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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s918917106 | p03555 | u189479417 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 79 | You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise. | A = input()
B = input()
if A == B[::-1]:
print('Yes')
else:
print('No') | s605811963 | Accepted | 17 | 2,940 | 79 | A = input()
B = input()
if A == B[::-1]:
print('YES')
else:
print('NO') |
s605154916 | p03494 | u439392790 | 2,000 | 262,144 | Wrong Answer | 19 | 2,940 | 221 | There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform. | n=int(input())
a=list(map(int,input().split()))
loop=True
cnt=0
while loop:
for i in range(n):
if a[i]%2==0:
a[i]=a[i]/2
else:
loop=False
break
cnt=cnt+1
print(cnt)
| s891646272 | Accepted | 19 | 2,940 | 223 | n=int(input())
a=list(map(int,input().split()))
loop=True
cnt=0
while loop:
for i in range(n):
if a[i]%2==0:
a[i]=a[i]/2
else:
loop=False
break
cnt=cnt+1
print(cnt-1)
|
s021594971 | p03448 | u267577436 | 2,000 | 262,144 | Wrong Answer | 48 | 3,060 | 257 | You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different. | a = int(input())
b = int(input())
c = int(input())
x = int(input())
rCnt = 0
ttla = 0
ttlb = 0
for ia in range(a):
ttla = 500 * ia
for ib in range(b):
ttlb = 100 * ib
for ic in range(c):
if ttla + ttlb + 50 * ic:
rCnt = rCnt + 1
| s406719977 | Accepted | 49 | 3,064 | 344 | a = int(input())
b = int(input())
c = int(input())
x = int(input())
rCnt = 0
ttla = 0
ttlb = 0
ia = 0
while ia <= a:
ib = 0
ic = 0
ttla = 500 * ia
ia = ia + 1
while ib <= b:
ic = 0
ttlb = 100 * ib
ib = ib + 1
while ic <= c:
if x == ttla + ttlb + 50 * ic:
rCnt = rCnt + 1
ic = ic + 1
print(rCnt) |
s807332472 | p02401 | u142359205 | 1,000 | 131,072 | Wrong Answer | 20 | 5,592 | 203 | Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part. | a, op, b = input().split()
if(op == '+'):
print(int(a) + int(b))
elif(op == '-'):
print(int(a) - int(b))
elif(op == '*'):
print(int(a) * int(b))
elif(op == '/'):
print(int(a) / int(b))
| s193363035 | Accepted | 30 | 5,592 | 275 | while True:
a, op, b = input().split()
if(op == '+'):
print(int(a) + int(b))
elif(op == '-'):
print(int(a) - int(b))
elif(op == '*'):
print(int(a) * int(b))
elif(op == '/'):
print(int(a) // int(b))
else:
break
|
s725941574 | p03494 | u695261159 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 281 | There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform. | import sys
Row=int(input())
List = [int(j) for j in input().split()]
for i in range(len(List)):
ans = List[i] % 2
if ans == 1:
print(0)
sys.exit()
minVal = min(List)
count = 0
while minVal%2 == 0:
count = count + 1
minVal = minVal/2
print(count) | s738607031 | Accepted | 19 | 3,064 | 359 | import sys
Row = input()
List = [int(j) for j in input().split()]
for i in range(len(List)):
ans = List[i] % 2
if ans == 1:
print(0)
sys.exit()
count = 0
flg = 0
while flg == 0:
count = count + 1
for j in range(len(List)):
List[j] = List[j] / 2
if List[j]%2:
flg = 1
break
print(count)
|
s112135728 | p03474 | u222841610 | 2,000 | 262,144 | Wrong Answer | 19 | 2,940 | 208 | The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom. | a,b = map(int,input().split())
s = list(input())
ans = True
for i in range(a+b+1):
if i == a and s[i] != '-':
ans = 'No'
if s[i] == '-' and i != a:
ans ='No'
break
print (ans)
| s290768362 | Accepted | 17 | 2,940 | 301 | a,b = map(int,input().split())
s = list(input())
ans = 'Yes'
if len(s) != a+b+1:
ans = 'No'
else:
for i in range(a+b+1):
if i == a and s[i] != '-':
ans = 'No'
break
if (not s[i].isdigit()) and i != a:
ans ='No'
break
print (ans)
|
s485446600 | p02841 | u583276018 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 105 | In this problem, a date is written as Y-M-D. For example, 2019-11-30 means November 30, 2019. Integers M_1, D_1, M_2, and D_2 will be given as input. It is known that the date 2019-M_2-D_2 follows 2019-M_1-D_1. Determine whether the date 2019-M_1-D_1 is the last day of a month. | a, b = map(int, input().split())
c, d = map(int, input().split())
if(a != b):
print(1)
else:
print(0) | s137057568 | Accepted | 17 | 2,940 | 106 | a, b = map(int, input().split())
c, d = map(int, input().split())
if(a != c):
print(1)
else:
print(0)
|
s483054246 | p03449 | u976469393 | 2,000 | 262,144 | Wrong Answer | 358 | 21,280 | 350 | We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j). You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down. The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them. At most how many candies can you collect when you choose the best way to travel? | # -*- coding: utf-8 -*-
import numpy as np
N=int(input())
A=np.zeros((2,N))
X=np.zeros(N)
A[0]=list(map(int,input().split()))
A[1]=list(map(int,input().split()))
print(A)
for k in range(N):
j=0
drop=0
for i in range(N):
drop+=A[j,i]
if i==k:
j+=1
drop+=A[j,i]
X[k]=drop
print(max(X))
| s480984899 | Accepted | 300 | 20,764 | 346 | # -*- coding: utf-8 -*-
import numpy as np
N=int(input())
A=np.zeros((2,N))
X=np.zeros(N)
A[0]=list(map(int,input().split()))
A[1]=list(map(int,input().split()))
for k in range(N):
j=0
drop=0
for i in range(N):
drop+=A[j,i]
if i==k:
j+=1
drop+=A[j,i]
X[k]=drop
print(int(max(X)))
|
s913768054 | p03815 | u166696759 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 103 | Snuke has decided to play with a six-sided die. Each of its six sides shows an integer 1 through 6, and two numbers on opposite sides always add up to 7. Snuke will first put the die on the table with an arbitrary side facing upward, then repeatedly perform the following operation: * Operation: Rotate the die 90° toward one of the following directions: left, right, front (the die will come closer) and back (the die will go farther). Then, obtain y points where y is the number written in the side facing upward. For example, let us consider the situation where the side showing 1 faces upward, the near side shows 5 and the right side shows 4, as illustrated in the figure. If the die is rotated toward the right as shown in the figure, the side showing 3 will face upward. Besides, the side showing 4 will face upward if the die is rotated toward the left, the side showing 2 will face upward if the die is rotated toward the front, and the side showing 5 will face upward if the die is rotated toward the back. Find the minimum number of operation Snuke needs to perform in order to score at least x points in total. | n = int(input())
count = 0
count += n // 11
count *= 2
n = n % 11
if n:
count += 1
print(count)
| s980204794 | Accepted | 17 | 2,940 | 127 | n = int(input())
count = 0
count += n // 11
count *= 2
n = n % 11
if n>6:
count += 2
elif n:
count +=1
print(count)
|
s143247409 | p03399 | u925353288 | 2,000 | 262,144 | Wrong Answer | 27 | 8,932 | 70 | You planned a trip using trains and buses. The train fare will be A yen (the currency of Japan) if you buy ordinary tickets along the way, and B yen if you buy an unlimited ticket. Similarly, the bus fare will be C yen if you buy ordinary tickets along the way, and D yen if you buy an unlimited ticket. Find the minimum total fare when the optimal choices are made for trains and buses. |
A = input()
B = input()
C = input()
D = input()
print(min(A,B,C,D))
| s001997489 | Accepted | 26 | 9,004 | 98 |
A = int(input())
B = int(input())
C = int(input())
D = int(input())
print(min(A+C,A+D,B+C,B+D))
|
s317637801 | p03945 | u781262926 | 2,000 | 262,144 | Wrong Answer | 205 | 7,924 | 73 | Two foxes Jiro and Saburo are playing a game called _1D Reversi_. This game is played on a board, using black and white stones. On the board, stones are placed in a row, and each player places a new stone to either end of the row. Similarly to the original game of Reversi, when a white stone is placed, all black stones between the new white stone and another white stone, turn into white stones, and vice versa. In the middle of a game, something came up and Saburo has to leave the game. The state of the board at this point is described by a string S. There are |S| (the length of S) stones on the board, and each character in S represents the color of the i-th (1 ≦ i ≦ |S|) stone from the left. If the i-th character in S is `B`, it means that the color of the corresponding stone on the board is black. Similarly, if the i-th character in S is `W`, it means that the color of the corresponding stone is white. Jiro wants all stones on the board to be of the same color. For this purpose, he will place new stones on the board according to the rules. Find the minimum number of new stones that he needs to place. | from itertools import groupby
for g, n in groupby(input()):
print(g, n) | s990553246 | Accepted | 55 | 15,860 | 66 | from itertools import groupby
print(len(list(groupby(input())))-1) |
