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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s834908999
|
p03435
|
u721569287
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 594
|
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.
|
grid = list()
for i in range(3):
grid.append([int(k) for k in input().split(" ")])
def check_vert():
for i in range(2):
diffs = [grid[k][i] - grid[k][i+1]for k in range(3)]
print(diffs)
if not len(set(diffs)) == 1:
return False
else:
return True
def check_hor():
for i in range(2):
diffs = [grid[i][k] - grid[i+1][k] for k in range(3)]
print(diffs)
if not len(set(diffs)) == 1:
return False
else:
return True
if check_hor() and check_vert():
print("Yes")
else:
print("No")
|
s112009321
|
Accepted
| 17
| 3,064
| 552
|
grid = list()
for i in range(3):
grid.append([int(k) for k in input().split(" ")])
def check_vert():
for i in range(2):
diffs = [grid[k][i] - grid[k][i+1]for k in range(3)]
if not len(set(diffs)) == 1:
return False
else:
return True
def check_hor():
for i in range(2):
diffs = [grid[i][k] - grid[i+1][k] for k in range(3)]
if not len(set(diffs)) == 1:
return False
else:
return True
if check_hor() and check_vert():
print("Yes")
else:
print("No")
|
s665650476
|
p03910
|
u408375121
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 98
|
The problem set at _CODE FESTIVAL 20XX Finals_ consists of N problems. The score allocated to the i-th (1≦i≦N) problem is i points. Takahashi, a contestant, is trying to score exactly N points. For that, he is deciding which problems to solve. As problems with higher scores are harder, he wants to minimize the highest score of a problem among the ones solved by him. Determine the set of problems that should be solved.
|
n = int(input())
k = 1
while True:
if k*(k+1)//2 >= n:
ans = k
break
k += 1
print(ans)
|
s492685091
|
Accepted
| 23
| 3,316
| 324
|
n = int(input())
k = 1
while True:
if k*(k+1)//2 >= n:
ans = k
break
k += 1
if n == ans*(ans+1)//2:
i = 1
while i <= ans:
print(i)
i += 1
else:
i = 1
l = []
cnt = 1
while i < ans:
if cnt <= (ans-1)*(ans+2)//2 - n:
print(i)
cnt += 1
else:
print(i+1)
i += 1
|
s912583839
|
p03110
|
u889550481
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 171
|
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?
|
N=int(input())
listA=[list(input().split()) for _ in range(N)]
print(listA)
ans=0
for i in range(N):
ans+=float(listA[i][0])*(1,380000)[listA[i][1]=='BTC']
print(ans)
|
s557624633
|
Accepted
| 19
| 3,060
| 158
|
N=int(input())
listA=[list(input().split()) for _ in range(N)]
ans=0
for i in range(N):
ans+=float(listA[i][0])*(1,380000)[listA[i][1]=='BTC']
print(ans)
|
s309754841
|
p03591
|
u852113235
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 72
|
Ringo is giving a present to Snuke. Ringo has found out that Snuke loves _yakiniku_ (a Japanese term meaning grilled meat. _yaki_ : grilled, _niku_ : meat). He supposes that Snuke likes grilled things starting with `YAKI` in Japanese, and does not like other things. You are given a string S representing the Japanese name of Ringo's present to Snuke. Determine whether S starts with `YAKI`.
|
s = input()
if s[0:4] == 'YAKI':
print('YES')
else:
print('NO')
|
s313913059
|
Accepted
| 17
| 2,940
| 72
|
s = input()
if s[0:4] == 'YAKI':
print('Yes')
else:
print('No')
|
s214586853
|
p02261
|
u591875281
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,728
| 792
|
Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
def bubble(A, N):
flag = True
while flag:
flag = False
for j in range(N-1, 0, -1):
if int(A[j][-1]) < int(A[j-1][-1]):
A[j], A[j-1] = A[j-1], A[j]
flag = True
return A
def selection(A, N):
for i in range(0, N):
minj = i
for j in range(i, N):
if int(A[j][-1]) < int(A[minj][-1]):
minj = j
if i != minj:
A[i], A[minj] = A[minj], A[i]
return A
if __name__ == "__main__":
N = int(input())
A = input().split()
A_bubble = bubble(A[:], N)
print (*A_bubble)
print ("Stable")
A_selection = selection(A[:], N)
print (*A_selection)
if A_bubble == A_selection:
print ("Stable")
else:
print ("Not Stable")
|
s976892514
|
Accepted
| 20
| 7,780
| 792
|
def bubble(A, N):
flag = True
while flag:
flag = False
for j in range(N-1, 0, -1):
if int(A[j][-1]) < int(A[j-1][-1]):
A[j], A[j-1] = A[j-1], A[j]
flag = True
return A
def selection(A, N):
for i in range(0, N):
minj = i
for j in range(i, N):
if int(A[j][-1]) < int(A[minj][-1]):
minj = j
if i != minj:
A[i], A[minj] = A[minj], A[i]
return A
if __name__ == "__main__":
N = int(input())
A = input().split()
A_bubble = bubble(A[:], N)
print (*A_bubble)
print ("Stable")
A_selection = selection(A[:], N)
print (*A_selection)
if A_bubble == A_selection:
print ("Stable")
else:
print ("Not stable")
|
s370584636
|
p03555
|
u190178779
| 2,000
| 262,144
|
Wrong Answer
| 27
| 8,964
| 107
|
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.
|
import sys
C1 = list(input())
C2 = list(input())
if C2 == C1[::-1]:
print("Yes")
else:
print("No")
|
s715125320
|
Accepted
| 30
| 9,100
| 107
|
import sys
C1 = list(input())
C2 = list(input())
if C2 == C1[::-1]:
print("YES")
else:
print("NO")
|
s709790747
|
p03759
|
u138486156
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 92
|
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')
|
s994582243
|
Accepted
| 17
| 2,940
| 87
|
a,b,c = map(int, input().split())
if b-a == c-b:
print('YES')
else:
print('NO')
|
s070071042
|
p03698
|
u845427284
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 62
|
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
s = list(input())
print("yes" if s == list(set(s)) else "no")
|
s742882355
|
Accepted
| 19
| 3,060
| 72
|
s = list(input())
print("yes" if len(s) == len(list(set(s))) else "no")
|
s358753408
|
p03079
|
u853185302
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 146
|
You are given three integers A, B and C. Determine if there exists an equilateral triangle whose sides have lengths A, B and C.
|
list_n = list(map(int,input().split()))
list_n.sort(reverse=True)
if list_n[0]**2 == list_n[1]**2+list_n[2]**2:
print('Yes')
else:
print('No')
|
s469089809
|
Accepted
| 18
| 2,940
| 154
|
list_n = list(map(int,input().split()))
list_n.sort(reverse=True)
if list_n[0] == list_n[1] and list_n[1] == list_n[2]:
print('Yes')
else:
print('No')
|
s937138087
|
p02796
|
u603958124
| 2,000
| 1,048,576
|
Wrong Answer
| 355
| 25,492
| 993
|
In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep.
|
from math import ceil,floor,factorial,gcd,sqrt,log2,cos,sin,tan,acos,asin,atan,degrees,radians,pi,inf,comb
from itertools import accumulate,groupby,permutations,combinations,product,combinations_with_replacement
from collections import deque,defaultdict,Counter
from bisect import bisect_left,bisect_right
from operator import itemgetter
from heapq import heapify,heappop,heappush
from queue import Queue,LifoQueue,PriorityQueue
from copy import deepcopy
from time import time
import string
import sys
sys.setrecursionlimit(10 ** 7)
def input() : return sys.stdin.readline().strip()
def INT() : return int(input())
def MAP() : return map(int,input().split())
def LIST() : return list(MAP())
n = INT()
x = [[0]*2 for i in range(n)]
for i in range(n):
x[i][0], x[i][1] = MAP()
x.sort()
tmp = -inf
ans = n
for i in range(n):
if tmp > x[i][0] - x[i][1]:
ans -= 1
tmp = min(tmp, x[i][0] + x[i][1])
print(i)
else:
tmp = x[i][0] + x[i][1]
print(ans)
|
s963016257
|
Accepted
| 280
| 27,440
| 930
|
from math import ceil,floor,factorial,gcd,sqrt,log2,cos,sin,tan,acos,asin,atan,degrees,radians,pi,inf,comb
from itertools import accumulate,groupby,permutations,combinations,product,combinations_with_replacement
from collections import deque,defaultdict,Counter
from bisect import bisect_left,bisect_right
from operator import itemgetter
from heapq import heapify,heappop,heappush
from queue import Queue,LifoQueue,PriorityQueue
from copy import deepcopy
from time import time
import string
import sys
sys.setrecursionlimit(10 ** 7)
def input() : return sys.stdin.readline().strip()
def INT() : return int(input())
def MAP() : return map(int,input().split())
def LIST() : return list(MAP())
n = INT()
ls = []
for i in range(n):
x, l = MAP()
ls.append([x-l, x+l])
ls = sorted(ls, key=itemgetter(1))
ans = n
for i in range(1, n):
if ls[i][0] < ls[i-1][1]:
ls[i][1] = ls[i-1][1]
ans -= 1
print(ans)
|
s496723653
|
p03713
|
u899975427
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 205
|
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())
i = h//3
j = -(-h//3)
a = i*w
b = (h-i)*(w//2)
c = (h-i)*-(-w//2)
a2 = j*w
b2 = (h-j)*(w//2)
c2 = (h-j)*-(-w//2)
print(min(max(a,b,c)-min(a,b,c),max(a2,b2,c2)-min(a2,b2,c2)))
|
s434212215
|
Accepted
| 17
| 3,064
| 290
|
h,w = sorted(list(map(int,input().split())))
if h % 3 == 0 or w % 3 == 0:
ret = 0
else:
c1 = ((w//3+1)*h)-((w-1-w//3)*(h//2))
c2 = ((w-w//3)*(h-h//2))-(w//3*h)
c3 = h
c4 = ((h//3+1)*w)-((h-1-h//3)*(w//2))
c5 = ((h-h//3)*(w-w//2))-(h//3*w)
ret = min(c1,c2,c3,c4,c5)
print(ret)
|
s363993600
|
p03644
|
u289288647
| 2,000
| 262,144
|
Wrong Answer
| 32
| 9,092
| 95
|
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 = 1
count = 0
while 2*ans <= N:
ans *= 2
count += 1
print(count)
|
s882647256
|
Accepted
| 26
| 9,144
| 68
|
N = int(input())
ans = 1
while 2*ans <= N:
ans *= 2
print(ans)
|
s665702681
|
p03455
|
u982591663
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 92
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
A, B = map(int, input().split())
if (A*B) % 2 == 0:
print("Evem")
else:
print("Odd")
|
s824528586
|
Accepted
| 17
| 2,940
| 92
|
A, B = map(int, input().split())
if (A*B) % 2 == 0:
print("Even")
else:
print("Odd")
|
s911280967
|
p02276
|
u519227872
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,652
| 373
|
Quick sort is based on the Divide-and-conquer approach. In QuickSort(A, p, r), first, a procedure Partition(A, p, r) divides an array A[p..r] into two subarrays A[p..q-1] and A[q+1..r] such that each element of A[p..q-1] is less than or equal to A[q], which is, inturn, less than or equal to each element of A[q+1..r]. It also computes the index q. In the conquer processes, the two subarrays A[p..q-1] and A[q+1..r] are sorted by recursive calls of QuickSort(A, p, q-1) and QuickSort(A, q+1, r). Your task is to read a sequence A and perform the Partition based on the following pseudocode: Partition(A, p, r) 1 x = A[r] 2 i = p-1 3 for j = p to r-1 4 do if A[j] <= x 5 then i = i+1 6 exchange A[i] and A[j] 7 exchange A[i+1] and A[r] 8 return i+1 Note that, in this algorithm, Partition always selects an element A[r] as a pivot element around which to partition the array A[p..r].
