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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s685930405
|
p04029
|
u587295817
| 2,000
| 262,144
|
Wrong Answer
| 37
| 3,064
| 32
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
a=int(input())
print(a*(a+1)/2)
|
s571030175
|
Accepted
| 39
| 3,064
| 37
|
a=int(input())
print(int(a*(a+1)/2))
|
s801562561
|
p03698
|
u912650255
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 79
|
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
S = input()
if len(set(S)) == len(S) :
print('Yes')
else:
print('No')
|
s097191981
|
Accepted
| 17
| 2,940
| 79
|
S = input()
if len(set(S)) == len(S) :
print('yes')
else:
print('no')
|
s169122617
|
p03160
|
u362771726
| 2,000
| 1,048,576
|
Wrong Answer
| 48
| 5,824
| 1,188
|
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.
|
import sys
from io import StringIO
import unittest
def resolve():
n = int(input())
h = [0] + list(map(int, input().split()))
inf = n * 10 ** 4 + 1
ans = [inf for _ in range(n + 1)]
ans[0] = 0
ans[1] = h[0]
for i in range(2, n + 1):
ans[i] = min(ans[i], ans[i - 1] + abs(h[i] - h[i - 1]), ans[i - 2] + abs(h[i - 2] + h[i - 2]))
print(ans[-1])
class TestClass(unittest.TestCase):
def assertIO(self, input, output):
stdout, stdin = sys.stdout, sys.stdin
sys.stdout, sys.stdin = StringIO(), StringIO(input)
resolve()
sys.stdout.seek(0)
out = sys.stdout.read()[:-1]
sys.stdout, sys.stdin = stdout, stdin
self.assertEqual(out, output)
def test_入力例_1(self):
input = """4
10 30 40 20"""
output = """30"""
self.assertIO(input, output)
def test_入力例_2(self):
input = """2
10 10"""
output = """0"""
self.assertIO(input, output)
def test_入力例_3(self):
input = """6
30 10 60 10 60 50"""
output = """40"""
self.assertIO(input, output)
if __name__ == "__main__":
unittest.main()
resolve()
|
s756988338
|
Accepted
| 275
| 122,620
| 1,866
|
import sys
from io import StringIO
import unittest
sys.setrecursionlimit(10 ** 7)
def memoize(f):
cache={}
def helper(*args):
if args not in cache:
cache[args] = f(*args)
return cache[args]
return helper
def resolve():
n = int(input())
h = list(map(int, input().split()))
inf = n * 10 ** 7
memo = [inf for _ in range(n)]
def dp(n: int) -> int:
if n == 0:
return 0
elif n == 1:
return abs(h[1] - h[0])
elif memo[n] != inf:
return memo[n]
else:
memo[n] = min(dp(n - 1) + abs(h[n] - h[n - 1]), dp(n - 2) + abs(h[n] - h[n - 2]))
return memo[n]
print(dp(n - 1))
class TestClass(unittest.TestCase):
def assertIO(self, input, output):
stdout, stdin = sys.stdout, sys.stdin
sys.stdout, sys.stdin = StringIO(), StringIO(input)
resolve()
sys.stdout.seek(0)
out = sys.stdout.read()[:-1]
sys.stdout, sys.stdin = stdout, stdin
self.assertEqual(out, output)
def test_入力例_1(self):
input = """4
10 30 40 20"""
output = """30"""
self.assertIO(input, output)
def test_入力例_2(self):
input = """2
10 10"""
output = """0"""
self.assertIO(input, output)
def test_入力例_3(self):
input = """6
30 10 60 10 60 50"""
output = """40"""
self.assertIO(input, output)
if __name__ == "__main__":
#unittest.main()
resolve()
|
s041230140
|
p03861
|
u599547273
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 60
|
You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x?
|
a, b, x = map(int, input().split(" "))
print((b+1)//x-a//x)
|
s173125648
|
Accepted
| 17
| 2,940
| 60
|
a, b, x = map(int, input().split(" "))
print(b//x-(a-1)//x)
|
s978589525
|
p03672
|
u074220993
| 2,000
| 262,144
|
Wrong Answer
| 28
| 9,136
| 268
|
We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not. You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
|
s = input()
D = {}
for c in s:
if c in D.keys():
D[c] += 1
else:
D[c] = 1
for i in range(len(s)+1):
for k in D.values():
if k % 1:
break
else:
ans = len(s)-i
break
D[s[len(s)-1-i]] -= 1
print(ans)
|
s966356990
|
Accepted
| 28
| 9,068
| 467
|
class mystr:
def __init__(self,string):
self.value = string
def isEven(self):
l = len(self.value)
if (not l & 1) and self.value[:l//2] == self.value[l//2:]:
return True
else:
return False
def pop(self):
l = len(self.value)
self.value = self.value[:l-1]
s = mystr(input())
l = len(s.value)
for i in reversed(range(l)):
s.pop()
if s.isEven():
print(i)
break
|
s955563180
|
p03599
|
u929793345
| 3,000
| 262,144
|
Wrong Answer
| 2,839
| 9,172
| 700
|
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())
a *= 100
b *= 100
wat = 0
wat_list = []
for i in range(f//a+1):
for j in range(f//b+1):
wat = a*i+b*j
if 0 < wat <= f:
wat_list.append(wat)
print(wat_list)
sug = 0
conc = 0
sug_max = 0
conc_best = 0
sug_best = 0
sug_wat_best = 0
for k in wat_list:
sug_max = k//100 * e
for l in range(sug_max // c + 1):
for m in range(sug_max //d + 1):
sug = c*l+d*m
conc = sug/(sug+k)*100
if 0<= sug <= sug_max and k+sug <= f and conc_best < conc :
conc_best = conc
sug_wat_best = sug + k
sug_best = sug
print(sug_wat_best, sug_best)
|
s029408052
|
Accepted
| 435
| 9,192
| 651
|
a, b, c, d, e, f = list(map(int, input().rstrip().split()))
a *= 100
b *= 100
x, y = a, 0
conc_best = 0
for i in range(f // a + 1):
for j in range(((f - a * i) // b) + 1):
water = i * a + j * b
if water == 0:
continue
rest = f - water
sug_max = e * water / 100
for k in range(rest // c + 1):
for l in range((rest - c * k) // d + 1):
sugar = c * k + d * l
conc = sugar / (sugar + water)
if sugar <= sug_max and conc> conc_best:
conc_best = conc
x, y = sugar+water , sugar
print(x, y)
|
s414936929
|
p03556
|
u239342230
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 31
|
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.
|
print(int(int(input())**.5//1))
|
s416635664
|
Accepted
| 18
| 2,940
| 31
|
print(int(int(input())**.5)**2)
|
s824664798
|
p03379
|
u940102677
| 2,000
| 262,144
|
Wrong Answer
| 217
| 25,556
| 149
|
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.
|
n = int(input())
x = list(map(int, input().split()))
a = x[n//2-1]
b = x[n//2-1]
for i in range(n):
if x[i] <= a:
print(b)
else:
print(a)
|
s606824339
|
Accepted
| 304
| 26,772
| 161
|
n = int(input())
x = list(map(int, input().split()))
y = sorted(x)
a = y[n//2-1]
b = y[n//2]
for i in range(n):
if x[i] <= a:
print(b)
else:
print(a)
|
s560658985
|
p03545
|
u354126779
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 617
|
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
seq=input()
list=[]
for i in range(4):
list.append(int(seq[i]))
ans=[]
brk=0
for a in range (2):
for b in range (2):
for c in range (2):
if list[0]+(2*a-1)*list[1]+(2*b-1)*list[2]+(2*c-1)*list[2]==7:
ans.append(a)
ans.append(b)
ans.append(c)
brk+=1
break
if brk!=0:
break
if brk!=0:
break
A=""
B=""
C=""
if a==0:
A="-"
else:
A="+"
if b==0:
B="-"
else:
B="+"
if c==0:
C="-"
else:
C="+"
print(str(list[0])+A+str(list[1])+B+str(list[2])+C+str(list[3]))
|
s342461589
|
Accepted
| 17
| 3,064
| 622
|
seq=input()
list=[]
for i in range(4):
list.append(int(seq[i]))
ans=[]
brk=0
for a in range (2):
for b in range (2):
for c in range (2):
if list[0]+(2*a-1)*list[1]+(2*b-1)*list[2]+(2*c-1)*list[3]==7:
ans.append(a)
ans.append(b)
ans.append(c)
brk+=1
break
if brk!=0:
break
if brk!=0:
break
A=""
B=""
C=""
if a==0:
A="-"
else:
A="+"
if b==0:
B="-"
else:
B="+"
if c==0:
C="-"
else:
C="+"
print(str(list[0])+A+str(list[1])+B+str(list[2])+C+str(list[3])+"=7")
|
s233507029
|
p03141
|
u092650292
| 2,000
| 1,048,576
|
Wrong Answer
| 928
| 41,348
| 1,385
|
There are N dishes of cuisine placed in front of Takahashi and Aoki. For convenience, we call these dishes Dish 1, Dish 2, ..., Dish N. When Takahashi eats Dish i, he earns A_i points of _happiness_ ; when Aoki eats Dish i, she earns B_i points of happiness. Starting from Takahashi, they alternately choose one dish and eat it, until there is no more dish to eat. Here, both of them choose dishes so that the following value is maximized: "the sum of the happiness he/she will earn in the end" minus "the sum of the happiness the other person will earn in the end". Find the value: "the sum of the happiness Takahashi earns in the end" minus "the sum of the happiness Aoki earns in the end".
|
from fractions import gcd
from math import factorial as f
from math import ceil,floor,sqrt
from sys import exit
from copy import deepcopy
import numpy as np
import bisect
def main():
n = int(input())
a = []
b = []
dif = []
for i in range(n):
atmp,btmp = map(int,input().split())
a.append(atmp)
b.append(btmp)
dif.append([atmp-btmp,i])
adif = sorted(dif, key=lambda x: a[x[1]])
adif = list(reversed(adif))
adif = sorted(adif, key=lambda x: x[0])
bdif = sorted(dif, key=lambda x: b[x[1]], reverse=True)
bdif = sorted(bdif, key=lambda x: x[0])
#bdif = list(reversed(bdif))
rem = [1 for i in range(n)]
takahashi = 0
aoki = 0
tp = 0
ap = 0
for i in range(ceil(n/2)):
while tp < n:
if rem[adif[tp][1]]:
takahashi += a[adif[tp][1]]
rem[adif[tp][1]]=0
# print('tp is ' + str(adif[tp][1]))
tp +=1
break
else:
tp +=1
while ap < n:
if rem[bdif[ap][1]]:
aoki+= b[bdif[ap][1]]
rem[bdif[ap][1]]=0
# print('ap is ' + str(bdif[ap][1]))
ap+=1
break
else:
ap += 1
print(takahashi-aoki)
main()
|
s681116900
|
Accepted
| 447
| 29,940
| 505
|
def main():
n = int(input())
a = []
b = []
s = []
for i in range(n):
atmp,btmp = map(int,input().split())
a.append(atmp)
b.append(btmp)
s.append([atmp+btmp,i])
s = sorted(s, key=lambda x: x[0], reverse=True)
rem = [1 for i in range(n)]
takahashi = 0
aoki = 0
for e,i in enumerate(s):
if not e % 2:
takahashi += a[i[1]]
else:
aoki += b[i[1]]
print(takahashi-aoki)
main()
|
s887386972
|
p03679
|
u813174766
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 116
|
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
|
a,b,c=map(int,input().split())
if b<=c:
print("delicious")
elif b<=a+c:
print("safe")
else:
print("dangerous")
|
s412024555
|
Accepted
| 17
| 3,064
| 116
|
a,b,c=map(int,input().split())
if c<=b:
print("delicious")
elif c<=a+b:
print("safe")
else:
print("dangerous")
|
s257496483
|
p03998
|
u408958033
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 303
|
Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.
