H + 1):
for x in range(1, W + 1):
if s[y][x] == "#" and s[y + dy[0]][x + dx[0]] == "."\
and s[y + dy[1]][x + dx[1]] == "."\
and s[y + dy[2]][x + dx[2]] == "."\
and s[y + dy[3]][x + dx[3]] == ".":
print("No")
return
print("Yes")
return
main()
|
s069011288
|
Accepted
| 18
| 3,064
| 551
|
def main():
dyx = [(1, 0), (0, 1), (-1, 0), (0, -1)]
H, W = map(int, input().split())
s = ["." * (W + 2)]
s += ["." + input() + "." for i in range(H)]
s.append("." * (W + 2))
for y in range(1, H + 1):
for x in range(1, W + 1):
if "." == s[y][x]:
continue
for dy, dx in dyx:
if "#" == s[y + dy][x + dx]:
break
else:
print("No")
return
print("Yes")
return
main()
|
s594273143
|
p02927
|
u569185453
| 2,000
| 1,048,576
|
Wrong Answer
| 22
| 2,940
| 221
|
Today is August 24, one of the five Product Days in a year. A date m-d (m is the month, d is the date) is called a Product Day when d is a two-digit number, and all of the following conditions are satisfied (here d_{10} is the tens digit of the day and d_1 is the ones digit of the day): * d_1 \geq 2 * d_{10} \geq 2 * d_1 \times d_{10} = m Takahashi wants more Product Days, and he made a new calendar called Takahashi Calendar where a year consists of M month from Month 1 to Month M, and each month consists of D days from Day 1 to Day D. In Takahashi Calendar, how many Product Days does a year have?
|
M, D = map(int, input().split())
for i in range(M+1):
for j in range(D+1):
d10 = int(j /10)
d1 = int(j % 10)
if d10 >=2 and d1 >=2 and d10 * d1 ==i:
print("{}月{}日".format(i, j))
|
s272540094
|
Accepted
| 23
| 3,060
| 222
|
M, D = map(int, input().split())
count = 0
for i in range(M+1):
for j in range(D+1):
d10 = int(j /10)
d1 = int(j % 10)
if d10 >=2 and d1 >=2 and d10 * d1 ==i:
count += 1
print(count)
|
s868732204
|
p03814
|
u834301346
| 2,000
| 262,144
|
Wrong Answer
| 36
| 10,244
| 109
|
Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s). Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`.
|
import re
letter = str(input())
pattern = 'A(.*)Z'
target = re.search(pattern, letter).group()
print(target)
|
s874404614
|
Accepted
| 35
| 10,024
| 114
|
import re
letter = str(input())
pattern = 'A(.*)Z'
target = re.search(pattern, letter).group()
print(len(target))
|
s168232848
|
p02613
|
u928758473
| 2,000
| 1,048,576
|
Wrong Answer
| 151
| 9,728
| 968
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
import os
import sys
from collections import defaultdict, Counter
from itertools import product, permutations,combinations, accumulate
from operator import itemgetter
from bisect import bisect_left,bisect
from heapq import heappop,heappush,heapify
from math import ceil, floor, sqrt, gcd
from copy import deepcopy
from functools import reduce
def main():
n = int(input())
lists = []
ans = [0]*4
for i in range(n):
s = input()
if not s in lists:
lists.append(s)
if s == "AC":
ans[0] += 1
elif s == "WA" :
ans[1] += 1
elif s == "TLE":
ans[2] += 1
else:
ans[3] += 1
for i in ans:
if i == "AC":
print("AC x", i)
elif i == "WA" :
print("WA x", i)
elif i == "TLE":
print("TLE x", i)
else:
print("RE x", i)
if __name__ == "__main__":
main()
|
s171238034
|
Accepted
| 149
| 9,736
| 978
|
import os
import sys
from collections import defaultdict, Counter
from itertools import product, permutations,combinations, accumulate
from operator import itemgetter
from bisect import bisect_left,bisect
from heapq import heappop,heappush,heapify
from math import ceil, floor, sqrt, gcd
from copy import deepcopy
from functools import reduce
def main():
n = int(input())
lists = []
ans = [0]*4
for i in range(n):
s = input()
if not s in lists:
lists.append(s)
if s == "AC":
ans[0] += 1
elif s == "WA" :
ans[1] += 1
elif s == "TLE":
ans[2] += 1
else:
ans[3] += 1
for i in range(len(ans)):
if i == 0:
print("AC x", ans[i])
elif i == 1 :
print("WA x", ans[i])
elif i == 2:
print("TLE x", ans[i])
else:
print("RE x", ans[i])
if __name__ == "__main__":
main()
|
s476629988
|
p03626
|
u721316601
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 404
|
We have a board with a 2 \times N grid. Snuke covered the board with N dominoes without overlaps. Here, a domino can cover a 1 \times 2 or 2 \times 1 square. Then, Snuke decided to paint these dominoes using three colors: red, cyan and green. Two dominoes that are adjacent by side should be painted by different colors. Here, it is not always necessary to use all three colors. Find the number of such ways to paint the dominoes, modulo 1000000007. The arrangement of the dominoes is given to you as two strings S_1 and S_2 in the following manner: * Each domino is represented by a different English letter (lowercase or uppercase). * The j-th character in S_i represents the domino that occupies the square at the i-th row from the top and j-th column from the left.
