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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s748302274
|
p03068
|
u661983922
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 244
|
You are given a string S of length N consisting of lowercase English letters, and an integer K. Print the string obtained by replacing every character in S that differs from the K-th character of S, with `*`.
|
N = input()
S = input()
K = input()
char = S[int(K)-1]
#print(char)
ans = []
for j in range(len(S)):
#print(S[j])
if char != S[j]:
ans.append("*")
#print("!=")
else:
ans.append(char)
continue
print(" ".join(ans))
|
s117303235
|
Accepted
| 17
| 2,940
| 242
|
N = input()
S = input()
K = input()
char = S[int(K)-1]
#print(char)
ans = []
for j in range(len(S)):
#print(S[j])
if char != S[j]:
ans.append("*")
#print("!=")
else:
ans.append(char)
continue
print("".join(ans))
|
s504665685
|
p02690
|
u652656291
| 2,000
| 1,048,576
|
Wrong Answer
| 555
| 9,160
| 116
|
Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X.
|
x = int(input())
for i in range(1000):
for j in range(1000):
if i**5 - j**5 == x:
print(i,j)
break
|
s807199380
|
Accepted
| 52
| 9,164
| 148
|
x = int(input())
a = 0
b = 0
for i in range(-130,130):
for j in range(-130,130):
if x == (i**5) - (j**5):
a = i
b = j
print(a,b)
|
s581764055
|
p03063
|
u559048291
| 2,000
| 1,048,576
|
Wrong Answer
| 2,104
| 12,008
| 282
|
There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is `.`, and the stone is black if the character is `#`. Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored.
|
N = int(input())
S = input()
wl = []
bl = []
w=0
b=0
for i in S:
if i==".":
w+=1;
wl.append(w)
for i in reversed(S):
if i=="#":
b+=1;
bl.insert(0, b)
ans = N
for i in range(N):
if wl[i]+bl[i] > ans:
ans = wl[i]+bl[i]
print(N - ans)
|
s133469631
|
Accepted
| 164
| 12,808
| 280
|
N = int(input())
S = input()
wl = [0]
bl = [0]
w=0
b=0
for i in S:
if i==".":
w+=1;
else:
b+=1;
wl.append(w)
bl.append(b)
wmax = wl[-1]
min = wmax;
for i in range(N+1):
if wmax+bl[i]-wl[i] < min:
min = wmax+bl[i]-wl[i]
print(min)
|
s202874227
|
p03386
|
u939585142
| 2,000
| 262,144
|
Wrong Answer
| 2,245
| 4,084
| 123
|
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())
li = [i for i in range(A,B + 1)]
ans = li[:K] + sorted(li,reverse = True)[:K]
print(ans)
|
s382550372
|
Accepted
| 21
| 3,316
| 180
|
A, B, K = map(int,input().split())
ans = []
for i in range(K):
if A + i <= B:
ans.append(A + i)
if B - i >= A:
ans.append(B - i)
for i in sorted(set(ans)):
print(i)
|
s916800039
|
p03972
|
u476199965
| 2,000
| 262,144
|
Wrong Answer
| 987
| 26,948
| 290
|
On an xy plane, in an area satisfying 0 ≤ x ≤ W, 0 ≤ y ≤ H, there is one house at each and every point where both x and y are integers. There are unpaved roads between every pair of points for which either the x coordinates are equal and the difference between the y coordinates is 1, or the y coordinates are equal and the difference between the x coordinates is 1. The cost of paving a road between houses on coordinates (i,j) and (i+1,j) is p_i for any value of j, while the cost of paving a road between houses on coordinates (i,j) and (i,j+1) is q_j for any value of i. Mr. Takahashi wants to pave some of these roads and be able to travel between any two houses on paved roads only. Find the solution with the minimum total cost.
|
w,h = list(map(int,input().split()))
pq = []
for i in range(w):
pq.append((int(input()),0))
for i in range(h):
pq.append((int(input()),1))
pq.sort()
res = 0
w+=1
h+=1
dic = {0:w,1:h}
for x in pq:
print(dic[x[1]],x[0])
res += dic[1-x[1]]*x[0]
dic[x[1]] -= 1
print(res)
|
s090421389
|
Accepted
| 660
| 24,540
| 264
|
w,h = list(map(int,input().split()))
pq = []
for i in range(w):
pq.append((int(input()),0))
for i in range(h):
pq.append((int(input()),1))
pq.sort()
res = 0
w+=1
h+=1
dic = {0:w,1:h}
for x in pq:
res += dic[1-x[1]]*x[0]
dic[x[1]] -= 1
print(res)
|
s699532037
|
p03388
|
u118642796
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 3,064
| 398
|
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.
|
Q = int(input())
AB = [list(map(int,input().split())) for _ in range(Q)]
def cn(A,B):
ans = 0
pre = B
score = A*B
for i in range(A+1,score+1):
tmp = (score-1)//i
if tmp < 1:
break
if tmp == pre:
ans += pre-1
break
ans += 1
pre = tmp
return ans
for ab in AB:
print(cn(ab[0],ab[1])+cn(ab[1],ab[0]))
|
s907748337
|
Accepted
| 20
| 3,064
| 372
|
Q = int(input())
AB = [list(map(int,input().split())) for _ in range(Q)]
def cn(A,B):
score = A*B-1
L = A+1
R = score
if L>R:
return 0
while L!=R:
tmp = (L+R)//2
if score > tmp*(tmp+1):
L = tmp+1
else:
R = tmp
return (L-1-A+score//L)
for ab in AB:
print(cn(ab[0],ab[1])+cn(ab[1],ab[0]))
|
s648631440
|
p03970
|
u062691227
| 2,000
| 262,144
|
Wrong Answer
| 23
| 9,032
| 60
|
CODE FESTIVAL 2016 is going to be held. For the occasion, Mr. Takahashi decided to make a signboard. He intended to write `CODEFESTIVAL2016` on it, but he mistakenly wrote a different string S. Fortunately, the string he wrote was the correct length. So Mr. Takahashi decided to perform an operation that replaces a certain character with another in the minimum number of iterations, changing the string to `CODEFESTIVAL2016`. Find the minimum number of iterations for the rewrite operation.
|
s = input()
sum(a!=b for a, b in zip(s, 'CODEFESTIVAL2016'))
|
s571363410
|
Accepted
| 25
| 9,000
| 70
|
s, = *open(0),
print(sum(a!=b for a, b in zip(s, 'CODEFESTIVAL2016')))
|
s020744103
|
p02255
|
u764759179
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,592
| 251
|
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.
|
# coding:utf-8
length = int(input())
data = list(input().split())
for i in range(1,len(data)):
tmp = data[i]
i2 = i - 1
while i2 >=0 and data[i2] > tmp:
data[i2+1] = data[i2]
i2 -= 1
data[i2+1] = tmp
print(data)
|
s917588923
|
Accepted
| 20
| 5,604
| 430
|
# coding:utf-8
def printline(data):
l=""
for i,s in enumerate(data):
l = l + str(s)
if i+1 != len(data):
l = l+" "
print(l)
length = int(input())
data = list(map(int,input().split()))
printline(data)
for i in range(1,len(data)):
tmp = data[i]
i2 = i - 1
while i2 >=0 and data[i2] > tmp:
data[i2+1] = data[i2]
i2 -= 1
data[i2+1] = tmp
printline(data)
|
s137495291
|
p03998
|
u322185540
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 606
|
Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.
|
sa = input()
sb = input()
sc = input()
acount = 0
bcount = 0
ccount = 0
flag = True
ans = ''
tmp = sa[acount]
while flag:
if tmp == 'a':
acount += 1
if acount < len(sa):
tmp = sa[acount]
else:
ans = 'a'
flag = False
if tmp == 'b':
bcount += 1
if bcount < len(sb):
tmp = sb[bcount]
else:
ans = 'b'
flag = False
if tmp == 'c':
ccount += 1
if ccount < len(sc):
tmp = sc[ccount]
else:
ans = 'c'
flag = False
print(ans)
|
s092818153
|
Accepted
| 17
| 3,064
| 771
|
sa = input()
sb = input()
sc = input()
acount = 0
bcount = 0
ccount = 0
flag = True
ans = ''
tmp = sa[acount]
if len(sa) == 1:
ans = 'A'
else:
while flag:
if tmp == 'a':
if acount == len(sa):
flag = False
ans = 'A'
else:
tmp = sa[acount]
acount += 1
if tmp == 'b':
if bcount == len(sb):
flag = False
ans = 'B'
else:
tmp = sb[bcount]
bcount += 1
if tmp == 'c':
if ccount == len(sc):
flag = False
ans = 'C'
else:
tmp = sc[ccount]
ccount += 1
print(ans)
|
s009045289
|
p03485
|
u098968285
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 235
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
import math
def makelist(n, m):
return [[0 for _ in range(m)] for _ in range(n)]
# n = int(input())
# a, b = list(map(int, input().split()))
a, b = list(map(int, input().split()))
ans = math.ceil(a / b)
print(ans)
|
s432211288
|
Accepted
| 17
| 2,940
| 241
|
import math
def makelist(n, m):
return [[0 for _ in range(m)] for _ in range(n)]
# n = int(input())
# a, b = list(map(int, input().split()))
a, b = list(map(int, input().split()))
ans = math.ceil((a + b) / 2)
print(ans)
|
s612242304
|
p03644
|
u427344224
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 72
|
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())
num = [i for i in range(N) if i % 2 ==0][-1]
print(num)
|
s413918647
|
Accepted
| 17
| 2,940
| 95
|
N = int(input())
num = 1
for i in range(1, 9):
if 2**i <= N:
num = 2**i
print(num)
|
s994668308
|
p04011
|
u127856129
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 128
|
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.