|
a = input()
b = input()
c = input()
a = list(a)
b = list(b)
c = list(c)
x = 'a'
while True:
if x=='a':
x = a.pop(0)
elif x=='b':
x = b.pop(0)
else:
x = c.pop(0)
if len(a)==0:
print("A")
break
if len(b)==0:
print("B")
break
if len(c)==0:
print("C")
break
|
s309467686
|
Accepted
| 17
| 3,064
| 288
|
ss = [input() for _ in range(3)]
i = 0
while ss[i % 3] != "":
card = ss[i % 3][0]
ss[i % 3] = ss[i % 3][1:]
if card == "a":
i = 0
elif card == "b":
i = 1
else:
i = 2
if i == 0:
print("A")
elif i == 1:
print("B")
else:
print("C")
|
s787283137
|
p03477
|
u802977614
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 118
|
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.
|
a,b,c,d=map(int,input().split())
if a+b<c+d:
print("Left")
elif a+b==c+d:
print("Balanced")
else:
print("Right")
|
s555927501
|
Accepted
| 17
| 2,940
| 118
|
a,b,c,d=map(int,input().split())
if a+b>c+d:
print("Left")
elif a+b==c+d:
print("Balanced")
else:
print("Right")
|
s387855567
|
p02398
|
u498511622
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,492
| 89
|
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
|
x,y,z = map(int,input().split())
i=0
for s in range(x,y):
if z%s==0:
i += 1
print(i)
|
s923225933
|
Accepted
| 20
| 7,648
| 93
|
a,b,c = map(int,input().split())
i=0
for s in range(a,b+1):
if c % s == 0:
i += 1
print(i)
|
s876419156
|
p03657
|
u960513073
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 119
|
Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.
|
a,b = list(map(int, input().split()))
if a%3 ==0 or b%3==0 or a+b%3==0:
print("Possible")
else:
print("Impossible")
|
s503105319
|
Accepted
| 17
| 2,940
| 121
|
a,b = list(map(int, input().split()))
if a%3 ==0 or b%3==0 or (a+b)%3==0:
print("Possible")
else:
print("Impossible")
|
s794930503
|
p02415
|
u634490486
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,560
| 206
|
Write a program which converts uppercase/lowercase letters to lowercase/uppercase for a given string.
|
from sys import stdin
s = list(stdin.readline())
print(s)
for i in range(len(s)):
if s[i].islower():
s[i] = s[i].upper()
elif s[i].isupper():
s[i] = s[i].lower()
print(*s, sep="")
|
s523573350
|
Accepted
| 20
| 5,544
| 72
|
from sys import stdin
s = stdin.readline().swapcase()
print(s, end="")
|
s484189335
|
p03814
|
u925406312
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,516
| 76
|
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`.
|
s=input()
i=s.find('A')
j=s.rfind('Z')
print(i)
print(j)
print(abs(i-j+1))
|
s278550666
|
Accepted
| 18
| 3,512
| 57
|
s=input()
i=s.find('A')
j=s.rfind('Z')
print(abs(i-j)+1)
|
s407511093
|
p03385
|
u807772568
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 90
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
a = list(input())
if "a" in a and "b" in a and "c" in a:
print("YES")
else:
print("NO")
|
s109855260
|
Accepted
| 17
| 2,940
| 199
|
d = list(input())
a = 0
b = 0
c = 0
for i in range(3):
if d[i] == "a":
a = 1
if d[i] == "b":
b = 1
if d[i] == "c":
c = 1
if a*b*c == 1:
print("Yes")
else:
print("No")
|
s704731052
|
p02743
|
u870262604
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 151
|
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
|
a, b, c = map(int,input().split())
left = a+b-c
if left <= 0:
print("Yes")
else:
if (a*b - left**2) > 0:
print("Yes")
else:
print("No")
|
s740863697
|
Accepted
| 17
| 2,940
| 155
|
a, b, c = map(int,input().split())
right = c-a-b
if right <= 0:
print("No")
else:
if (right**2 - 4*a*b) > 0:
print("Yes")
else:
print("No")
|
s572953741
|
p03698
|
u363118893
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 88
|
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
S = str(input())
if len(list(S)) == len(set(S)):
print("Yes")
else:
print("No")
|
s345271157
|
Accepted
| 17
| 2,940
| 88
|
S = str(input())
if len(list(S)) == len(set(S)):
print("yes")
else:
print("no")
|
s392063078
|
p04031
|
u576917603
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 779
|
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal **integers** by transforming some of them. He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2). Find the minimum total cost to achieve his objective.
|
n=int(input())
a=list(map(int,input().split()))
def custom_round(number, ndigits=0):
if type(number) == int:
return number
d_point = len(str(number).split('.')[1])
if ndigits >= d_point:
return number
c = (10 ** d_point) * 2
return round((number * c + 1) / c, ndigits)
ave=custom_round(sum(a)/n)
ans=0
for i in a:
ans+=(ave-i)**2
print(ans)
|
s691924474
|
Accepted
| 19
| 3,188
| 784
|
n=int(input())
a=list(map(int,input().split()))
def custom_round(number, ndigits=0):
if type(number) == int:
return number
d_point = len(str(number).split('.')[1])
if ndigits >= d_point:
return number
c = (10 ** d_point) * 2
return round((number * c + 1) / c, ndigits)
ave=custom_round(sum(a)/n)
ans=0
for i in a:
ans+=(ave-i)**2
print(int(ans))
|
s079840962
|
p02401
|
u885889402
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,736
| 265
|
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) = [s for s in input().split()]
a = int(a)
b = int(b)
if op == "+":
print(a+b)
elif op == "-":
print(a-b)
elif op == "//":
print(a/b)
elif op == "*":
print(a*b)
else:
break
|
s717740300
|
Accepted
| 20
| 7,676
| 265
|
while(True):
(a,op,b) = [s for s in input().split()]
a = int(a)
b = int(b)
if op == "+":
print(a+b)
elif op == "-":
print(a-b)
elif op == "/":
print(a//b)
elif op == "*":
print(a*b)
else:
break
|
s781943764
|
p02612
|
u096025032
| 2,000
| 1,048,576
|
Wrong Answer
| 30
| 9,152
| 77
|
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.
|
A = int(input())
if A % 1000 == 0:
print(A/1000)
else:
print(A/1000 + 1)
|
s491498889
|
Accepted
| 29
| 9,152
| 62
|
import math
A = int(input())
print(math.ceil(A/1000)*1000-A)
|
s001415526
|
p02271
|
u301461168
| 5,000
| 131,072
|
Wrong Answer
| 30
| 7,560
| 395
|
Write a program which reads a sequence _A_ of _n_ elements and an integer _M_ , and outputs "yes" if you can make _M_ by adding elements in _A_ , otherwise "no". You can use an element only once. You are given the sequence _A_ and _q_ questions where each question contains _M i_.
|
numA = int(input().rstrip())
listA = list(map(int,input().rstrip().split(" ")))
numQ = int(input().rstrip())
listQ = list(map(int,input().rstrip().split(" ")))
sumA = []
cnt = 0
for i in range(numA):
for j in range(i+1, numA):
sumA.append(listA[i]+listA[j])
for i in range(numQ):
if sumA.count(listQ[i]) > 0:
#exsits
print("yes")
else:
print("no")
|
s889154247
|
Accepted
| 630
| 58,252
| 441
|
import itertools
numA = int(input().rstrip())
listA = list(map(int,input().rstrip().split(" ")))
numQ = int(input().rstrip())
listQ = list(map(int,input().rstrip().split(" ")))
sumA = set([])
cnt = 0
for i in range(1,numA+1):
tmpComb = list(itertools.combinations(listA, i))
for nums in tmpComb:
sumA.add(sum(nums))
for i in range(numQ):
if listQ[i] in sumA:
print("yes")
else:
print("no")
|
s521213664
|
p02692
|
u858742833
| 2,000
| 1,048,576
|
Wrong Answer
| 24
| 9,148
| 11
|
There is a game that involves three variables, denoted A, B, and C. As the game progresses, there will be N events where you are asked to make a choice. Each of these choices is represented by a string s_i. If s_i is `AB`, you must add 1 to A or B then subtract 1 from the other; if s_i is `AC`, you must add 1 to A or C then subtract 1 from the other; if s_i is `BC`, you must add 1 to B or C then subtract 1 from the other. After each choice, none of A, B, and C should be negative. Determine whether it is possible to make N choices under this condition. If it is possible, also give one such way to make the choices.
|
print('No')
|
s807207970
|
Accepted
| 181
| 10,716
| 731
|
def main():
N, A, B, C = list(map(int, input().split()))
S = [{'AB':(0, 1), 'BC': (1, 2), 'AC': (0, 2)}[input()] for _ in range(N)]
ABC = [A, B, C]
RR = ['A', 'B', 'C']
R = []
for i, (s1, s2) in enumerate(S):
if ABC[s1] == 0 and ABC[s2] == 0:
return False, None
if ABC[s1] < ABC[s2]:
s1, s2 = s2, s1
if A + B + C == 2 and ABC[s1] == 1 and ABC[s2] == 1 and i < N - 1 and S[i] != S[i + 1]:
s1n, s2n = S[i + 1]
if s1 in [s1n, s2n]:
s1, s2 = s2, s1
ABC[s1] -= 1
ABC[s2] += 1
R.append(RR[s2])
return True, R
a, s = main()
if not a:
print('No')
else:
print('Yes')
print('\n'.join(s))
|
s477144474
|
p03644
|
u943057856
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 2,940
| 128
|
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
n=int(input())
ans=0
for i in range(n):
a=0
while int(i%2)==0:
a+=1
i=i//2
ans=max(ans,a)
print(ans)
|
s123232155
|
Accepted
| 17
| 2,940
| 103
|
n=int(input())
l=[2**i for i in range(8)]
for i in l[::-1]:
if i<=n:
print(i)
break
|
s797122524
|
p03493
|
u390958150
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 99
|
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.
|
masu = list(input())
count = 0
for i in range(3):
if masu[0] == '1':
count += 1
print(count)
|
s443210295
|
Accepted
| 17
| 2,940
| 100
|
masu = list(input())
count = 0
for i in range(3):
if masu[i] == '1':
count += 1
print(count)
|
s251497128
|
p02420
|
u025362139
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,592
| 202
|
Your task is to shuffle a deck of n cards, each of which is marked by a alphabetical letter. A single shuffle action takes out h cards from the bottom of the deck and moves them to the top of the deck. The deck of cards is represented by a string as follows. abcdeefab The first character and the last character correspond to the card located at the bottom of the deck and the card on the top of the deck respectively. For example, a shuffle with h = 4 to the above deck, moves the first 4 characters "abcd" to the end of the remaining characters "eefab", and generates the following deck: eefababcd You can repeat such shuffle operations. Write a program which reads a deck (a string) and a sequence of h, and prints the final state (a string).