|
MOD = 1000000007
N = int(input())
S1 = input()
S2 = input()
if S1[0] == S2[0]:
ans, idx = 3, 1
else:
ans, idx = 6, 2
while idx < N:
print(S1[idx])
if S1[idx] == S2[idx]:
if S1[idx-1] == S2[idx-1]:
ans *= 2
idx += 1
else:
if S1[idx-1] == S2[idx-1]:
ans *= 2
else:
ans *= 3
idx += 2
ans %= MOD
print(ans)
|
s458062050
|
Accepted
| 17
| 3,064
| 385
|
MOD = 1000000007
N = int(input())
S1 = input()
S2 = input()
if S1[0] == S2[0]:
ans, idx = 3, 1
else:
ans, idx = 6, 2
while idx < N:
if S1[idx] == S2[idx]:
if S1[idx-1] == S2[idx-1]:
ans *= 2
idx += 1
else:
if S1[idx-1] == S2[idx-1]:
ans *= 2
else:
ans *= 3
idx += 2
ans %= MOD
print(ans)
|
s556547924
|
p02690
|
u465101448
| 2,000
| 1,048,576
|
Wrong Answer
| 64
| 9,132
| 153
|
Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X.
|
X=int(input())
for a in range(-125,125):
for b in range(-125,125):
i = a**5-b**5
if i == X:
print(a,b)
break
|
s730243022
|
Accepted
| 62
| 9,184
| 171
|
X=int(input())
ans=''
for a in range(-125,125):
for b in range(-125,125):
i = a**5-b**5
if i == X:
ans=[a,b]
print(' '.join(map(str,ans)))
|
s025722276
|
p02259
|
u841567836
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,596
| 419
|
Write a program of the Bubble Sort algorithm which sorts a sequence _A_ in ascending order. The algorithm should be based on the following pseudocode: BubbleSort(A) 1 for i = 0 to A.length-1 2 for j = A.length-1 downto i+1 3 if A[j] < A[j-1] 4 swap A[j] and A[j-1] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 4 of the pseudocode.
|
def bubble(l):
N = len(l)
flag = 1
while flag:
flag = 0
for j in range(N - 1, 1 - 1, -1):
if l[j] < l[j - 1]:
l[j], l[j - 1] = l[j - 1], l[j]
flag = 1
return l
if __name__ == '__main__':
n = int(input())
l = input().split()
for i in range(n):
l[i] = int(l[i])
l = bubble(l)
count = 0
leng = len(l)
for i in l:
count += 1
if count < leng:
print(i, end =' ')
else:
print(i)
|
s654927356
|
Accepted
| 20
| 5,608
| 467
|
def bubble(l):
count = 0
N = len(l)
flag = 1
while flag:
flag = 0
for j in range(N - 1, 1 - 1, -1):
if l[j] < l[j - 1]:
l[j], l[j - 1] = l[j - 1], l[j]
flag = 1
count += 1
return l, count
if __name__ == '__main__':
n = int(input())
l = input().split()
for i in range(n):
l[i] = int(l[i])
l, c = bubble(l)
count = 0
leng = len(l)
for i in l:
count += 1
if count < leng:
print(i, end =' ')
else:
print(i)
print(c)
|
s561836733
|
p02393
|
u656153606
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,732
| 306
|
Write a program which reads three integers, and prints them in ascending order.
|
a, b, c = [int(i) for i in input().split()]
if a > b:
if b > c:
print(a, b, c)
else:
print(a, c, b)
elif b > a:
if a > c:
print(b, a, c)
else:
print(b, c, a)
elif c > b:
if b > a:
print(c, b, a)
else:
print(c, a, b)
|
s037232553
|
Accepted
| 20
| 7,728
| 91
|
input = [int(i) for i in input().split()]
input.sort()
print(input[0], input[1], input[2])
|
s309964843
|
p02397
|
u648117624
| 1,000
| 131,072
|
Time Limit Exceeded
| 9,990
| 5,896
| 129
|
Write a program which reads two integers x and y, and prints them in ascending order.
|
x,y = map(int, input().split())
while True:
if x == 0 and y == 0 : break
list = [x, y]
print(list.sort())
|
s420239211
|
Accepted
| 60
| 5,608
| 193
|
while True:
x,y = map(int, input().split())
if x == 0 and y == 0:
break
elif x <= y:
print(str(x) + " " + str(y))
else:
print(str(y) + " " + str(x))
|
s207260306
|
p03130
|
u151625340
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 157
|
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads. Determine if we can visit all the towns by traversing each of the roads exactly once.