|
n = int(input())
A = list(map(int, input().split()))
def partition(A, p, r):
x = A[r]
i = p - 1
for j in range(p, r):
if A[j] <= x:
i += 1
A[i], A[j] = A[j], A[i]
A[i + 1], A[r] = A[r], A[i + 1]
return i + 1
i = partition(A, 0, n-1)
print(' '.join(map(str, A[:i])) + ' [' + str(A[i]) + '] ' + ' '.join(map(str, A[i:])))
|
s531932459
|
Accepted
| 80
| 18,936
| 375
|
n = int(input())
A = list(map(int, input().split()))
def partition(A, p, r):
x = A[r]
i = p - 1
for j in range(p, r):
if A[j] <= x:
i += 1
A[i], A[j] = A[j], A[i]
A[i + 1], A[r] = A[r], A[i + 1]
return i + 1
i = partition(A, 0, n-1)
print(' '.join(map(str, A[:i])) + ' [' + str(A[i]) + '] ' + ' '.join(map(str, A[i+1:])))
|
s057296038
|
p03854
|
u161442663
| 2,000
| 262,144
|
Wrong Answer
| 20
| 3,636
| 375
|
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=str(input())
Smojisuu=len(S)
c=S.count("eraser")
S=S.replace("eraser","xxxxxx",c)
print(S)
d=S.count("erase")
S=S.replace("erase","xxxxx",d)
print(S)
e=S.count("dreamer")
S=S.replace("dreamer","xxxxxxx",e)
f=S.count("dream")
S=S.replace("dream","xxxxx",f)
print(S)
g=S.count("x")
if len(S)>g:
print("NO")
elif len(S)==g:
print("YES")
else:
print("error")
|
s251254982
|
Accepted
| 20
| 3,188
| 348
|
S=str(input())
Smojisuu=len(S)
c=S.count("eraser")
S=S.replace("eraser","xxxxxx",c)
d=S.count("erase")
S=S.replace("erase","xxxxx",d)
e=S.count("dreamer")
S=S.replace("dreamer","xxxxxxx",e)
f=S.count("dream")
S=S.replace("dream","xxxxx",f)
g=S.count("x")
if len(S)>g:
print("NO")
elif len(S)==g:
print("YES")
else:
print("error")
|
s097821704
|
p02255
|
u936051377
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,664
| 390
|
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.
|
def insertion_sort(A, N):
for i in range(N):
v = A[i]
j = i - 1
while j >= 0 and A[j] > v:
A[j + 1] = A[j]
j -= 1
A[j + 1] = v
def main():
N = int(input().rstrip())
A = list(map(int, input().rstrip().split()))
insertion_sort(A, N)
print(' '.join(map(str, A)))
if __name__ == '__main__':
main()
|
s638058985
|
Accepted
| 40
| 7,656
| 390
|
def insertion_sort(A, N):
for i in range(N):
v = A[i]
j = i - 1
while j >= 0 and A[j] > v:
A[j + 1] = A[j]
j -= 1
A[j + 1] = v
print(' '.join(map(str, A)))
def main():
N = int(input().rstrip())
A = list(map(int, input().rstrip().split()))
insertion_sort(A, N)
if __name__ == '__main__':
main()
|
s467628748
|
p03659
|
u982762220
| 2,000
| 262,144
|
Wrong Answer
| 77
| 24,812
| 293
|
Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer a_i written on it. They will share these cards. First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards. Here, both Snuke and Raccoon have to take at least one card. Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively. They would like to minimize |x-y|. Find the minimum possible value of |x-y|.
|
N = int(input())
A = list(map(int, input().split()))
tot = sum(A)
res = 0
s, t = A[0], sum(A[1:])
min_dif = abs(s - t)
for a in A[1:-1]:
s += a
t -= a
dif = abs(s - t)
if min_dif > dif:
min_dif = dif
else:
print(dif)
exit()
print(abs(s - A[-1]))
|
s683615592
|
Accepted
| 146
| 24,832
| 245
|
N = int(input())
A = list(map(int, input().split()))
tot = sum(A)
half = tot / 2
res = 0
s, t = A[0], sum(A[1:])
min_v = abs(s - t)
for a in A[1:-1]:
s += a
tmp = abs(tot - 2 * s)
if tmp < min_v:
min_v = tmp
print(min_v)
|
s818614323
|
p02843
|
u391157755
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 263
|
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())
x -= min(x // 105, 1000000) * 105
x -= min(x // 104, 1000000) * 104
x -= min(x // 103, 1000000) * 103
x -= min(x // 102, 1000000) * 102
x -= min(x // 101, 1000000) * 101
x -= min(x // 100, 1000000) * 100
if x == 0:
print(1)
else:
print(0)
|
s258331982
|
Accepted
| 212
| 3,876
| 248
|
x = int(input())
a = list(False for i in range(x + 1))
a[0] = True
for i in range(1, x + 1):
for j in range(6):
if (i - 100 - j) >= 0:
if a[(i - 100 - j)]:
a[i] = True
if a[x]:
print(1)
else:
print(0)
|
s054146683
|
p03080
|
u575101291
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 98
|
There are N people numbered 1 to N. Each person wears a red hat or a blue hat. You are given a string s representing the colors of the people. Person i wears a red hat if s_i is `R`, and a blue hat if s_i is `B`. Determine if there are more people wearing a red hat than people wearing a blue hat.
|
N=input()
s=list(input())
r=s.count("r")
b=s.count("b")
if r>b:
print("Yes")
else:
print("No")
|
s022480131
|
Accepted
| 17
| 2,940
| 98
|
N=input()
s=list(input())
r=s.count("R")
b=s.count("B")
if r>b:
print("Yes")
else:
print("No")
|
s215951623
|
p03644
|
u940780117
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 215
|
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())
n=0
ans = 0
ans2 =0
for i in range(1,N+1):
n=i
res = 0
while n % 2 !=0:
n=n//2
res = res + 1
if ans < res:
ans = res
ans2 =i+1
print(ans2)
|
s002381845
|
Accepted
| 17
| 3,060
| 242
|
N = int(input())
n=0
ans = 0
ans2 =0
for i in range(1,N+1):
n=i
res = 0
while n % 2 ==0:
n=n//2
res = res + 1
if ans < res:
ans = res
ans2 =i
if ans2==0:
print(1)
else:
print(ans2)
|
s471299167
|
p02972
|
u761471989
| 2,000
| 1,048,576
|
Wrong Answer
| 969
| 22,312
| 585
|
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
N = int(input())
a_list = list(map(int, input().split()))
b_list = [0]*(N//2)+a_list[N//2:]
for i, a in enumerate(range(N//2)):
j = N//2 - i
j_list = []
k = j
while(1):
j = j + k
if j > N:
break
else:
j_list.append(j)
s = 0
for j in j_list:
s += b_list[j-1]
if s % 2 == a_list[k-1]:
b_list[k-1] = 0
else:
b_list[k-1] = 1
if len([str(i+1) for i, b in enumerate(b_list) if b == 1]) == 0:
print(0)
else:
print(' '.join([str(i+1) for i, b in enumerate(b_list) if b == 1]))
|
s891192578
|
Accepted
| 1,047
| 22,572
| 650
|
N = int(input())
a_list = list(map(int, input().split()))
b_list = [0]*(N//2)+a_list[N//2:]
for i, a in enumerate(range(N//2)):
j = N//2 - i
j_list = []
k = j
while(1):
j = j + k
if j > N:
break
else:
j_list.append(j)
#print(j_list)
s = 0
for j in j_list:
s += b_list[j-1]
if s % 2 == a_list[k-1]:
b_list[k-1] = 0
else:
b_list[k-1] = 1
print(len([str(i+1) for i, b in enumerate(b_list) if b == 1]))
print(' '.join([str(i+1) for i, b in enumerate(b_list) if b == 1]))
|
s119531995
|
p03485
|
u703890795
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 48
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
a, b = map(int, input().split())
print((a+b)//1)
|
s172779760
|
Accepted
| 19
| 2,940
| 50
|
a, b = map(int, input().split())
print((a+b+1)//2)
|
s305514614
|
p03719
|
u287880059
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 72
|
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
a,b,c = map(int,input().split())
print("YES" if a>=c and a<=b else "NO")
|
s709810806
|
Accepted
| 17
| 2,940
| 79
|
a,b,c = map(int,input().split())
if a<=c<=b:
print("Yes")
else:
print("No")
|
s568205404
|
p03457
|
u294721290
| 2,000
| 262,144
|
Wrong Answer
| 907
| 3,316
| 245
|
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
N = int(input())
count = 0
T, X, Y = 0, 0, 0
for i in range(N):
t, x, y = map(int, input().split())
if abs(x-X)+abs(y-Y) <= t-T and t % 2 == (x+y) % 2:
count += 1
T, X, Y = t, x, y
print("Yes" if count == N else "No")
|
s462304747
|
Accepted
| 383
| 3,064
| 237
|
N = int(input())
count = 0
T, X, Y = 0, 0, 0
for i in range(N):
t, x, y = map(int, input().split())
if abs(x-X)+abs(y-Y) <= t-T and t % 2 == (x+y) % 2:
count += 1
T, X, Y = t, x, y
print("Yes" if count == N else "No")
|
s110472077
|
p02843
|
u951401193
| 2,000
| 1,048,576
|
Wrong Answer
| 33
| 3,064
| 407
|
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.)