|
#coding: UTF-8
while True:
str = input()
if str == '-':
break
m = int(input())
for num in range(m):
h = int(input())
str = str[h:] + str[:h]
print(str)
|
s880353808
|
Accepted
| 20
| 5,604
| 272
|
#coding: UTF-8
shuffle = []
while True:
str = input()
if str == '-':
break
m = int(input())
for num in range(m):
h = int(input())
str = str[h:] + str[:h]
shuffle.append(str)
for i in range(len(shuffle)):
print(shuffle[i])
|
s677000984
|
p00102
|
u811773570
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,564
| 353
|
Your task is to develop a tiny little part of spreadsheet software. Write a program which adds up columns and rows of given table as shown in the following figure:
|
def main():
while True:
n = int(input())
if n == 0:
break
for i in range(n):
num = 0
m = map(int, input().split())
for j in m:
print("{:5}".format(j), end='')
num += j
print("{:5}".format(num))
if __name__ == '__main__':
main()
|
s820275768
|
Accepted
| 40
| 7,920
| 647
|
from copy import deepcopy
def main():
while True:
n = int(input())
if n == 0:
break
tate = [0 for i in range(n)]
for i in range(n):
num = 0
m = map(int, input().split())
hoge = deepcopy(m)
tate = [x + y for (x, y) in zip(tate, hoge)]
for j in m:
print("{:5}".format(j), end='')
num += j
print("{:5}".format(num))
num = 0
for j in tate:
print("{:5}".format(j), end='')
num += j
print("{:5}".format(num))
if __name__ == '__main__':
main()
|
s479167029
|
p03943
|
u760831084
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 97
|
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
a, b, c = sorted(map(int, input().split()))
if a == b + c:
print('Yes')
else:
print('No')
|
s809029206
|
Accepted
| 17
| 2,940
| 97
|
a, b, c = sorted(map(int, input().split()))
if a + b == c:
print('Yes')
else:
print('No')
|
s869750453
|
p02419
|
u607723579
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,460
| 1
|
Write a program which reads a word W and a text T, and prints the number of word W which appears in text T T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.
|
s239421697
|
Accepted
| 20
| 5,572
| 328
|
import sys
BIG_NUM = 2000000000
MOD = 1000000007
EPS = 0.000000001
P = str(input()).upper()
line = []
while True:
tmp_line = str(input())
if tmp_line == 'END_OF_TEXT':
break
line += tmp_line.upper().split()
ans = 0
for tmp_word in line:
if tmp_word == P:
ans += 1
print("%d"%(ans))
|
|
s184692562
|
p02401
|
u982632052
| 1,000
| 131,072
|
Wrong Answer
| 40
| 7,300
| 83
|
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:
s = input()
if (s == '0 ? 0'):
break
print(eval(s))
|
s692607101
|
Accepted
| 30
| 7,544
| 112
|
import re
while True:
s = input()
if (re.match(r'.+\s\?\s.+', s)):
break
print(int(eval(s)))
|
s309848053
|
p03712
|
u934442292
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 157
|
You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1.
|
H, W = map(int, input().split())
a = [""] * H
for _a in a:
_a = input()
print("#" * (W + 2))
for _a in a:
print("#{}#".format(_a))
print("#" * (W + 2))
|
s622208223
|
Accepted
| 18
| 3,060
| 166
|
H, W = map(int, input().split())
a = [""] * H
for i in range(H):
a[i] = input()
print("#" * (W+2))
for i in range(H):
print("#" + a[i] + "#")
print("#" * (W+2))
|
s266600972
|
p03139
|
u814986259
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 76
|
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())
print(max(0,min(A,B) -(N-max(A,B))),min(A,B))
|
s617706794
|
Accepted
| 17
| 2,940
| 76
|
N,A,B=map(int,input().split())
print(min(A,B),max(0,min(A,B) -(N-max(A,B))))
|
s546169506
|
p02612
|
u015647294
| 2,000
| 1,048,576
|
Wrong Answer
| 25
| 9,072
| 48
|
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())
L = N // 1000
print((L + 1)-N)
|
s800832529
|
Accepted
| 29
| 9,160
| 91
|
N = int(input())
if (N % 1000) != 0:
ans = 1000 - N % 1000
else:
ans = 0
print(ans)
|
s688129548
|
p03196
|
u340249922
| 2,000
| 1,048,576
|
Wrong Answer
| 417
| 3,060
| 226
|
There are N integers a_1, a_2, ..., a_N not less than 1. The values of a_1, a_2, ..., a_N are not known, but it is known that a_1 \times a_2 \times ... \times a_N = P. Find the maximum possible greatest common divisor of a_1, a_2, ..., a_N.
|
N, P = [int(_c) for _c in input().split(" ")]
if N == 1:
print(P)
else:
_max = -1
M = int(P ** (1 / N))
for i in range(1, M):
var = i ** N
if P % var == 0:
_max = i
print(_max)
|
s844548102
|
Accepted
| 383
| 3,060
| 226
|
N, P = [int(_c) for _c in input().split(" ")]
if N == 1:
print(P)
else:
_max = 1
M = int(P ** (1 / N) + 1.0 * 1e-9)
for i in range(1, M + 1):
if P % (i ** N) == 0:
_max = i
print(_max)
|
s127506155
|
p03853
|
u856232850
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,188
| 167
|
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).
|
n,m =list(map(int,input().split()))
a =[]
for i in range(n):
a.append(list(input()))
b = []
for j in a:
b.append(j)
b.append(j)
for k in b:
print(k)
|
s773566975
|
Accepted
| 18
| 3,060
| 97
|
n,m =list(map(int,input().split()))
for i in range(n):
a = input()
print(a)
print(a)
|
s864402019
|
p02613
|
u875756191
| 2,000
| 1,048,576
|
Wrong Answer
| 151
| 9,136
| 319
|
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())
a = 0
w = 0
t = 0
r = 0
for _ in range(n):
s = input()
if s == 'AC':
a += 1
elif s == 'WA':
w += 1
elif s == 'TLE':
t += 1
elif s == 'RE':
r += 1
print('AC × ' + str(a))
print('WA × ' + str(w))
print('TLE × ' + str(t))
print('RE × ' + str(r))
|
s605782671
|
Accepted
| 146
| 9,204
| 315
|
n = int(input())
a = 0
w = 0
t = 0
r = 0
for _ in range(n):
s = input()
if s == 'AC':
a += 1
elif s == 'WA':
w += 1
elif s == 'TLE':
t += 1
elif s == 'RE':
r += 1
print('AC x ' + str(a))
print('WA x ' + str(w))
print('TLE x ' + str(t))
print('RE x ' + str(r))
|
s892929017
|
p03555
|
u778700306
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 85
|
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
a = input()
b = reversed(input())
if a == b:
print("YES")
else:
print("NO")
|
s489049564
|
Accepted
| 18
| 2,940
| 81
|
a = input()
b = input()[::-1]
if a == b:
print("YES")
else:
print("NO")
|
s276729936
|
p03606
|
u492447501
| 2,000
| 262,144
|
Wrong Answer
| 20
| 2,940
| 125
|
Joisino is working as a receptionist at a theater. The theater has 100000 seats, numbered from 1 to 100000. According to her memo, N groups of audiences have come so far, and the i-th group occupies the consecutive seats from Seat l_i to Seat r_i (inclusive). How many people are sitting at the theater now?
|
N = int(input())
count = 0
for i in range(N):
l, r = map(int, input().split())
count = count + (r-l)
print(count)
|
s522825399
|
Accepted
| 21
| 2,940
| 129
|
N = int(input())
count = 0
for i in range(N):
l, r = map(int, input().split())
count = count + (r-l) + 1
print(count)
|
s057543311
|
p02747
|
u892308039
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 109
|
A Hitachi string is a concatenation of one or more copies of the string `hi`. For example, `hi` and `hihi` are Hitachi strings, while `ha` and `hii` are not. Given a string S, determine whether S is a Hitachi string.
|
S = input()
S_size = len(S)
S_hi = S.count('hi')
if S_size == S_hi:
print('Yes')
else:
print('No')
|
s412822681
|
Accepted
| 18
| 2,940
| 110
|
S = input()
S_size = len(S)
S_hi = S.count('hi')
if S_size == S_hi*2:
print('Yes')
else:
print('No')
|
s096604072
|
p03386
|
u379692329
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 193
|
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
A, B, K = map(int, input().split())
select = []
for i in range(A, min(B, A+K)):
select.append(i)
for i in range(max(A, B-K+1), B+1):
select.append(i)
for i in set(select):
print(i)
|
s895520808
|
Accepted
| 17
| 3,060
| 201
|
A, B, K = map(int, input().split())
select = []
for i in range(A, min(B, A+K)):
select.append(i)
for i in range(max(A, B-K+1), B+1):
select.append(i)
for i in sorted(set(select)):
print(i)
|
s944173668
|
p03695
|
u013408661
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 458
|
In AtCoder, a person who has participated in a contest receives a _color_ , which corresponds to the person's rating as follows: * Rating 1-399 : gray * Rating 400-799 : brown * Rating 800-1199 : green * Rating 1200-1599 : cyan * Rating 1600-1999 : blue * Rating 2000-2399 : yellow * Rating 2400-2799 : orange * Rating 2800-3199 : red Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not. Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i. Find the minimum and maximum possible numbers of different colors of the users.
|
N=int(input())
rate=[0]*8
tourist=0
a=list(map(int,input().split()))
for i in range(N):
if 1<=i<399:
rate[0]+=1
elif 400<=i<=799:
rate[1]+=1
elif 800<=i<=1199:
rate[2]+=1
elif 1200<=i<=1599:
rate[3]+=1
elif 1600<=i<=1999:
rate[4]+=1
elif 2000<=i<=2399:
rate[5]+=1
elif 2400<=i<=2799:
rate[6]+=1
elif 2800<=i<=3199:
rate[7]+=1
elif i<=3200:
tourist+=1
print(max(8-rate.count(0),1),8-rate.count(0)+tourist)
|
s992678396
|
Accepted
| 17
| 3,064
| 400
|
N=int(input())
rate=[0]*8
tourist=0
a=list(map(int,input().split()))
for i in a:
if i<400:
rate[0]+=1
elif i<800:
rate[1]+=1
elif i<1200:
rate[2]+=1
elif i<1600:
rate[3]+=1
elif i<2000:
rate[4]+=1
elif i<2400:
rate[5]+=1
elif i<2800:
rate[6]+=1
elif i<3200:
rate[7]+=1
else:
tourist+=1
print(max(8-rate.count(0),1),max(1,8-rate.count(0)+tourist))
|
s307392789
|
p03455
|
u354126779
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 81
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a,b=map(int,input().split())
c=a*b
if c%2==0:
print("Odd")
else:
print("Even")
|
s111596529
|
Accepted
| 17
| 2,940
| 83
|
a,b=map(int,input().split())
c=a*b
if c%2==0:
print("Even")
else:
print("Odd")
|
s293582592
|
p03494
|
u642120132
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 2,940
| 144
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
_ = input()
a = list(map(int, input().split()))
cnt = 0
while all(i % 2 == 0 for i in a):
for j in a:
j /= 2
cnt += 1
print(cnt)
|
s910500292
|
Accepted
| 18
| 3,060
| 140
|
_ = input()
a = list(map(int, input().split()))
cnt = 0
while all(i % 2 == 0 for i in a):
a = [j / 2 for j in a]
cnt += 1
print(cnt)
|
s083683202
|
p03827
|
u963903527
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 125
|
You have an integer variable x. Initially, x=0. Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`. Find the maximum value taken by x during the operations (including before the first operation, and after the last operation).