s506070556 | p03487 | u188827677 | 2,000 | 262,144 | Wrong Answer | 131 | 23,072 | 113 | You are given a sequence of positive integers of length N, a = (a_1, a_2, ..., a_N). Your objective is to remove some of the elements in a so that a will be a **good sequence**. Here, an sequence b is a **good sequence** when the following condition holds true: * For each element x in b, the value x occurs exactly x times in b. For example, (3, 3, 3), (4, 2, 4, 1, 4, 2, 4) and () (an empty sequence) are good sequences, while (3, 3, 3, 3) and (2, 4, 1, 4, 2) are not. Find the minimum number of elements that needs to be removed so that a will be a good sequence. | s = "".join(sorted(input()))
t = "".join(sorted(input(), reverse=True))
if s < t: print("Yes")
else: print("No") | s437831496 | Accepted | 72 | 22,304 | 195 | from collections import Counter
n = int(input())
a = list(map(int, input().split()))
count = Counter(a)
ans = 0
for k,v in count.items():
if k < v: ans += v-k
elif v < k: ans += v
print(ans) |
s525179450 | p03737 | u543954314 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 87 | You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words. | s = input().split()
ans = ""
for i in range(3):
ans += s[i][0]
ans.upper()
print(ans) | s885481149 | Accepted | 18 | 2,940 | 83 | s = input().split()
ans = ""
for i in range(3):
ans += s[i][0]
print(ans.upper()) |
s807809097 | p03957 | u300538442 | 1,000 | 262,144 | Wrong Answer | 24 | 3,316 | 109 | This contest is `CODEFESTIVAL`, which can be shortened to the string `CF` by deleting some characters. Mr. Takahashi, full of curiosity, wondered if he could obtain `CF` from other strings in the same way. You are given a string s consisting of uppercase English letters. Determine whether the string `CF` can be obtained from the string s by deleting some characters. | import re
S1 = input()
p = r'C*F'
result = re.match(p, S1)
if result:
print('yes')
else:
print('No')
| s180616303 | Accepted | 30 | 3,316 | 114 | import re
S1 = input()
p = r'C.+F|CF'
result = re.search(p, S1)
if result:
print('Yes')
else:
print('No')
|
s052310994 | p00002 | u002010345 | 1,000 | 131,072 | Wrong Answer | 20 | 5,536 | 170 | Write a program which computes the digit number of sum of two integers a and b. | def main():
while True:
try:
a, b = (int(x) for x in input().split())
print(len(str(a+b)))
except EOFError:
break
| s615163678 | Accepted | 20 | 5,600 | 177 | def main():
while True:
try:
a, b = (int(x) for x in input().split())
print(len(str(a+b)))
except EOFError:
break
main()
|
s335832244 | p03455 | u952396514 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 128 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | import math
num = int(input().replace(" ", ""))
squreRoot = math.sqrt(num)
print('Yes' if squreRoot == int(squreRoot) else 'No') | s951725124 | Accepted | 18 | 2,940 | 77 | a, b = map(int, input().split())
print('Odd' if (a * b) % 2 == 1 else 'Even') |
s345879025 | p02601 | u923662841 | 2,000 | 1,048,576 | Wrong Answer | 30 | 9,200 | 379 | M-kun has the following three cards: * A red card with the integer A. * A green card with the integer B. * A blue card with the integer C. He is a genius magician who can do the following operation at most K times: * Choose one of the three cards and multiply the written integer by 2. His magic is successful if both of the following conditions are satisfied after the operations: * The integer on the green card is **strictly** greater than the integer on the red card. * The integer on the blue card is **strictly** greater than the integer on the green card. Determine whether the magic can be successful. | A,B,C = map(int, input().split())
K = int(input())
d = 0
for i in range(K):
for j in range(K - i):
for k in range(K - i - j+1):
a = A*2**i
b = B*2**j
c = C*2**k
print("a:",a, ",b:",b, ",c:",c)
if a<b and b<c:
d = 1
break
if d ==0:
print("No") | s168430136 | Accepted | 28 | 9,200 | 452 | A,B,C = map(int, input().split())
K = int(input())
def F(d):
for i in range(K):
for j in range(K - i):
for k in range(K - i - j+1):
a = A*2**i
b = B*2**j
c = C*2**k
#print("a:",a, ",b:",b, ",c:",c)
if a<b and b<c:
d = 1
return "Yes"
if d ==0:
return "No"
print(F(0)) |
s259624726 | p04043 | u799691369 | 2,000 | 262,144 | Wrong Answer | 16 | 2,940 | 228 | 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. | import sys
args = sys.argv
count_5 = 0
count_7 = 0
for i in args:
print(i)
if i == '5':
count_5+=1
if i == '7':
count_7+=1
if count_5 == 2 and count_7 == 1:
print('YES')
else:
print('NO')
| s550922753 | Accepted | 50 | 3,060 | 288 | A, B, C = map(int, input().split())
l = [5, 5, 7]
def main(A=A, B=B, C=C):
if A in l:
l.remove(A)
if B in l:
l.remove(B)
if C in l:
return 'YES'
return 'NO'
if __name__ == '__main__':
result = main()
print(result) |
s563646702 | p02842 | u148423304 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 95 | Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact. | n = int(input())
x = int(n / 1.08)
n2 = x * 1.08
if x == n2:
print(x)
else:
print(":(") | s787322566 | Accepted | 17 | 3,060 | 174 | n = int(input())
x1 = int(n / 1.08)
x2 = int(n / 1.08) + 1
n1 = int(x1 * 1.08)
n2 = int(x2 * 1.08)
if n == n1:
print(x1)
elif n == n2:
print(x2)
else:
print(":(") |
s028713750 | p03455 | u311919427 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 103 | 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())
ans = a * b
if ans % 2 == 0:
print("Odd")
else:
print("Even") | s811985026 | Accepted | 18 | 2,940 | 103 | a, b = map(int, input().split())
ans = a * b
if ans % 2 == 0:
print("Even")
else:
print("Odd") |
s400529849 | p03860 | u365254117 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 81 | Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name. | a, s, c=input().split()
s.lower()
s[0].upper()
print("{} {} {}".format(a, s, c))
| s054618873 | Accepted | 17 | 2,940 | 54 | a, s, c=input().split()
print(a[0]+s[0].upper()+c[0])
|
s527980604 | p02607 | u036340997 | 2,000 | 1,048,576 | Wrong Answer | 25 | 9,120 | 127 | We have N squares assigned the numbers 1,2,3,\ldots,N. Each square has an integer written on it, and the integer written on Square i is a_i. How many squares i satisfy both of the following conditions? * The assigned number, i, is odd. * The written integer is odd. | n = int(input())
a = list(map(int, input().split()))
ans = 0
for i in range(0,n,2):
if a[i] % 2 == 0:
ans += 1
print(ans) | s904551458 | Accepted | 24 | 9,096 | 128 | n = int(input())
a = list(map(int, input().split()))
ans = 0
for i in range(0,n,2):
if a[i] % 2 == 1:
ans += 1
print(ans)
|
s464190058 | p03388 | u319818856 | 2,000 | 262,144 | Wrong Answer | 19 | 3,064 | 1,251 | 10^{10^{10}} participants, including Takahashi, competed in two programming contests. In each contest, all participants had distinct ranks from first through 10^{10^{10}}-th. The _score_ of a participant is the product of his/her ranks in the two contests. Process the following Q queries: * In the i-th query, you are given two positive integers A_i and B_i. Assuming that Takahashi was ranked A_i-th in the first contest and B_i-th in the second contest, find the maximum possible number of participants whose scores are smaller than Takahashi's. | def upper_square(A: int, B: int) -> int:
l, r = min(A, B), max(A, B)
while r - l > 1:
m = (l + r) // 2
if m * m < A * B:
l = m
else:
r = m
return l
def query(A: int, B: int) -> int:
A, B = min(A, B), max(A, B)
if A == B:
# (1, 2A-1), ..., (A-1, A+1), (A+1, A-1), ..., (2A-1, 1)
return 2 * A - 2
if B == A + 1:
# (1, 2A), ..., (A-1, A+2), (A+2, A-1), ..., (2A, 1)
return 2 * A - 2
# C = A
# while (C + 1) ** 2 < A * B:
# C += 1
C = upper_square(A, B)
if A * B <= C * (C + 1):
return 2 * C - 2
return 2*C - 1
def worst_case(Q: int, queries: list) -> list:
return [query(A, B) for A, B in queries]
if __name__ == "__main__":
Q = int(input())
queries = [tuple(map(int, input().split())) for _ in range(Q)]
ans = worst_case(Q, queries)
print(ans)
| s464631561 | Accepted | 18 | 3,064 | 1,271 | def upper_square(A: int, B: int) -> int:
l, r = min(A, B), max(A, B)
while r - l > 1:
m = (l + r) // 2
if m * m < A * B:
l = m
else:
r = m
return l
def query(A: int, B: int) -> int:
A, B = min(A, B), max(A, B)
if A == B:
# (1, 2A-1), ..., (A-1, A+1), (A+1, A-1), ..., (2A-1, 1)
return 2 * A - 2
if B == A + 1:
# (1, 2A), ..., (A-1, A+2), (A+2, A-1), ..., (2A, 1)
return 2 * A - 2
# C = A
# while (C + 1) ** 2 < A * B:
# C += 1
C = upper_square(A, B)
if A * B <= C * (C + 1):
return 2 * C - 2
return 2*C - 1
def worst_case(Q: int, queries: list) -> list:
return [query(A, B) for A, B in queries]
if __name__ == "__main__":
Q = int(input())
queries = [tuple(map(int, input().split())) for _ in range(Q)]
ans = worst_case(Q, queries)
for a in ans:
print(a)
|
s222568233 | p03644 | u327248573 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 89 | 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())
power = [2,4,8,16,32,64]
print(max([i if i <= N else N for i in power])) | s600221674 | Accepted | 17 | 3,060 | 107 | N = int(input())
power = [2,4,8,16,32,64]
if N < 2: print(N)
else: print(max([i for i in power if i <= N])) |
s706600116 | p03795 | u642012866 | 2,000 | 262,144 | Wrong Answer | 26 | 8,980 | 39 | Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y. | N = int(input())
print(N*800-15//N*400) | s971480112 | Accepted | 31 | 9,064 | 39 | N = int(input())