|
n = int(input())
m = n%615
for i in range(7):
for j in range(7):
for k in range(7):
for l in range(7):
for m in range(7):
for n in range(7):
tmp = 100*i+101*j + 102*k + 103*l * 104*m + 105*n
if m == tmp:
print('1')
elif m < tmp:break
print('0')
|
s646802931
|
Accepted
| 17
| 2,940
| 133
|
x=int(input())
n=x//100
if x<100:
print('0')
else:
if 100*n<=x and 105*n>=x:
print('1')
else:
print('0')
|
s425027402
|
p03555
|
u943657163
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 90
|
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, b = input(), input()
print('Yes' if a[0]==b[2] and a[1]==b[1] and a[2]==b[0] else 'No')
|
s586074763
|
Accepted
| 17
| 2,940
| 62
|
a, b = input(), input()
print('YES' if a == b[::-1] else 'NO')
|
s423073700
|
p03485
|
u509094491
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 84
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
i = list(map(int, input().split()))
sum=i[0]+i[1]
if sum%2==1:
sum+=1
print(sum/2)
|
s820826273
|
Accepted
| 17
| 2,940
| 89
|
i = list(map(int, input().split()))
sum=i[0]+i[1]
if sum%2==1:
sum+=1
print(int(sum/2))
|
s253961841
|
p03005
|
u095021077
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 43
|
Takahashi is distributing N balls to K persons. If each person has to receive at least one ball, what is the maximum possible difference in the number of balls received between the person with the most balls and the person with the fewest balls?
|
N, K=map(int, input().split())
print(N-K-1)
|
s115945190
|
Accepted
| 17
| 2,940
| 69
|
N, K=map(int, input().split())
if K==1:
print(0)
else:
print(N-K)
|
s180005961
|
p03149
|
u373047809
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 48
|
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".
|
print("YNEOS"[sorted(input())!=list(" 1479")])
|
s421477404
|
Accepted
| 18
| 2,940
| 51
|
print("YNEOS"[sorted(input())!=list(" 1479")::2])
|
s174705513
|
p03545
|
u101350975
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 312
|
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.
|
from sys import stdin
l = list(stdin.readline().rstrip())
s = [int(i) for i in l]
def dfs(i, f, sum):
if i == 3:
if sum == 7:
print(f)
exit()
else:
dfs(i+1, f + '+' + str(s[i]), sum + s[i])
dfs(i+1, f + '-' + str(s[i]), sum - s[i])
dfs(0, str(s[0]), s[0])
|
s964339503
|
Accepted
| 17
| 3,060
| 327
|
from sys import stdin
l = list(stdin.readline().rstrip())
s = [int(i) for i in l]
def dfs(i, f, sum):
if i == 3:
if sum == 7:
print(f + '=7')
exit()
else:
dfs(i+1, f + '+' + str(s[i+1]), sum + s[i+1])
dfs(i+1, f + '-' + str(s[i+1]), sum - s[i+1])
dfs(0, str(s[0]), s[0])
|
s363775237
|
p03385
|
u045408189
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 58
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
s=input().split()
print('Yes' if len(set(s))==3 else 'No')
|
s717423421
|
Accepted
| 17
| 2,940
| 51
|
s=input()
print('Yes' if len(set(s))==3 else 'No')
|
s132155751
|
p03386
|
u076764813
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 104
|
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())
r=range(a, b+1)
for i in sorted(set(r[:k]) or set(r[-k:])):
print(i)
|
s525112527
|
Accepted
| 17
| 3,060
| 94
|
a,b,k=map(int,input().split())
r=range(a,b+1)
for i in sorted(set(r[:k])|set(r[-k:])):print(i)
|
s258073755
|
p03593
|
u652737716
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,188
| 1,437
|
We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter. Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition: * Every row and column in A' can be read as a palindrome. Determine whether he can create a matrix satisfying the condition.
|
H, W = [int(x) for x in input().split(" ")]
d = {}
for h in range(H):
row = input()
for a in list(row):
if a in d:
d[a] += 1
else:
d[a] = 1
print(d)
res = False
if H % 2 == 0 and W % 2 == 0:
for v in d.values():
if v % 4 != 0:
res = True
break
res = not res
elif H % 2 == 1 and W % 2 == 0:
cnt4 = (W // 2) * ((H - 1) // 2)
cnt2 = (W // 2)
temp4 = 0
temp2 = 0
f = True
for v in d.values():
if v % 2 != 0:
f = False
break
temp4 += v // 4
temp2 += 1 if v % 4 != 0 else 0
if f and cnt2 >= temp2 and (cnt2 - temp2) % 2 == 0:
res = True
elif H % 2 == 0 and W % 2 == 1:
cnt4 = (H // 2) * ((W - 1) // 2)
cnt2 = (H // 2)
temp4 = 0
temp2 = 0
f = True
for v in d.values():
if v % 2 != 0:
f = False
break
temp4 += v // 4
temp2 += 1 if v % 4 != 0 else 0
if f and cnt2 >= temp2 and (cnt2 - temp2) % 2 == 0:
res = True
else:
cnt4 = ((H - 1) // 2) * ((W - 1) // 2)
cnt2 = ((H - 1) // 2) + ((W - 1) // 2)
temp4 = 0
temp2 = 0
temp1 = 0
for v in d.values():
temp4 += v // 4
temp2 += (v % 4) // 2
temp1 += (v % 2)
if temp1 == 1 and cnt2 >= temp2 and (cnt2 - temp2) % 2 == 0:
res = True
print("Yes" if res else "No")
|
s799965989
|
Accepted
| 20
| 3,192
| 1,427
|
H, W = [int(x) for x in input().split(" ")]
d = {}
for h in range(H):
row = input()
for a in list(row):
if a in d:
d[a] += 1
else:
d[a] = 1
res = False
if H % 2 == 0 and W % 2 == 0:
for v in d.values():
if v % 4 != 0:
res = True
break
res = not res
elif H % 2 == 1 and W % 2 == 0:
cnt4 = (W // 2) * ((H - 1) // 2)
cnt2 = (W // 2)
temp4 = 0
temp2 = 0
f = True
for v in d.values():
if v % 2 != 0:
f = False
break
temp4 += v // 4
temp2 += 1 if v % 4 != 0 else 0
if f and cnt2 >= temp2 and (cnt2 - temp2) % 2 == 0:
res = True
elif H % 2 == 0 and W % 2 == 1:
cnt4 = (H // 2) * ((W - 1) // 2)
cnt2 = (H // 2)
temp4 = 0
temp2 = 0
f = True
for v in d.values():
if v % 2 != 0:
f = False
break
temp4 += v // 4
temp2 += 1 if v % 4 != 0 else 0
if f and cnt2 >= temp2 and (cnt2 - temp2) % 2 == 0:
res = True
else:
cnt4 = ((H - 1) // 2) * ((W - 1) // 2)
cnt2 = ((H - 1) // 2) + ((W - 1) // 2)
temp4 = 0
temp2 = 0
temp1 = 0
for v in d.values():
temp4 += v // 4
temp2 += (v % 4) // 2
temp1 += (v % 2)
if temp1 == 1 and cnt2 >= temp2 and (cnt2 - temp2) % 2 == 0:
res = True
print("Yes" if res else "No")
|
s646906786
|
p03486
|
u836737505
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 96
|
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.
|
a = "".join(sorted(input()))
b = "".join(sorted(input())[::-1])
print("YES" if a < b else "NO" )
|
s594154147
|
Accepted
| 17
| 2,940
| 77
|
s = sorted(input())
t = sorted(input())[::-1]
print("Yes" if s < t else "No")
|
s402727625
|
p04030
|
u301624971
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 364
|
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?
|
def myAnswer(s:list)->str:
ans = []
for string in s:
if(string == "B"):
if(ans == []):
continue
else:
ans.pop()
else:
ans.append(string)
print(ans)
return "".join(ans)
def modelAnswer():
return
def main():
s = list(input())
print(myAnswer(s))
if __name__ == '__main__':
main()
|
s343952897
|
Accepted
| 18
| 3,060
| 351
|
def myAnswer(s:list)->str:
ans = []
for string in s:
if(string == "B"):
if(ans == []):
continue
else:
ans.pop()
else:
ans.append(string)
return "".join(ans)
def modelAnswer():
return
def main():
s = list(input())
print(myAnswer(s))
if __name__ == '__main__':
main()
|
s629732366
|
p04011
|
u151625340
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 124
|
There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.
|
N = int(input())
K = int(input())
X = int(input())
Y = int(input())
if N <= K:
print(N*X)
else:
print((N-K)*Y+N*X)
|
s723410718
|
Accepted
| 17
| 2,940
| 126
|
N = int(input())
K = int(input())
X = int(input())
Y = int(input())
if N <= K:
print(N*X)
else:
print((N-K)*Y+K*X)
|
s449263646
|
p03251
|
u677705680
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 231
|
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
N, M, X, Y = map(int, input().split())
x_list = list(map(int, input().split()))
y_list = list(map(int, input().split()))
print(X)
print(x_list)
if X <= Y and max(x_list) <= min(y_list):
print("No War")
else:
print("War")
|
s623947459
|
Accepted
| 17
| 2,940
| 235
|
N, M, X, Y = map(int, input().split())
X_list = [X]
X_list.extend(list(map(int, input().split())))
Y_list = [Y]
Y_list.extend(list(map(int, input().split())))
if max(X_list) < min(Y_list):
print("No War")
else:
print("War")
|
s778758395
|
p02608
|
u379917971
| 2,000
| 1,048,576
|
Wrong Answer
| 2,205
| 8,760
| 234
|
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
N=int(input())
sum=[0]*N
for x in range(1,N):
for y in range(1,N):
for z in range(1,N):
a=x*x+y*y+z*z+x*y+y*z+x*z
if a<=N:
sum[a-1]=sum[a-1]+1
for i in range(N):
print(sum[i-1])
|
s078935107
|
Accepted
| 446
| 9,320
| 228
|
N=int(input())
sum=[0]*N
for x in range(1,100):
for y in range(1,100):
for z in range(1,100):
a=x*x+y*y+z*z+x*y+y*z+x*z
if a<=N:
sum[a-1]=sum[a-1]+1
for i in sum:
print(i)
|
s629518697
|
p03730
|
u016881126
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 302
|
We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
import sys
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
read = sys.stdin.buffer.read
sys.setrecursionlimit(10 ** 7)
INF = float('inf')
A, B, C = map(int, readline().split())
for i in range(1, B+1):
if A * i % B == C:
print('Yes')
quit()
print('No')
|
s593089800
|
Accepted
| 18
| 3,060
| 302
|
import sys
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
read = sys.stdin.buffer.read
sys.setrecursionlimit(10 ** 7)
INF = float('inf')
A, B, C = map(int, readline().split())
for i in range(1, B+1):
if A * i % B == C:
print('YES')
quit()
print('NO')
|
s770190034
|
p03160
|
u357230322
| 2,000
| 1,048,576
|
Wrong Answer
| 157
| 13,928
| 246
|
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
N = int(input())
*h, = map(int, input().split())
dp = [float("inf") for _ in range(N)]
dp[0] = 0
for i in range(N - 1):
dp[i + 1] = dp[i] + abs(h[i + 1] - h[i])
if i < N - 2:
dp[i + 2] =dp[i + 2], dp[i] + abs(h[i + 2] - h[i])
print(dp[N - 1])
|
s694025638
|
Accepted
| 125
| 13,980
| 181
|
n = int(input())
h=list(map(int,input().split()))
dp=[0]*(n)
dp[1]=abs(h[1]-h[0])
for i in range(2,n):
dp[i]=min(dp[i-1]+abs(h[i]-h[i-1]),dp[i-2]+abs(h[i]-h[i-2]))
print(dp[n-1])
|
s829410209
|
p03729
|
u064963667
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 99
|
You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`.