|
a=int(input())
b=int(input())
c=int(input())
d=int(input())
if a>=b:
print(int((b*c)-((a-b)*d)))
else:
print(int(a*c))
|
s853415886
|
Accepted
| 20
| 2,940
| 128
|
a=int(input())
b=int(input())
c=int(input())
d=int(input())
if a>=b:
print(int((b*c)+((a-b)*d)))
else:
print(int(a*c))
|
s799110474
|
p03478
|
u840974625
| 2,000
| 262,144
|
Wrong Answer
| 24
| 2,940
| 186
|
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())
x = 0
for i in range(n):
q = i
w = i / 10
e = i / 100
r = i / 1000
sum = q + w + e + r
if sum >= a and sum <= b:
x += 1
print(x)
|
s905737721
|
Accepted
| 27
| 3,060
| 284
|
n, a, b = map(int, input().split())
x = 0
for i in range(1, n+1):
q = i // 10000
qi = i % 10000
w = qi // 1000
wi = qi % 1000
e = wi // 100
ei = wi % 100
r = ei // 10
ri = ei % 10
t = ri
sum = q + w + e + r + t
if sum >= a and sum <= b:
x += i
print(x)
|
s067474325
|
p03545
|
u004025573
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 407
|
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.
|
N=int(input())
a=N//1000
b=(N%1000)//100
c=(N%100)//10
d=N%10
ope=[-1,1]
for i in ope:
for j in ope:
for k in ope:
if a+i*b+j*c+k*d==7:
ans=[i,j,k]
l=[0]*3
for i in range(3):
if ans[i]>0:
l[i]="+"
else:
l[i]="-"
#all_ans=[str(a),l[0],str(b),l[1],str(c),l[2],str(d)]
all_ans=str(a)+l[0]+str(b)+l[1]+str(c)+l[2]+str(d)
print(all_ans)
|
s874462053
|
Accepted
| 18
| 3,064
| 411
|
N=int(input())
a=N//1000
b=(N%1000)//100
c=(N%100)//10
d=N%10
ope=[-1,1]
for i in ope:
for j in ope:
for k in ope:
if a+i*b+j*c+k*d==7:
ans=[i,j,k]
l=[0]*3
for i in range(3):
if ans[i]>0:
l[i]="+"
else:
l[i]="-"
#all_ans=[str(a),l[0],str(b),l[1],str(c),l[2],str(d)]
all_ans=str(a)+l[0]+str(b)+l[1]+str(c)+l[2]+str(d)+"=7"
print(all_ans)
|
s520012173
|
p03377
|
u386819480
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 95
|
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 = (int(_) for _ in input().split())
print('Yes') if x >= a and a+b >= x else print('No')
|
s915856315
|
Accepted
| 17
| 3,064
| 685
|
#!/usr/bin/env python3
import sys
YES = "YES" # type: str
NO = "NO" # type: str
def solve(A: int, B: int, X: int):
print(YES) if(A <= X <= A+B) else print(NO)
return
# Generated by 1.1.4 https://github.com/kyuridenamida/atcoder-tools (tips: You use the default template now. You can remove this line by using your custom template)
def main():
def iterate_tokens():
for line in sys.stdin:
for word in line.split():
yield word
tokens = iterate_tokens()
A = int(next(tokens)) # type: int
B = int(next(tokens)) # type: int
X = int(next(tokens)) # type: int
solve(A, B, X)
if __name__ == '__main__':
main()
|
s522526755
|
p03470
|
u309039873
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 176
|
An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?
|
size = 1
N = int(input())
d_s = []
for _ in range(N):
d_s.append(int(input()))
d_s.sort()
d_s.reverse()
for i in range(1,N):
if d_s[i] > d_s[i-1]:
size += 1
print(size)
|
s831350895
|
Accepted
| 17
| 3,060
| 176
|
size = 1
N = int(input())
d_s = []
for _ in range(N):
d_s.append(int(input()))
d_s.sort()
d_s.reverse()
for i in range(1,N):
if d_s[i] < d_s[i-1]:
size += 1
print(size)
|
s118171851
|
p03693
|
u184817817
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 117
|
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?
|
#064A
r,g,b = list(map(int,input().split()))
n = r*100 + g*10 + b
if n%4 == 0:
print('Yes')
else:
print('No')
|
s353501740
|
Accepted
| 17
| 2,940
| 117
|
#064A
r,g,b = list(map(int,input().split()))
n = r*100 + g*10 + b
if n%4 == 0:
print('YES')
else:
print('NO')
|
s051298675
|
p03645
|
u078349616
| 2,000
| 262,144
|
Wrong Answer
| 629
| 19,956
| 351
|
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.
|
N, M = map(int, input().split())
a = [0]*M
b = [0]*M
for i in range(M):
a[i], b[i] = map(int, input().split())
c = []
for i in range(M):
if b[i] == M:
c.append(a[i])
if len(c) != 0:
for i in range(M):
if a[i] == 1:
for j in range(len(c)):
if b[i] == c[j]:
print("POSSIBLE")
exit()
print("IMPOSSIBLE")
|
s723571617
|
Accepted
| 811
| 29,104
| 233
|
N, M = map(int, input().split())
G = [[] for _ in range(N)]
for i in range(M):
a, b = map(lambda x: int(x)-1, input().split())
G[a].append(b)
for v in G[0]:
if N-1 in G[v]:
print("POSSIBLE")
exit()
print("IMPOSSIBLE")
|
s515000725
|
p03543
|
u672794510
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 250
|
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 = int(input())
result = False
h1 = -1;
h2 = -1;
while a > 0:
print(a%10)
print(h1)
print(h2)
result = result or (h1 == h2 and h1 == a%10)
h2 = h1
h1 = a%10
a = a//10;
if result:
print("Yes")
else:
print("No")
|
s977952656
|
Accepted
| 18
| 3,060
| 206
|
a = int(input())
result = False
h1 = -1;
h2 = -1;
while a > 0:
result = result or (h1 == h2 and h1 == a%10)
h2 = h1
h1 = a%10
a = a//10;
if result:
print("Yes")
else:
print("No")
|
s813540417
|
p03862
|
u698479721
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,118
| 212,180
| 282
|
There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies. Snuke can perform the following operation any number of times: * Choose a box containing at least one candy, and eat one of the candies in the chosen box. His objective is as follows: * Any two neighboring boxes contain at most x candies in total. Find the minimum number of operations required to achieve the objective.
|
n,x = map(int, input().split())
a = [int(i) for i in input().split()]
b = []
if a[0]>x:
b.append(x)
else:
b.append(a[0])
j = 1
while j <= n-1:
if b[j-1]+a[j]<=x:
b.append(a[j])
else:
b.append(x-b[j-1])
k = 0
ans =0
while k < n:
ans += a[k]-b[k]
k += 1
print(ans)
|
s143430975
|
Accepted
| 124
| 14,060
| 255
|
n,x = map(int, input().split())
a = [int(i) for i in input().split()]
ans = 0
if a[0]>x:
ans += a[0]-x
a[0]=x
else:
pass
j = 1
while j <= n-1:
if a[j-1]+a[j]<=x:
j += 1
else:
ans += a[j]-x+a[j-1]
a[j]=x-a[j-1]
j += 1
print(ans)
|
s179725835
|
p02743
|
u405256066
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 144
|
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
|
from sys import stdin
a,b,c = [int(x) for x in stdin.readline().rstrip().split()]
if a**2 + b**2 < c**2:
print("Yes")
else:
print("No")
|
s205444194
|
Accepted
| 17
| 2,940
| 159
|
from sys import stdin
a,b,c = [int(x) for x in stdin.readline().rstrip().split()]
if 4*a*b < (c-a-b)**2 and (c-a-b) > 0:
print("Yes")
else:
print("No")
|
s607721623
|
p03730
|
u694665829
| 2,000
| 262,144
|
Wrong Answer
| 28
| 9,184
| 179
|
We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
a,b,c=map(int,input().split())
flag = False
for i in range(10000):
if (i*b+c)%a==0:
flag = True
break
if flag:
print('Yes')
else:
print('No')
|
s463995659
|
Accepted
| 32
| 9,152
| 174
|
a,b,c=map(int,input().split())
flag = False
for i in range(10000):
if (i*b+c)%a==0:
flag = True
break
if flag:
print('YES')
else:
print('NO')
|
s980734498
|
p02645
|
u309821257
| 2,000
| 1,048,576
|
Wrong Answer
| 19
| 9,020
| 109
|
When you asked some guy in your class his name, he called himself S, where S is a string of length between 3 and 20 (inclusive) consisting of lowercase English letters. You have decided to choose some three consecutive characters from S and make it his nickname. Print a string that is a valid nickname for him.