print(N*800-N//15*200) |
s543946122 | p03478 | u505420467 | 2,000 | 262,144 | Wrong Answer | 38 | 2,940 | 146 | Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive). | n,a,b=map(int,input().split())
ans=0
for i in range(n):
tmp=sum([int(j) for j in list(str(i+1))])
if a<=tmp<=b:
ans+=1
print(ans)
| s485071691 | Accepted | 38 | 2,940 | 148 | n,a,b=map(int,input().split())
ans=0
for i in range(n):
tmp=sum([int(j) for j in list(str(i+1))])
if a<=tmp<=b:
ans+=i+1
print(ans)
|
s039650162 | p03433 | u835353642 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 92 | E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins. | N = int(input())
A = int(input())
if (N%500) <= A:
print ("YES")
else:
print ("NO") | s556073210 | Accepted | 17 | 2,940 | 92 | N = int(input())
A = int(input())
if (N%500) <= A:
print ("Yes")
else:
print ("No") |
s080698528 | p02399 | u566311709 | 1,000 | 131,072 | Wrong Answer | 20 | 5,596 | 77 | Write a program which reads two integers a and b, and calculates the following values: * a ÷ b: d (in integer) * remainder of a ÷ b: r (in integer) * a ÷ b: f (in real number) | a, b = map(int, input().split())
print(a // b, a % b, "{:.5}".format(a / b))
| s702652533 | Accepted | 20 | 5,600 | 79 | a, b = map(int, input().split())
print(a // b, a % b, "{0:.5f}".format(a / b))
|
s346627555 | p03023 | u095756391 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 155 | Given an integer N not less than 3, find the sum of the interior angles of a regular polygon with N sides. Print the answer in degrees, but do not print units. | s = input()
win = 0
for i in s:
if i == 'o':
win += 1
if (15 - len(s)) + win >= 8:
print('YES')
else:
print('NO') | s012967169 | Accepted | 17 | 2,940 | 35 | n = int(input())
print(180*(n-2))
|
s993715459 | p02795 | u341267151 | 2,000 | 1,048,576 | Wrong Answer | 29 | 9,204 | 329 | We have a grid with H rows and W columns, where all the squares are initially white. You will perform some number of painting operations on the grid. In one operation, you can do one of the following two actions: * Choose one row, then paint all the squares in that row black. * Choose one column, then paint all the squares in that column black. At least how many operations do you need in order to have N or more black squares in the grid? It is guaranteed that, under the conditions in Constraints, having N or more black squares is always possible by performing some number of operations. | n=int(input())
a=list(map(int,input().split()))
t=sum(a)
m=10**9
flag=0
for i in range(len(a)-1):
t-=2*a[i]
if t<=0 :
if i==0:
print(abs(t))
flag = 1
break
else:
print(min(abs(t+2*a[i]),abs(t)))
flag=1
break
if flag==0:
print(t) | s312236759 | Accepted | 27 | 9,032 | 87 | import math
h=int(input())
w=int(input())
n=int(input())
print(math.ceil(n/max(h,w)))
|
s048459499 | p02401 | u825994660 | 1,000 | 131,072 | Wrong Answer | 20 | 7,392 | 354 | Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part. | import sys
while True:
x = sys.stdin.readline().strip()
if "?" in x:
break
else:
a = list(x.split())
if a[1] == "+":
print(a[0] + a[2])
elif a[1] == "-":
print(a[0] - a[2])
elif a[1] == "*":
print(a[0] * a[2])
elif a[1] == "/":
print(a[0] / a[2]) | s776942566 | Accepted | 30 | 7,768 | 397 | import sys
import math
while True:
x = sys.stdin.readline().strip()
a = list(x.split())
if a[1] == "?":
break
else:
x = int(a[0])
y = int(a[2])
if a[1] == "+":
print(x + y)
elif a[1] == "-":
print(x - y)
elif a[1] == "*":
print(x * y)
elif a[1] == "/":
print(math.floor(x / y)) |
s298491484 | p03486 | u233720439 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 156 | You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order. | S = str(input())
T = str(input())
sl = [c for c in S]
tl = [c for c in T]
sl.sort()
tl.sort()
print("YES" if sl[0] < [c for c in reversed(tl)][0] else "NO") | s550685905 | Accepted | 17 | 2,940 | 84 | S = sorted(input())
N = sorted(input())
N.reverse()
print("Yes" if S < N else "No") |
s908421847 | p02259 | u918276501 | 1,000 | 131,072 | Wrong Answer | 20 | 7,640 | 228 | Write a program of the Bubble Sort algorithm which sorts a sequence _A_ in ascending order. The algorithm should be based on the following pseudocode: BubbleSort(A) 1 for i = 0 to A.length-1 2 for j = A.length-1 downto i+1 3 if A[j] < A[j-1] 4 swap A[j] and A[j-1] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 4 of the pseudocode. | cnt = 0
n = int(input())
lst = list(map(int, input().split()))
for i in range(n):
for j in range(n-1,i,-1):
if lst[j] < lst[j-1]:
lst[j-1:j+1] = lst[j], lst[j-1]
cnt += 1
print(lst)
print(cnt) | s238941092 | Accepted | 30 | 7,696 | 229 | cnt = 0
n = int(input())
lst = list(map(int, input().split()))
for i in range(n):
for j in range(n-1,i,-1):
if lst[j] < lst[j-1]:
lst[j-1:j+1] = lst[j], lst[j-1]
cnt += 1
print(*lst)
print(cnt) |
s131705125 | p03731 | u906017074 | 2,000 | 262,144 | Wrong Answer | 164 | 25,200 | 249 | In a public bath, there is a shower which emits water for T seconds when the switch is pushed. If the switch is pushed when the shower is already emitting water, from that moment it will be emitting water for T seconds. Note that it does not mean that the shower emits water for T additional seconds. N people will push the switch while passing by the shower. The i-th person will push the switch t_i seconds after the first person pushes it. How long will the shower emit water in total? | n, T = [int(i) for i in input().split()]
t = [int(i) for i in input().split()]
print(n, T, t)
sum = 0
for i in range(1,n):
interval = t[i] - t[i-1]
if interval < T:
sum += interval
else:
sum += T
sum += T
print(sum)
| s822666149 | Accepted | 139 | 25,200 | 251 | n, T = [int(i) for i in input().split()]
t = [int(i) for i in input().split()]
# print(n, T, t)
sum = 0
for i in range(1,n):
interval = t[i] - t[i-1]
if interval < T:
sum += interval
else:
sum += T
sum += T
print(sum)
|
s156963827 | p02396 | u260980560 | 1,000 | 131,072 | Wrong Answer | 30 | 6,776 | 126 | 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. | ans = []
for i, x in enumerate(open(0).readlines()):
ans.append("Case %d: %s" % (i, x))
open(1,'w').writelines(ans[:-1])
| s439941146 | Accepted | 30 | 6,768 | 128 | ans = []
for i, x in enumerate(open(0).readlines()):
ans.append("Case %d: %s" % (i+1, x))
open(1,'w').writelines(ans[:-1])
|
s868194663 | p04029 | u339503988 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 26 | There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total? | sum(range(int(input())+1)) | s627713823 | Accepted | 17 | 2,940 | 33 | print(sum(range(int(input())+1))) |
s468196192 | p02613 | u507456172 | 2,000 | 1,048,576 | Wrong Answer | 146 | 9,212 | 337 | Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format. | N = int(input())
result=[0,0,0,0]
for i in range(N):
S=input()
if S == "AC":
result[0] += 1
elif S == "WA":
result[1] += 1
elif S == "TLE":
result[2] += 1
else:
result[3] += 1
print("AC × " + str(result[0]))
print("WA × " + str(result[1]))
print("TLE × " + str(result[2]))
print("RE × " + str(result[3])) | s270317249 | Accepted | 149 | 9,156 | 333 | N = int(input())
result=[0,0,0,0]
for i in range(N):
S=input()
if S == "AC":
result[0] += 1
elif S == "WA":
result[1] += 1
elif S == "TLE":
result[2] += 1
else:
result[3] += 1
print("AC x " + str(result[0]))
print("WA x " + str(result[1]))
print("TLE x " + str(result[2]))
print("RE x " + str(result[3])) |
s335534277 | p02646 | u485172913 | 2,000 | 1,048,576 | Wrong Answer | 22 | 9,108 | 132 | 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())
s=int(input())
if(((b*w)-(a*v))>(a*v)):
print("Yes")
else:
print("No") | s554082070 | Accepted | 21 | 9,108 | 129 | a,v=map(int,input().split())
b,w=map(int,input().split())
s=int(input())
if(abs(a-b)>(v-w)*s):
print("NO")
else:
print("YES") |
s637809448 | p02972 | u747602774 | 2,000 | 1,048,576 | Wrong Answer | 296 | 8,784 | 269 | 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(map(int,input().split()))
b = [0 for i in range(N//2)]+a[N//2:]
for i in range(N//2-1,-1,-1):
if sum(b[i::i+1])%2 == 0:
b[i] = a[i]
else:
b[i] = 1-a[i]
for i in range(N):
if b[i] == 1:
print(i+1,end=' ')
| s461821223 | Accepted | 189 | 20,276 | 322 | N = int(input())
a = list(map(int,input().split()))
b = [0 for i in range(N//2)]+a[N//2:]
for i in range(N//2-1,-1,-1):
if sum(b[i::i+1])%2 == 0:
b[i] = a[i]
else:
b[i] = 1-a[i]
ans = []
for i in range(N):
if b[i] == 1:
ans.append(i+1)
print(len(ans))
print(' '.join(map(str,ans)))
|
s735439401 | p03943 | u218984487 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 118 | Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students. | a, b, c = map(int, input().split())
if a + b == c or a == c + b or b == a + c:
print("YES")
else:
print("NO")
| s613480505 | Accepted | 18 | 2,940 | 118 | a, b, c = map(int, input().split())
if a + b == c or a == c + b or b == a + c:
print("Yes")
else:
print("No")
|