|
p = [0,0,0,0]
for i in range(3):
ai, bi = map(int, input().split())
p[ai-1] += 1
p[bi-1] += 1
if 3 in p:
print('YES')
else:
print('NO')
|
s551219618
|
Accepted
| 17
| 2,940
| 157
|
p = [0,0,0,0]
for i in range(3):
ai, bi = map(int, input().split())
p[ai-1] += 1
p[bi-1] += 1
if 3 in p:
print('NO')
else:
print('YES')
|
s261011879
|
p03079
|
u456595418
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 136
|
You are given three integers A, B and C. Determine if there exists an equilateral triangle whose sides have lengths A, B and C.
|
l = []
l= list(map(int,input().split()))
ls = sorted(l)
a = ls[0]
b = ls[1]
c = ls[2]
if a + b <= c:
print("Yes")
else:
print("No")
|
s801355725
|
Accepted
| 17
| 2,940
| 125
|
l = []
l= list(map(int,input().split()))
a = l[0]
b = l[1]
c = l[2]
if a == b and a == c:
print("Yes")
else:
print("No")
|
s646676357
|
p03575
|
u111508936
| 2,000
| 262,144
|
Wrong Answer
| 152
| 12,504
| 1,426
|
You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges. The i-th edge (1 \leq i \leq M) connects Vertex a_i and Vertex b_i. An edge whose removal disconnects the graph is called a _bridge_. Find the number of the edges that are bridges among the M edges.
|
import numpy as np
import math
# K = int(input())
# N, K = map(int, input().split())
# A = list(map(int, input().split()))
N, M = map(int, input().split())
E = np.zeros((N, N), dtype=int)
for m in range(M):
s, t = map(int, input().split())
s = s-1
t = t-1
E[s][t] = 1
E[t][s] = 1
# print(E)
bridges = set()
TIME = 0
def bridgeN(n, visited, parent, low, dis):
global TIME
visited[n] = 1
dis[n] = TIME
low[n] = TIME
TIME += 1
# print('time', TIME)
# print('n=', n)
# print('visited=', visited)
for v in range(N):
if E[n][v] == 0:
continue
if visited[v] == 1:
if v != parent[n]:
low[n] = min(low[n], dis[v])
continue
print('visited[', v, ']=', visited[v])
parent[v] = n
bridgeN(v, visited, parent, low, dis)
low[n] = min(low[n], low[v])
if low[v] > dis[n]:
if v <= n:
bridges.add((v, n))
else:
bridges.add((n, v))
def bridge(E):
visited = np.zeros(N, dtype=int)
dis = np.full_like(visited, N*N*10)
low = np.full_like(visited, N*N*10)
parent = np.full_like(visited, -1)
for n in range(N):
if visited[n] == 0:
bridgeN(n, visited, parent, low, dis)
bridge(E)
nBridges = len(bridges)
# print(bridges)
print(nBridges)
|
s318847440
|
Accepted
| 152
| 12,512
| 1,428
|
import numpy as np
import math
# K = int(input())
# N, K = map(int, input().split())
# A = list(map(int, input().split()))
N, M = map(int, input().split())
E = np.zeros((N, N), dtype=int)
for m in range(M):
s, t = map(int, input().split())
s = s-1
t = t-1
E[s][t] = 1
E[t][s] = 1
# print(E)
bridges = set()
TIME = 0
def bridgeN(n, visited, parent, low, dis):
global TIME
visited[n] = 1
dis[n] = TIME
low[n] = TIME
TIME += 1
# print('time', TIME)
# print('n=', n)
# print('visited=', visited)
for v in range(N):
if E[n][v] == 0:
continue
if visited[v] == 1:
if v != parent[n]:
low[n] = min(low[n], dis[v])
continue
# print('visited[', v, ']=', visited[v])
parent[v] = n
bridgeN(v, visited, parent, low, dis)
low[n] = min(low[n], low[v])
if low[v] > dis[n]:
if v <= n:
bridges.add((v, n))
else:
bridges.add((n, v))
def bridge(E):
visited = np.zeros(N, dtype=int)
dis = np.full_like(visited, N*N*10)
low = np.full_like(visited, N*N*10)
parent = np.full_like(visited, -1)
for n in range(N):
if visited[n] == 0:
bridgeN(n, visited, parent, low, dis)
bridge(E)
nBridges = len(bridges)
# print(bridges)
print(nBridges)
|
s286898627
|
p03853
|
u292978925
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 124
|
There is an image with a height of H pixels and a width of W pixels. Each of the pixels is represented by either `.` or `*`. The character representing the pixel at the i-th row from the top and the j-th column from the left, is denoted by C_{i,j}. Extend this image vertically so that its height is doubled. That is, print a image with a height of 2H pixels and a width of W pixels where the pixel at the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division is rounded down).
|
h, w = map(int, input().split())
in1 = []
for idx1 in range(h):
in1.append(input())
for item1 in in1:
print(item1)
|
s168222173
|
Accepted
| 17
| 3,060
| 225
|
#a = '2 2'
#b = ['*.', '.*']
h, w = map(int, input().split())
#h, w = map(int, a.split())
in1 = []
for idx1 in range(h):
in1.append(input())
# in1.append(b[idx1])
for item1 in in1:
print(item1)
print(item1)
|
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