|
n = input()
s = input()
li = list(s)
x = 0
for argv in li:
if argv == "I":
x += 1
elif argv == "D":
x -= 1
print(x)
|
s436171933
|
Accepted
| 19
| 3,060
| 217
|
n = input()
s = input()
li = list(s)
x = 0
num_list = list()
num_list.append(0)
for argv in li:
if argv == "I":
x += 1
num_list.append(x)
elif argv == "D":
x -= 1
num_list.append(x)
print(max(num_list))
|
s735408853
|
p03909
|
u902151549
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 196
|
There is a grid with H rows and W columns. The square at the i-th row and j-th column contains a string S_{i,j} of length 5. The rows are labeled with the numbers from 1 through H, and the columns are labeled with the uppercase English letters from `A` through the W-th letter of the alphabet. Exactly one of the squares in the grid contains the string `snuke`. Find this square and report its location. For example, the square at the 6-th row and 8-th column should be reported as `H6`.
|
h,w=map(int,input().split())
A=[input().split() for _ in range(h)]
for a in range(h):
for b in range(w):
if A[a][b]=="snuke":
print("{}{}".format(a+1,chr(ord("A")+b)))
break
|
s071064858
|
Accepted
| 17
| 3,060
| 193
|
h,w=map(int,input().split())
A=[input().split() for _ in range(h)]
for a in range(h):
for b in range(w):
if A[a][b]=="snuke":
print("{1}{0}".format(a+1,chr(ord("A")+b)))
break
|
s662713812
|
p02612
|
u331640179
| 2,000
| 1,048,576
|
Wrong Answer
| 2,205
| 9,116
| 56
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n = int(input())
while n < 1000:
n //= 1000
print(n)
|
s260496376
|
Accepted
| 29
| 9,156
| 62
|
n = int(input())
while n > 1000:
n -= 1000
print(1000 - n)
|
s264702627
|
p03971
|
u062691227
| 2,000
| 262,144
|
Wrong Answer
| 69
| 9,244
| 318
|
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these. Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests. * A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B. * An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students. A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these. Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
|
x, s = *open(0),
n, a, b = map(int, x.split())
p = 0
wp = 0
for char in s:
if char == 'c' or p >= a+b:
print('No')
continue
if char == 'a':
p += 1
print('Yes')
continue
if wp < b:
p += 1
wp += 1
print('Yes')
else:
print('No')
|
s556362937
|
Accepted
| 72
| 9,192
| 323
|
x = input()
s = input()
n, a, b = map(int, x.split())
p = 0
wp = 0
for char in s:
if char == 'c' or p >= a+b:
print('No')
continue
if char == 'a':
p += 1
print('Yes')
continue
if wp < b:
p += 1
wp += 1
print('Yes')
else:
print('No')
|
s478794359
|
p03682
|
u711238850
| 2,000
| 262,144
|
Wrong Answer
| 2,113
| 147,592
| 3,980
|
There are N towns on a plane. The i-th town is located at the coordinates (x_i,y_i). There may be more than one town at the same coordinates. You can build a road between two towns at coordinates (a,b) and (c,d) for a cost of min(|a-c|,|b-d|) yen (the currency of Japan). It is not possible to build other types of roads. Your objective is to build roads so that it will be possible to travel between every pair of towns by traversing roads. At least how much money is necessary to achieve this?
|
import heapq
class Graph:
def __init__(self,v,edgelist,w_v = None,directed = False):
super().__init__()
self.v = v
self.w_e = [{} for _ in [0]*self.v]
self.neighbor = [[] for _ in [0]*self.v]
self.w_v = w_v
self.directed = directed
for i,j,w in edgelist:
self.w_e[i][j] = w
self.neighbor[i].append(j)
def dijkstra(self,v_n):
d = [float('inf')]*self.v
d[v_n] = 0
prev = [-1]*self.v
queue = []
for i,d_i in enumerate(d): heapq.heappush(queue,(d_i,i))
while len(queue)>0:
d_u,u = queue.pop()
if d[u]<d_u :continue
for v in self.neighbor[u]:
alt = d[u]+self.w_e[u][v]
if d[v]>alt:
d[v] = alt
prev[v] = u
heapq.heappush(queue,(alt,v))
return d,prev
def warshallFloyd(self):
d = [[10**18]*self.v for _ in [0]*self.v]
for i in range(self.v):
d[i][i] = 0
for i in range(self.v):
for j in self.neighbor[i]:
d[i][j] = self.w_e[i][j]
for i in range(self.v):
for j in self.neighbor[i]:
d[i][j] = self.w_e[i][j]
for k in range(self.v):
for i in range(self.v):
for j in range(self.v):
check = d[i][k] + d[k][j]
if d[i][j] > check:
d[i][j] = check
return d
def prim(self):
gb = GraphBuilder(self.v,self.directed)
queue = []
for i,w in self.w_e[0].items(): heapq.heappush(queue,(w,0,i))
rest = [True]*self.v
rest[0] = False
while len(queue)>0:
w,i,j = heapq.heappop(queue)
if rest[j]:
gb.addEdge(i,j,w)
rest[j] = False
for k,w in self.w_e[j].items():
heapq.heappush(queue,(w,j,k))
return gb
class Tree():
def __init__(self,v,e):
pass
class GraphBuilder():
def __init__(self,v,directed = False):
self.v = v
self.directed = directed
self.edge = []
def addEdge(self,i,j,w=1):
if not self.directed:
self.edge.append((j,i,w))
self.edge.append((i,j,w))
def addEdges(self,edgelist,weight = True):
if weight:
if self.directed:
for i,j,w in edgelist:
self.edge.append((i,j,w))
else:
for i,j,w in edgelist:
self.edge.append((i,j,w))
self.edge.append((j,i,w))
else:
if self.directed:
for i,j,w in edgelist:
self.edge.append((i,j,1))
else:
for i,j,w in edgelist:
self.edge.append((i,j,1))
self.edge.append((j,i,1))
def addAdjMat(self, mat):
for i,mat_i in enumerate(mat):
for j,w in enumerate(mat_i):
self.edge.append((i,j,w))
def buildTree(self):
pass
def buildGraph(self):
return Graph(self.v,self.edge,directed=self.directed)
def main():
n= int(input())
edge = []
for i in range(n):
x,y = list(map(int,input().split()))
edge.append((x,y,i))
xs = sorted(edge,key=lambda a:a[0])
ys = sorted(edge,key=lambda a:a[1])
edge = []
for i in range(n-1):
x1,y1,p1 = xs[i]
x2,y2,p2 = xs[i+1]
edge.append((p1,p2,min(x2-x1,y2-y1)))
x1,y1,p1 = ys[i]
x2,y2,p2 = ys[i+1]
edge.append((p1,p2,y2-y1))
gb = GraphBuilder(n)
gb.addEdges(edge)
mint = gb.buildGraph().prim()
print(mint.edge)
ans = 0
for a in mint.edge:
ans+=a[2]
print(ans//2)
if __name__ == "__main__":
main()
|
s583514615
|
Accepted
| 1,236
| 49,016
| 1,630
|
class UnionFind:
def __init__(self,n):
super().__init__()
self.par = [-1]*n
self.rank = [0]*n
self.tsize = [1]*n
def root(self,x):
if self.par[x] == -1:
return x
else:
self.par[x] = self.root(self.par[x])
return self.par[x]
def unite(self, x, y):
x_r = self.root(x)
y_r = self.root(y)
if self.rank[x_r]>self.rank[y_r]:
self.par[y_r] = x_r
elif self.rank[x_r]<self.rank[y_r]:
self.par[x_r] = y_r
elif x_r != y_r:
self.par[y_r] = x_r
self.rank[x_r] += 1
if x_r != y_r:
size = self.tsize[x_r]+self.tsize[y_r]
self.tsize[x_r] = size
self.tsize[y_r] = size
def isSame(self,x,y):
return self.root(x) == self.root(y)
def size(self,x):
return self.tsize[self.root(x)]
def main():
n= int(input())
edge = []
for i in range(n):
x,y = list(map(int,input().split()))
edge.append((x,y,i))
xs = sorted(edge,key=lambda a:a[0])
ys = sorted(edge,key=lambda a:a[1])
edge = []
for i in range(n-1):
x1,y1,p1 = xs[i]
x2,y2,p2 = xs[i+1]
edge.append((p1,p2,min(x2-x1,abs(y2-y1))))
x1,y1,p1 = ys[i]
x2,y2,p2 = ys[i+1]
edge.append((p1,p2,min(abs(x2-x1),y2-y1)))
edge.sort(key = lambda a:a[2])
uf = UnionFind(n)
ans = 0
for x,y,w in edge:
if uf.isSame(x,y):continue
uf.unite(x,y)
ans+=w
print(ans)
if __name__ == "__main__":
main()
|
s924851114
|
p03385
|
u702208001
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 49
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
print('YES' if len(set(input())) == 3 else 'No')
|
s185403068
|
Accepted
| 17
| 2,940
| 49
|
print('Yes' if len(set(input())) == 3 else 'No')
|
s098182958
|
p04044
|
u681323954
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 43
|
Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.
|
s=input()
print(s[0]+str((len(s)-2))+s[-1])
|
s132647828
|
Accepted
| 17
| 3,060
| 89
|
n,l = map(int, input().split())
s = sorted([input() for i in range(n)])
print(*s, sep="")
|
s692745959
|
p03433
|
u936047618
| 2,000
| 262,144
|
Wrong Answer
| 24
| 9,156
| 108
|
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
n = int(input("N--->"))
a = int(input("A--->"))
b = n%500
if a > b:
print("Yes")
else:
print("No")
|
s456381059
|
Accepted
| 23
| 9,024
| 95
|
N = int(input())
A = int(input())
b = N%500
if A >= b:
print("Yes")
else:
print("No")
|
s860239899
|
p03998
|
u136395536
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 7,660
| 147
|
Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.
|
A = input()
B = input()
C = input()
moji = A[0]
while(len(A)!=0 or len(B)!=0 or len(C)!=0):
if moji == "a":
A = A[2:]
print(A)
|
s125375615
|
Accepted
| 17
| 3,064
| 466
|
A = input()
B = input()
C = input()
throw = "a"
a = 0
b = 0
c = 0
while True:
if throw == "a":
if a == len(A):
print("A")
break
throw = A[a]
a += 1
elif throw == "b":
if b == len(B):
print("B")
break
throw = B[b]
b += 1
elif throw == "c":
if c == len(C):
print("C")
break
throw = C[c]
c += 1
|
s839210880
|
p03407
|
u007550226
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 65
|
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')
|
s858484192
|
Accepted
| 17
| 2,940
| 65
|
A,B,C = map(int,input().split())
print('Yes' if A+B>=C else 'No')
|
s412390440
|
p03574
|
u576917603
| 2,000
| 262,144
|
Wrong Answer
| 36
| 3,188
| 542
|
You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process.