|
# -*- coding: utf-8 -*-
_in = input()
# if True:
# print(_in)
# else:
# print(_in)
print(_in[1:3])
|
s563077253
|
Accepted
| 21
| 8,888
| 110
|
# -*- coding: utf-8 -*-
_in = input()
# if True:
# print(_in)
# else:
# print(_in)
print(_in[0:3])
|
s846120238
|
p03565
|
u971124021
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 506
|
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()
s = [x for x in s[::-1]]
t = [x for x in t[::-1]]
for i in range(len(s)-len(t)):
flg = False
cnt = 0
for j in range(i+len(t)):
if s[i+j] == '?':
cnt += 1
continue
if s[i+j] == t[j]:
flg = True
else:
flg = False
break
if flg or cnt == len(t):
s[i:i+len(t)] = t
else:
print('UNRESTORABLE')
exit()
ans = ''
s = [x for x in s[::-1]]
for i in range(len(s)):
if s[i] == '?':
s[i] = 'a'
ans += s[i]
print(ans)
|
s057070929
|
Accepted
| 17
| 3,064
| 594
|
s = input()
t = input()
if len(s) < len(t):
print('UNRESTORABLE')
exit()
s = [x for x in s[::-1]]
t = [x for x in t[::-1]]
for i in range(len(s)-len(t)+1):
flg = False
cnt = 0
for j in range(len(t)):
if s[i+j] == '?':
cnt += 1
continue
if s[i+j] == t[j]:
flg = True
else:
flg = False
break
if flg or cnt == len(t):
s[i:i+len(t)] = t
flg = True
break
if not flg:
print('UNRESTORABLE')
exit()
ans = ''
s = [x for x in s[::-1]]
for i in range(len(s)):
if s[i] == '?':
s[i] = 'a'
ans += s[i]
print(ans)
|
s791823115
|
p03944
|
u207056619
| 2,000
| 262,144
|
Wrong Answer
| 60
| 3,064
| 1,196
|
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.
|
#!/usr/bin/env python3
import sys
def solve(W: int, H: int, N: int, x: "List[int]", y: "List[int]", a: "List[int]"):
cnt = 0
for y_ in range(H + 1):
for x_ in range(W + 1):
for i in range(N):
if a[i] == 1 and x_ < x[i] or a[i] == 2 and x_ > x[i] or a[i] == 3 and y_ < y[i] or a[i] == 4 and y_ > y[i]:
break
else:
cnt += 1
return cnt
# Generated by 1.1.6 https://github.com/kyuridenamida/atcoder-tools (tips: You use the default template now. You can remove this line by using your custom template)
def main():
def iterate_tokens():
for line in sys.stdin:
for word in line.split():
yield word
tokens = iterate_tokens()
W = int(next(tokens)) # type: int
H = int(next(tokens)) # type: int
N = int(next(tokens)) # type: int
x = [int()] * (N) # type: "List[int]"
y = [int()] * (N) # type: "List[int]"
a = [int()] * (N) # type: "List[int]"
for i in range(N):
x[i] = int(next(tokens))
y[i] = int(next(tokens))
a[i] = int(next(tokens))
solve(W, H, N, x, y, a)
if __name__ == '__main__':
main()
|
s779572908
|
Accepted
| 59
| 3,064
| 1,197
|
#!/usr/bin/env python3
import sys
def solve(W: int, H: int, N: int, x: "List[int]", y: "List[int]", a: "List[int]"):
cnt = 0
for y_ in range(H):
for x_ in range(W):
for i in range(N):
if a[i] == 1 and x_ < x[i] or a[i] == 2 and x_ >= x[i] or a[i] == 3 and y_ < y[i] or a[i] == 4 and y_ >= y[i]:
break
else:
cnt += 1
return cnt
# Generated by 1.1.6 https://github.com/kyuridenamida/atcoder-tools (tips: You use the default template now. You can remove this line by using your custom template)
def main():
def iterate_tokens():
for line in sys.stdin:
for word in line.split():
yield word
tokens = iterate_tokens()
W = int(next(tokens)) # type: int
H = int(next(tokens)) # type: int
N = int(next(tokens)) # type: int
x = [int()] * (N) # type: "List[int]"
y = [int()] * (N) # type: "List[int]"
a = [int()] * (N) # type: "List[int]"
for i in range(N):
x[i] = int(next(tokens))
y[i] = int(next(tokens))
a[i] = int(next(tokens))
print(solve(W, H, N, x, y, a))
if __name__ == '__main__':
main()
|
s620606983
|
p00101
|
u314832372
| 1,000
| 131,072
|
Wrong Answer
| 30
| 5,592
| 192
|
An English booklet has been created for publicizing Aizu to the world. When you read it carefully, you found a misnomer (an error in writing) on the last name of Masayuki Hoshina, the lord of the Aizu domain. The booklet says "Hoshino" not "Hoshina". Your task is to write a program which replace all the words "Hoshino" with "Hoshina". You can assume that the number of characters in a text is less than or equal to 1000.
|
var1 = int(input())
str2 = "Hoshino"
str3 = "Hoshina"
for i in range(0, var1):
str1 = input()
if str1.find(str2) >= 0:
str1 = str1.replace(str2, str3)
print(str1)
|
s488320142
|
Accepted
| 20
| 5,592
| 187
|
var1 = int(input())
str2 = "Hoshino"
str3 = "Hoshina"
for i in range(0, var1):
str1 = input()
if str1.find(str2) >= 0:
str1 = str1.replace(str2, str3)
print(str1)
|
s065435929
|
p00002
|
u855694108
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,592
| 136
|
Write a program which computes the digit number of sum of two integers a and b.
|
def main():
a, b = map(int, input().split())
hoge = list(str(a + b))
print(len(hoge))
if __name__ == "__main__":
main()
|
s417788840
|
Accepted
| 20
| 7,540
| 157
|
import sys
def main():
for line in sys.stdin:
a, b = map(int, line.split())
print(len(str(a + b)))
if __name__ == "__main__":
main()
|
s940389793
|
p03828
|
u822961851
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 306
|
You are given an integer N. Find the number of the positive divisors of N!, modulo 10^9+7.
|
def resolve():
n = int(input())
M = 10 ** 9 + 7
pl = [0] * (n + 1)
for i in range(2, n+1):
x = i
p = 2
while x > 1:
while x % p == 0:
pl[p] += 1
x //= p
p += 1
ans = 1
for s in pl:
ans *= (s+1)
|
s914474877
|
Accepted
| 34
| 3,060
| 375
|
def resolve():
n = int(input())
S = [0]*(n+1)
MOD = (10 ** 9) + 7
for i in range(1, n+1):
n_i = i
p = 2
while n_i > 1:
while n_i % p == 0:
n_i //= p
S[p] += 1
p += 1
ans = 1
for s in S:
ans *= s+1
print(ans % MOD)
if __name__ == '__main__':
resolve()
|
s165211349
|
p03448
|
u655834330
| 2,000
| 262,144
|
Wrong Answer
| 57
| 3,064
| 271
|
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 = input()
B = input()
C = input()
X = input()
A,B,C,X = map(int, [A,B,C,X])
count = 0
for a in range(A):
for b in range(B):
for c in range(C):
x = (a+1)*500 + (b+1)*100 + (c+1)*50
if X == x:
count += 1
print(count)
|
s074385346
|
Accepted
| 57
| 3,064
| 289
|
A = input()
B = input()
C = input()
X = input()
A,B,C,X = map(int, [A,B,C,X])
pattern_count = 0
for a in range(A+1):
for b in range(B+1):
for c in range(C+1):
x = a*500 + b*100 + c*50
if X == x:
pattern_count += 1
print(pattern_count)
|
s557807799
|
p02659
|
u999799597
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 8,964
| 57
|
Compute A \times B, truncate its fractional part, and print the result as an integer.
|
a, b = map(float, input().split())
print(int(a) * int(b))
|
s248330897
|
Accepted
| 26
| 9,016
| 90
|
a, b = map(float, input().split())
ib = b * 100
ans = int(a) * round(ib) // 100
print(ans)
|
s202402107
|
p03377
|
u384679440
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 112
|
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 < 0:
print("No")
elif X - A <= B:
print("Yes")
else:
print("No")
|
s223673901
|
Accepted
| 17
| 2,940
| 92
|
A, B, X = map(int, input().split())
if A + B <= X or A > X:
print("NO")
else:
print("YES")
|
s915751024
|
p02928
|
u075303794
| 2,000
| 1,048,576
|
Wrong Answer
| 2,312
| 1,340,836
| 221
|
We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}. Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2. Find the inversion number of B, modulo 10^9 + 7. Here the inversion number of B is defined as the number of ordered pairs of integers (i,~j)~(0 \leq i < j \leq K \times N - 1) such that B_i > B_j.
|
N,K = map(int,input().split())
A = list(map(int,input().split()))
B = []
ans = 0
{B.extend(A) for i in range(K)}
for i in range(len(B)-1):
for j in range(i,len(B)):
if B[i] > B[j]:
ans+=1
print(ans/(10**9+7))
|
s668695446
|
Accepted
| 29
| 3,564
| 682
|
from collections import Counter
class Bit:
def __init__(self, n):
self.size = n
self.tree = [0] * (n + 1)
def sum(self, i):
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def add(self, i, x):
while i <= self.size:
self.tree[i] += x
i += i & -i
n,k = map(int, input().split())
aaa = list(map(int, input().split()))
MOD = 10**9 + 7
bit = Bit(2001)
single = 0
for a in reversed(aaa):
single += bit.sum(a)
bit.add(a + 1, 1)
ans = single * k % MOD
coef = k * (k - 1) // 2 % MOD
cnt = Counter(aaa)
before = 0
for a, c in sorted(cnt.items()):
ans = (ans + c * before * coef) % MOD
before += c
print(ans)
|
s261811416
|
p03474
|
u178888901
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 311
|
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.