|
a,b,c = input().split()
if a[-1] == b[0] and b[-1] == c[0]:
print('Yes')
else:
print('No')
|
s981160604
|
Accepted
| 17
| 2,940
| 100
|
a,b,c = input().split()
if a[-1] == b[0] and b[-1] == c[0]:
print('YES')
else:
print('NO')
|
s554934036
|
p02422
|
u855694108
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,916
| 1,114
|
Write a program which performs a sequence of commands to a given string $str$. The command is one of: * print a b: print from the a-th character to the b-th character of $str$ * reverse a b: reverse from the a-th character to the b-th character of $str$ * replace a b p: replace from the a-th character to the b-th character of $str$ with p Note that the indices of $str$ start with 0.
|
def main():
m = list(input())
q = int(input())
A = []
for _ in range(q):
A.append(input().split())
for x in range(q):
if A[x][0] == "print":
A[x][1] = int(A[x][1])
A[x][2] = int(A[x][2])
for y in range(A[x][1], A[x][2] + 1):
if y != A[x][2]:
print(m[y], end = "")
else:
print(m[y])
elif A[x][0] == "reverse":
A[x][1] = int(A[x][1])
A[x][2] = int(A[x][2])
hoge = []
for y in range(A[x][1], A[x][2] + 1):
hoge.append(m[y])
hoge = hoge[::-1]
i = 0
for y in range(A[x][1], A[x][2] + 1):
m[y] = hoge[i]
i += 1
elif A[x][0] == "replace":
A[x][1] = int(A[x][1])
A[x][2] = int(A[x][2])
fuga = list(A[x][3])
print(fuga)
j = 0
for y in range(A[x][1], A[x][2] + 1):
m[y] = fuga[j]
j += 1
if __name__ == "__main__":
main()
|
s408637492
|
Accepted
| 70
| 9,140
| 1,090
|
def main():
m = list(input())
q = int(input())
A = []
for _ in range(q):
A.append(input().split())
for x in range(q):
if A[x][0] == "print":
A[x][1] = int(A[x][1])
A[x][2] = int(A[x][2])
for y in range(A[x][1], A[x][2] + 1):
if y != A[x][2]:
print(m[y], end = "")
else:
print(m[y])
elif A[x][0] == "reverse":
A[x][1] = int(A[x][1])
A[x][2] = int(A[x][2])
hoge = []
for y in range(A[x][1], A[x][2] + 1):
hoge.append(m[y])
hoge = hoge[::-1]
i = 0
for y in range(A[x][1], A[x][2] + 1):
m[y] = hoge[i]
i += 1
elif A[x][0] == "replace":
A[x][1] = int(A[x][1])
A[x][2] = int(A[x][2])
fuga = list(A[x][3])
j = 0
for y in range(A[x][1], A[x][2] + 1):
m[y] = fuga[j]
j += 1
if __name__ == "__main__":
main()
|
s932699318
|
p03478
|
u144072139
| 2,000
| 262,144
|
Wrong Answer
| 22
| 3,060
| 273
|
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
N, A, B = map(int, input().split())
sum = 0
def FindSumOfDigits(x):
count = 0
while x>0:
count += x%10
x /= 10
return count
for n in range(1, N+1):
count= FindSumOfDigits(n)
if A <= count <= B:
sum += n
print(sum)
|
s248481773
|
Accepted
| 25
| 2,940
| 263
|
N, A, B = map(int, input().split())
ans = 0
def FindSumOfDigits(x):
count = 0
while x>0:
count += x%10
x = x//10
return count
for n in range(1, N+1):
count = FindSumOfDigits(n)
if A<= count <=B:
ans += n
print(ans)
|
s939089726
|
p03360
|
u210827208
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 77
|
There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations?
|
x=list(map(int,input().split()))
k=int(input())
print(max(x)*k+sum(x)-max(x))
|
s626145915
|
Accepted
| 17
| 2,940
| 82
|
x=list(map(int,input().split()))
k=int(input())
print(max(x)*(2**k)+sum(x)-max(x))
|
s907208714
|
p02916
|
u094932051
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 388
|
Takahashi went to an all-you-can-eat buffet with N kinds of dishes and ate all of them (Dish 1, Dish 2, \ldots, Dish N) once. The i-th dish (1 \leq i \leq N) he ate was Dish A_i. When he eats Dish i (1 \leq i \leq N), he gains B_i satisfaction points. Additionally, when he eats Dish i+1 just after eating Dish i (1 \leq i \leq N - 1), he gains C_i more satisfaction points. Find the sum of the satisfaction points he gained.
|
while True:
try:
N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
C = list(map(int, input().split()))
res = B[A[0]-1]
for i in range(1, len(A)):
res += B[A[i]-1]
if A[i-1]-1 < len(C):
res += C[A[i-1]-1]
print(res)
except:
break
|
s414525199
|
Accepted
| 17
| 3,064
| 450
|
while True:
try:
N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
C = list(map(int, input().split()))
A.insert(0, 0)
B.insert(0, 0)
C.insert(0, 0)
res = B[A[1]]
for i in range(2, len(A)):
res += B[A[i]]
if A[i-1]+1 == A[i]:
res += C[A[i-1]]
print(res)
except:
break
|
s342128631
|
p03672
|
u478266845
| 2,000
| 262,144
|
Wrong Answer
| 293
| 18,844
| 260
|
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.
|
import numpy as np
S = str(input())
import numpy as np
for i in range(int(np.floor(len(S)/2))):
j = int(np.floor(len(S)/2)) - i
first_str = S[:j]
second_str = S[j:2*j]
if first_str == second_str:
num = 2*j
break
print(num)
|
s024779331
|
Accepted
| 149
| 12,424
| 339
|
import numpy as np
S = str(input())
import numpy as np
for i in range(1,int(np.floor(len(S)/2))):
j = int(np.floor(len(S)/2)) - i
first_str = S[:j]
second_str = S[j:2*j]
if first_str == second_str:
num = 2*j
break
elif (first_str != second_str) & (j==1):
num = 0
break
print(num)
|
s380055462
|
p02747
|
u801009312
| 2,000
| 1,048,576
|
Wrong Answer
| 20
| 3,188
| 230
|
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.
|
# coding: utf-8
# Your code here!
import re
input = input()
pattern = '^(hi)+$'
repatter = re.compile(pattern)
result = repatter.fullmatch(input)
if result is not None:
print("yes")
else:
print("no")
|
s487481709
|
Accepted
| 22
| 3,316
| 230
|
# coding: utf-8
# Your code here!
import re
input = input()
pattern = '^(hi)+$'
repatter = re.compile(pattern)
result = repatter.fullmatch(input)
if result is not None:
print("Yes")
else:
print("No")
|
s631726958
|
p03612
|
u723583932
| 2,000
| 262,144
|
Wrong Answer
| 78
| 14,008
| 161
|
You are given a permutation p_1,p_2,...,p_N consisting of 1,2,..,N. You can perform the following operation any number of times (possibly zero): Operation: Swap two **adjacent** elements in the permutation. You want to have p_i ≠ i for all 1≤i≤N. Find the minimum required number of operations to achieve this.
|
n=int(input())
p=list(map(int,input().split()))
cnt=0
for i in range(n-1):
if p[i]==(i+1):
p[i],p[i+1]=p[i+1],p[i]
cnt+=1
print(cnt)
print(p)
|
s378043059
|
Accepted
| 74
| 14,008
| 191
|
n=int(input())
p=list(map(int,input().split()))
cnt=0
for i in range(n-1):
if p[i]==(i+1):
p[i],p[i+1]=p[i+1],p[i]
cnt+=1
if p[-1]==n:
cnt+=1
print(cnt)
|
s482843397
|
p03407
|
u771167374
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 88
|
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 c == a or c== b or c ==a+b else 'No')
|
s833988601
|
Accepted
| 17
| 2,940
| 69
|
a, b, c = map(int, input().split())
print('Yes' if c <=a+b else 'No')
|
s848741086
|
p02390
|
u579699497
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,668
| 128
|
Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively.
|
import math
S = int(input())
h = math.floor(S/3600)
m = math.floor((S - h*3600)/60)
s = (S - (h*3600 + m*60))%60
print(h, m, s)
|
s634492061
|
Accepted
| 30
| 7,756
| 149
|
import math
S = int(input())
h = math.floor(S/3600)
m = math.floor((S - h*3600)/60)
s = (S - (h*3600 + m*60))%60
print(str(h)+":"+str(m)+":"+str(s))
|
s368822237
|
p03339
|
u985596066
| 2,000
| 1,048,576
|
Wrong Answer
| 2,104
| 5,796
| 284
|
There are N people standing in a row from west to east. Each person is facing east or west. The directions of the people is given as a string S of length N. The i-th person from the west is facing east if S_i = `E`, and west if S_i = `W`. You will appoint one of the N people as the leader, then command the rest of them to face in the direction of the leader. Here, we do not care which direction the leader is facing. The people in the row hate to change their directions, so you would like to select the leader so that the number of people who have to change their directions is minimized. Find the minimum number of people who have to change their directions.
|
N=int(input())
H=list(str(input()))
m = 0
for i in range(N):
c = 0
for s in range(N):
if i > s and H[s] == 'W':
c += 1
if i < s and H[s] == 'E':
c += 1
if m == 0 or m >c:
n = i
m = c
print(n)
|
s277901831
|
Accepted
| 119
| 16,160
| 211
|
from itertools import accumulate
N=int(input())
s=str(input())
a = accumulate(c=='W' for c in 'X'+s[:-1])
b = reversed(list(accumulate(c=='E' for c in 'X'+s[-1:0:-1])))
print(min(x + y for x, y in zip(a, b)))
|
s360745940
|
p02612
|
u600261652
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 9,140
| 30
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
print(N%1000)
|
s566046898
|
Accepted
| 25
| 9,180
| 94
|
def resolve():
N = int(input())
print("0" if N%1000 == 0 else 1000-(N%1000))
resolve()
|
s575671815
|
p02613
|
u081141316
| 2,000
| 1,048,576
|
Wrong Answer
| 139
| 16,152
| 229
|
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.