|
dx=[-1,0,1,-1,1,-1,0,1]
dy=[-1,-1,-1,0,0,1,1,1]
h,w=map(int,input().split())
field = []
for i in range(h):
field.append(list(input()))
print(field)
for y in range(h):
for x in range(w):
if field[y][x]=="#":
continue
cnt=0
for k in range(8):
nx=x+dx[k]
ny=y+dy[k]
if nx<0 or nx>=w or ny<0 or ny>=h:
continue
if field[ny][nx]=="#":
cnt+=1
field[y][x]=str(cnt)
for i in range(h):
print("".join(field[i]))
|
s690134319
|
Accepted
| 31
| 3,188
| 524
|
dx=[-1,0,1,-1,1,-1,0,1]
dy=[-1,-1,-1,0,0,1,1,1]
h,w=map(int,input().split())
field = []
for i in range(h):
field.append(list(input()))
for y in range(h):
for x in range(w):
if field[y][x]=="#":
continue
cnt=0
for k in range(8):
nx=x+dx[k]
ny=y+dy[k]
if nx<0 or nx>=w or ny<0 or ny>=h:
continue
if field[ny][nx]=="#":
cnt+=1
field[y][x]=str(cnt)
for i in range(h):
print("".join(field[i]))
|
s518526979
|
p02612
|
u257332942
| 2,000
| 1,048,576
|
Wrong Answer
| 35
| 9,076
| 58
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n = int(input())
while n >= 1000:
n -= 1000
print(n)
|
s310163142
|
Accepted
| 27
| 9,152
| 79
|
n = int(input())
a = n % 1000
if a == 0:
print(0)
else:
print(1000-a)
|
s637826951
|
p02975
|
u712397093
| 2,000
| 1,048,576
|
Wrong Answer
| 114
| 14,212
| 137
|
Snuke has N hats. The i-th hat has an integer a_i written on it. There are N camels standing in a circle. Snuke will put one of his hats on each of these camels. If there exists a way to distribute the hats to the camels such that the following condition is satisfied for every camel, print `Yes`; otherwise, print `No`. * The bitwise XOR of the numbers written on the hats on both adjacent camels is equal to the number on the hat on itself. What is XOR? The bitwise XOR x_1 \oplus x_2 \oplus \ldots \oplus x_n of n non- negative integers x_1, x_2, \ldots, x_n is defined as follows: - When x_1 \oplus x_2 \oplus \ldots \oplus x_n is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if the number of integers among x_1, x_2, \ldots, x_n whose binary representations have 1 in the 2^k's place is odd, and 0 if that count is even. For example, 3 \oplus 5 = 6.
|
N = int(input())
a = list(map(int, input().split()))
sum = 0
for i in range(N):
sum ^= a[i]
print('Yes' if sum == 0 else 'No')
|
s259490528
|
Accepted
| 58
| 14,116
| 133
|
N = int(input())
a = list(map(int, input().split()))
sum = 0
for i in range(N):
sum ^= a[i]
print('Yes' if sum == 0 else 'No')
|
s815858999
|
p03729
|
u830054172
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 107
|
You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`.
|
a, b, c = list(input().split())
if a[-1] == b[0] and b[-1] == c[0]:
print("Yes")
else:
print("No")
|
s750049035
|
Accepted
| 17
| 2,940
| 107
|
a, b, c = list(input().split())
if a[-1] == b[0] and b[-1] == c[0]:
print("YES")
else:
print("NO")
|
s690954974
|
p03657
|
u701318346
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 125
|
Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.
|
A, B = map(int, input().split())
if A % 3 == 0 or B % 3 == 0 or (A + B) == 0:
print('Possible')
else:
print('Impossible')
|
s347788193
|
Accepted
| 24
| 3,316
| 132
|
A, B = map(int, input().split())
if A % 3 == 0 or B % 3 == 0 or (A + B) % 3 == 0:
print('Possible')
else:
print('Impossible')
|
s519717271
|
p03854
|
u110927356
| 2,000
| 262,144
|
Wrong Answer
| 1,655
| 21,708
| 461
|
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
import math
import numpy as np
#n, a, b = map(int, input().split())
#print(len(z))
# print(i)
#print(sum(a[::2]) - sum(a[1::2]))
s = input()
s.replace("eraser","").replace("erase","").replace("dreamer","").replace("dream","")
if s:
print("NO")
else:
print("YES")
|
s826395133
|
Accepted
| 1,349
| 21,876
| 466
|
import math
import numpy as np
#n, a, b = map(int, input().split())
#print(len(z))
# print(i)
#print(sum(a[::2]) - sum(a[1::2]))
s = input()
s = s.replace("eraser","").replace("erase","").replace("dreamer","").replace("dream","")
if s:
print("NO")
else:
print("YES")
|
s446234187
|
p03545
|
u468206018
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 435
|
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
s = list(input())
def calc(i, a, b,tag):
a ,b = int(a), int(b)
if i==0:
c = a+b
tag.append('+')
elif i==1:
c = a-b
tag.append('-')
return c, tag
for i in range(2):
tag = []
c ,tag= calc(i,s[0],s[1],tag)
for j in range(2):
d ,tag = calc(j,c,s[2],tag)
for k in range(2):
e ,tag = calc(k,d,s[3],tag)
if e == 7:
print(s[0]+tag[0]+s[1]+tag[1]+s[2]+tag[2]+s[3])
exit()
|
s968412314
|
Accepted
| 18
| 3,064
| 522
|
s = list(input())
def calc(i, a, b):
a ,b = int(a), int(b)
if i==0:
c = a+b
elif i==1:
c = a-b
return c
def tag(t_list):
t = []
for p in t_list:
if p==0:
t.append('+')
else:
t.append('-')
return(t)
for i in range(2):
c = calc(i,s[0],s[1])
for j in range(2):
d = calc(j,c,s[2])
for k in range(2):
e = calc(k,d,s[3])
if e == 7:
t_list = [i,j,k]
t = tag(t_list)
print(s[0]+t[0]+s[1]+t[1]+s[2]+t[2]+s[3]+'=7')
exit()
|
s694050670
|
p03369
|
u331036636
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 103
|
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
S = input()
price = 700
for i in range(len(S)):
if S[i] == "○":
price += 100
print(price)
|
s154745777
|
Accepted
| 17
| 2,940
| 107
|
S = list(input())
price = 700
for i in range(len(S)):
if S[i] == "o":
price += 100
print(price)
|
s562956888
|
p03486
|
u429319815
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 195
|
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
|
s = input()
S = "".join(sorted(s))
t = input()
T = "".join(sorted(t))
if S < T:
print("Yes")
else:
print("No")
|
s352114037
|
Accepted
| 17
| 2,940
| 176
|
s = sorted(list(input()))
t = sorted(list(input()), reverse=True)
s_sorted = "".join(s)
t_sorted = "".join(t)
if s_sorted < t_sorted:
print("Yes")
else:
print("No")
|
s649634201
|
p03610
|
u525117558
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,444
| 52
|
You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.
|
s=[input().split()]
for i in s[::2]:
print(s[::2])
|
s789365121
|
Accepted
| 17
| 3,188
| 23
|
s=input()
print(s[::2])
|
s227439201
|
p00003
|
u585035894
| 1,000
| 131,072
|
Wrong Answer
| 40
| 5,592
| 146
|
Write a program which judges wheather given length of three side form a right triangle. Print "YES" if the given sides (integers) form a right triangle, "NO" if not so.
|
for _ in range(int(input())):
l = list(map(int, input().split()))
l.sort()
print('YES' if l[0]**2 + l[1] ** 2 == l[2] ** 2 else 'No')
|
s816231591
|
Accepted
| 50
| 5,604
| 169
|
for _ in range(int(input())):
a = [int(i) for i in input().split()]
print('YES') if a[0]**2 + a[1]**2 + a[2] ** 2 - max(a) **2 == max(a) ** 2 else print('NO')
|
s841627228
|
p03351
|
u333731247
| 2,000
| 1,048,576
|
Wrong Answer
| 21
| 2,940
| 104
|
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a,b,c,d=map(int,input().split())
if abs(a-b)<d and abs(b-c)<d:
print('Yes')
else :
print('No')
|
s248730098
|
Accepted
| 17
| 2,940
| 141
|
a,b,c,d=map(int,input().split())
if abs(a-b)<=d and abs(b-c)<=d:
print('Yes')
elif abs(a-c)<=d:
print('Yes')
else :
print('No')
|
s649068366
|
p02389
|
u095590628
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,588
| 60
|
Write a program which calculates the area and perimeter of a given rectangle.
|
ab = [int(n) for n in input().split()]
print(ab[0]*ab[1])
|
s004258845
|
Accepted
| 20
| 5,608
| 113
|
ab = [int(n) for n in input().split()]
S = ab[0]*ab[1]
L = 2*ab[0] + 2*ab[1]
print(" ".join(map(str,[S,L])))
|
s084660637
|
p03386
|
u901582103
| 2,000
| 262,144
|
Wrong Answer
| 2,217
| 1,839,640
| 105
|
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
a,b,k=map(int,input().split())
l=[i for i in range(a,b+1)]
for s in list(set(l[:k]+l[-k:])):
print(s)
|
s671409885
|
Accepted
| 19
| 3,060
| 158
|
a,b,k=map(int,input().split())
s=[i for i in range(a,min(a+k,b+1))]
l=[i for i in range(b,max(b-k,a-1),-1)]
L=list(set(s+l))
L.sort()
for i in L:
print(i)
|
s298577420
|
p03456
|
u710921979
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 100
|
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.
|
a,b=map(int,input().split())
n=int(a+b)
if n==int(n**0.5)**2:
print('YES')
else:
print('NO')
|
s289605471
|
Accepted
| 17
| 3,060
| 100
|
a,b=map(str,input().split())
n=int(a+b)
if n==int(n**0.5)**2:
print("Yes")
else:
print("No")
|
s380498849
|
p03610
|
u479638406
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 3,956
| 94
|
You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.
|
s = list(input())
for i in range(len(s)):
if i%2 == 0:
odd = ''.join(s)
print(odd)
|
s049498051
|
Accepted
| 38
| 4,388
| 122
|
s = list(input())
t = []
for i in range(len(s)):
if i%2 == 0:
t.append(s[i])
odd = ''.join(t)
print(odd)
|
s347166281
|
p03149
|
u078349616
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 172
|
You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974".