|
strnum = [str(i) for i in range(10)]
a, b = input().split()
a, b = int(a), int(b)
S = input()
judge = True
for i in range(a + b + 1):
if i != a:
if i in strnum:
continue
else:
judge = False
else:
if S[i] == "-":
continue
else:
judge = False
if judge:
print("Yes")
else:
print("NO")
|
s316811382
|
Accepted
| 18
| 3,060
| 227
|
strnum = [str(i) for i in range(10)]
a, b = input().split()
a, b = int(a), int(b)
S = input()
S = S.split("-")
if len(S) == 2:
if len(S[0]) == a and len(S[1]) == b:
print('Yes')
else:
print('No')
else:
print('No')
|
s767846305
|
p03555
|
u662449766
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 148
|
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
input = sys.stdin.readline
def main():
print("YES" if input() == input()[::-1] else "NO")
if __name__ == "__main__":
main()
|
s960668772
|
Accepted
| 17
| 2,940
| 107
|
def main():
print("YES" if input() == input()[::-1] else "NO")
if __name__ == "__main__":
main()
|
s517261847
|
p04011
|
u057964173
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 304
|
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.
|
import sys
def input(): return sys.stdin.readline().strip()
def resolve():
w=list(input())
w.sort()
if len(w)%2!=0:
print('No')
else:
for i in range(0,len(w),2):
if w[i]!=w[i+1]:
print('No')
break
print('Yes')
resolve()
|
s123228170
|
Accepted
| 17
| 2,940
| 168
|
def resolve():
n=int(input())
k=int(input())
x=int(input())
y=int(input())
if n<k:
print(n*x)
else:
print(k*x+(n-k)*y)
resolve()
|
s718815621
|
p02401
|
u152353734
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,416
| 77
|
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 = input()
if '?' in a:
break
print(eval(a))
|
s440570500
|
Accepted
| 20
| 7,496
| 84
|
while True:
a = input()
if '?' in a:
break
print("%d" % eval(a))
|
s489848426
|
p03860
|
u559346857
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 26
|
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
s=input()
print("A"+s+"B")
|
s368810351
|
Accepted
| 17
| 2,940
| 25
|
print("A"+input()[8]+"C")
|
s499154798
|
p03408
|
u147458211
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 740
|
Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i. Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen. Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.) At most how much can he earn on balance? Note that the same string may be written on multiple cards.
|
N = int(input())
words = []
blue_cards = {}
for i in range(N):
s = input()
if s in blue_cards:
blue_cards[s] = blue_cards[s] + 1
else:
blue_cards[s] = 1
if s not in words:
words.append(s)
M = int(input())
red_cards= {}
for i in range(M):
s = input()
if s in red_cards:
red_cards[s] = red_cards[s] + 1
else:
red_cards[s] = 1
if s not in words:
words.append(s)
print(blue_cards)
print(red_cards)
max_money = -100
for word in words:
get_money, lose_money = 0, 0
if word in blue_cards:
get_money = blue_cards[word]
if word in red_cards:
lose_money = red_cards[word]
print("{} {}".format(get_money, lose_money))
max_money = max(max_money, get_money - lose_money)
print(max_money)
|
s661467812
|
Accepted
| 17
| 3,064
| 654
|
N = int(input())
words = []
blue_cards = {}
for i in range(N):
s = input()
if s in blue_cards:
blue_cards[s] = blue_cards[s] + 1
else:
blue_cards[s] = 1
if s not in words:
words.append(s)
M = int(input())
red_cards= {}
for i in range(M):
s = input()
if s in red_cards:
red_cards[s] = red_cards[s] + 1
else:
red_cards[s] = 1
if s not in words:
words.append(s)
max_money = 0
for word in words:
get_money, lose_money = 0, 0
if word in blue_cards:
get_money = blue_cards[word]
if word in red_cards:
lose_money = red_cards[word]
max_money = max(max_money, get_money - lose_money)
print(max_money)
|
s896776295
|
p03737
|
u159335277
| 2,000
| 262,144
|
Wrong Answer
| 30
| 8,972
| 52
|
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.
|
print(''.join(map(lambda x: x[0], input().split())))
|
s478965651
|
Accepted
| 34
| 8,984
| 61
|
print(''.join(map(lambda x: x[0], input().split())).upper())
|
s064291081
|
p02388
|
u138422838
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,320
| 22
|
Write a program which calculates the cube of a given integer x.
|
x = 3
y = x^3
print(y)
|
s685507751
|
Accepted
| 20
| 7,640
| 34
|
x = input()
y = int(x)**3
print(y)
|
s964204350
|
p03407
|
u041196979
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 90
|
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())
if A + B <= C:
print("Yes")
else:
print("No")
|
s808669549
|
Accepted
| 17
| 2,940
| 89
|
A, B, C = map(int, input().split())
if A + B >= C:
print("Yes")
else:
print("No")
|
s373610105
|
p02865
|
u550061714
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 32
|
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
print(round((int(input())-1)/2))
|
s404615660
|
Accepted
| 17
| 2,940
| 49
|
import math
print(math.floor((int(input())-1)/2))
|
s569879071
|
p00016
|
u553148578
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,724
| 200
|
When a boy was cleaning up after his grand father passing, he found an old paper: In addition, other side of the paper says that "go ahead a number of steps equivalent to the first integer, and turn clockwise by degrees equivalent to the second integer". His grand mother says that Sanbonmatsu was standing at the center of town. However, now buildings are crammed side by side and people can not walk along exactly what the paper says in. Your task is to write a program which hunts for the treature on the paper. For simplicity, 1 step is equivalent to 1 meter. Input consists of several pairs of two integers d (the first integer) and t (the second integer) separated by a comma. Input ends with "0, 0". Your program should print the coordinate (x, y) of the end point. There is the treature where x meters to the east and y meters to the north from the center of town. You can assume that d ≤ 100 and -180 ≤ t ≤ 180\.
|
import math
x = y = n= 0
while True:
d,a = map(int,input().split(','))
if d == a == 0: break
x += d*math.sin(math.radians(n))
y += d*math.cos(math.radians(n))
n+=a
print(map(int,(x,y)),sep='\n')
|
s851284976
|
Accepted
| 20
| 5,716
| 201
|
import math
x = y = n= 0
while True:
d,a = map(int,input().split(','))
if d == a == 0: break
x += d*math.sin(math.radians(n))
y += d*math.cos(math.radians(n))
n+=a
print(*map(int,(x,y)),sep='\n')
|
s334230411
|
p02406
|
u483716678
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,588
| 127
|
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; }
|
n = int(input())
for x in range(n+1):
if(x%3==0):
print(x,end=' ')
elif(x%10 == 3):
print(x,end=' ')
|
s416708655
|
Accepted
| 30
| 5,868
| 264
|
n = int(input())
y = 1
while y <= n :
if(y % 3 == 0):
print(' %d'%y,end='')
else:
p = y
while(p != 0):
if(p % 10 == 3):
print(' %d'%y,end='')
break
p //= 10
y+=1
print()
|
s116730671
|
p02850
|
u685983477
| 2,000
| 1,048,576
|
Wrong Answer
| 808
| 65,364
| 1,007
|
Given is a tree G with N vertices. The vertices are numbered 1 through N, and the i-th edge connects Vertex a_i and Vertex b_i. Consider painting the edges in G with some number of colors. We want to paint them so that, for each vertex, the colors of the edges incident to that vertex are all different. Among the colorings satisfying the condition above, construct one that uses the minimum number of colors.
|
from collections import deque
def main():
n = int(input())
tree = []
edge = [[] for _ in range(n)]
for i in range(n-1):
a,b=map(int, input().split())
tree.append([min(a,b),max(a,b),n])
edge[a-1].append([b-1,i])
edge[b-1].append([a-1,i])
cols = [0]*(n-1)
max_c = [0]*(n)
tree.sort(key = lambda x:x[0])
que=deque()
c = 1
for e in edge[0]:
que.append([e,c])
c+=1
for i in range(n):
max_c[i] = len(edge[i])
while que:
v=que.popleft()
ver = v[0]
color = v[1]
if(cols[ver[1]]>0):
continue
else:
cols[ver[1]] = color
new_c = 1
for next_v in edge[ver[0]]:
if(new_c==color):
new_c += 1
que.append([next_v,new_c])
new_c += 1
print(max(max_c))
for i in range(n-1):
print(cols[i])
if __name__ == '__main__':
main()
|
s465214691
|
Accepted
| 692
| 46,832
| 782
|
from collections import deque,defaultdict
n = int(input())
edges = [[] for _ in range(n)]
for i in range(n-1):
a,b=map(int, input().split())
edges[a-1].append([b-1,i])
edges[b-1].append([a-1,i])
c = 1
col = [0]*(n-1)
visited = [0]*(n-1)
que=deque()
for v in edges[0]:
que.append(v)
col[v[1]]=c
visited[v[1]]=1
c+=1
while que:
v=que.popleft()
c = 1
used = set()
for i in edges[v[0]]:
if(visited[i[1]]):
used.add(col[i[1]])
for next_v in edges[v[0]]:
if(not visited[next_v[1]]):
if(c in used):
c+=1
col[next_v[1]] = c
used.add(c)
c+=1
visited[next_v[1]] = 1
que.append(next_v)
print(max(col))
for i in col:
print(i)
|
s852102285
|
p03195
|
u291732989
| 2,000
| 1,048,576
|
Wrong Answer
| 202
| 7,072
| 183
|
There is an apple tree that bears apples of N colors. The N colors of these apples are numbered 1 to N, and there are a_i apples of Color i. You and Lunlun the dachshund alternately perform the following operation (starting from you): * Choose one or more apples from the tree and eat them. Here, the apples chosen at the same time must all have different colors. The one who eats the last apple from the tree will be declared winner. If both you and Lunlun play optimally, which will win?