s610605224 | p00008 | u362104929 | 1,000 | 131,072 | Wrong Answer | 190 | 7,644 | 569 | Write a program which reads an integer n and identifies the number of combinations of a, b, c and d (0 ≤ a, b, c, d ≤ 9) which meet the following equality: a + b + c + d = n For example, for n = 35, we have 4 different combinations of (a, b, c, d): (8, 9, 9, 9), (9, 8, 9, 9), (9, 9, 8, 9), and (9, 9, 9, 8). | for _ in range(50):
try:
n = int(input()) # 1 <= n <= 50
ans =set()
for a in range(10):
if (n - a) >= 1:
for b in range(10):
if(n - a -b) >= 1:
for c in range(10):
if(n - a - b - c) >= 1:
for d in range(10):
if(n - a - b - c - d) == 0:
ans.add("{},{},{},{}".format(a,b,c,d))
print(len(ans))
except EOFError:
break | s768894653 | Accepted | 220 | 7,656 | 806 | for _ in range(50):
try:
n = int(input()) # 1 <= n <= 50
ans =set()
for a in range(10):
if (n - a) >= 1:
for b in range(10):
if(n - a -b) >= 1:
for c in range(10):
if(n - a - b - c) >= 1:
for d in range(10):
if(n - a - b - c - d) == 0:
ans.add("{},{},{},{}".format(a,b,c,d))
ans.add("{},{},{},{}".format(b,c,d,a))
ans.add("{},{},{},{}".format(c,d,a,b))
ans.add("{},{},{},{}".format(d,a,b,c))
print(len(ans))
except EOFError:
break |
s496215524 | p03543 | u484856305 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 111 | We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**? | a=input()
if a[0]==a[1]==a[3] or a[1]==a[2]==a[3] or a[0]==a[1]==a[2]==a[3]:
print("Yes")
else:
print("No") | s694013485 | Accepted | 17 | 2,940 | 73 | a,b,c,d=input()
if a==b==c or b==c==d:
print("Yes")
else:
print("No") |
s310119032 | p03992 | u751047721 | 2,000 | 262,144 | Wrong Answer | 22 | 3,068 | 11 | This contest is `CODE FESTIVAL`. However, Mr. Takahashi always writes it `CODEFESTIVAL`, omitting the single space between `CODE` and `FESTIVAL`. So he has decided to make a program that puts the single space he omitted. You are given a string s with 12 letters. Output the string putting a single space between the first 4 letters and last 8 letters in the string s. | print("No") | s601001725 | Accepted | 22 | 3,192 | 32 | i=input()
print(i[:4]+" "+i[4:]) |
s191856725 | p02402 | u767850279 | 1,000 | 131,072 | Wrong Answer | 20 | 7,620 | 160 | Write a program which reads a sequence of $n$ integers $a_i (i = 1, 2, ... n)$, and prints the minimum value, maximum value and sum of the sequence. | # coding : utf-8
a = int(input())
n1 = input().split()
n1 = [float(i) for i in n1]
print(min(n1), max(n1), sum(n1)) | s467819058 | Accepted | 20 | 8,600 | 158 | # coding : utf-8
a = int(input())
n1 = input().split()
n1 = [int(i) for i in n1]
print(min(n1), max(n1), sum(n1)) |
s141949647 | p02261 | u301461168 | 1,000 | 131,072 | Wrong Answer | 30 | 7,800 | 1,358 | 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 is_stable(output_str, same_num_str):
if len(same_num_str)>0:
for i in range(len(same_num_str)):
if output_str.find(same_num_str[i]) == -1:
return "Not Stable"
return "Stable"
data_num = int(input())
cards = input().split(" ")
cards2 = cards[:]
cnt = 0
num_1 = 0
num_2 = 0
same_num_str = []
output_str = ""
judge_stbl = ""
for i in range(data_num):
for j in range(data_num-1, i,-1):
if int(cards[j][1:]) < int(cards[j-1][1:]):
cards[j], cards[j-1] = cards[j-1], cards[j]
elif int(cards[j][1:]) == int(cards[j-1][1:]):
same_num_str.append(cards[j-1]+" "+cards[j])
output_str = " ".join(cards)
judge_stbl = is_stable(output_str, same_num_str)
print(output_str + "\n" + judge_stbl)
same_num_str = []
output_str = ""
judge_stbl = ""
for i in range(data_num):
min_idx = i
for j in range(i, data_num):
if int(cards2[j][1:]) < int(cards2[min_idx][1:]):
min_idx = j
elif int(cards2[j][1:]) == int(cards2[min_idx][1:]) and j != min_idx:
same_num_str.append(cards2[min_idx]+" "+cards2[j])
if i != min_idx:
cards2[i], cards2[min_idx] = cards2[min_idx], cards2[i]
output_str = " ".join(cards2)
judge_stbl = is_stable(output_str, same_num_str)
print(output_str + "\n" + judge_stbl) | s440762062 | Accepted | 20 | 7,764 | 1,358 | def is_stable(output_str, same_num_str):
if len(same_num_str)>0:
for i in range(len(same_num_str)):
if output_str.find(same_num_str[i]) == -1:
return "Not stable"
return "Stable"
data_num = int(input())
cards = input().split(" ")
cards2 = cards[:]
cnt = 0
num_1 = 0
num_2 = 0
same_num_str = []
output_str = ""
judge_stbl = ""
for i in range(data_num):
for j in range(data_num-1, i,-1):
if int(cards[j][1:]) < int(cards[j-1][1:]):
cards[j], cards[j-1] = cards[j-1], cards[j]
elif int(cards[j][1:]) == int(cards[j-1][1:]):
same_num_str.append(cards[j-1]+" "+cards[j])
output_str = " ".join(cards)
judge_stbl = is_stable(output_str, same_num_str)
print(output_str + "\n" + judge_stbl)
same_num_str = []
output_str = ""
judge_stbl = ""
for i in range(data_num):
min_idx = i
for j in range(i, data_num):
if int(cards2[j][1:]) < int(cards2[min_idx][1:]):
min_idx = j
elif int(cards2[j][1:]) == int(cards2[min_idx][1:]) and j != min_idx:
same_num_str.append(cards2[min_idx]+" "+cards2[j])
if i != min_idx:
cards2[i], cards2[min_idx] = cards2[min_idx], cards2[i]
output_str = " ".join(cards2)
judge_stbl = is_stable(output_str, same_num_str)
print(output_str + "\n" + judge_stbl) |
s733795553 | p03645 | u970267139 | 2,000 | 262,144 | Wrong Answer | 575 | 67,668 | 629 | In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands. For convenience, we will call them Island 1, Island 2, ..., Island N. There are M kinds of regular boat services between these islands. Each service connects two islands. The i-th service connects Island a_i and Island b_i. Cat Snuke is on Island 1 now, and wants to go to Island N. However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services. Help him. | from collections import deque
n, m = map(int, input().split())
p = [tuple(map(int, input().split())) for _ in range(m)]
d = dict()
for a, b in p:
if a not in d.keys():
d[a] = [b]
else:
d[a].append(b)
if b not in d.keys():
d[b] = []
node = deque()
node.append((1, 0))
visit = set()
while node:
(s, count) = node.popleft()
if s in visit:
continue
if s == n:
print(count)
if count == 2:
print('POSSIBLE')
exit()
else:
break
for i in d[s]:
node.append((i, count+1))
visit.add(s)
print('IMPOSSIBLE')
| s880523670 | Accepted | 607 | 67,784 | 608 | from collections import deque
n, m = map(int, input().split())
p = [tuple(map(int, input().split())) for _ in range(m)]
d = dict()
for a, b in p:
if a not in d.keys():
d[a] = [b]
else:
d[a].append(b)
if b not in d.keys():
d[b] = []
node = deque()
node.append((1, 0))
visit = set()
while node:
(s, count) = node.popleft()
if s in visit:
continue
if s == n:
if count == 2:
print('POSSIBLE')
exit()
else:
break
for i in d[s]:
node.append((i, count+1))
visit.add(s)
print('IMPOSSIBLE')
|
s562678585 | p03044 | u711539583 | 2,000 | 1,048,576 | Wrong Answer | 768 | 42,668 | 456 | We have a tree with N vertices numbered 1 to N. The i-th edge in the tree connects Vertex u_i and Vertex v_i, and its length is w_i. Your objective is to paint each vertex in the tree white or black (it is fine to paint all vertices the same color) so that the following condition is satisfied: * For any two vertices painted in the same color, the distance between them is an even number. Find a coloring of the vertices that satisfies the condition and print it. It can be proved that at least one such coloring exists under the constraints of this problem. | n = int(input())
d = {}
for i in range(n - 1):
u, v, w = map(int, input().split())
if u in d:
d[u].append([v, w])
else:
d[u] = [[v, w]]
if v in d:
d[v] = [[u, w]]
else:
d[v] = [[u, w]]
start = list(d.keys())[0]
ans = [0 for _ in range(n)]
def walk(poj):
ans[poj - 1] = 1
for i, way in enumerate(d[poj]):
if way[1] % 2 == 0:
d[poj][i][1] = 1
walk(way[0])
walk(start)
for res in ans:
print(res) | s372592465 | Accepted | 915 | 99,740 | 562 | import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline
n = int(input())
d = {}
for i in range(n - 1):
u, v, w = map(int, input().split())
u -= 1
v -= 1
if u in d:
d[u].append([v, w])
else:
d[u] = [[v, w]]
if v in d:
d[v].append([u, w])
else:
d[v] = [[u, w]]
rem = {}
for i in range(n):
rem[i] = 1
ans = [0 for _ in range(n)]
def check(p, tmp):
rem.pop(p)
for n, w in d[p]:
if n in rem:
ntmp = (tmp + w) % 2
ans[n] = ntmp
check(n, ntmp)
check(0, 0)
for res in ans:
print(res)
|
s538297769 | p03229 | u607865971 | 2,000 | 1,048,576 | Wrong Answer | 517 | 9,492 | 589 | You are given N integers; the i-th of them is A_i. Find the maximum possible sum of the absolute differences between the adjacent elements after arranging these integers in a row in any order you like. | import collections
N = int(input())
A = [0] * N
for i in range(N):
A[i] = int(input())
A = sorted(A)
if N % 2 == 1:
d1 = A[0:N // 2 + 1]
u1 = A[N // 2 + 1:N + 1]
else:
d1 = A[0:N // 2]
u1 = A[N // 2:N + 1]
d1 = collections.deque(d1)
u1 = collections.deque(u1)
d1.rotate(1)
u1.rotate(1)
sum1 = 0
i = 0
while (True):
print("{},{},{}".format(i, len(d1), len(u1)))
if i >= len(d1) or i >= len(u1):
break
sum1 += abs(u1[i] - d1[i])
if i+1 >= len(d1) or i >= len(u1):
break
sum1 += abs(u1[i] - d1[i + 1])
i += 1
print(sum1)
| s767441878 | Accepted | 1,056 | 10,516 | 1,575 | import collections
N = int(input())
A = [0] * N
for i in range(N):
A[i] = int(input())
A = sorted(A)
if N % 2 == 1:
d1 = A[0:N // 2 + 1]
u1 = A[N // 2 + 1:N + 1]
else:
d1 = A[0:N // 2]
u1 = A[N // 2:N + 1]
d1 = collections.deque(d1)
u1 = collections.deque(u1)
d1.rotate(1)
u1.rotate(-1)
sum1 = 0
i = 0
while (True):
if i >= len(d1) or i >= len(u1):
break
sum1 += abs(u1[i] - d1[i])
if i+1 >= len(d1) or i >= len(u1):
break
sum1 += abs(u1[i] - d1[i + 1])
i += 1
sum2 = 0
i = 0
while (True):
if i >= len(d1) or i >= len(u1):
break
sum2 += abs(u1[i] - d1[i])
if i >= len(d1) or i+1 >= len(u1):
break
sum2 += abs(u1[i+1] - d1[i])
i += 1
###########################
if N % 2 == 1:
d1 = A[0:N // 2]
u1 = A[N // 2:N + 1]
d1 = collections.deque(d1)
u1 = collections.deque(u1)
d1.rotate(1)
u1.rotate(-1)
sum3 = 0
i = 0
while (True):
if i >= len(d1) or i >= len(u1):
break
sum3 += abs(u1[i] - d1[i])
if i+1 >= len(d1) or i >= len(u1):
break
sum3 += abs(u1[i] - d1[i + 1])
i += 1
sum4 = 0
i = 0
while (True):
if i >= len(d1) or i >= len(u1):
break
sum4 += abs(u1[i] - d1[i])
if i >= len(d1) or i+1 >= len(u1):
break
sum4 += abs(u1[i+1] - d1[i])
i += 1
print(max(sum1, sum2, sum3, sum4))
|
s739635435 | p02388 | u885889402 | 1,000 | 131,072 | Wrong Answer | 30 | 7,340 | 28 | Write a program which calculates the cube of a given integer x. | s=input()
print(float(s)**3) | s946007192 | Accepted | 20 | 7,684 | 26 | s=input()
print(int(s)**3) |
s257680098 | p03711 | u472721500 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 190 | Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group. | a, b = map(int, input().split())
group1 = [1,3,5,7,8,10,12]
group2 = [4,6,9,11]
if a in group1 and b in group1: print("YES")
elif a in group2 and b in group2: print("YES")
else: print("NO") | s497283900 | Accepted | 18 | 3,060 | 172 | a, b = map(int, input().split())
x = [1,3,5,7,8,10,12]
y = [4,6,9,11]
if a in x and b in x:
print("Yes")
elif a in y and b in y:
print("Yes")
else:
print("No") |
s837544131 | p03150 | u328131364 | 2,000 | 1,048,576 | Wrong Answer | 98 | 3,700 | 347 | A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string. | import copy
S = input()
ja = 0
sremove = ""
S_tmp = ""
for i in range(len(S)):
for j in range(len(S)- i):
S_tmp = S
sremove = S_tmp[i:]
sremove = sremove[j:]
S_tmp.replace(sremove, "")
if S_tmp == "keyence":
ja = 1
break
if ja == 0:
print("NO")
else:
print("YES") | s580905779 | Accepted | 319 | 3,444 | 307 | import copy
S = list(input())
ja = 0
sremove = ""
S_tmp = []
for i in range(len(S)):
for j in range(i,len(S)+1):
S_tmp = copy.deepcopy(S)
del S_tmp[i:j]
if "".join(S_tmp) == "keyence":
ja = 1
break
if ja == 1:
print("YES")
else:
print("NO") |
s000732947 | p03693 | u379424722 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 115 | AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4? | n = input().split(" ")
n = int(n[0] + n[1] + n[2])
print(n)
if n % 4 == 0:
print("YES")
else:
print("NO") | s042497699 | Accepted | 17 | 3,064 | 106 | n = input().split(" ")
n = int(n[0] + n[1] + n[2])
if n % 4 == 0:
print("YES")
else:
print("NO") |
s428298580 | p03658 | u740284863 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 92 | Snuke has N sticks. The length of the i-th stick is l_i. Snuke is making a snake toy by joining K of the sticks together. The length of the toy is represented by the sum of the individual sticks that compose it. Find the maximum possible length of the toy. | n,k = map(int,input().split())
x = sorted(list(map(int,input().split())))
print(sum(x[k:]))
| s763298246 | Accepted | 18 | 2,940 | 106 | n,k = map(int,input().split())
x = sorted(list(map(int,input().split())),reverse=True)
print(sum(x[:k]))
|
s707991421 | p02406 | u316584871 | 1,000 | 131,072 | Wrong Answer | 20 | 5,588 | 142 | In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; } | i = int(input())
for n in range(1,i+1):
if (n%3==0):
print(' {}'.format(n))
elif(n%10 == 3):
print(' {}'.format(n))
| s344236924 | Accepted | 20 | 5,876 | 319 | i = int(input())
for n in range(1,i+1):
if (n%3==0):
print(' {}'.format(n),end="")
else:
x = n
while(x>0):
if (x%10 == 3):
print(' {}'.format(n),end="")
break
else:
x = x//10
print()
|
s603253166 | p02612 | u275242455 | 2,000 | 1,048,576 | Wrong Answer | 32 | 9,140 | 43 | 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())
ans = N % 1000
print(ans)
| s485474301 | Accepted | 32 | 9,184 | 76 | N = int(input())
ans = N % 1000
if ans != 0:
ans = 1000 - ans
print(ans) |
s571534474 | p03720 | u445380615 | 2,000 | 262,144 | Wrong Answer | 20 | 2,940 | 259 | There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city? | wlist = []
city, road = map(int, input().split())
for i in range(road):
wlist.append(input())
dst = []
for i in range(city):
count = 0
for k in wlist:
if str(i) in k:
count += 1
dst.append(count)
for d in dst:
print(d)
| s991629809 | Accepted | 18 | 2,940 | 280 | wlist = []
city, road = map(int, input().split())
for i in range(road):
wlist.append(list(map(int, input().split())))
dst = []
for i in range(city):
count = 0
for k in wlist:
if i+1 in k:
count += 1
dst.append(count)
for d in dst:
print(d)
|
s447955831 | p03494 | u925836726 | 2,000 | 262,144 | Time Limit Exceeded | 2,104 | 3,064 | 385 | There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform. | def main():
n = int(input())
a = list(map(int, input().split()))
flag = True
count = 0
for i in a:
if(i%2 == 1):
flag = False
while flag == True:
count += 1
a = list(map( lambda x: int(x/2), a))
for i in a:
if(i%2 != 0):
False
print(count)
if __name__ == '__main__':
main() | s871645077 | Accepted | 19 | 3,064 | 392 | def main():
n = int(input())
a = list(map(int, input().split()))
flag = True
count = 0
for i in a:
if(i%2 == 1):
flag = False
while flag == True:
count += 1
a = list(map( lambda x: int(x/2), a))
for i in a:
if(i%2 != 0):
flag = False
print(count)
if __name__ == '__main__':
main() |
s950891436 | p02389 | u721446434 | 1,000 | 131,072 | Wrong Answer | 30 | 7,608 | 63 | Write a program which calculates the area and perimeter of a given rectangle. | nums = list(map(int, input().split()))
print(nums[0] * nums[1]) | s603613641 | Accepted | 20 | 7,572 | 88 | nums = list(map(int, input().split()))
print(nums[0] * nums[1], (nums[0] + nums[1]) * 2) |
s732001192 | p03673 | u811528179 | 2,000 | 262,144 | Wrong Answer | 2,104 | 26,176 | 125 | You are given an integer sequence of length n, a_1, ..., a_n. Let us consider performing the following n operations on an empty sequence b. The i-th operation is as follows: 1. Append a_i to the end of b. 2. Reverse the order of the elements in b. Find the sequence b obtained after these n operations. | n=int(input())
a=list(map(int,input().split()))
ans=[]
for i in range(1,n+1):
ans.append(i)
ans=ans[::-1]
print(ans)
| s349152621 | Accepted | 181 | 29,252 | 232 | from collections import deque
n=int(input())
a=list(map(int,input().split()))
ans=deque([a[0]])
for i in range(1,n):
if (n-1)%2==i%2:
ans.appendleft(a[i])
else:
ans.append(a[i])
print(" ".join(map(str,ans)))
|
s333700754 | p02927 | u167908302 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 181 | Today is August 24, one of the five Product Days in a year. A date m-d (m is the month, d is the date) is called a Product Day when d is a two-digit number, and all of the following conditions are satisfied (here d_{10} is the tens digit of the day and d_1 is the ones digit of the day): * d_1 \geq 2 * d_{10} \geq 2 * d_1 \times d_{10} = m Takahashi wants more Product Days, and he made a new calendar called Takahashi Calendar where a year consists of M month from Month 1 to Month M, and each month consists of D days from Day 1 to Day D. In Takahashi Calendar, how many Product Days does a year have? | #coding:utf-8
m, d = map(int, input().split())
ans = 0
for i in range(10, d+1):
d1 = d % 10
d10 = d // 10
if d1 >= 2 and d10 >= 2 and d1*d10 == m:
ans += 1
print(ans) | s812112819 | Accepted | 17 | 2,940 | 180 | #coding:utf-8
m, d = map(int, input().split())
ans = 0
for i in range(10, d+1):
i1 = i % 10
i10 = i // 10
if i1 >= 2 and i10 >= 2 and i1*i10 <= m:
ans += 1
print(ans)
|
s451770194 | p03024 | u026788530 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 78 | 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=[x for x in input()]
if(s.count('x')>7):
print("No")
else:
print("Yes") | s287674206 | Accepted | 17 | 2,940 | 78 | s=[x for x in input()]
if(s.count('x')>7):
print("NO")
else:
print("YES") |
s135335613 | p02833 | u871841829 | 2,000 | 1,048,576 | Wrong Answer | 18 | 3,060 | 137 | For an integer n not less than 0, let us define f(n) as follows: * f(n) = 1 (if n < 2) * f(n) = n f(n-2) (if n \geq 2) Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N). | N = int(input())
import sys
if N & 1:
print(0)
sys.exit()
ans = 0
N /= 2
while N != 0:
ans += N // 5
N /= 5
print(ans) | s010572982 | Accepted | 17 | 2,940 | 139 | N = int(input())
import sys
if N & 1:
print(0)
sys.exit()
ans = 0
N //= 2
while N != 0:
ans += N // 5
N //= 5
print(ans) |
s142453794 | p03624 | u789290859 | 2,000 | 262,144 | Wrong Answer | 51 | 9,760 | 313 | You are given a string S consisting of lowercase English letters. Find the lexicographically (alphabetically) smallest lowercase English letter that does not occur in S. If every lowercase English letter occurs in S, print `None` instead. | str1 = input()