|
A = list(map(int, input().split()))
A1 = A.count("1")
A9 = A.count("9")
A7 = A.count("7")
A4 = A.count("4")
if A1 == A9 == A7 == A4 == 1:
print("YES")
else:
print("NO")
|
s956326800
|
Accepted
| 17
| 3,060
| 164
|
A = list(map(int, input().split()))
A1 = A.count(1)
A9 = A.count(9)
A7 = A.count(7)
A4 = A.count(4)
if A1 == A9 == A7 == A4 == 1:
print("YES")
else:
print("NO")
|
s964706722
|
p03759
|
u090225501
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 89
|
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')
|
s108253002
|
Accepted
| 17
| 2,940
| 89
|
a, b, c = map(int, input().split())
if b - a == c - b:
print('YES')
else:
print('NO')
|
s624743000
|
p02742
|
u268792407
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 57
|
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())
print(((h+1)//2)*((w+1)//2))
|
s970710133
|
Accepted
| 18
| 2,940
| 171
|
h,w=map(int,input().split())
if h==1 or w==1:
print(1)
exit()
if h%2==1 and w%2==1:
a1=((h+1)//2)*(w+1)//2
a2=(h//2)*(w//2)
print(a1+a2)
else:
print((h*w)//2)
|
s079294836
|
p03433
|
u057415180
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 84
|
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
n = int(input())
a = list(map(int,input().split()))
print(sum(a[::2])-sum(a[1::2]))
|
s923695169
|
Accepted
| 17
| 2,940
| 88
|
n = int(input())
a = int(input())
if n%500 <= a:
print('Yes')
else:
print('No')
|
s762037547
|
p02607
|
u124212130
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 9,104
| 168
|
We have N squares assigned the numbers 1,2,3,\ldots,N. Each square has an integer written on it, and the integer written on Square i is a_i. How many squares i satisfy both of the following conditions? * The assigned number, i, is odd. * The written integer is odd.
|
N = int(input())
Mylist = input().split()
Mylist = [int(s) for s in Mylist]
ANS = 0
for i in range(N):
if i+1 % 2 == 1 and Mylist[i] % 2 == 1:
ANS += 1
print(ANS)
|
s684776705
|
Accepted
| 31
| 9,160
| 173
|
N = int(input())
Mylist = input().split()
Mylist = [int(s) for s in Mylist]
ANS = 0
for i in range(1,N+1):
if i % 2 == 1 and Mylist[i-1] % 2 == 1:
ANS += 1
print(ANS)
|
s898989213
|
p03854
|
u460878159
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,188
| 953
|
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()
l=len(S)
r=0
amari = [1,2,3,4]
list5 = ['dream', 'erase']
list81 = ['dreamdre', 'dreamera', 'erasedre', 'eraseera']
list82 = ['dreamerd', 'dreamere', 'eraserdre', 'eraserera']
if l<=4:
print(False)
while l>=5:
if S[r:r+5] in list5:
if r+5==l:
print(True)
break
elif r+5 == l-1:
if S[r:r+6] == 'eraser':
print(True)
else:
print(False)
break
elif r+5 == l-2:
if S[r:r+7] == 'dreamer':
print(True)
else:
print(False)
break
elif r+5 == l-3 or r+5 == l-4:
print(False)
break
else:
if S[r:r+8] in list81:
r=r+5
elif S[r:r+8] in list82:
r=r+7
else:
print(False)
break
else:
print(False)
break
|
s678410282
|
Accepted
| 40
| 3,316
| 1,000
|
S=input()
l=len(S)
r=0
list5 = ['dream', 'erase']
list81 = ['dreamdre', 'dreamera', 'erasedre', 'eraseera']
list82 = ['dreamerd', 'dreamere']
list83 = ['eraserdr', 'eraserer']
if l<=4:
print('NO')
while l>=5:
if S[r:r+5] in list5:
if r+5==l:
print('YES')
break
elif r+5 == l-1:
if S[r:r+6] == 'eraser':
print('YES')
else:
print('NO')
break
elif r+5 == l-2:
if S[r:r+7] == 'dreamer':
print('YES')
else:
print('NO')
break
elif r+5 == l-3 or r+5 == l-4:
print('NO')
break
else:
if S[r:r+8] in list81:
r=r+5
elif S[r:r+8] in list82:
r=r+7
elif S[r:r+8] in list83:
r=r+6
else:
print('NO')
break
else:
print('NO')
break
|
s811603379
|
p03493
|
u266874640
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 83
|
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
S = input()
result =0
for i in S:
if i == 1:
result += 1
print(result)
|
s375047803
|
Accepted
| 20
| 2,940
| 88
|
S = input()
result =0
for i in S:
if i == str(1):
result += 1
print(result)
|
s411642124
|
p03449
|
u917558625
| 2,000
| 262,144
|
Wrong Answer
| 29
| 9,144
| 228
|
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j). You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down. The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them. At most how many candies can you collect when you choose the best way to travel?
|
N=int(input())
p=[0]
q=[0]
A1=list(map(int,input().split()))
A2=list(map(int,input().split()))
for i in range(N):
p.append(p[i]+A1[N-i-1])
q.append(q[i]+A2[i])
ans=0
for j in range(N):
ans=max(ans,p[j+1]+q[j+1])
print(ans)
|
s353854854
|
Accepted
| 32
| 9,144
| 228
|
N=int(input())
p=[0]
q=[0]
A1=list(map(int,input().split()))
A2=list(map(int,input().split()))
for i in range(N):
p.append(p[i]+A1[i])
q.append(q[i]+A2[N-i-1])
ans=0
for j in range(N):
ans=max(ans,p[j+1]+q[N-j])
print(ans)
|
s563762869
|
p03493
|
u903005414
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 50
|
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
s = input()
print(sum([1 for i in s if s == '1']))
|
s116132254
|
Accepted
| 18
| 2,940
| 45
|
print(sum([1 for i in input() if i == '1']))
|
s404051727
|
p03698
|
u136869985
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 53
|
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
s=input()
print(['no', 'ues'][len(set(s)) == len(s)])
|
s219525263
|
Accepted
| 17
| 2,940
| 53
|
s=input()
print(['no', 'yes'][len(set(s)) == len(s)])
|
s420949430
|
p03796
|
u450904670
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 100
|
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.
|
n = int(input())
res = 1
for num in (1, n+1):
res = res * num
res = res % (10**9 + 7)
print(res)
|
s422534863
|
Accepted
| 40
| 2,940
| 110
|
n = int(input())
res = 1
for num in range(1, n+1):
res = res * num
res = res % (10**9 + 7)
print(res)
|
s773738886
|
p02618
|
u595833382
| 2,000
| 1,048,576
|
Wrong Answer
| 151
| 27,304
| 613
|
AtCoder currently hosts three types of contests: ABC, ARC, and AGC. As the number of users has grown, in order to meet the needs of more users, AtCoder has decided to increase the number of contests to 26 types, from AAC to AZC. For convenience, we number these 26 types as type 1 through type 26. AtCoder wants to schedule contests for D days so that user satisfaction is as high as possible. For every day, AtCoder will hold exactly one contest, and each contest will end on that day. The satisfaction is calculated as follows. * The satisfaction at the beginning of day 1 is 0. Satisfaction can be negative. * Holding contests increases satisfaction. The amount of increase will vary depending on a variety of factors. Specifically, we know in advance that holding a contest of type i on day d will increase the satisfaction by s_{d,i}. * If a particular type of contest is not held for a while, the satisfaction decreases. Each contest type i has an integer c_i, and at the end of each day d=1,2,...,D, the satisfaction decreases as follows. Let \mathrm{last}(d,i) be the last day before day d (including d) on which a contest of type i was held. If contests of type i have never been held yet, we define \mathrm{last}(d,i)=0. At the end of day d, the satisfaction decreases by \sum _{i=1}^{26}c_i \times (d-\mathrm{last}(d,i)). Please schedule contests on behalf of AtCoder. If the satisfaction at the end of day D is S, you will get a score of \max(10^6 + S, 0). There are 50 test cases, and the score of a submission is the total scores for each test case. You can make submissions multiple times, and the highest score among your submissions will be your score.
|
import numpy as np
D = int(input())
C = ([int(x) for x in input().split()])
for d in range(D):
exec('s{} = ([int(x) for x in input().split()])'.format(d))
S = 0
d = 0
lday = np.array([0] * 26)
C = np.array(C)
while d != D:
exec("curs = np.array(s{})".format(d))
pos_max_i = np.argmax(curs)
neg_min_i = np.argmin(C * ( np.array([d]*26 ) * lday ))
if np.abs(curs[pos_max_i]) > np.abs(curs[neg_min_i]):
S += curs[pos_max_i]
lday[pos_max_i] = d
print(pos_max_i)
else:
S += curs[neg_min_i]
lday[neg_min_i] = d
print(neg_min_i)
d += 1
|
s119649056
|
Accepted
| 148
| 27,400
| 569
|
import numpy as np
D = int(input())
C = ([int(x) for x in input().split()])
for d in range(D):
exec('s{} = ([int(x) for x in input().split()])'.format(d))
d = 0
lday = np.array([0] * 26)
C = np.array(C)
while d != D:
exec("curs = np.array(s{})".format(d))
pos_max_i = np.argmax(curs)
decl = C * ( np.array([d]*26 ) - lday )
neg_min_i = np.argmin(decl)
if np.abs(curs[pos_max_i]) > np.abs(decl[neg_min_i]):
lday[pos_max_i] = d
print(pos_max_i+1)
else:
lday[neg_min_i] = d
print(neg_min_i+1)
d += 1
|
s077701312
|
p03438
|
u867069435
| 2,000
| 262,144
|
Wrong Answer
| 1,087
| 21,492
| 272
|
You are given two integer sequences of length N: a_1,a_2,..,a_N and b_1,b_2,..,b_N. Determine if we can repeat the following operation zero or more times so that the sequences a and b become equal. Operation: Choose two integers i and j (possibly the same) between 1 and N (inclusive), then perform the following two actions **simultaneously** : * Add 2 to a_i. * Add 1 to b_j.
|
import numpy as np
n = int(input())
a = np.array(list(map(int, input().split())))
b = np.array(list(map(int, input().split())))
if np.sum(b[a < b] - a[a < b]) > np.sum(a[b < a] - b[b < a]):
print("No")
elif np.sum(a) > np.sum(b):
print("No")
else:
print("Yes")
|
s466596303
|
Accepted
| 30
| 4,596
| 562
|
import math
n = int(input())
lis1 = list(map(int,input().split()))
lis2 = list(map(int,input().split()))
k = 0
c = 0
s = 0
cou = sum(lis2)-sum(lis1)
if cou < 0:
print("No")
else:
for i in range(n):
if lis1[i] < lis2[i]:
c += math.ceil((lis2[i] - lis1[i]) / 2)
if (lis2[i] - lis1[i])%2 == 0:
k += 1
else:
s += 1
if c > cou:
print("No")
elif c == cou:
print("Yes")
else:
if k >= 1:
print("Yes")
else:
print("No")
|
s373192185
|
p03435
|
u641722141
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 237
|
We have a 3 \times 3 grid. A number c_{i, j} is written in the square (i, j), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left. According to Takahashi, there are six integers a_1, a_2, a_3, b_1, b_2, b_3 whose values are fixed, and the number written in the square (i, j) is equal to a_i + b_j. Determine if he is correct.
|
l =[list(map(int, input().split())) for i in range(3)]
total1 = total2 = 0
for i in range(3):
total1 += sum(l[i])
total2 += l[i][i]
print(total1, total2)
if total1 == total2 * 3:
print('Yes')
else:
print('No')
|
s353498167
|
Accepted
| 17
| 2,940
| 214
|
l = [[int(i) for i in input().split()] for x in range(3)]
total1 = 0
total2 = 0
for i in range(3):
total1 += sum(l[i])
total2 += l[i][i]
if total1 == total2 * 3:
print('Yes')
else:
print('No')
|
s722244719
|
p03165
|
u102461423
| 2,000
| 1,048,576
|
Wrong Answer
| 531
| 82,760
| 843
|
You are given strings s and t. Find one longest string that is a subsequence of both s and t.