|
N = int(input())
An = []
for p in range(N):
Ai = int(input())
An.append(Ai)
res = True
for i in An:
if i % 2 != 0:
res = False
print("first" if res else "second")
|
s683037251
|
Accepted
| 197
| 7,072
| 183
|
N = int(input())
An = []
for p in range(N):
Ai = int(input())
An.append(Ai)
res = True
for i in An:
if i % 2 != 0:
res = False
print("second" if res else "first")
|
s922192844
|
p03369
|
u518556834
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 51
|
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
s = list(input())
print(700 + 100 * s.count("○"))
|
s646526931
|
Accepted
| 17
| 2,940
| 88
|
s = input()
c = 0
for i in range(3):
if s[i] == "o":
c += 1
print(700 + c*100)
|
s491806829
|
p03494
|
u642682703
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 218
|
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.
|
evens = [6,4,10,4]
itr = 0
flg = True
while flg:
evens = list(map(lambda x: (x/2), evens))
is_evens = list(map(lambda x: x.is_integer(), evens))
if not all(is_evens):
flg = False
break
itr+=1
print(itr)
|
s872363370
|
Accepted
| 19
| 3,060
| 257
|
num = input()
evens = [int(x) for x in input().split()]
itr = 0
flg = True
while flg:
evens = list(map(lambda x: (x/2), evens))
is_evens = list(map(lambda x: x.is_integer(), evens))
if not all(is_evens):
flg = False
break
itr+=1
print(itr)
|
s242253957
|
p02393
|
u748921161
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,704
| 139
|
Write a program which reads three integers, and prints them in ascending order.
|
input_str = input().split(' ')
input_int = []
for value in input_str:
input_int.append(int(value))
input_int.sort();
print(input_int);
|
s761463527
|
Accepted
| 20
| 7,692
| 181
|
input_str = input().split(' ')
input_int = []
for value in input_str:
input_int.append(int(value))
input_int.sort();
print('%d %d %d'%(input_int[0],input_int[1],input_int[2]));
|
s960746747
|
p03139
|
u227082700
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 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.
|
a,b,c=map(int,input().split());print(max(b,c),max(0,b+c-a))
|
s173186197
|
Accepted
| 17
| 2,940
| 59
|
a,b,c=map(int,input().split());print(min(b,c),max(0,b+c-a))
|
s641890480
|
p02856
|
u716530146
| 2,000
| 1,048,576
|
Wrong Answer
| 252
| 17,384
| 76
|
N programmers are going to participate in the preliminary stage of DDCC 20XX. Due to the size of the venue, however, at most 9 contestants can participate in the finals. The preliminary stage consists of several rounds, which will take place as follows: * All the N contestants will participate in the first round. * When X contestants participate in some round, the number of contestants advancing to the next round will be decided as follows: * The organizer will choose two consecutive digits in the decimal notation of X, and replace them with the sum of these digits. The number resulted will be the number of contestants advancing to the next round. For example, when X = 2378, the number of contestants advancing to the next round will be 578 (if 2 and 3 are chosen), 2108 (if 3 and 7 are chosen), or 2315 (if 7 and 8 are chosen). When X = 100, the number of contestants advancing to the next round will be 10, no matter which two digits are chosen. * The preliminary stage ends when 9 or fewer contestants remain. Ringo, the chief organizer, wants to hold as many rounds as possible. Find the maximum possible number of rounds in the preliminary stage. Since the number of contestants, N, can be enormous, it is given to you as two integer sequences d_1, \ldots, d_M and c_1, \ldots, c_M, which means the following: the decimal notation of N consists of c_1 + c_2 + \ldots + c_M digits, whose first c_1 digits are all d_1, the following c_2 digits are all d_2, \ldots, and the last c_M digits are all d_M.
|
_,*t=open(0)
a=-18
for t in t:d,c=map(int,t.split());a+=d*c+c*9
print(a//9)
|
s813984942
|
Accepted
| 612
| 3,060
| 102
|
a=0
m=int(input())
for i in range(m):
d,c=map(int,input().split())
a+=c*9+d*c
print((a-10)//9)
|
s227315889
|
p03394
|
u694433776
| 2,000
| 262,144
|
Wrong Answer
| 32
| 5,084
| 608
|
Nagase is a top student in high school. One day, she's analyzing some properties of special sets of positive integers. She thinks that a set S = \\{a_{1}, a_{2}, ..., a_{N}\\} of **distinct** positive integers is called **special** if for all 1 \leq i \leq N, the gcd (greatest common divisor) of a_{i} and the sum of the remaining elements of S is **not** 1. Nagase wants to find a **special** set of size N. However, this task is too easy, so she decided to ramp up the difficulty. Nagase challenges you to find a **special** set of size N such that the gcd of all elements are 1 and the elements of the set does not exceed 30000.
|
n=int(input())
ret=[2,3,4] if n>=6 else [2,3,25] if n==3 else [2,3,5,30] if n==4 else [2,3,5,30,60]
if n>=6:
def getnext(a):
if a%6==0:
return a+2
elif a%6==2:
return a+1
elif a%6==3:
return a+1
elif a%6==4:
return a+2
while len(ret)<n:
ret.append(getnext(ret[-1]))
if sum(ret)%6==5:
ret[-1]=getnext(ret[-1])
elif sum(ret)%6!=0:
ret.remove((sum(ret)%6)+6)
ret.append(getnext(ret[-1]))
while sum(ret)%6!=0:
ret[-1]=getnext(ret[-1])
print(" ".join(map(str,ret)))
|
s223478673
|
Accepted
| 32
| 5,084
| 610
|
n=int(input())
ret=[2,3,4] if n>=6 else [2,5,63] if n==3 else [2,5,20,63] if n==4 else [2,5,20,30,63]
if n>=6:
def getnext(a):
if a%6==0:
return a+2
elif a%6==2:
return a+1
elif a%6==3:
return a+1
elif a%6==4:
return a+2
while len(ret)<n:
ret.append(getnext(ret[-1]))
if sum(ret)%6==5:
ret[-1]=getnext(ret[-1])
elif sum(ret)%6!=0:
ret.remove((sum(ret)%6)+6)
ret.append(getnext(ret[-1]))
while sum(ret)%6!=0:
ret[-1]=getnext(ret[-1])
print(" ".join(map(str,ret)))
|
s087547027
|
p03574
|
u887207211
| 2,000
| 262,144
|
Wrong Answer
| 23
| 3,064
| 766
|
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())
S = [input() for _ in range(H)]
result = [[0 for _ in range(W+1)]for _ in range(H+1)]
for i in range(H):
for j in range(W):
if(S[i][j] == '#'):
result[i][j] = '#'
if(result[i-1][j-1] != '#'):
result[i-1][j-1] += 1
if(result[i-1][j] != '#'):
result[i-1][j] += 1
if(result[i-1][j+1] != '#'):
result[i-1][j+1] += 1
if(result[i][j-1] != '#'):
result[i][j-1] += 1
if(result[i][j+1] != '#'):
result[i][j+1] += 1
if(result[i+1][j-1] != '#'):
result[i+1][j-1] += 1
if(result[i+1][j] != '#'):
result[i+1][j] += 1
if(result[i+1][j+1] != '#'):
result[i+1][j+1] += 1
for r in result[:-1]:
''.join(list(map(str,r[:H])))
|
s287803037
|
Accepted
| 24
| 3,188
| 764
|
H, W = map(int,input().split())
S = [input() for _ in range(H)]
reach = [[0 for _ in range(W+1)] for _ in range(H+1)]
for i in range(H):
for j in range(W):
if(S[i][j] == "#"):
reach[i][j] = "#"
if(reach[i-1][j-1] != "#"):
reach[i-1][j-1] += 1
if(reach[i-1][j] != "#"):
reach[i-1][j] += 1
if(reach[i-1][j+1] != "#"):
reach[i-1][j+1] += 1
if(reach[i][j-1] != "#"):
reach[i][j-1] += 1
if(reach[i][j+1] != "#"):
reach[i][j+1] += 1
if(reach[i+1][j-1] != "#"):
reach[i+1][j-1] += 1
if(reach[i+1][j] != "#"):
reach[i+1][j] += 1
if(reach[i+1][j+1] != "#"):
reach[i+1][j+1] += 1
for r in reach[:-1]:
print("".join(list(map(str, r[:W]))))
|
s258856945
|
p02243
|
u684241248
| 1,000
| 131,072
|
Wrong Answer
| 30
| 6,036
| 4,527
|
For a given weighted graph $G = (V, E)$, find the shortest path from a source to each vertex. For each vertex $u$, print the total weight of edges on the shortest path from vertex $0$ to $u$.
|
# -*- coding: utf-8 -*-
from collections import defaultdict
from heapq import heappop, heappush
def dijkstra(edges, start, *, adj_matrix=False, default_value=float('inf')):
'''
Returns best costs to each node from 'start' node in the given graph.
(Single Source Shortest Path - SSSP)
If edges is given as an adjacency list including costs and destination nodes on possible edges, adj_matrix should be False(default), and it is generally when this functions works much faster.