li = list(str1)
anli = []
for i in range(26):
anli.append(0)
for i in range(len(li)):
anli[ord(li[i]) - 97] += 1
flag = True
mi = 27
print(anli)
for i in range(len(anli)):
if anli[i] == 0:
flag = False
mi = min(i,mi)
if flag:
print('None')
else:
print(chr(mi+97)) | s810018667 | Accepted | 50 | 9,700 | 301 | str1 = input()
li = list(str1)
anli = []
for i in range(26):
anli.append(0)
for i in range(len(li)):
anli[ord(li[i]) - 97] += 1
flag = True
mi = 27
for i in range(len(anli)):
if anli[i] == 0:
flag = False
mi = min(i,mi)
if flag:
print('None')
else:
print(chr(mi+97)) |
s332652602 | p02393 | u548155360 | 1,000 | 131,072 | Wrong Answer | 20 | 7,528 | 255 | Write a program which reads three integers, and prints them in ascending order. | nums=list(map(int,input().split()))
if nums[2] >= nums[1]:
a=nums[1]
b=nums[2]
else:
b=nums[1]
a=nums[2]
if nums[0] <= a:
min=nums[0]
mid=a
max=b
elif nums[0] < b:
min=a
mid=nums[0]
max=b
else:
min=a
mid=b
max=nums[0]
print("min,mid,max") | s569806674 | Accepted | 20 | 7,624 | 254 | nums=list(map(int,input().split()))
if nums[2] >= nums[1]:
a=nums[1]
b=nums[2]
else:
b=nums[1]
a=nums[2]
if nums[0] <= a:
min=nums[0]
mid=a
max=b
elif nums[0] <= b:
min=a
mid=nums[0]
max=b
else:
min=a
mid=b
max=nums[0]
print(min,mid,max) |
s977039250 | p03377 | u914797917 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 89 | There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals. | A,B,X = map(int,input().split())
if X>A+B or A>X:
print('No')
else:
print('Yes')
| s006046528 | Accepted | 17 | 2,940 | 89 | A,B,X = map(int,input().split())
if X>A+B or A>X:
print('NO')
else:
print('YES')
|
s286820692 | p03097 | u675317308 | 2,000 | 1,048,576 | Wrong Answer | 1,094 | 22,836 | 865 | You are given integers N,\ A and B. Determine if there exists a permutation (P_0,\ P_1,\ ...\ P_{2^N-1}) of (0,\ 1,\ ...\ 2^N-1) that satisfies all of the following conditions, and create one such permutation if it exists. * P_0=A * P_{2^N-1}=B * For all 0 \leq i < 2^N-1, the binary representations of P_i and P_{i+1} differ by exactly one bit. | def Double(vec):
m=len(vec)
res=[]
for i in range(0,m,2):
res.extend([vec[i],vec[i]^m,vec[i+1]^m,vec[i+1]])
return res
def Greedy(n):
m=1<<n
res=[0]*m
vis=[False]*m
vis[0]=True
for i in range(1,m):
x=res[i-1]
for j in range(n):
if not vis[x^(1<<j)]:
x^=1<<j
break
vis[x]=True
res[i]=x
return res
def Perfect(n):
if n==1:
return [0,1]
m=1<<n
A=Greedy(n-1)
B=Double(Perfect(n-2))
B.reverse()
for i in range(m>>1):
B[i]^=m-1
A.extend(B)
return A
n,a,b=map(int,input().split())
m=1<<n
c=bin(a^b).count('1')
if not c&1:
print('NO')
exit()
tmp=Perfect(c)
for i in range(c,n):
tmp=Double(tmp)
p=0
q=c
ans=[0]*m
for i in range(n):
if (a^b)>>i&1:
for j in range(m):
ans[j]^=(((tmp[j]>>p)^(a>>i))&1)<<i
p+=1
else:
for j in range(m):
ans[j]^=(((tmp[j]>>q)^(a>>i))&1)<<i
q+=1
print(' '.join(str(i) for i in ans)) | s431711792 | Accepted | 1,146 | 22,840 | 880 | def Double(vec):
m=len(vec)
res=[]
for i in range(0,m,2):
res.extend([vec[i],vec[i]^m,vec[i+1]^m,vec[i+1]])
return res
def Greedy(n):
m=1<<n
res=[0]*m
vis=[False]*m
vis[0]=True
for i in range(1,m):
x=res[i-1]
for j in range(n):
if not vis[x^(1<<j)]:
x^=1<<j
break
vis[x]=True
res[i]=x
return res
def Perfect(n):
if n==1:
return [0,1]
m=1<<n
A=Greedy(n-1)
B=Double(Perfect(n-2))
B.reverse()
for i in range(m>>1):
B[i]^=m-1
A.extend(B)
return A
n,a,b=map(int,input().split())
m=1<<n
c=bin(a^b).count('1')
if not c&1:
print('NO')
exit()
tmp=Perfect(c)
for i in range(c,n):
tmp=Double(tmp)
p=0
q=c
ans=[0]*m
for i in range(n):
if (a^b)>>i&1:
for j in range(m):
ans[j]^=(((tmp[j]>>p)^(a>>i))&1)<<i
p+=1
else:
for j in range(m):
ans[j]^=(((tmp[j]>>q)^(a>>i))&1)<<i
q+=1
print('YES\n',' '.join(str(i) for i in ans),sep='') |
s600017869 | p03693 | u736479342 | 2,000 | 262,144 | Wrong Answer | 24 | 9,092 | 99 | AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4? | r, g, b = map(int, input().split())
if (100*r + 10*b + b)%4==0:
print('YES')
else:
print('NO') | s237994542 | Accepted | 30 | 9,040 | 99 | r, g, b = map(int, input().split())
if (100*r + 10*g + b)%4==0:
print('YES')
else:
print('NO') |
s441908199 | p02843 | u580049189 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 7 | 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.) | input() | s940028604 | Accepted | 18 | 2,940 | 146 | def solve(target):
n = int(target / 100)
if n * 100 <= target <= n * 105:
print(1)
else:
print(0)
solve(int(input())) |
s202502377 | p04044 | u628018347 | 2,000 | 262,144 | Wrong Answer | 37 | 3,060 | 492 | Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m. | num_list = list(map(int, input().split()))
N = num_list[0]
L = num_list[1]
S = [input(i) for i in range(N)]
for i in range(N):
for j in range(N):
if i < j:
for k in range(L):
if S[i][k] > S[j][k]:
S[i], S[j] = S[j], S[i]
break
else:
continue
print(''.join(S))
| s853887270 | Accepted | 17 | 3,060 | 118 | num_list = list(map(int, input().split()))
S = [input() for i in range(num_list[0])]
S2 = sorted(S)
print(*S2, sep='') |
s859561541 | p03474 | u667949809 | 2,000 | 262,144 | Wrong Answer | 19 | 3,060 | 126 | The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom. | A, B = map(int,input().split())
S = list(input())
if S.count("-") == 1 and S[A] == "-":
print("yes")
else:
print("No") | s800272568 | Accepted | 18 | 3,064 | 113 | a,b = map(int,input().split())
s=input()
if s[a]=="-" and s.count("-")==1:
print("Yes")
else:
print("No") |
s698509731 | p03024 | u657208344 | 2,000 | 1,048,576 | Wrong Answer | 32 | 9,024 | 68 | 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. | n = input()
if n.count('o') >= 8:
print("YES")
else:
print("NO") | s142410121 | Accepted | 26 | 9,064 | 81 | n = input()
if 15-len(n) + n.count('o') >= 8:
print("YES")
else:
print("NO")
|
s732697821 | p02401 | u391228754 | 1,000 | 131,072 | Wrong Answer | 20 | 7,624 | 263 | Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part. | while True:
a, b, c = input().split()
a = int(a)
c = int(c)
if b == "?":
break
elif b == "-":
d = a - c
elif b == "+":
d = a + c
elif b == "*":
d = a * c
elif b == "/":
d = a / c
print(d) | s912753781 | Accepted | 30 | 7,656 | 264 | while True:
a, b, c = input().split()
a = int(a)
c = int(c)
if b == "?":
break
elif b == "-":
d = a - c
elif b == "+":
d = a + c
elif b == "*":
d = a * c
elif b == "/":
d = a // c
print(d) |
s484251380 | p03578 | u075304271 | 2,000 | 262,144 | Wrong Answer | 2,207 | 45,216 | 385 | Rng is preparing a problem set for a qualification round of CODEFESTIVAL. He has N candidates of problems. The difficulty of the i-th candidate is D_i. There must be M problems in the problem set, and the difficulty of the i-th problem must be T_i. Here, one candidate of a problem cannot be used as multiple problems. Determine whether Rng can complete the problem set without creating new candidates of problems. | import math
import collections
import fractions
import itertools
def solve():
n = int(input())
d = list(map(int, input().split()))
m = int(input())
t = list(map(int, input().split()))
t = collections.Counter(t)
for i in t:
if t[i] > d.count(i):
print("YES")
else:
print("NO")
return 0
if __name__ == "__main__":
solve() | s166405264 | Accepted | 250 | 56,608 | 428 | import math
import collections
import fractions
import itertools
def solve():
n = int(input())
d = list(map(int, input().split()))
m = int(input())
t = list(map(int, input().split()))
t = collections.Counter(t)
d = collections.Counter(d)
for i in t:
if t[i] > d[i]:
print("NO")
break
else:
print("YES")
return 0
if __name__ == "__main__":
solve() |
s786690014 | p02865 | u967789504 | 2,000 | 1,048,576 | Wrong Answer | 20 | 2,940 | 70 | How many ways are there to choose two distinct positive integers totaling N, disregarding the order? | n = int(input())
if n % 2:
print(n // 2)
else:
print((n / 2) - 1)
| s623989345 | Accepted | 17 | 3,064 | 30 | n=int(input())
print((n-1)//2) |
s845737433 | p03719 | u765590009 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 180 | You are given three integers A, B and C. Determine whether C is not less than A and not greater than B. |
a, b, c = map(int, input().split())
if (c >= a) and (c <= b) :
print("YES")
else :
print("NO") | s506191230 | Accepted | 17 | 2,940 | 180 |
a, b, c = map(int, input().split())
if (c >= a) and (c <= b) :
print("Yes")
else :
print("No") |
s376499357 | p03729 | u785205215 | 2,000 | 262,144 | Wrong Answer | 18 | 3,064 | 165 | 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`. | def read_str_list():
return list(str(_) for _ in input().split())
a,b,c = read_str_list()
if a[-1]==b[0] and b[-1]==c[0]:
print("Yes")
else:
print("No") | s666598442 | Accepted | 17 | 2,940 | 165 | def read_str_list():
return list(str(s) for s in input().split())
a,b,c = read_str_list()
if a[-1]==b[0] and b[-1]==c[0]:
print("YES")
else:
print("NO") |