|
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
import numpy as np
S = np.frombuffer(readline().rstrip(),'S1')
T = np.frombuffer(readline().rstrip(),'S1')
S,T
def LCS(S,T):
LS = len(S); LT = len(T)
dp = np.zeros((LS+1,LT+1),np.int64)
for n,s in enumerate(S,1):
equal = (s == T)
np.maximum(dp[n,1:],dp[n-1,:-1]+equal*1,out=dp[n,1:])
np.maximum.accumulate(dp[n],out=dp[n])
return dp
dp = LCS(S,T)
answer = []
max_len = dp[-1,-1]
for n in range(len(S),0,-1):
if max_len == 0:
break
equal = (S[n-1]==T)
ind = np.where((dp[n,1:]*equal)==max_len)[0]
if len(ind):
i = ind[0]
answer.append(T[i].decode('utf-8'))
dp = dp[:,:i+1]; T = T[:i]
max_len -= 1
print(''.join(answer[::-1]))
|
s474530848
|
Accepted
| 273
| 61,968
| 1,049
|
import sys
import numpy as np
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
def longest_common_subsequence(S, T):
LS, LT = len(S), len(T)
dp = np.zeros((LS + 1, LT + 1), np.int32)
for n in range(1, LS + 1):
dp[n, 1:] = dp[n - 1, :-1] + (S[n - 1] == T)
np.maximum(dp[n], dp[n - 1], out=dp[n])
np.maximum.accumulate(dp[n], out=dp[n])
return dp
def reconstruct_LCS(S, T, dp):
tmp = []
i, j = len(S), len(T)
while i > 0 and j > 0:
if S[i - 1] == T[j - 1]:
i, j = i - 1, j - 1
tmp.append(S[i])
elif dp[i, j] == dp[i - 1, j]:
i -= 1
else:
j -= 1
return ''.join(reversed(tmp))
S = np.array(list(readline().decode()), 'U1')[:-1]
T = np.array(list(readline().decode()), 'U1')[:-1]
dp = longest_common_subsequence(S, T)
print(reconstruct_LCS(S, T, dp))
|
s187305385
|
p03157
|
u547167033
| 2,000
| 1,048,576
|
Wrong Answer
| 671
| 7,028
| 823
|
There is a grid with H rows and W columns, where each square is painted black or white. You are given H strings S_1, S_2, ..., S_H, each of length W. If the square at the i-th row from the top and the j-th column from the left is painted black, the j-th character in the string S_i is `#`; if that square is painted white, the j-th character in the string S_i is `.`. Find the number of pairs of a black square c_1 and a white square c_2 that satisfy the following condition: * There is a path from the square c_1 to the square c_2 where we repeatedly move to a vertically or horizontally adjacent square through an alternating sequence of black and white squares: black, white, black, white...
|
h,w=map(int,input().split())
grid =[list(input()) for i in range(h)]
cntW =[[0]*w for i in range(h)]
visited=[[False]*w for i in range(h)]
for i in range(h):
for j in range(w):
if grid[i][j]=='#':
if j<w-1 and grid[i][j+1]=='.':cntW[i][j]+=1
if j>0 and grid[i][j-1]=='.':cntW[i][j]+=1
if i<h-1 and grid[i+1][j]=='.':cntW[i][j]+=1
if i>0 and grid[i-1][j]=='.':cntW[i][j]+=1
ans=0
for i in range(h):
for j in range(w):
if visited[i][j]:
continue
q=[(i,j)]
ans+=cntW[i][j]
visited[i][j]=True
while q:
cy,cx=q.pop(0)
for nx,ny in ((cy+1,cx+1),(cy-1,cx-1),(cy-1,cx+1),(cy+1,cx-1)):
if 0<=nx<w and 0<=ny<h and grid[ny][nx]=='#' and not(visited[ny][nx]):
ans+=cntW[ny][nx]
visited[ny][nx]=True
q.append((ny,nx))
print(ans)
|
s778382629
|
Accepted
| 604
| 5,748
| 540
|
h,w=map(int,input().split())
grid =[list(input()) for i in range(h)]
visited=[[False]*w for i in range(h)]
ans=0
for i in range(h):
for j in range(w):
if visited[i][j]:
continue
q=[(i,j)]
d={'#':0,'.':0}
while q:
cy,cx=q.pop(0)
for ny,nx in ((cy+1,cx),(cy-1,cx),(cy,cx+1),(cy,cx-1)):
if not(0<=ny<h and 0<=nx<w) or grid[cy][cx]==grid[ny][nx] or visited[ny][nx]:
continue
d[grid[ny][nx]]+=1
visited[ny][nx]=True
q.append((ny,nx))
ans+=d['.']*d['#']
print(ans)
|
s140564345
|
p03054
|
u788137651
| 2,000
| 1,048,576
|
Wrong Answer
| 162
| 3,784
| 705
|
We have a rectangular grid of squares with H horizontal rows and W vertical columns. Let (i,j) denote the square at the i-th row from the top and the j-th column from the left. On this grid, there is a piece, which is initially placed at square (s_r,s_c). Takahashi and Aoki will play a game, where each player has a string of length N. Takahashi's string is S, and Aoki's string is T. S and T both consist of four kinds of letters: `L`, `R`, `U` and `D`. The game consists of N steps. The i-th step proceeds as follows: * First, Takahashi performs a move. He either moves the piece in the direction of S_i, or does not move the piece. * Second, Aoki performs a move. He either moves the piece in the direction of T_i, or does not move the piece. Here, to move the piece in the direction of `L`, `R`, `U` and `D`, is to move the piece from square (r,c) to square (r,c-1), (r,c+1), (r-1,c) and (r+1,c), respectively. If the destination square does not exist, the piece is removed from the grid, and the game ends, even if less than N steps are done. Takahashi wants to remove the piece from the grid in one of the N steps. Aoki, on the other hand, wants to finish the N steps with the piece remaining on the grid. Determine if the piece will remain on the grid at the end of the game when both players play optimally.
|
H, W, N = map(int, input().split())
sr, sc = map(int, input().split())
S = input()
T = input()
left = 1
right = W
up = 1
down = H
for i in reversed(range(N)):
if i != N - 1:
if T[i] == "U":
if down != H:
down -= 1
elif T[i] == "D":
if up != 1:
up += 1
elif T[i] == "L":
if right != W:
right += 1
else:
if left != 1:
left -= 1
if S[i] == "U":
up -= 1
elif S[i] == "D":
down += 1
elif S[i] == "L":
left += 1
else:
right -= 1
if left <= sr <= right and down <= sc <= up:
print("YES")
else:
print("NO")
|
s187430729
|
Accepted
| 183
| 3,784
| 731
|
H, W, N = map(int, input().split())
sr, sc = map(int, input().split())
S = input()
T = input()
left = 1
right = W
up = 1
down = H
for i in reversed(range(N)):
if T[i] == "U":
down = min(H, down+1)
elif T[i] == "D":
up = max(1, up-1)
elif T[i] == "L":
right = min(W, right+1)
else:
left = max(1, left-1)
if S[i] == "U":
up += 1
elif S[i] == "D":
down -= 1
elif S[i] == "L":
left += 1
else:
right -= 1
#print(left, right, " ", up, down)
if left > right or up > down:
print("NO")
exit()
#print(left, right)
#print(up, down)
if left <= sc <= right and up <= sr <= down:
print("YES")
else:
print("NO")
|
s143142159
|
p03543
|
u441320782
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 123
|
We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**?
|
x = [int(i) for i in input()]
x.sort()
print(x)
if x[0]==x[1]==x[2] or x[1]==x[2]==x[3]:
print("Yes")
else:
print("No")
|
s513861361
|
Accepted
| 17
| 2,940
| 106
|
x = [int(i) for i in input()]
if x[0]==x[1]==x[2] or x[1]==x[2]==x[3]:
print("Yes")
else:
print("No")
|
s501194304
|
p03713
|
u057586044
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 406
|
There is a bar of chocolate with a height of H blocks and a width of W blocks. Snuke is dividing this bar into exactly three pieces. He can only cut the bar along borders of blocks, and the shape of each piece must be a rectangle. Snuke is trying to divide the bar as evenly as possible. More specifically, he is trying to minimize S_{max} \- S_{min}, where S_{max} is the area (the number of blocks contained) of the largest piece, and S_{min} is the area of the smallest piece. Find the minimum possible value of S_{max} - S_{min}.
|
H,W = map(int,input().split())
hr,ht,vr,vt = 0,0,0,0
hr,vr = H%3*W,W%3*H
def t(x,y):
if x%3==0:
return 0
elif x%3==1:
a = ((x//3)*2+1)*(y//2)
b = x//3*y
c = x*y-a-b
return max(a,b,c)-min(a,b,c)
else:
a = ((x//3)*2+1)*(y//2)
b = (x//3+1)*y
c = x*y-a-b
return max(a,b,c)-min(a,b,c)
ht = t(H,W)
vt = t(W,H)
print(hr,ht,vr,vt)
|
s976021234
|
Accepted
| 17
| 3,064
| 430
|
H,W = map(int,input().split())
res = min(H,W)
hr,ht,vr,vt = 0,0,0,0
hr,vr = H%3*W,W%3*H
def t(x,y):
if x%3==0:
return 0
elif x%3==1:
a = ((x//3)*2+1)*(y//2)
b = x//3*y
c = x*y-a-b
return max(a,b,c)-min(a,b,c)
else:
a = ((x//3)*2+1)*(y//2)
b = (x//3+1)*y
c = x*y-a-b
return max(a,b,c)-min(a,b,c)
ht = t(H,W)
vt = t(W,H)
print(min(res,hr,ht,vr,vt))
|
s384493853
|
p02694
|
u344232182
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 9,164
| 93
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
x = int(input())
m = 100
count = 1
while x > m:
m = m + m/100
count += 1
print(count)
|
s519998211
|
Accepted
| 21
| 9,108
| 109
|
x = int(input())
m = 100
count = 0
while x > m:
r = int(m*0.01)
m = m + r
count += 1
print(count)
|
s175641440
|
p03644
|
u113255362
| 2,000
| 262,144
|
Wrong Answer
| 25
| 9,092
| 122
|
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
A = int(input())
chek = 1
res = 0
for i in range(10):
chek = 2**i
if chek < A:
res = 2**(i-1)
break
print(res)
|
s237610656
|
Accepted
| 30
| 9,020
| 122
|
A = int(input())
chek = 1
res = 0
for i in range(10):
chek = 2**i
if chek > A:
res = 2**(i-1)
break
print(res)
|
s970784622
|
p03643
|
u893209854
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 24
|
This contest, _AtCoder Beginner Contest_ , is abbreviated as _ABC_. When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example, ABC680 is the 680th round of ABC. What is the abbreviation for the N-th round of ABC? Write a program to output the answer.
|
N = input()
print(N[3:])
|
s573018644
|
Accepted
| 19
| 2,940
| 28
|
N = input()
print('ABC' + N)
|
s436633604
|
p03050
|
u476435125
| 2,000
| 1,048,576
|
Wrong Answer
| 140
| 2,940
| 121
|
Snuke received a positive integer N from Takahashi. A positive integer m is called a _favorite number_ when the following condition is satisfied: * The quotient and remainder of N divided by m are equal, that is, \lfloor \frac{N}{m} \rfloor = N \bmod m holds. Find all favorite numbers and print the sum of those.