|
def count_data(s):
judge_list = ['AC', 'WA', 'TEL', 'RE']
for judge in judge_list:
ans = s.count(judge)
print('{} x {}'.format(judge, ans))
n = int(input())
s = []
for i in range(n):
s.append(input())
|
s207694999
|
Accepted
| 148
| 16,240
| 245
|
def count_data(s):
judge_list = ['AC', 'WA', 'TLE', 'RE']
for judge in judge_list:
ans = s.count(judge)
print('{} x {}'.format(judge, ans))
n = int(input())
s = []
for i in range(n):
s.append(input())
count_data(s)
|
s994428641
|
p03719
|
u689835643
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 91
|
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
x, y, z = map(int, input().split())
if x <= y <= z:
print("Yes")
else:
print("No")
|
s981619720
|
Accepted
| 19
| 3,060
| 91
|
x, y, z = map(int, input().split())
if x <= z <= y:
print("Yes")
else:
print("No")
|
s667219366
|
p03351
|
u905582793
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 110
|
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-c) <= d or abs(a-b)+abs(b-c)<= d:
print('Yes')
else:
print('No')
|
s812463762
|
Accepted
| 17
| 2,940
| 119
|
a,b,c,d=map(int,input().split())
if abs(a-c) <= d or (abs(a-b)<=d and abs(b-c)<= d):
print('Yes')
else:
print('No')
|
s478659594
|
p03131
|
u610042046
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 9,032
| 140
|
Snuke has one biscuit and zero Japanese yen (the currency) in his pocket. He will perform the following operations exactly K times in total, in the order he likes: * Hit his pocket, which magically increases the number of biscuits by one. * Exchange A biscuits to 1 yen. * Exchange 1 yen to B biscuits. Find the maximum possible number of biscuits in Snuke's pocket after K operations.
|
k, a, b = map(int, input().split())
if 2 >= b-a or a >= k:
print(k+1)
else:
print((k-a+1)//2 + ((k-a+1)//2-1)*(b-a) + ((k-a+1)%2))
|
s418365716
|
Accepted
| 28
| 9,060
| 131
|
k, a, b = map(int, input().split())
if 2 >= b-a or a >= k:
print(k+1)
else:
print(b + ((k-a+1)//2-1)*(b-a) + ((k-a+1)%2))
|
s062334973
|
p02842
|
u422242927
| 2,000
| 1,048,576
|
Wrong Answer
| 30
| 9,156
| 149
|
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
N = int(input())
if int(N // 1.08 * 1.08) == N:
print(N // 1.08)
elif int((N // 1.08 + 1) * 1.08) == N:
print(N // 1.08 + 1)
else:
print(":(")
|
s272450720
|
Accepted
| 28
| 9,072
| 117
|
N = int(input())
for i in range(1, N + 1):
if i + (i * 8) // 100 == N:
print(i)
break
else:
print(":(")
|
s157884657
|
p02697
|
u542541293
| 2,000
| 1,048,576
|
Wrong Answer
| 99
| 9,192
| 199
|
You are going to hold a competition of one-to-one game called AtCoder Janken. _(Janken is the Japanese name for Rock-paper-scissors.)_ N players will participate in this competition, and they are given distinct integers from 1 through N. The arena has M playing fields for two players. You need to assign each playing field two distinct integers between 1 and N (inclusive). You cannot assign the same integer to multiple playing fields. The competition consists of N rounds, each of which proceeds as follows: * For each player, if there is a playing field that is assigned the player's integer, the player goes to that field and fight the other player who comes there. * Then, each player adds 1 to its integer. If it becomes N+1, change it to 1. You want to ensure that no player fights the same opponent more than once during the N rounds. Print an assignment of integers to the playing fields satisfying this condition. It can be proved that such an assignment always exists under the constraints given.
|
N, M = map(int, input().split())
for i in range(M):
L = []
if M % 2 == 0:
L.append(i+2)
L.append(N-i-1)
print(*L)
if M % 2 == 1:
L.append(i+1)
L.append(N-i-1)
print(*L)
|
s906462005
|
Accepted
| 99
| 9,204
| 542
|
N, M = map(int, input().split())
if M % 2 == 0:
l = 1
for i in range(M//2):
L = []
L.append(l)
L.append(2*(M//2-i)+l)
print(*L)
l += 1
l = 2*(M//2)+2
for i in range(M//2):
L = []
L.append(l)
L.append(2*(M//2-i)+l-1)
print(*L)
l += 1
elif M % 2 == 1:
l = 1
for i in range(M//2):
L = []
L.append(l)
L.append(2*(M//2-i)+l)
print(*L)
l += 1
l = 2*(M//2)+2
for i in range(M//2+1):
L = []
L.append(l)
L.append(2*(M//2+1-i)+l-1)
print(*L)
l += 1
|
s037027844
|
p03024
|
u366959492
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 108
|
Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.
|
s=list(input())
c=0
for i in s:
if i=="o":
c+=1
if c>=8:
print("YES")
else:
print("NO")
|
s957764713
|
Accepted
| 17
| 2,940
| 152
|
s=list(input())
c=0
cc=0
for i in s:
if i=="o":
c+=1
if i=="x":
cc+=1
if c>=8 or cc<=7:
print("YES")
else:
print("NO")
|
s514154949
|
p02540
|
u104282757
| 2,000
| 1,048,576
|
Wrong Answer
| 1,138
| 56,764
| 1,900
|
There are N cities on a 2D plane. The coordinate of the i-th city is (x_i, y_i). Here (x_1, x_2, \dots, x_N) and (y_1, y_2, \dots, y_N) are both permuations of (1, 2, \dots, N). For each k = 1,2,\dots,N, find the answer to the following question: Rng is in City k. Rng can perform the following move arbitrarily many times: * move to another city that has a smaller x-coordinate and a smaller y-coordinate, or a larger x-coordinate and a larger y-coordinate, than the city he is currently in. How many cities (including City k) are reachable from City k?
|
from heapq import heappop, heappush
class UnionFind:
def __init__(self, n):
self.par = [i for i in range(n+1)]
self.rank = [0] * (n+1)
# search
def find(self, x):
if self.par[x] == x:
return x
else:
self.par[x] = self.find(self.par[x])
return self.par[x]
# unite
def union(self, x, y):
x = self.find(x)
y = self.find(y)
if self.rank[x] < self.rank[y]:
self.par[x] = y
else:
self.par[y] = x
if self.rank[x] == self.rank[y]:
self.rank[x] += 1
def same_check(self, x, y):
return self.find(x) == self.find(y)
def solve(n, xy_list):
# sort y by x
xy_list_s = sorted(xy_list, key=lambda x: x[0])
y_list_s = [xy[1] for xy in xy_list_s]
# union_find
uf = UnionFind(n)
# heap
h = []
for y in y_list_s:
yp_list = [y]
while len(h):
p = heappop(h)
if p < y:
yp_list.append(p)
uf.union(p, y)
else:
heappush(h, p)
break
heappush(h, min(yp_list))
# unite
parent_n = [0] * n
parent_cnt = [0] * (n + 1)
for i in range(n):
p = uf.find(xy_list[i][1])
parent_n[i] = p
parent_cnt[p] += 1
return [parent_cnt[parent_n[i]] for i in range(n)]
def main():
n = int(input())
xy_list = [[0] * 2 for _ in range(n)]
for i in range(n):
x, y = map(int, input().split())
xy_list[i][0] = x
xy_list[i][1] = y
res = solve(n, xy_list)
print(res)
def test():
assert solve(4, [[1, 4], [2, 3], [3, 1], [4, 2]]) == [1, 1, 2, 2]
assert solve(7, [[6, 4], [4, 3], [3, 5], [7, 1], [2, 7], [5, 2], [1, 6]]) == [3, 3, 1, 1, 2, 3, 2]
if __name__ == "__main__":
test()
main()
|
s282814846
|
Accepted
| 1,085
| 56,872
| 1,920
|
from heapq import heappop, heappush
class UnionFind:
def __init__(self, n):
self.par = [i for i in range(n+1)]
self.rank = [0] * (n+1)
# search
def find(self, x):
if self.par[x] == x:
return x
else:
self.par[x] = self.find(self.par[x])
return self.par[x]
# unite
def union(self, x, y):
x = self.find(x)
y = self.find(y)
if self.rank[x] < self.rank[y]:
self.par[x] = y
else:
self.par[y] = x
if self.rank[x] == self.rank[y]:
self.rank[x] += 1
def same_check(self, x, y):
return self.find(x) == self.find(y)
def solve(n, xy_list):
# sort y by x
xy_list_s = sorted(xy_list, key=lambda x: x[0])
y_list_s = [xy[1] for xy in xy_list_s]
# union_find
uf = UnionFind(n)
# heap
h = []
for y in y_list_s:
yp_list = [y]
while len(h):
p = heappop(h)
if p < y:
yp_list.append(p)
uf.union(p, y)
else:
heappush(h, p)
break
heappush(h, min(yp_list))
# unite
parent_n = [0] * n
parent_cnt = [0] * (n + 1)
for i in range(n):
p = uf.find(xy_list[i][1])
parent_n[i] = p
parent_cnt[p] += 1
return [parent_cnt[parent_n[i]] for i in range(n)]
def main():
n = int(input())
xy_list = [[0] * 2 for _ in range(n)]
for i in range(n):
x, y = map(int, input().split())
xy_list[i][0] = x
xy_list[i][1] = y
res = solve(n, xy_list)
for r in res:
print(r)
def test():
assert solve(4, [[1, 4], [2, 3], [3, 1], [4, 2]]) == [1, 1, 2, 2]
assert solve(7, [[6, 4], [4, 3], [3, 5], [7, 1], [2, 7], [5, 2], [1, 6]]) == [3, 3, 1, 1, 2, 3, 2]
if __name__ == "__main__":
test()
main()
|
s251256700
|
p03251
|
u434822333
| 2,000
| 1,048,576
|
Wrong Answer
| 19
| 3,060
| 236
|
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
N, M, X, Y = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
A = [a for a in range(101) if (X < a <= Y) and (max(x) < a <= min(y))]
if len(A) is not 0: print('No war')
else: print('War')
|
s492094537
|
Accepted
| 18
| 3,060
| 242
|
N, M, X, Y = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
Z = [z for z in range(-100, 101) if (X < z <= Y) and (max(x) < z <= min(y))]
if len(Z) is not 0: print('No War')
else: print('War')
|
s226374124
|
p03351
|
u183481524
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 175
|
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 or abs(b - a) < d:
if abs(b - c) < d or abs(c - b) < d:
print('Yes')
else:print('No')
else:print('No')
|
s597932612
|
Accepted
| 17
| 3,060
| 197
|
a, b, c, d = map(int,input().split())
if (abs(a - b) <= d or abs(b - a) <= d) and (abs(b - c) <= d or abs(c - b) <= d) or (abs(c - a) <= d or abs(a - c) <= d):
print('Yes')
else:print('No')
|
s085096302
|
p02972
|
u751724075
| 2,000
| 1,048,576
|
Wrong Answer
| 2,109
| 99,940
| 715
|
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
import math
N = int(input())
a_list = [-1] + list(map(int, input().split()))
cache = {}
def decompose(n):
if n in cache:
return cache[n]
if n % 2 == 0:
return [2] + decompose(n // 2)
for i in range(3, math.floor(math.sqrt(n)), 2):
if n % i == 0:
return [i] + decompose(n // i)
return [n]
rels = [set() for _ in range(N + 1)]
for i in range(2, N + 1):
for d in decompose(i):
if d != i:
rels[d].add(i)
rels[1].add(i)
b_list = [None] * (N + 1)
for i in range(N, 0, -1):
b = a_list[i]
for j in rels[i]:
b = b ^ b_list[j]
b_list[i] = b
count = 0
b_list2 = []
for i in range(1, N + 1):
if b_list[i] == 1:
count += 1
b_list2.append(str(i + 1))
print(count)
print(" ".join(b_list2))
|
s600115873
|
Accepted
| 549
| 17,196
| 362
|
import math
N = int(input())
a_list = [-1] + list(map(int, input().split()))
b_list = [None] * (N + 1)
for i in range(N, 0, -1):
b = a_list[i]
for j in range(i * 2, N + 1, i):
b = b ^ b_list[j]
b_list[i] = b
count = 0
b_list2 = []
for i in range(1, N + 1):
if b_list[i] == 1:
count += 1
b_list2.append(str(i))
print(count)
print(" ".join(b_list2))
|
s270124475
|
p03386
|
u679089074
| 2,000
| 262,144
|
Wrong Answer
| 32
| 9,068
| 267
|
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.