Note that costs in edges should follow the rules below
1. The cost from a node itself should be 0,
2. If there is no edge between nodes, the cost between them should be default_value
3. The costs can't be negative
# sample input when given as adjacency list (adj_matrix=False)
edges = [[(1, 2), (2, 5), (3, 4)], # node 0
[(0, 2), (3, 3), (4, 6)], # node 1
[(0, 5), (3, 2), (5, 6)], # node 2
[(0, 4), (1, 3), (2, 2), (4, 2)], # node 3
[(1, 6), (3, 2), (5, 4)], # node 4
[(2, 6), (4, 4)]] # node 5
# sample input when given as adjacency matrix (adj_matrix=True)
edges = [[0, 2, 5, 4, inf, inf], # node 0
[2, 0, inf, 3, 6, inf], # node 1
[5, inf, 0, 2, inf, 6], # node 2
[4, 3, 2, 0, 2, inf], # node 3
[inf, 6, inf, 2, 0, 4], # node 4
[inf, inf, 6, inf, 4, 0]] # node 5
'''
n = len(edges)
inf = float('inf')
# costs = [inf] * n
costs = defaultdict(lambda: inf)
costs[start] = 0
pq, rem = [(0, start)], n - 1
while pq and rem:
tmp_cost, tmp_node = heappop(pq)
if costs[tmp_node] < tmp_cost:
continue
rem -= 1
nxt_edges = ((node, cost) for node, cost in enumerate(edges[tmp_node]) if cost != default_value) if adj_matrix else\
edges[tmp_node]
for nxt_node, nxt_cost in nxt_edges:
new_cost = tmp_cost + nxt_cost
if costs[nxt_node] > new_cost:
costs[nxt_node] = new_cost
heappush(pq, (new_cost, nxt_node))
return costs
def dijkstra_route(edges,
start,
goal,
*,
adj_matrix=False,
default_value=float('inf'),
verbose=False):
'''
Trys to find the best route to the 'goal' from the 'start'
'''
n = len(edges)
inf = float('inf')
# costs = [inf] * n
costs = defaultdict(lambda: inf)
costs[start] = 0
pq, rem = [(0, start)], n - 1
# prevs = [-1 for _ in [None] * n]
prevs = defaultdict(lambda: -1)
while pq and rem:
tmp_cost, tmp_node = heappop(pq)
if costs[tmp_node] < tmp_cost:
continue
rem -= 1
nxt_edges = ((node, cost) for node, cost in enumerate(edges[tmp_node]) if cost != default_value) if adj_matrix else\
edges[tmp_node]
for nxt_node, nxt_cost in nxt_edges:
new_cost = tmp_cost + nxt_cost
if costs[nxt_node] > new_cost:
costs[nxt_node] = new_cost
heappush(pq, (new_cost, nxt_node))
prevs[nxt_node] = tmp_node
min_route = []
prev = goal
cnt = 0
while prev != start:
min_route.append(prev)
prev = prevs[prev]
cnt += 1
if prev == -1 or cnt > n:
raise NoRouteError(
'There is no possible route in this graph \nedges: {} \nstart: {} \ngoal: {}'.
format(edges, start, goal))
else:
min_route.append(prev)
min_route = min_route[::-1]
min_cost = costs[goal]
if verbose:
print('---route---')
for node in min_route[:-1]:
print('{} -> '.format(node), end='')
else:
print(min_route[-1])
print('---distance---')
print(min_cost)
return costs, min_route
class NoRouteError(Exception):
pass
if __name__ == '__main__':
'''
an example of using this function
AIZU ONLINE JUDGE - ALDS_1_12_C
'''
n = int(input())
edges = [[] for i in range(n)]
for i in range(n):
i, k, *kedges = map(int, input().split())
for edge in zip(kedges[::2], kedges[1::2]):
edges[i].append(edge)
for i, cost in enumerate(dijkstra(edges, 0)):
print(i, cost)
|
s889552347
|
Accepted
| 460
| 68,708
| 4,374
|
# -*- coding: utf-8 -*-
from heapq import heappop, heappush
def dijkstra(edges, start, *, adj_matrix=False, default_value=float('inf')):
'''
Returns best costs to each node from 'start' node in the given graph.
(Single Source Shortest Path - SSSP)
If edges is given as an adjacency list including costs and destination nodes on possible edges, adj_matrix should be False(default), and it is generally when this functions works much faster.
Note that costs in edges should follow the rules below
1. The cost from a node itself should be 0,
2. If there is no edge between nodes, the cost between them should be default_value
3. The costs can't be negative
# sample input when given as adjacency list (adj_matrix=False)
edges = [[(1, 2), (2, 5), (3, 4)], # node 0
[(0, 2), (3, 3), (4, 6)], # node 1
[(0, 5), (3, 2), (5, 6)], # node 2
[(0, 4), (1, 3), (2, 2), (4, 2)], # node 3
[(1, 6), (3, 2), (5, 4)], # node 4
[(2, 6), (4, 4)]] # node 5
# sample input when given as adjacency matrix (adj_matrix=True)
edges = [[0, 2, 5, 4, inf, inf], # node 0
[2, 0, inf, 3, 6, inf], # node 1
[5, inf, 0, 2, inf, 6], # node 2
[4, 3, 2, 0, 2, inf], # node 3
[inf, 6, inf, 2, 0, 4], # node 4
[inf, inf, 6, inf, 4, 0]] # node 5
'''
n = len(edges)
costs = [float('inf')] * n
costs[start] = 0
pq = [(0, start)]
rem = n - 1
while pq and rem:
tmp_cost, tmp_node = heappop(pq)
if costs[tmp_node] < tmp_cost:
continue
rem -= 1
nxt_edges = ((node, cost) for node, cost in enumerate(edges[tmp_node]) if cost != default_value) if adj_matrix else\
edges[tmp_node]
for nxt_node, nxt_cost in nxt_edges:
new_cost = tmp_cost + nxt_cost
if costs[nxt_node] > new_cost:
costs[nxt_node] = new_cost
heappush(pq, (new_cost, nxt_node))
return costs
def dijkstra_route(edges,
start,
goal,
*,
adj_matrix=False,
default_value=float('inf'),
verbose=False):
'''
Trys to find the best route to the 'goal' from the 'start'
'''
n = len(edges)
costs = [float('inf')] * n
costs[start] = 0
pq = [(0, start)]
rem = n - 1
prevs = [-1 for _ in [None] * n]
while pq and rem:
tmp_cost, tmp_node = heappop(pq)
if costs[tmp_node] < tmp_cost:
continue
rem -= 1
nxt_edges = ((node, cost) for node, cost in enumerate(edges[tmp_node]) if cost != default_value) if adj_matrix else\
edges[tmp_node]
for nxt_node, nxt_cost in nxt_edges:
new_cost = tmp_cost + nxt_cost
if costs[nxt_node] > new_cost:
costs[nxt_node] = new_cost
heappush(pq, (new_cost, nxt_node))
prevs[nxt_node] = tmp_node
min_route = []
prev = goal
cnt = 0
while prev != start:
min_route.append(prev)
prev = prevs[prev]
cnt += 1
if prev == -1 or cnt > n:
raise NoRouteError(
'There is no possible route in this graph \nedges: {} \nstart: {} \ngoal: {}'.
format(edges, start, goal))
else:
min_route.append(prev)
min_route = min_route[::-1]
min_cost = costs[goal]
if verbose:
print('---route---')
for node in min_route[:-1]:
print('{} -> '.format(node), end='')
else:
print(min_route[-1])
print('---distance---')
print(min_cost)
return costs, min_route
class NoRouteError(Exception):
pass
if __name__ == '__main__':
'''
an example of using this function
AIZU ONLINE JUDGE - ALDS_1_12_C
'''
n = int(input())
edges = [[] for i in range(n)]
for i in range(n):
i, k, *kedges = map(int, input().split())
for edge in zip(kedges[::2], kedges[1::2]):
edges[i].append(edge)
costs = dijkstra(edges, 0)
for i, cost in enumerate(costs):
print(i, cost)
|
s000556123
|
p02853
|
u732870425
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 170
|
We held two competitions: Coding Contest and Robot Maneuver. In each competition, the contestants taking the 3-rd, 2-nd, and 1-st places receive 100000, 200000, and 300000 yen (the currency of Japan), respectively. Furthermore, a contestant taking the first place in both competitions receives an additional 400000 yen. DISCO-Kun took the X-th place in Coding Contest and the Y-th place in Robot Maneuver. Find the total amount of money he earned.
|
X, Y = map(int, input().split())
ans = 0
if X >= 3:
ans += (4 - X) * 100000
if Y >= 3:
ans += (4 - Y) * 100000
if X == 1 and Y == 1:
ans += 400000
print(ans)
|
s970646842
|
Accepted
| 17
| 2,940
| 170
|
X, Y = map(int, input().split())
ans = 0
if X <= 3:
ans += (4 - X) * 100000
if Y <= 3:
ans += (4 - Y) * 100000
if X == 1 and Y == 1:
ans += 400000
print(ans)
|
s743337277
|
p04029
|
u876742094
| 2,000
| 262,144
|
Wrong Answer
| 28
| 9,008
| 32
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
n=int(input())
print(n*(n+1)/2)
|
s581540154
|
Accepted
| 26
| 9,140
| 33
|
n=int(input())
print(n*(n+1)//2)
|
s908666455
|
p03696
|
u329865314
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 223
|
You are given a string S of length N consisting of `(` and `)`. Your task is to insert some number of `(` and `)` into S to obtain a _correct bracket sequence_. Here, a correct bracket sequence is defined as follows: * `()` is a correct bracket sequence. * If X is a correct bracket sequence, the concatenation of `(`, X and `)` in this order is also a correct bracket sequence. * If X and Y are correct bracket sequences, the concatenation of X and Y in this order is also a correct bracket sequence. * Every correct bracket sequence can be derived from the rules above. Find the shortest correct bracket sequence that can be obtained. If there is more than one such sequence, find the lexicographically smallest one.