s443906523 | p03845 | u084320347 | 2,000 | 262,144 | Wrong Answer | 32 | 3,572 | 217 | Joisino is about to compete in the final round of a certain programming competition. In this contest, there are N problems, numbered 1 through N. Joisino knows that it takes her T_i seconds to solve problem i(1≦i≦N). Also, there are M kinds of drinks offered to the contestants, numbered 1 through M. If Joisino takes drink i(1≦i≦M), her brain will be stimulated and the time it takes for her to solve problem P_i will become X_i seconds. It does not affect the time to solve the other problems. A contestant is allowed to take exactly one of the drinks before the start of the contest. For each drink, Joisino wants to know how many seconds it takes her to solve all the problems if she takes that drink. Here, assume that the time it takes her to solve all the problems is equal to the sum of the time it takes for her to solve individual problems. Your task is to write a program to calculate it instead of her. | import copy
n = int(input())
t = list(map(int,input().split()))
m = int(input())
for i in range(m):
c_t = copy.deepcopy(t)
p,x = list(map(int,input().split()))
c_t[p-1]=x
print(c_t)
print(sum(c_t)) | s953165616 | Accepted | 31 | 3,444 | 202 | import copy
n = int(input())
t = list(map(int,input().split()))
m = int(input())
for i in range(m):
c_t = copy.deepcopy(t)
p,x = list(map(int,input().split()))
c_t[p-1]=x
print(sum(c_t)) |
s155999915 | p02263 | u650790815 | 1,000 | 131,072 | Wrong Answer | 20 | 7,620 | 239 | An expression is given in a line. Two consequtive symbols (operand or operator) are separated by a space character. You can assume that +, - and * are given as the operator and an operand is a positive integer less than 106 | q = []
for e in input().strip().split():
if e == '+':
q.append(q.pop() + q.pop())
elif e == '-':
q.append(-q.pop() + q.pop())
elif e == '*':
q.append(q.pop() * q.pop())
else:
q.append(int(e)) | s239903523 | Accepted | 20 | 7,636 | 258 | f = input().strip().split()
q = []
for e in f:
if e == '+':
q.append(q.pop() + q.pop())
elif e == '-':
q.append(-q.pop() + q.pop())
elif e == '*':
q.append(q.pop() * q.pop())
else:
q.append(int(e))
print(q[0]) |
s066884029 | p02274 | u918276501 | 1,000 | 131,072 | Wrong Answer | 20 | 7,704 | 684 | For a given sequence $A = \\{a_0, a_1, ... a_{n-1}\\}$, the number of pairs $(i, j)$ where $a_i > a_j$ and $i < j$, is called the number of inversions. The number of inversions is equal to the number of swaps of Bubble Sort defined in the following program: bubbleSort(A) cnt = 0 // the number of inversions for i = 0 to A.length-1 for j = A.length-1 downto i+1 if A[j] < A[j-1] swap(A[j], A[j-1]) cnt++ return cnt For the given sequence $A$, print the number of inversions of $A$. Note that you should not use the above program, which brings Time Limit Exceeded. | def merge(A, left, mid, right):
global count
# count += (right - left)
L = A[left:mid] + [int(1e9)]
R = A[mid:right] + [int(1e9)]
i,j = 0,0
for k in range(left, right):
if L[i] <= R[j]:
A[k] = L[i]
i += 1
else:
A[k] = R[j]
j += 1
count += 1
def merge_sort(A, left, right):
if left+1 < right:
mid = (left + right) // 2
merge_sort(A, left, mid)
merge_sort(A, mid, right)
merge(A, left, mid, right)
if __name__ == "__main__":
n = int(input())
A = list(map(int, input().strip().split()))
count = 0
merge_sort(A, 0, n)
print(count) | s403222715 | Accepted | 1,600 | 30,444 | 695 | def merge(A, left, mid, right):
global count
# count += (right - left)
L = A[left:mid] + [int(1e9)]
R = A[mid:right] + [int(1e9)]
i = j = 0
for k in range(left, right):
if L[i] <= R[j]:
A[k] = L[i]
i += 1
else:
A[k] = R[j]
j += 1
count += (mid-left-i)
def merge_sort(A, left, right):
if left+1 < right:
mid = (left + right) // 2
merge_sort(A, left, mid)
merge_sort(A, mid, right)
merge(A, left, mid, right)
if __name__ == "__main__":
n = int(input())
A = list(map(int, input().strip().split()))
count = 0
merge_sort(A, 0, n)
print(count) |
s548856929 | p03555 | u021548497 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 109 | 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. | s = input()
t = input()
if s[0] == t[2] and s[1] == t[1] and s[2] == t[0]:
print("Yes")
else:
print("No") | s619521226 | Accepted | 17 | 2,940 | 109 | s = input()
t = input()
if s[0] == t[2] and s[1] == t[1] and s[2] == t[0]:
print("YES")
else:
print("NO") |
s405715401 | p03486 | u392361133 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 72 | You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order. | s = sorted(input())
t = sorted(input())
print("Yes" if s < t else "No")
| s712584688 | Accepted | 17 | 2,940 | 78 | s = sorted(input())
t = sorted(input())[::-1]
print("Yes" if s < t else "No")
|
s503589893 | p02422 | u613534067 | 1,000 | 131,072 | Wrong Answer | 20 | 5,596 | 396 | 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. | t = input()
for i in range(int(input())):
cmd = list(input().split())
cmd[1] = int(cmd[1])
cmd[2] = int(cmd[2])
if cmd[0] == "print":
print(t[cmd[1]-1 : cmd[2]])
elif cmd[0] == "reverse":
tmp = t[cmd[1]-1 : cmd[2]]
tmp = tmp[::-1]
t = t[:cmd[1]-1] + tmp + t[cmd[2]:]
elif cmd[0] == "replace":
t = t[:cmd[1]-1] + cmd[3] + t[cmd[2]:]
| s285011673 | Accepted | 20 | 5,620 | 482 |
t = input()
for i in range(int(input())):
cmd = list(input().split())
cmd[1] = int(cmd[1])
cmd[2] = int(cmd[2])
if cmd[0] == "print":
print(t[cmd[1] : cmd[2]] + t[cmd[2]])
elif cmd[0] == "reverse":
tmp = t[cmd[1] : cmd[2]] + t[cmd[2]]
tmp = tmp[::-1]
t = t[:cmd[1]] + tmp + t[cmd[2]+1:]
elif cmd[0] == "replace":
t = t[:cmd[1]] + cmd[3] + t[cmd[2]+1:]
|
s764087811 | p03555 | u346474533 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 96 | You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise. | a = input()
b = input()
if a == b[::-1] and b == a[::-1]:
print('Yes')
else:
print('No') | s338664623 | Accepted | 17 | 2,940 | 79 | a = input()
b = input()
if a == b[::-1]:
print('YES')
else:
print('NO') |
s092726660 | p03610 | u197300260 | 2,000 | 262,144 | Wrong Answer | 18 | 3,188 | 543 | You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1. |
# Python 2nd Try
class Problem:
def __init__(self, problemstring):
self.problemstring = problemstring
self.length = len(self.problemstring)
def solve(self):
result = ''
resultlist = []
for j in range(0, self.length):
if j % 2 == 0:
resultlist.append(self.problemstring[j])
result = ''.join(resultlist)
return result
if __name__ == "__main__":
S = input()
print(Problem(S).solve) | s416647096 | Accepted | 22 | 3,700 | 475 |
# Python 3rd Try
import copy
class Problem:
def __init__(self, inputstring):
self.inputstring = inputstring
def __str__(self):
return self.inputstring
def solver(self):
result = ''
string = copy.copy(self.inputstring)
result = string[0:len(string):2]
return result
if __name__ == "__main__":
s = input()
x = Problem(s)
print(x.solver())
|
s720256613 | p04043 | u919978322 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 201 | 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. | ip = input().split()
A = int(ip[0])
B = int(ip[1])
C = int(ip[2])
boolean = False
if A+B+C==19:
if A==5 or B==5 or C==5:
boolean = True
if boolean:
print("YES")
else:
print("NO")
| s302140215 | Accepted | 17 | 2,940 | 201 | ip = input().split()
A = int(ip[0])
B = int(ip[1])
C = int(ip[2])
boolean = False
if A+B+C==17:
if A==5 or B==5 or C==5:
boolean = True
if boolean:
print("YES")
else:
print("NO")
|
s441871880 | p02694 | u242580186 | 2,000 | 1,048,576 | Wrong Answer | 31 | 9,200 | 472 | 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? | import sys
read = sys.stdin.readline
import time
import math
import itertools as it
def inp():
return int(input())
def inpl():
return list(map(int, input().split()))
start_time = time.perf_counter()
# ------------------------------
X = inp()
yen = 100
for i in range(10**4):
yen += (yen // 100)
if yen >= X:
print(i)
break
# -----------------------------
end_time = time.perf_counter()
print('time:', end_time-start_time, file=sys.stderr) | s029894427 | Accepted | 30 | 9,188 | 474 | import sys
read = sys.stdin.readline
import time
import math
import itertools as it
def inp():
return int(input())
def inpl():
return list(map(int, input().split()))
start_time = time.perf_counter()
# ------------------------------
X = inp()
yen = 100
for i in range(10**4):
yen += (yen // 100)
if yen >= X:
print(i+1)
break
# -----------------------------
end_time = time.perf_counter()
print('time:', end_time-start_time, file=sys.stderr) |
s946080084 | p03635 | u001495709 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 52 | In _K-city_ , there are n streets running east-west, and m streets running north-south. Each street running east-west and each street running north-south cross each other. We will call the smallest area that is surrounded by four streets a block. How many blocks there are in K-city? | n, m = map(int, input().split())
print = (n-1)*(m-1) | s584504113 | Accepted | 18 | 2,940 | 52 | n, m = map(int, input().split())
print((n-1)*(m-1))
|
s244543428 | p03457 | u316603606 | 2,000 | 262,144 | Wrong Answer | 261 | 9,156 | 384 | 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 ())
t2 = 0