|
#56
a,b,k = map(int,input().split())
num = []
if b-a <= k:
for i in range(a,b+1):
print(i)
else:
for i in range(a,a+k):
num.append(i)
for i in range(b-k+1,b+1):
num.append(i)
num = set(num)
for i in num:
print(i)
|
s436342041
|
Accepted
| 28
| 9,036
| 300
|
#56
a,b,k = map(int,input().split())
num = []
if b-a <= k:
for i in range(a,b+1):
print(i)
else:
for i in range(a,a+k):
num.append(i)
for i in range(b-k+1,b+1):
num.append(i)
num = list(set(num))
num.sort()
for i in num:
print(i)
|
s528985423
|
p02659
|
u586832145
| 2,000
| 1,048,576
|
Wrong Answer
| 25
| 9,928
| 118
|
Compute A \times B, truncate its fractional part, and print the result as an integer.
|
from decimal import Decimal
from decimal import getcontext
s = input();
a,b = s.split();
print(Decimal(a)*Decimal(b));
|
s799923173
|
Accepted
| 27
| 10,052
| 123
|
from decimal import Decimal
from decimal import getcontext
s = input();
a,b = s.split();
print(int(Decimal(a)*Decimal(b)));
|
s453762067
|
p03089
|
u981767024
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 944
|
Snuke has an empty sequence a. He will perform N operations on this sequence. In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1). You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
|
# 2019/03/23
# AtCoder Grand Contest 032 - A
# Input
n = int(input())
b = list(map(int, input().split()))
a = [0] * n
anslist = list()
ans = 0
# Possible Check
for idx in range(n):
if b[idx] > (idx + 1):
ans = -1
break
# fixcnt = 0 // (n-1) - idx2
# Make list
if ans == 0:
fixcnt = 0
# Loop from n-1, to -1, by -1
for idx2 in range(n-1, -1, -1):
if a[idx2] == 0:
if b[idx2] <= fixcnt + 1:
a[idx2] = b[idx2]
anslist.append(b[idx2])
else:
for idx3 in range(idx2-1, -1, -1):
if b[idx3] <= fixcnt + 1:
a[idx3] = b[idx3]
anslist.append(b[idx3])
fixcnt += 1
break
anslist.append(b[idx2])
a[idx2] = b[idx2]
fixcnt += 1
for idx4 in range(n):
print(anslist[idx4])
|
s719446322
|
Accepted
| 18
| 3,064
| 492
|
# 2019/03/23
# AtCoder Grand Contest 032 - A
# Input
n = int(input())
b = list(map(int, input().split()))
anslist = list()
ngflg = False
for idx6 in range(n):
maxidx = -1
for idx7 in range(len(b)):
if b[idx7] == idx7 + 1:
maxidx = idx7
if maxidx >= 0:
anslist.append(b.pop(maxidx))
else:
ngflg = True
if ngflg:
print(-1)
break
if ngflg == False:
for idx4 in range(n-1, -1, -1):
print(anslist[idx4])
|
s979520927
|
p03251
|
u122195031
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 279
|
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
n,m,X,Y = map(int,input().split())
x = list(map(int,input().split()))
y = list(map(int,input().split()))
x_max = max(x)
y_min = min(y)
for i in range(-100,101):
if X <= i <= Y and x_max < i and y_min >= i:
print("No War")
print(i)
exit()
print("War")
|
s813817978
|
Accepted
| 18
| 3,064
| 261
|
n,m,X,Y = map(int,input().split())
x = list(map(int,input().split()))
y = list(map(int,input().split()))
x_max = max(x)
y_min = min(y)
for i in range(-100,101):
if X < i <= Y and x_max < i and y_min >= i:
print("No War")
exit()
print("War")
|
s958528057
|
p04043
|
u759518460
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 115
|
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
k = input().split()
k.sort()
if k[0] == '5' and k[1] == '5' and k[2] == '7':
print('Yes')
else:
print('No')
|
s756622347
|
Accepted
| 16
| 2,940
| 115
|
k = input().split()
k.sort()
if k[0] == '5' and k[1] == '5' and k[2] == '7':
print('YES')
else:
print('NO')
|
s396345655
|
p02256
|
u209940149
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,604
| 194
|
Write a program which finds the greatest common divisor of two natural numbers _a_ and _b_
|
def gcd(a, b):
c = max([a, b])
d = min([a, b])
if c % d == 0:
return d
else:
gcd(d, c % d)
nums = input().split()
print(gcd(int(nums[0]), int(nums[1])))
|
s147069234
|
Accepted
| 20
| 5,604
| 201
|
def gcd(a, b):
c = max([a, b])
d = min([a, b])
if c % d == 0:
return d
else:
return gcd(d, c % d)
nums = input().split()
print(gcd(int(nums[0]), int(nums[1])))
|
s654121777
|
p02602
|
u380854465
| 2,000
| 1,048,576
|
Wrong Answer
| 2,207
| 53,472
| 310
|
M-kun is a student in Aoki High School, where a year is divided into N terms. There is an exam at the end of each term. According to the scores in those exams, a student is given a grade for each term, as follows: * For the first through (K-1)-th terms: not given. * For each of the K-th through N-th terms: the multiplication of the scores in the last K exams, including the exam in the graded term. M-kun scored A_i in the exam at the end of the i-th term. For each i such that K+1 \leq i \leq N, determine whether his grade for the i-th term is **strictly** greater than the grade for the (i-1)-th term.
|
import sys
import numpy as np
N, K = list(map(int, input().split()))
A = list(map(int, input().split()))
B = []
for i in range(N-K+1):
s = 1
for j in range(K):
s *= A[i+j]
B.append(s)
print(B)
for i in range(N-K):
if B[i] < B[i+1]:
print('Yes')
else:
print('No')
|
s529086157
|
Accepted
| 267
| 50,148
| 302
|
import sys
import numpy as np
N, K = list(map(int, input().split()))
A = list(map(int, input().split()))
B = 0
s = 1
for i, a in enumerate(A):
if i >= K:
B = A[i-K]
s = a
if B != 0 and B < s:
print('Yes')
elif B != 0 and B >= s:
print('No')
|
s909468723
|
p03067
|
u223904637
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 105
|
There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.
|
a,b,c=map(int,input().split())
if (a<b and b<c) or (c<b and b<a):
print('Yes')
else:
print('No')
|
s630754569
|
Accepted
| 18
| 2,940
| 105
|
a,b,c=map(int,input().split())
if (a<c and c<b) or (b<c and c<a):
print('Yes')
else:
print('No')
|
s934206264
|
p03699
|
u962330718
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 245
|
You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well. However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade?
|
n=int(input())
s=[int(input()) for i in range(n)]
a=[0]
for i in range(n):
if s[i]%10!=0:
a.append(s[i])
a.pop(0)
if sum(s)%10==0:
if 0 in a:
print(0)
else:
print(sum(s)-min(a))
else:
print(sum(s))
|
s885483354
|
Accepted
| 17
| 3,060
| 257
|
n=int(input())
s=[int(input()) for i in range(n)]
a=[]
for i in range(n):
if s[i]%10!=0:
a.append(s[i])
if len(a)==0:
a.append(0)
if sum(s)%10==0:
if 0 in a:
print(0)
else:
print(sum(s)-min(a))
else:
print(sum(s))
|
s020037206
|
p02389
|
u362104929
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,612
| 200
|
Write a program which calculates the area and perimeter of a given rectangle.
|
def main(arg):
numbers = [int(x) for x in arg.split(" ")]
s = numbers[0] * numbers[1]
p = (numbers[0] + numbers[1]) * 2
return s,p
if __name__ == '__main__':
print(main(input()))
|
s099037363
|
Accepted
| 20
| 7,624
| 236
|
def main(arg):
numbers = [int(x) for x in arg.split(" ")]
s = numbers[0] * numbers[1]
p = (numbers[0] + numbers[1]) * 2
result = "{} {}".format(s,p)
return result
if __name__ == '__main__':
print(main(input()))
|
s454896497
|
p02646
|
u725993280
| 2,000
| 1,048,576
|
Wrong Answer
| 26
| 9,056
| 510
|
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
a,v = map(int,input().split())
b,w = map(int,input().split())
t = int(input())
daplus = a + v*t
daminus = a - v*t
if daplus < daminus:
tmp = daplus
daplus = daminus
daminus = tmp
dbplus = b + w*t
dbminus = b - w*t
if dbplus < dbminus:
tmp = dbplus
dbplus = dbminus
dbminus = tmp
flag = 0
if(daminus <= dbplus & dbplus <= daplus):
flag += 1
if(daminus <= dbminus & dbminus <= daplus):
flag += 1
print(flag)
if flag == 2:
print("YES")
else:
print("NO")
|
s312311329
|
Accepted
| 24
| 9,188
| 497
|
a,v = map(int,input().split())
b,w = map(int,input().split())
t = int(input())
daplus = a + v*t
daminus = a - v*t
if daplus < daminus:
tmp = daplus
daplus = daminus
daminus = tmp
dbplus = b + w*t
dbminus = b - w*t
if dbplus < dbminus:
tmp = dbplus
dbplus = dbminus
dbminus = tmp
flag = 0
if(daminus <= dbplus & dbplus <= daplus):
flag += 1
if(daminus <= dbminus & dbminus <= daplus):
flag += 1
if flag == 2:
print("YES")
else:
print("NO")
|
s136968240
|
p03565
|
u108617242
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 718
|
E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One more thing he found is a sheet of paper with the following facts written on it: * Condition 1: The string S contains a string T as a contiguous substring. * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1. Print the string S. If such a string does not exist, print `UNRESTORABLE`.