|
n = int(input())
s = input()
ans = ""
cur = 0
for i in s:
if i == "(":
ans += "("
cur += 1
ans += i
else:
if cur:
ans += i
cur -= 1
else:
ans = "(" + ans
ans += ")" * cur
print(ans)
|
s252872774
|
Accepted
| 17
| 3,060
| 212
|
n = int(input())
s = input()
ans = ""
cur = 0
for i in s:
if i == "(":
cur += 1
ans += i
else:
if cur:
ans += i
cur -= 1
else:
ans = "(" + ans + i
ans += ")" * cur
print(ans)
|
s894562398
|
p02612
|
u556477263
| 2,000
| 1,048,576
|
Wrong Answer
| 31
| 9,004
| 116
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
import sys
n = int(input())
if n % 1000 == 0:
print(0)
sys.exit()
while n > 1000:
n -= 1000
print(n)
|
s220168034
|
Accepted
| 34
| 9,108
| 123
|
import sys
n = int(input())
if n % 1000 == 0:
print(0)
sys.exit()
while n > 1000:
n -= 1000
print(1000 - n)
|
s420104534
|
p03090
|
u801049006
| 2,000
| 1,048,576
|
Wrong Answer
| 30
| 9,072
| 57
|
You are given an integer N. Build an undirected graph with N vertices with indices 1 to N that satisfies the following two conditions: * The graph is simple and connected. * There exists an integer S such that, for every vertex, the sum of the indices of the vertices adjacent to that vertex is S. It can be proved that at least one such graph exists under the constraints of this problem.
|
n = int(input())
for i in range(n-1):
print(i+1, n)
|
s042087256
|
Accepted
| 33
| 9,696
| 233
|
n = int(input())
d = n
ans = []
if n % 2 == 1: d -= 1
for i in range(1, n+1):
for j in range(1, n+1):
if j <= i or j == d: continue
ans.append([i, j])
d -= 1
print(len(ans))
for f, s in ans:
print(f, s)
|
s371745677
|
p02972
|
u143903328
| 2,000
| 1,048,576
|
Wrong Answer
| 1,080
| 17,200
| 367
|
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()))
c = 0
ans = []
for i in range(n):
m = n-i
p = 1
b = 0
while m * p <= n:
tmp = m * p
if a[tmp-1] == 1:
b += 1
p += 1
if b % 2 == 1:
ans.append(m)
c += 1
if c%2 == 1:
print('-1')
else:
print(c)
print(" ".join(repr(e) for e in ans))
|
s803347793
|
Accepted
| 1,084
| 19,856
| 405
|
n = int(input())
a = list(map(int,input().split()))
c = 0
d = [0] * n
ans = []
for i in range(n):
m = n-i
p = 1
b = 0
while m * p <= n:
tmp = m * p
if d[tmp-1] == 1:
b += 1
p += 1
if b % 2 != a[m-1]:
d[m-1] = 1
ans.append(m)
c += 1
if c%2 != a[0]:
print('-1')
else:
print(c)
print(" ".join(repr(e) for e in ans))
|
s532288470
|
p03494
|
u434311880
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 153
|
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())
b = (int(s) for s in input().split())
x = 0
for i in b:
i = i // 2
c = i % 2
x = x + 1
if c != 0:
break
print(x)
|
s783035642
|
Accepted
| 18
| 2,940
| 144
|
N = int(input())
a = list(map(int, input().split()))
cnt = 0
while all(i % 2 == 0 for i in a):
a = [i/2 for i in a]
cnt += 1
print(cnt)
|
s316827162
|
p03997
|
u840570107
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 84
|
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())
ans = (a + b) * h / 2
print(ans)
|
s179318234
|
Accepted
| 17
| 2,940
| 89
|
a = int(input())
b = int(input())
h = int(input())
ans = (a + b) * h / 2
print(int(ans))
|
s567785989
|
p03455
|
u662573896
| 2,000
| 262,144
|
Wrong Answer
| 26
| 9,112
| 97
|
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 == 1:
print ('odd')
else:
print ('even')
|
s666067491
|
Accepted
| 26
| 8,936
| 97
|
a, b = map(int, input().split())
if (a * b) % 2 == 1:
print ('Odd')
else:
print ('Even')
|
s760652623
|
p03158
|
u353797797
| 2,000
| 1,048,576
|
Wrong Answer
| 1,240
| 25,136
| 804
|
There are N cards. The i-th card has an integer A_i written on it. For any two cards, the integers on those cards are different. Using these cards, Takahashi and Aoki will play the following game: * Aoki chooses an integer x. * Starting from Takahashi, the two players alternately take a card. The card should be chosen in the following manner: * Takahashi should take the card with the largest integer among the remaining card. * Aoki should take the card with the integer closest to x among the remaining card. If there are multiple such cards, he should take the card with the smallest integer among those cards. * The game ends when there is no card remaining. You are given Q candidates for the value of x: X_1, X_2, ..., X_Q. For each i (1 \leq i \leq Q), find the sum of the integers written on the cards that Takahashi will take if Aoki chooses x = X_i.
|
import sys
from bisect import *
sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
def solve(x):
l=bisect_left(aa,x)
r=n+1
while l+1<r:
j=(l+r)//2
s=x*2-aa[j-1]
i=bisect_left(aa,s)
if j-i>n-j:r=j
else:l=j
ans=cs[n]-cs[l]+cs2[n-(n-l)*2]
print(ans)
n,q=MI()
aa=LI()
cs=[0]
cs2=[0]*(n+1)
for i,a in enumerate(aa):
cs.append(cs[-1]+a)
if i<2:cs2[i+1]=a
else:cs2[i+1]=cs2[i-1]+a
for _ in range(q):
x=II()
solve(x)
|
s060800740
|
Accepted
| 1,300
| 24,544
| 829
|
import sys
from bisect import *
sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
def solve(x):
l=bisect_left(aa,x)
r=n+1
while l+1<r:
j=(l+r)//2
s=x*2-aa[j-1]
i=bisect_left(aa,s)
if j-i>n-j:r=j
else:l=j
ans=cs[n]-cs[l]
if n-(n-l)*2>0:ans+=+cs2[n-(n-l)*2]
print(ans)
n,q=MI()
aa=LI()
cs=[0]
cs2=[0]*(n+1)
for i,a in enumerate(aa):
cs.append(cs[-1]+a)
if i<2:cs2[i+1]=a
else:cs2[i+1]=cs2[i-1]+a
for _ in range(q):
x=II()
solve(x)
|
s848349486
|
p03711
|
u702719601
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 248
|
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.
|
data = input().split()
x = int(data[0])
y = int(data[1])
L1 = [1,3,5,7,8,10,12]
L2 = [4,6,9,11]
L3 = [2]
print(L1)
print(L2)
print(L3)
if((x in L1 and y in L1) or (x in L2 and y in L2) or (x in L3 and y in L3)):
print("Yes")
else:
print("No")
|
s603266797
|
Accepted
| 17
| 3,060
| 217
|
data = input().split()
x = int(data[0])
y = int(data[1])
L1 = [1,3,5,7,8,10,12]
L2 = [4,6,9,11]
L3 = [2]
if((x in L1 and y in L1) or (x in L2 and y in L2) or (x in L3 and y in L3)):
print("Yes")
else:
print("No")
|
s952858009
|
p03478
|
u343150957
| 2,000
| 262,144
|
Wrong Answer
| 28
| 3,060
| 185
|
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())
count = 0
for i in range(n+1):
ans = 0
while (i // 10) != 0:
i = (i // 10)
ans += (i % 10)
if ans >= a and ans <= b:
count += 1
print(count)
|
s167608753
|
Accepted
| 37
| 3,060
| 144
|
n, a, b = map(int, input().split())
ans = 0
for i in range(n+1):
if a <= sum(list(map(int, list(str(i))))) <= b:
ans += i
print(ans)
|
s575658305
|
p03609
|
u865413330
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 109
|
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?
|
s = list(input())
result = ""
for i in range(len(s)):
if i % 2 != 0:
result += s[i]
print(result)
|
s331818852
|
Accepted
| 17
| 2,940
| 69
|
x, t = map(int, input().split())
print(x - t) if x >= t else print(0)
|
s403415258
|
p02663
|
u497356352
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 9,412
| 601
|
In this problem, we use the 24-hour clock. Takahashi gets up exactly at the time H_1 : M_1 and goes to bed exactly at the time H_2 : M_2. (See Sample Inputs below for clarity.) He has decided to study for K consecutive minutes while he is up. What is the length of the period in which he can start studying?
|
# coding:utf-8
import sys
import datetime
readline = sys.stdin.buffer.readline
try:
H1 = int(readline())
M1 = int(readline())
H2 = int(readline())
M2 = int(readline())
K = int(readline())
while not (0 <= H1 < 23 and 0 <= M1 < 59 and 0 <= H2 < 23 and 0 <= M2 < 59):
H1 = int(readline())
M1 = int(readline())
H2 = int(readline())
M2 = int(readline())
K = int(readline())
if M2 > M1:
a = (H2 - H1) * 60 + M2 - M1
print(a - K)
if M2 < M1:
b = (H2 - H1) * 60 + M1 - M2
print(b - K)
except:
print(1)
|
s004133242
|
Accepted
| 20
| 9,196
| 132
|
import sys
readline = sys.stdin.buffer.readline
h1, m1, h2, m2, k = map(int, readline().split())
print(60 * (h2 - h1) + m2 - m1 - k)
|
s039610041
|
p04029
|
u658978681
| 2,000
| 262,144
|
Wrong Answer
| 20
| 3,060
| 33
|
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?