x2 = 0
y2 = 0
for i in range (N):
t,x,y = map(int, input().split())
if abs (t-t2) >= abs ((x-x2)+(y-y2)):
if (round (((t-t2)-((x-x2)+(y-y2)))/2))*2 ==((t-t2)-((x-x2)+(y-y2))):
ans = 'YES'
else:
ans = 'NO'
break
else:
ans = 'NO'
break
t2 = t
x2 = x
y2 = y
print (ans) | s164804494 | Accepted | 240 | 9,160 | 316 | N = int (input ())
t2 = 0
x2 = 0
y2 = 0
ans = 'Yes'
for i in range (N):
t,x,y = map(int, input().split())
if abs(x+y-x2-y2) <= t-t2:
if ((abs(x2+y2-x-y))-(t-t2)) % 2 == 1:
ans = 'No'
break
else:
ans = 'No'
break
t2 = t
x2 = x
y2 = y
print (ans) |
s839915871 | p03351 | u869728296 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 171 | 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=input().strip().split(" ")
b=[int(i) for i in a]
if(((abs(b[0]-b[1])<b[3]) and (abs(b[2]-b[1])<b[3])) or abs(b[0]-b[2])<b[3] ):
print("Yes")
else:
print("No")
| s327271501 | Accepted | 18 | 2,940 | 174 | a=input().strip().split(" ")
b=[int(i) for i in a]
if(((abs(b[0]-b[1])<=b[3]) and (abs(b[2]-b[1])<=b[3])) or abs(b[0]-b[2])<=b[3] ):
print("Yes")
else:
print("No")
|
s990377150 | p02255 | u342125850 | 1,000 | 131,072 | Wrong Answer | 20 | 5,596 | 271 | 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. | num = int(input())
cards = [int(i) for i in input().split()]
def insertion_sort(num, cards):
for i in range(num):
v = cards[i]
j = i - 1
while j >= 0 and cards[j] > v:
cards[j+1] = cards[j]
j -= 1
cards[j+1] = v
print(cards)
insertion_sort(num, cards)
| s733344277 | Accepted | 20 | 5,608 | 353 | num = int(input())
cards = [int(i) for i in input().split()]
def insertion_sort(num, cards):
for i in range(num):
v = cards[i]
j = i - 1
while j >= 0 and cards[j] > v:
cards[j+1] = cards[j]
j -= 1
cards[j+1] = v
output(cards)
def output(cards):
sorting = [str(i) for i in cards]
print(" ".join(sorting))
insertion_sort(num, cards)
|
s193644064 | p02578 | u198065632 | 2,000 | 1,048,576 | Wrong Answer | 123 | 32,264 | 170 | N persons are standing in a row. The height of the i-th person from the front is A_i. We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person: Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool. Find the minimum total height of the stools needed to meet this goal. | n = int(input())
a = list(map(int, input().split()))
s = 0
for i in range(0, n - 1):
h = a[i] - a[i+1]
if h > 0:
s += h
a[i] += h
print(str(s))
| s105169478 | Accepted | 141 | 32,172 | 196 | n = int(input())
a = list(map(int, input().split()))
s = 0
for i in range(0, n - 1):
h = a[i] - a[i+1]
if h > 0:
s += h
a[i+1] += h
# a[i+1] = a[i]
print(str(s))
|
s958597704 | p02612 | u729805651 | 2,000 | 1,048,576 | Wrong Answer | 25 | 8,948 | 32 | 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())
a=N%1000
print(a) | s562442416 | Accepted | 27 | 9,108 | 69 | N=int(input())
a=N%1000
if a==0:
print(a)
else:
print(1000-a) |
s503430462 | p03609 | u965436898 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 65 | We have a sandglass that runs for X seconds. The sand drops from the upper bulb at a rate of 1 gram per second. That is, the upper bulb initially contains X grams of sand. How many grams of sand will the upper bulb contains after t seconds? | n = list(input())
if "9" in n:
print("Yes")
else:
print("No") | s045039962 | Accepted | 17 | 2,940 | 77 | x,t = map(int,input().split())
if x - t >= 0:
print(x - t)
else:
print(0) |
s266233559 | p02771 | u528594564 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 205 | A triple of numbers is said to be _poor_ when two of those numbers are equal but the other number is different from those two numbers. You will be given three integers A, B, and C. If this triple is poor, print `Yes`; otherwise, print `No`. | from sys import stdin,stdout
def main():
for x in stdin:
A = list(map(int,x.split()))
A.sort()
if A[0]==A[1] and A[1]!=A[2]:print('Yes')
else:print('No')
if __name__ == "__main__":
main()
| s814908133 | Accepted | 17 | 3,064 | 328 | from sys import stdin,stdout
def f(i,j,n,mat):
ans = -1
for k in range(n):
ans += mat[i][k]+mat[k][j]
return ans
def main():
for x in stdin:
A = list(map(int,x.split()))
A.sort()
if (A[0]==A[1] and A[1]!=A[2]) or (A[0]!=A[1] and A[1]==A[2]):print('Yes')
else:print('No')
if __name__ == "__main__":
main()
|
s012017733 | p03487 | u875291233 | 2,000 | 262,144 | Time Limit Exceeded | 2,132 | 532,388 | 335 | You are given a sequence of positive integers of length N, a = (a_1, a_2, ..., a_N). Your objective is to remove some of the elements in a so that a will be a **good sequence**. Here, an sequence b is a **good sequence** when the following condition holds true: * For each element x in b, the value x occurs exactly x times in b. For example, (3, 3, 3), (4, 2, 4, 1, 4, 2, 4) and () (an empty sequence) are good sequences, while (3, 3, 3, 3) and (2, 4, 1, 4, 2) are not. Find the minimum number of elements that needs to be removed so that a will be a good sequence. | N=int(input())
a=[int(i) for i in input().split()]
#print(a)
counter = [0 for i in range(10**9+3)]
#print(counter[a[N-1]])
#print(counter[6])
for i in range(N):
counter[a[i]] += 1
ans = 0
for i in range(N):
if counter[i] < i:
ans += counter[i]
else:
ans += counter[i] - i
print(ans)
| s853671785 | Accepted | 90 | 14,024 | 470 | N=int(input())
a=[int(i) for i in input().split()]
counter = [0]*(10**5+10)
ans = 0
for i in range(N):
# print('a[i]=')
# print(a[i])
if a[i] < 10**5 + 3:
counter[a[i]] += 1
# print('hoge')
else:
ans += 1
# print(ans)
for i in range(10**5+3):
if counter[i] < i:
ans += counter[i]
# print(counter[i])
else:
ans += counter[i] - i
# print(counter[i]-i)
# print(ans)
print(ans)
|
s914720847 | p03555 | u197300773 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 49 | 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. | print("Yes" if input()==input()[::-1] else "No" ) | s438739998 | Accepted | 17 | 2,940 | 49 | print("YES" if input()==input()[::-1] else "NO" ) |
s086993124 | p03658 | u729133443 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 58 | Snuke has N sticks. The length of the i-th stick is l_i. Snuke is making a snake toy by joining K of the sticks together. The length of the toy is represented by the sum of the individual sticks that compose it. Find the maximum possible length of the toy. | t,l=open(0);print(sorted(map(int,l.split()))[-int(t[2:])]) | s543193519 | Accepted | 17 | 2,940 | 64 | t,l=open(0);print(sum(sorted(map(int,l.split()))[-int(t[2:]):])) |
s223886275 | p03694 | u735008991 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 80 | It is only six months until Christmas, and AtCoDeer the reindeer is now planning his travel to deliver gifts. There are N houses along _TopCoDeer street_. The i-th house is located at coordinate a_i. He has decided to deliver gifts to all these houses. Find the minimum distance to be traveled when AtCoDeer can start and end his travel at any positions. | N = input()
A = sorted(list(map(int, input().split())))
print(A[-1] - A[0] + 1)
| s108299202 | Accepted | 17 | 2,940 | 76 | N = input()
A = sorted(list(map(int, input().split())))
print(A[-1] - A[0])
|
s925685968 | p03545 | u094778153 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 599 | 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. | if __name__ == '__main__':
num = input()
num_list = [int(i) for i in str(num)]
A = num_list[0]
B = num_list[1]
C = num_list[2]
D = num_list[3]
if A + B + C + D == 7:
print("A+B+C+D")
elif A + B + C - D == 7:
print("A+B+C-D")
elif A + B - C + D == 7:
print("A+B-C+D")
elif A + B - C - D == 7:
print("A+B-C-D")
elif A - B + C + D == 7:
print("A-B+C+D")
elif A - B + C - D == 7:
print("A-B+C-D")
elif A - B - C + D == 7:
print("A-B-C+D")
elif A - B - C - D == 7:
print("A-B-C-D")
| s047900819 | Accepted | 17 | 3,064 | 831 | if __name__ == '__main__':
num = input()
num_list = [int(i) for i in str(num)]
A = num_list[0]
B = num_list[1]
C = num_list[2]
D = num_list[3]
if A + B + C + D == 7:
print("{0}+{1}+{2}+{3}=7".format(A, B, C, D))
elif A + B + C - D == 7:
print("{0}+{1}+{2}-{3}=7".format(A, B, C, D))
elif A + B - C + D == 7:
print("{0}+{1}-{2}+{3}=7".format(A, B, C, D))
elif A + B - C - D == 7:
print("{0}+{1}-{2}-{3}=7".format(A, B, C, D))
elif A - B + C + D == 7:
print("{0}-{1}+{2}+{3}=7".format(A, B, C, D))
elif A - B + C - D == 7:
print("{0}-{1}+{2}-{3}=7".format(A, B, C, D))
elif A - B - C + D == 7:
print("{0}-{1}-{2}+{3}=7".format(A, B, C, D))
elif A - B - C - D == 7:
print("{0}-{1}-{2}-{3}=7".format(A, B, C, D))
|
s735525818 | p03963 | u816631826 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 99 | There are N balls placed in a row. AtCoDeer the deer is painting each of these in one of the K colors of his paint cans. For aesthetic reasons, any two adjacent balls must be painted in different colors. Find the number of the possible ways to paint the balls. | a=list(map(int,input().split()))
if a[0]==1:
print(a[1])
exit(0)
print(a[1]*(a[1]**(a[0]-1))) | s993391819 | Accepted | 17 | 2,940 | 72 | # your code goes here
n,k=map(int,input().split())
print(k*pow(k-1,n-1)) |
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