|
S = input()
T = input()
len_S = len(S)
len_T = len(T)
qc = 0
ans = ''
for i in range(len_S-1, -1, -1):
print(i, S[i])
si = S[i]
if si == '?':
qc += 1
else:
qc = 0
if qc == len_T:
ans = ans = S[:i] + T + S[i+len_T:]
break
if si == T[0] and len_S - i >= len_T:
print(T)
flag = True
for ii in range(1, len_T):
print(ii, S[i+ii])
if T[ii] == S[i+ii] or S[i+ii] == '?':
continue
else:
flag = False
break
if flag:
ans = S[:i] + T + S[i+len_T:]
break
if ans:
print(ans.replace('?', 'a'))
else:
print('UNRESTORABLE')
|
s371298593
|
Accepted
| 17
| 3,064
| 525
|
S = input()
T = input()
len_S = len(S)
len_T = len(T)
qc = 0
ans = ''
for i in range(len_S-1, -1, -1):
si = S[i]
if (si == '?' or si == T[0]) and len_S - i >= len_T:
flag = True
for ii in range(1, len_T):
if T[ii] == S[i+ii] or S[i+ii] == '?':
continue
else:
flag = False
break
if flag:
ans = S[:i] + T + S[i+len_T:]
break
if ans:
print(ans.replace('?', 'a'))
else:
print('UNRESTORABLE')
|
s106889848
|
p03433
|
u113971909
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 51
|
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
print('YNeos'[(int(input())%500)<=int(input())::2])
|
s745904779
|
Accepted
| 17
| 2,940
| 52
|
print('NYoe s'[(int(input())%500)<=int(input())::2])
|
s069809598
|
p03471
|
u191635495
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 15,348
| 866
|
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.
|
N, Y = map(int, input().split())
a = 10000
b = 5000
c = 1000
x = -1 # 10000
y = -1 # 5000
z = -1 # 1000
if N * a >= Y:
x = N
y += 1
z += 1
diff = N*a - Y
while diff >= 0:
print(diff)
if diff == 0:
break
else:
if diff > 9000:
# exchange 10000 to 1000
diff -= 9000
x -= 1
z += 1
elif diff > 5000:
# exchange 10000 to 5000
diff -= 5000
x -= 1
y += 1
elif diff % 4000 == 0 and diff // 4000 == y:
# exchange 5000 to 1000
diff = 0
y -= 1
z += 1
elif diff <= 4000:
diff = 0
x = -1
y = -1
z = -1
print(x, y, z)
|
s387171735
|
Accepted
| 1,356
| 3,064
| 376
|
N, Y = map(int, input().split())
A = 10000
B = 5000
C = 1000
x = -1
y = -1
z = -1
for a in range(N+1):
for b in range(N+1):
if a + b > N:
continue
else:
total = A*a + B*b + C*(N - (a + b))
if total == Y:
x = a
y = b
z = N - (a + b)
break
print(x, y, z)
|
s827235923
|
p02678
|
u222668979
| 2,000
| 1,048,576
|
Wrong Answer
| 834
| 76,592
| 732
|
There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.
|
from collections import deque
import sys
input = sys.stdin.readline
def bfs(NEAR, S, N):
pas = [-1 for _ in range(N)]
pas[S] = 's'
frag = set([S])
que = deque([S])
while len(que) > 0:
q = que.popleft()
for i in NEAR[q]:
if i in frag:
continue
pas[i]=q
que.append(i)
frag.add(i)
return pas
n, m = map(int, input().split())
ab = [list(map(int, input().split())) for _ in range(m)]
near = [[] for _ in range(n)]
for a, b in ab:
near[a - 1].append(b - 1)
near[b - 1].append(a - 1)
print(near)
pas = bfs(near, 0, n)
for i in range(1, n):
print(pas[i] + 1)
|
s595898965
|
Accepted
| 734
| 76,312
| 806
|
from collections import deque
import sys
input = sys.stdin.readline
def bfs(NEAR, S, N):
pas = [-1 for _ in range(N)]
pas[S] = 's'
frag = set([S])
que = deque([S])
while len(que) > 0:
q = que.popleft()
for i in NEAR[q]:
if i in frag:
continue
pas[i]=q
que.append(i)
frag.add(i)
return pas
n, m = map(int, input().split())
ab = [list(map(int, input().split())) for _ in range(m)]
near = [[] for _ in range(n)]
for a, b in ab:
near[a - 1].append(b - 1)
near[b - 1].append(a - 1)
pas = bfs(near, 0, n)
if all(pas[i] != -1 for i in range(n)):
print('Yes')
for i in range(1, n):
print(pas[i] + 1)
else:
print('No')
|
s857912563
|
p03944
|
u075595666
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 387
|
There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white. Snuke plotted N points into the rectangle. The coordinate of the i-th (1 ≦ i ≦ N) point was (x_i, y_i). Then, he created an integer sequence a of length N, and for each 1 ≦ i ≦ N, he painted some region within the rectangle black, as follows: * If a_i = 1, he painted the region satisfying x < x_i within the rectangle. * If a_i = 2, he painted the region satisfying x > x_i within the rectangle. * If a_i = 3, he painted the region satisfying y < y_i within the rectangle. * If a_i = 4, he painted the region satisfying y > y_i within the rectangle. Find the area of the white region within the rectangle after he finished painting.
|
W,H,N = map(int,input().split())
grid = []
for i in range(N):
array = list(map(int,input().split()))
grid.append(array)
x1 = 0
xW = W
y1 = 0
yH = H
for i in grid:
if i[2] == 1 and x1 < i[0]:
x1 = i[0]
elif i[2] == 2 and xW > i[0]:
xH = i[0]
elif i[2] == 3 and y1 < i[1]:
y1 == i[1]
elif i[2] == 4 and yH > i[1]:
yH == i[1]
else:
continue
print(max(0,xW-x1)*max(0,yH-y1))
|
s502183491
|
Accepted
| 18
| 3,064
| 367
|
W,H,N = map(int,input().split())
grid = []
for i in range(N):
array = list(map(int,input().split()))
grid.append(array)
x1 = 0
xW = W
y1 = 0
yH = H
for i in grid:
if i[2] == 1 and x1 < i[0]:
x1 = i[0]
elif i[2] == 2 and xW > i[0]:
xW = i[0]
elif i[2] == 3 and y1 < i[1]:
y1 = i[1]
elif i[2] == 4 and yH > i[1]:
yH = i[1]
print(max(0,xW-x1)*max(0,yH-y1))
|
s503799119
|
p03048
|
u430937688
| 2,000
| 1,048,576
|
Wrong Answer
| 1,933
| 3,060
| 222
|
Snuke has come to a store that sells boxes containing balls. The store sells the following three kinds of boxes: * Red boxes, each containing R red balls * Green boxes, each containing G green balls * Blue boxes, each containing B blue balls Snuke wants to get a total of exactly N balls by buying r red boxes, g green boxes and b blue boxes. How many triples of non-negative integers (r,g,b) achieve this?
|
r, g, b, n = map(int, input().split())
cnt = 0
nr = 0
while nr <= n:
m = n - nr * r
ng = 0
while ng <= m:
l = m - ng * g
if not l % b:
cnt += 1
ng += 1
nr += 1
print(cnt)
|
s137680785
|
Accepted
| 1,949
| 2,940
| 230
|
r, g, b, n = map(int, input().split())
cnt = 0
nr = 0
while nr * r <= n:
m = n - nr * r
ng = 0
while ng * g <= m:
l = m - ng * g
if not l % b:
cnt += 1
ng += 1
nr += 1
print(cnt)
|
s820963474
|
p02694
|
u406405116
| 2,000
| 1,048,576
|
Wrong Answer
| 24
| 9,232
| 118
|
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())
y = 100
risi = 0
result = 0
while y <= x:
y *= 1.01
y = y // 1
result += 1
print(int(result))
|
s915490117
|
Accepted
| 24
| 9,232
| 141
|
x = int(input())
y = 100
risi = 0
result = 0
while y <= x:
y *= 1.01
y = y // 1
result += 1
if y >= x:
break
print(int(result))
|
s549930695
|
p03387
|
u090225501
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 557
|
You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order: * Choose two among A, B and C, then increase both by 1. * Choose one among A, B and C, then increase it by 2. It can be proved that we can always make A, B and C all equal by repeatedly performing these operations.