|
N=int(input())
sqrtn=int(N**0.5)
ans=0
for r in range(1,sqrtn):
if N%r==0:
ans+=(int(N/r)-1)
print(int(ans))
|
s749915079
|
Accepted
| 145
| 3,060
| 135
|
N=int(input())
sqrtn=int(N**0.5)
ans=0
for r in range(1,sqrtn+1):
if N%r==0 and N/r-1>r:
ans+=(int(N/r)-1)
print(int(ans))
|
s563027181
|
p03478
|
u094191970
| 2,000
| 262,144
|
Wrong Answer
| 37
| 3,628
| 229
|
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
n,a,b=map(int,input().split())
ans=0
for i in range(1,n+1):
n=i
cnt=0
while i//10 != 0:
cnt+=i%10
i=i//10
else:
cnt+=i
if a<=cnt and cnt<=b:
print(cnt)
ans+=n
print(ans)
|
s177560039
|
Accepted
| 29
| 3,060
| 210
|
n,a,b=map(int,input().split())
ans=0
for i in range(1,n+1):
n=i
cnt=0
while i//10 != 0:
cnt+=i%10
i=i//10
else:
cnt+=i
if a<=cnt and cnt<=b:
ans+=n
print(ans)
|
s135027647
|
p00460
|
u797673668
| 1,000
| 131,072
|
Time Limit Exceeded
| 40,000
| 7,588
| 472
|
あるプログラミングコンテストでは,競技後の懇親会でビンゴゲームをする習わしがある．しかし,このビンゴゲームで使うビンゴカードは少々特殊で,以下の条件に従って作成される． * ビンゴカードは N 行 N 列のマス目に区切られており，各マス目には正整数が１つずつ書かれている．それらの整数は全て異なる． * マス目に書かれている整数は 1 以上 M 以下である． * ビンゴカードに書かれているN×N個の整数の合計は S である． * どの列を見たときも，上から下に向かって整数は昇順に並んでいる． * どのマス目の整数も，そのマス目より左の列のどの整数よりも大きい． 以下は， N = 5, M = 50, S = 685 のときのビンゴカードの例である． 懇親会のために上の条件を満たすビンゴカードをできるだけたくさん作りたい．ただし，同一のカードを2枚以上作ってはならない．作ることができるビンゴカードの枚数の最大値を 100000 で割った余りを出力するプログラムを作成せよ．
|
def allocate(remains, limit_h, limit_w):
global m
if not remains:
return 1
if limit_w == 1:
return int(remains <= limit_h)
return sum(allocate(remains - i, i, limit_w - 1) for i in
range(min(limit_h, remains, m - limit_w), (remains - 1) // limit_w, -1))
while True:
n, m, s = map(int, input().split())
if not n:
break
remains = s - sum(range(1, n ** 2 + 1))
print(allocate(remains, remains, n ** 2))
|
s178889445
|
Accepted
| 210
| 7,756
| 427
|
while True:
n, m, s = map(int, input().split())
if not n:
break
n2 = n ** 2
dpp = [0] * (s + 1)
dpp[0] = 1
for i in range(1, n2 + 1):
dpn = [0] * (s + 1)
for j in range(i * (i + 1) // 2, s + 1):
dpn[j] += dpp[j - i] + dpn[j - i]
if j - m - 1 >= 0:
dpn[j] -= dpp[j - m - 1]
dpn[j] %= 100000
dpp = dpn
print(dpp[s])
|
s517559739
|
p03737
|
u127856129
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 49
|
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
|
a,b,c=input().split()
print(a[0]+b[0]+c[0])
|
s478464839
|
Accepted
| 17
| 2,940
| 73
|
a,b,c=input().split()
print(a[0].upper()+b[0].upper()+c[0].upper())
|
s075756066
|
p02262
|
u742013327
| 6,000
| 131,072
|
Wrong Answer
| 20
| 7,628
| 1,347
|
Shell Sort is a generalization of [Insertion Sort](http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_1_A) to arrange a list of $n$ elements $A$. 1 insertionSort(A, n, g) 2 for i = g to n-1 3 v = A[i] 4 j = i - g 5 while j >= 0 && A[j] > v 6 A[j+g] = A[j] 7 j = j - g 8 cnt++ 9 A[j+g] = v 10 11 shellSort(A, n) 12 cnt = 0 13 m = ? 14 G[] = {?, ?,..., ?} 15 for i = 0 to m-1 16 insertionSort(A, n, G[i]) A function shellSort(A, n) performs a function insertionSort(A, n, g), which considers every $g$-th elements. Beginning with large values of $g$, it repeats the insertion sort with smaller $g$. Your task is to complete the above program by filling ?. Write a program which reads an integer $n$ and a sequence $A$, and prints $m$, $G_i (i = 0, 1, ..., m − 1)$ in the pseudo code and the sequence $A$ in ascending order. The output of your program must meet the following requirements: * $1 \leq m \leq 100$ * $0 \leq G_i \leq n$ * cnt does not exceed $\lceil n^{1.5}\rceil$
|
#http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_2_D
#???????????????????????????????????????????????????????????????????????????
cnt = 0
def insertion_sort(target_list,g):
maxlen = len(target_list)
global cnt
for focus_index in range(g, maxlen):
target = target_list[focus_index]
if target < target_list[focus_index - g]:
compare_index = focus_index - g
while compare_index >= 0 and target_list[compare_index] > target:
target_list[compare_index + g] = target_list[compare_index]
compare_index = compare_index - g;
cnt += 1
target_list[compare_index + g] = target
return target_list
def shell_sort(target_list):
g = []
g_value = 1
while g_value < len(target_list):
g.append(g_value)
g_value = 3 * g_value + 1
g = g[::-1]
print(*g)
for i in range(0,len(g)):
insertion_sort(target_list, g[i])
def main():
n_list = int(input())
target_list = [int(input()) for n in range(n_list)]
cc = shell_sort(target_list)
print(cnt)
print("\n".join([str(n) for n in target_list]))
if __name__ == "__main__":
main()
|
s418374401
|
Accepted
| 22,080
| 127,948
| 1,371
|
#http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_2_D
#???????????????????????????????????????????????????????????????????????????
cnt = 0
def insertion_sort(target_list,g):
maxlen = len(target_list)
global cnt
for focus_index in range(g, maxlen):
target = target_list[focus_index]
if target < target_list[focus_index - g]:
compare_index = focus_index - g
while compare_index >= 0 and target_list[compare_index] > target:
target_list[compare_index + g] = target_list[compare_index]
compare_index = compare_index - g;
cnt += 1
target_list[compare_index + g] = target
return target_list
def shell_sort(target_list):
g = []
g_value = 1
while g_value <= len(target_list):
g.append(g_value)
g_value = 3 * g_value + 1
g = g[::-1]
print(len(g))
print(*g)
for i in range(0,len(g)):
insertion_sort(target_list, g[i])
def main():
n_list = int(input())
target_list = [int(input()) for n in range(n_list)]
cc = shell_sort(target_list)
print(cnt)
print("\n".join([str(n) for n in target_list]))
if __name__ == "__main__":
main()
|
s250073480
|
p02843
|
u760802228
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 901
|
AtCoder Mart sells 1000000 of each of the six items below: * Riceballs, priced at 100 yen (the currency of Japan) each * Sandwiches, priced at 101 yen each * Cookies, priced at 102 yen each * Cakes, priced at 103 yen each * Candies, priced at 104 yen each * Computers, priced at 105 yen each Takahashi wants to buy some of them that cost exactly X yen in total. Determine whether this is possible. (Ignore consumption tax.)
|
x = int(input())
total = 0
if x % 10 == 9:
total += 104 + 105
elif x % 10 == 8:
total += 104 + 104
elif x % 10 == 7:
total += 103 + 104
elif x % 10 == 6:
total += 105 + 101
elif x % 10 == 5:
total += 105
elif x % 10 == 4:
total += 104
elif x % 10 == 3:
total += 103
elif x % 10 == 2:
total += 102
elif x % 10 == 1:
total += 101
if x % 100 // 10 == 1:
total += 105 + 105
elif x % 100 // 10 == 2:
total += 105 * 4
elif x % 100 // 10 == 3:
total += 105 * 6
elif x % 100 // 10 == 4:
total += 105 * 8
elif x % 100 // 10 == 5:
total += 105 * 10
elif x % 100 // 10 == 6:
total += 105 * 12
elif x % 100 // 10 == 7:
total += 105 * 14
elif x % 100 // 10 == 8:
total += 105 * 16
elif x % 100 // 10 == 9:
total += 105 * 18
print(total)
if total <= x:
print(1)
|
s783690454
|
Accepted
| 17
| 3,064
| 911
|
x = int(input())
total = 0
if x % 10 == 9:
total += 104 + 105
elif x % 10 == 8:
total += 104 + 104
elif x % 10 == 7:
total += 103 + 104
elif x % 10 == 6:
total += 105 + 101
elif x % 10 == 5:
total += 105
elif x % 10 == 4:
total += 104
elif x % 10 == 3:
total += 103
elif x % 10 == 2:
total += 102
elif x % 10 == 1:
total += 101
if x % 100 // 10 == 1:
total += 105 + 105
elif x % 100 // 10 == 2:
total += 105 * 4
elif x % 100 // 10 == 3:
total += 105 * 6
elif x % 100 // 10 == 4:
total += 105 * 8
elif x % 100 // 10 == 5:
total += 105 * 10
elif x % 100 // 10 == 6:
total += 105 * 12
elif x % 100 // 10 == 7:
total += 105 * 14
elif x % 100 // 10 == 8:
total += 105 * 16
elif x % 100 // 10 == 9:
total += 105 * 18
if total <= x:
print(1)
else:
print(0)
|
s120026651
|
p03592
|
u057993957
| 2,000
| 262,144
|
Wrong Answer
| 33
| 9,112
| 360
|
We have a grid with N rows and M columns of squares. Initially, all the squares are white. There is a button attached to each row and each column. When a button attached to a row is pressed, the colors of all the squares in that row are inverted; that is, white squares become black and vice versa. When a button attached to a column is pressed, the colors of all the squares in that column are inverted. Takahashi can freely press the buttons any number of times. Determine whether he can have exactly K black squares in the grid.
|
n, m, k = list(map(int, input().split()))
flag = False
for y in range(m):
if n - 2 * y == 0:
continue
x = (k - m * y) / (n - 2 * y)
if (k - m * y) % (n - 2 * y) == 0 and (0 <= x and x <= m):
print(y, (k - m * y) / (n - 2 * y), (n - 2 * y), (k - m * y))
flag = True
break
print("Yes" if flag else "No")
|
s138800140
|
Accepted
| 29
| 9,132
| 292
|
n, m, k = list(map(int, input().split()))
flag = False
for y in range(n+1):
if n - 2 * y == 0:
continue
x = (k - m * y) / (n - 2 * y)
if (k - m * y) % (n - 2 * y) == 0 and (0 <= x and x <= m):
flag = True
break
print("Yes" if flag else "No")
|
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