|
def main():
a, b, c = map(int, input().split())
print(solve(a, b, c))
def solve(a, b, c):
a, b, c = sorted([a, b, c])
x = a % 2
y = b % 2
z = c % 2
print(a, b, c)
if x == y == z:
return (c - a) // 2 + (c - b) // 2
if x == y:
return solve(a + 1, b + 1, c) + 1
if x == z:
return solve(a + 1, b, c + 1) + 1
if y == z:
return solve(a, b + 1, c + 1) + 1
def evens(*args):
n = 0
for i in args:
if i % 2 == 0:
n += 1
return n
main()
|
s474940539
|
Accepted
| 17
| 3,064
| 541
|
def main():
a, b, c = map(int, input().split())
print(solve(a, b, c))
def solve(a, b, c):
a, b, c = sorted([a, b, c])
x = a % 2
y = b % 2
z = c % 2
if x == y == z:
return ((c - a) // 2) + ((c - b) // 2)
if x == y:
return solve(a + 1, b + 1, c) + 1
if x == z:
return solve(a + 1, b, c + 1) + 1
if y == z:
return solve(a, b + 1, c + 1) + 1
def evens(*args):
n = 0
for i in args:
if i % 2 == 0:
n += 1
return n
main()
|
s530352519
|
p03457
|
u578850957
| 2,000
| 262,144
|
Wrong Answer
| 340
| 3,060
| 231
|
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
N = int(input())
x_before = 0
y_before = 0
ans = 'YES'
for i in range(N):
t, x, y = map(int, input().split())
abs_distace = abs(x) + abs(y)
if((t-abs_distace)%2!=0):
ans = 'NO'
break
print(ans)
|
s377361789
|
Accepted
| 383
| 3,064
| 608
|
N = int(input())
x_before = 0
y_before = 0
t_before = 0
ans = 'Yes'
for i in range(N):
t, x, y = map(int, input().split())
distance = abs(x-x_before) + abs(y-y_before)
time = t-t_before
if time-distance<0:
ans = 'No'
break
if time%2 != distance%2:
ans = 'No'
break
x_before = x
y_before = y
t_before = t
print(ans)
|
s699233797
|
p03385
|
u668726177
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 67
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
s = input()
if sorted(s)=='abc':
print('Yes')
else:
print('No')
|
s406887348
|
Accepted
| 17
| 2,940
| 76
|
s = input()
if ''.join(sorted(s))=='abc':
print('Yes')
else:
print('No')
|
s594924722
|
p03997
|
u733321071
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 72
|
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a = int(input())
b = int(input())
h = int(input())
print((a+b) * h / 2)
|
s508125316
|
Accepted
| 17
| 2,940
| 70
|
a = int(input())
b = int(input())
h = int(input())
print((a+b)*h // 2)
|
s569278172
|
p03644
|
u699944218
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 73
|
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 = 64
if ans > N:
ans /= 2
print(int(ans))
|
s689744934
|
Accepted
| 18
| 3,060
| 76
|
N = int(input())
ans = 64
while ans > N:
ans /= 2
print(int(ans))
|
s321058913
|
p03370
|
u546440137
| 2,000
| 262,144
|
Wrong Answer
| 28
| 9,156
| 109
|
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts. Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition: * For each of the N kinds of doughnuts, make at least one doughnut of that kind. At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
|
N,X=map(int,input().split())
m=[]
for i in range(N):
m.append(int(input()))
print(X-sum(m)//min(m)+N)
|
s910754675
|
Accepted
| 25
| 9,156
| 111
|
N,X=map(int,input().split())
m=[]
for i in range(N):
m.append(int(input()))
print((X-sum(m))//min(m)+N)
|
s642250946
|
p03485
|
u111365362
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 63
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
#18:50
a,b = map(int,input().split())
print( (a+b+1) // 2 + 1 )
|
s449452087
|
Accepted
| 17
| 2,940
| 63
|
#18:50
a,b = map(int,input().split())
print( (a+b-1) // 2 + 1 )
|
s044497429
|
p02396
|
u098047375
| 1,000
| 131,072
|
Wrong Answer
| 90
| 5,904
| 155
|
In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem.
|
a = []
while True:
n = int(input())
if n == 0:
break
a.append(n)
for i in range(len(a)):
print('Case '+ str(i) + ': ' + str(a[i]))
|
s835294056
|
Accepted
| 80
| 5,908
| 157
|
a = []
while True:
n = int(input())
if n == 0:
break
a.append(n)
for i in range(len(a)):
print('Case '+ str(i+1) + ': ' + str(a[i]))
|
s162188089
|
p03370
|
u469063372
| 2,000
| 262,144
|
Wrong Answer
| 194
| 2,940
| 164
|
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts. Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition: * For each of the N kinds of doughnuts, make at least one doughnut of that kind. At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
|
N, X = map(int, input().split())
ms = []
for i in range(N):
ms.append(int(input()))
count = len(ms)
while X >= 0:
count += 1
X -= min(ms)
print(count-1)
|
s722389849
|
Accepted
| 197
| 3,060
| 177
|
N, X = map(int, input().split())
ms = []
for i in range(N):
ms.append(int(input()))
X -= sum(ms)
count = len(ms)
while X >= 0:
count += 1
X -= min(ms)
print(count-1)
|
s928809453
|
p03574
|
u256464928
| 2,000
| 262,144
|
Wrong Answer
| 23
| 3,572
| 649
|
You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process.
|
h,w = map(int,input().split())
count = 0
X = []
p = []
X.append(["@"]*(w+2))
for _ in range(h):
p = []
x = input()
for i in range(w):
p.append(x[i])
X.append(["@"]+p+["@"])
X.append(["@"]*(w+2))
for i in range(1,h+1):
for j in range(1,w+1):
count=0
if X[i][j]==".":
if X[i-1][j-1]=="#":count+=1
if X[i-1][j]=="#":count+=1
if X[i-1][j+1]=="#":count+=1
if X[i][j-1]=="#":count+=1
if X[i][j+1]=="#":count+=1
if X[i+1][j-1]=="#":count+=1
if X[i+1][j]=="#":count+=1
if X[i+1][j+1]=="#":count+=1
count=str(count)
X[i][j]=count
for k in range(1,h+1):
print(*X[k][1:w+1])
|
s442568823
|
Accepted
| 22
| 3,572
| 656
|
h,w = map(int,input().split())
count = 0
X = []
p = []
X.append(["@"]*(w+2))
for _ in range(h):
p = []
x = input()
for i in range(w):
p.append(x[i])
X.append(["@"]+p+["@"])
X.append(["@"]*(w+2))
for i in range(1,h+1):
for j in range(1,w+1):
count=0
if X[i][j]==".":
if X[i-1][j-1]=="#":count+=1
if X[i-1][j]=="#":count+=1
if X[i-1][j+1]=="#":count+=1
if X[i][j-1]=="#":count+=1
if X[i][j+1]=="#":count+=1
if X[i+1][j-1]=="#":count+=1
if X[i+1][j]=="#":count+=1
if X[i+1][j+1]=="#":count+=1
count=str(count)
X[i][j]=count
for k in range(1,h+1):
print(*X[k][1:w+1],sep="")
|
s142505484
|
p03455
|
u734252743
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,036
| 95
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b =map(int, input().split())
if ( a*b % 2 == 0 ):
print("EVEN")
else:
print("ODD")
|
s311457842
|
Accepted
| 27
| 9,080
| 95
|
a, b =map(int, input().split())
if ( a*b % 2 == 0 ):
print("Even")
else:
print("Odd")
|
s940004397
|
p02408
|
u796301295
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,596
| 223
|
Taro is going to play a card game. However, now he has only n cards, even though there should be 52 cards (he has no Jokers). The 52 cards include 13 ranks of each of the four suits: spade, heart, club and diamond.
|
cards = []
for i in set({'S','H','C','D'}):
for j in range(1, 14):
cards.append(i + " " + str(j))
N = int(input())
for i in range(N):
cards.remove(input())
for i in range(len(cards)):
print (cards[i])
|
s693747853
|
Accepted
| 20
| 5,600
| 159
|
c = [x+' '+y for x in ['S','H','C','D'] for y in [str(i) for i in range(1,14)]]
for i in range(int(input())):
c.remove(input())
for i in c:
print (i)
|
s759569818
|
p02612
|
u804711544
| 2,000
| 1,048,576
|
Wrong Answer
| 32
| 9,140
| 30
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n = int(input())
print(n%1000)
|
s332881402
|
Accepted
| 28
| 9,140
| 76
|
n = int(input())
if n%1000 == 0:
print(0)
else:
print(1000-(n%1000))
|
s983702406
|
p02534
|
u973972117
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 9,016
| 73
|
You are given an integer K. Print the string obtained by repeating the string `ACL` K times and concatenating them. For example, if K = 3, print `ACLACLACL`.
|
K = int(input())
Ans = 'ACL'
for i in range(K-1):
Ans += Ans
print(Ans)
|
s122130696
|
Accepted
| 20
| 9,020
| 77
|
K = int(input())
Ans = 'ACL'
for i in range(K-1):
Ans += 'ACL'
print(Ans)
|
s764726168
|
p03697
|
u319612498
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 61
|
You are given two integers A and B as the input. Output the value of A + B. However, if A + B is 10 or greater, output `error` instead.
|
a,b=map(int,input().split())
print(a+b if a+b<9 else "error")
|
s119807881
|
Accepted
| 17
| 2,940
| 63
|
a,b=map(int,input().split())
print(a+b if a+b<10 else "error")
|
s148347575
|
p03139
|
u278670845
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 59
|
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(min(a,b), abs(a-b))
|
s844465376
|
Accepted
| 17
| 2,940
| 95
|
n,a,b = map(int, input().split())
if a+b >n:
print(min(a,b), a+b-n)
else:
print(min(a,b), 0)
|
s277518597
|
p03139
|
u569742427
| 2,000
| 1,048,576
|
Wrong Answer
| 20
| 3,060
| 299
|
We conducted a survey on newspaper subscriptions. More specifically, we asked each of the N respondents the following two questions: * Question 1: Are you subscribing to Newspaper X? * Question 2: Are you subscribing to Newspaper Y? As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2. What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y? Write a program to answer this question.
|
N,A,B=[int(_) for _ in input().split() ]
if A>=B:
MAX=str(A+B)
MIN=str(B)
if A+B>=N:
print(str(N)+' '+MIN)
exit(0)
print(MAX+" "+MIN)
exit(0)
if B>A:
MAX=str(A+B)
MIN=str(B)
if A+B>=N:
print(str(N)+' '+MIN)
print(MAX+" "+MIN)
exit(0)
|
s126048621
|
Accepted
| 17
| 3,060
| 231
|
N,A,B=[int(_) for _ in input().split() ]
if A>=B:
BIG=A
SMALL=B
if B>A:
BIG=B
SMALL=A
MAX=SMALL
if BIG+SMALL<=N:
MIN=SMALL
print(str(MAX)+' 0')
exit(0)
MIN=BIG+SMALL-N
print(str(MAX)+' '+str(MIN))
|
s555476689
|
p04030
|
u477216059
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 228
|
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?
|
if __name__ == "__main__":
sl = list(input())
print(sl)
text = ""
for s in sl:
if s == "0" or s == "1":
text = text + s
elif s == "B":
text = text[:-1]
print(text)
|
s586927923
|
Accepted
| 17
| 2,940
| 244
|
if __name__ == "__main__":
sl = list(input())
text = ""
for s in sl:
if s == "0" or s == "1":
text = text + s
elif s == "B":
if text != "":
text = text[:-1]
print(text)
|
s075054760
|
p03359
|
u951601135
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 61
|
In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year. For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year. In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For example, 5-5 is Takahashi. How many days from 2018-1-1 through 2018-a-b are Takahashi?
|
a,b=map(int,input().split())
print(a if(a<b) else print(a-1))
|
s441857797
|
Accepted
| 18
| 2,940
| 56
|
a,b=map(int,input().split())
print(a if (a<=b) else a-1)
|
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