|
n = int(input())
print(n*(n+1)/2)
|
s089529459
|
Accepted
| 18
| 2,940
| 34
|
n = int(input())
print(n*(n+1)//2)
|
s294477566
|
p03555
|
u725757838
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 69
|
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 b[::-1]==a:
print("Yes")
else:
print("NO")
|
s694373802
|
Accepted
| 17
| 2,940
| 69
|
a=input()
b=input()
if b[::-1]==a:
print("YES")
else:
print("NO")
|
s262485123
|
p02744
|
u896741788
| 2,000
| 1,048,576
|
Wrong Answer
| 20
| 3,184
| 420
|
In this problem, we only consider strings consisting of lowercase English letters. Strings s and t are said to be **isomorphic** when the following conditions are satisfied: * |s| = |t| holds. * For every pair i, j, one of the following holds: * s_i = s_j and t_i = t_j. * s_i \neq s_j and t_i \neq t_j. For example, `abcac` and `zyxzx` are isomorphic, while `abcac` and `ppppp` are not. A string s is said to be in **normal form** when the following condition is satisfied: * For every string t that is isomorphic to s, s \leq t holds. Here \leq denotes lexicographic comparison. For example, `abcac` is in normal form, but `zyxzx` is not since it is isomorphic to `abcac`, which is lexicographically smaller than `zyxzx`. You are given an integer N. Print all strings of length N that are in normal form, in lexicographically ascending order.
|
from time import time
n=int(input())
from itertools import combinations as pe
al=[]
for i in range(n):
for tu in pe(range(n-1),i):
if tu:
pre=tu[0]
ans="a"*(pre+1)
for d,v in enumerate(tu[1:]):
ans+=chr(98+d)*(v-pre);pre=v
ans+=chr(ord(ans[-1])+1)*(n-len(ans))
al.append(ans)
else:print("a"*n)
print(*sorted(al),sep="\n")
|
s006517301
|
Accepted
| 152
| 4,404
| 162
|
li="abcdefghijklmn"
n=int(input())
def f(s,l,k):
global li,n
if l==n:print(s);return
for i in range(2+k):
f(s+li[i],l+1,max(i,k))
f("a",1,0)
|
s117010763
|
p03693
|
u632413369
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,096
| 71
|
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?
|
t = sum(map(int, input().split()))
print('YES' if t % 4 == 0 else 'NO')
|
s348639278
|
Accepted
| 23
| 9,160
| 93
|
a, b, c = map(int, input().split())
print('YES' if (a * 100 + b * 10 + c) % 4 == 0 else 'NO')
|
s347939393
|
p03080
|
u899929023
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 103
|
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=int(input())
st=input()
R_num = st.count('R')
if R_num*2 > N:
print('YES')
else:
print('NO')
|
s564687742
|
Accepted
| 17
| 2,940
| 104
|
N=int(input())
st=input()
R_num = st.count('R')
if R_num*2 > N:
print('Yes')
else:
print('No')
|
s968734020
|
p03228
|
u077337864
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 173
|
In the beginning, Takahashi has A cookies, and Aoki has B cookies. They will perform the following operation alternately, starting from Takahashi: * If the number of cookies in his hand is odd, eat one of those cookies; if the number is even, do nothing. Then, give one-half of the cookies in his hand to the other person. Find the numbers of cookies Takahashi and Aoki respectively have after performing K operations in total.
|
A, B, K = map(int, input().strip().split())
for k in range(K):
if k % 2 == 0:
A = A // 2
B = B + A // 2
else:
B = B // 2
A = A + B // 2
print(A, B)
|
s455429653
|
Accepted
| 18
| 2,940
| 209
|
A, B, K = map(int, input().strip().split())
for k in range(K):
if k % 2 == 0:
_A = A // 2
_B = B + A // 2
A, B = _A, _B
else:
_B = B // 2
_A = A + B // 2
A, B = _A, _B
print(A, B)
|
s434498413
|
p03130
|
u131634965
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 188
|
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads. Determine if we can visit all the towns by traversing each of the roads exactly once.
|
city={1:0, 2:0, 3:0, 4:0}
for _ in range(3):
a,b=map(int, input().split())
city[a]+=1
city[b]+=1
if list(city.values()).count(2)==2:
print("Yes")
else:
print("No")
|
s242330009
|
Accepted
| 17
| 3,060
| 188
|
city={1:0, 2:0, 3:0, 4:0}
for _ in range(3):
a,b=map(int, input().split())
city[a]+=1
city[b]+=1
if list(city.values()).count(2)==2:
print("YES")
else:
print("NO")
|
s167732768
|
p03680
|
u689322583
| 2,000
| 262,144
|
Wrong Answer
| 224
| 7,084
| 273
|
Takahashi wants to gain muscle, and decides to work out at AtCoder Gym. The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it. Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up. Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.
|
#coding: utf-8
N = int(input())
a = []
for i in range(N):
a.append(int(input()))
start = 0
count = 0
for i in range(N):
count += 1
next = a[start]
start = a[next-1]
if(start == 2):
print(count)
break
start -= 1
else:
print(-1)
|
s693512344
|
Accepted
| 53
| 7,068
| 162
|
import sys
N = list(map(int, sys.stdin))
bot = N[1]
for i in range(N[0]):
if bot == 2:
print(i+1)
break
bot = N[bot]
else:
print(-1)
|
s638037713
|
p02418
|
u956226421
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,368
| 287
|
Write a program which finds a pattern $p$ in a ring shaped text $s$.
|
s = input()
p = input()
Ans = False
for i in range(len(s)):
for j in range(len(p)):
if i + j >= len(s):
i = -1
if s[i + j] != p[j]:
break
if j == len(p) - 1:
Ans = True
break
if Ans:
print('Yes')
else:
print('No')
|
s621562413
|
Accepted
| 60
| 7,528
| 322
|
s = input()
p = input()
Ans = False
cnt = 0
for i in range(len(s)):
for j in range(len(p)):
if i + j >= len(s):
i = -j
if s[i + j] != p[j]:
break
cnt += 1
if cnt == len(p):
Ans = True
break
cnt = 0
if Ans:
print('Yes')
else:
print('No')
|
s879638043
|
p03493
|
u962423738
| 2,000
| 262,144
|
Wrong Answer
| 24
| 8,948
| 50
|
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
n=list(map(int,input().split()))
print(n.count(1))
|
s681993158
|
Accepted
| 28
| 8,944
| 29
|
n=input()
print(n.count("1"))
|
s472435594
|
p03854
|
u738622346
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,940
| 830
|
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
s = input()
t = ""
while len(t) < len(s):
target_len = len(s) - len(t)
if s[len(t)] == 'd':
if target_len == 5:
t += "dream"
elif target_len == 7:
t += "dreamer"
elif target_len > 7:
if s[len(t) + 5] != 'd' and s[len(t) + 7] != 'a':
t += "dreamer"
else:
t += "dream"
else:
break
else:
if target_len == 5:
t += "erase"
elif target_len == 6:
t += "eraser"
elif target_len > 6:
if s[len(t) + 4] != 'd' and s[len(t) + 6] != 'a':
t += "eraser"
else:
t += "erase"
else:
break
print(s, t, sep=" ")
if s[0:len(t)] != t:
break
print("YES" if s == t else "NO")
|
s321967328
|
Accepted
| 50
| 3,188
| 440
|
s = input()
div = ["dream", "dreamer", "erase", "eraser"]
rev = []
t = ""
s_rev = s[::-1]
for d in div:
rev.append(d[::-1])
result = True
i = 0
while i < len(s):
can_divide = False
for d in rev:
if len(s_rev) - i >= len(d) and s_rev[i:i + len(d)] == d:
can_divide = True
i += len(d)
break
if not can_divide:
result = False
break
print("YES" if result else "NO")
|
s387112457
|
p03416
|
u090225501
| 2,000
| 262,144
|
Wrong Answer
| 96
| 3,060
| 134
|
Find the number of _palindromic numbers_ among the integers between A and B (inclusive). Here, a palindromic number is a positive integer whose string representation in base 10 (without leading zeros) reads the same forward and backward.
|
a, b = map(int, input().split())
c = 0
for i in range(a, b + 1):
x = list(str(i))
y = reversed(x)
if x == y:
c += 1
print(c)
|
s295796917
|
Accepted
| 103
| 2,940
| 137
|
a, b = map(int, input().split())
c = 0
for i in range(a, b + 1):
x = str(i)
y = ''.join(reversed(x))
if x == y:
c += 1
print(c)
|
s493484404
|
p02612
|
u277175276
| 2,000
| 1,048,576
|
Wrong Answer
| 31
| 9,128
| 28
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N=int(input())
print(N%1000)
|
s353094030
|
Accepted
| 29
| 9,124
| 65
|
N=int(input())
b=N%1000
if b==0:
print(0)
else:
print(1000-b)
|
s636447721
|
p03044
|
u188745744
| 2,000
| 1,048,576
|
Wrong Answer
| 339
| 28,892
| 406
|
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())
l=[0]*N
com_l=[[] for i in range(N)]
for i in range(N-1):
X,Y,Z=list(map(int,input().split()))
com_l[X-1].append((Y-1,Z%2))
for i in range(N-1):
if com_l[i] != []:
if l[i] == 0:
for j,k in com_l[i]:
if k ==1:
l[j]=1
else:
for j,k in com_l[i]:
if k==0:
l[j]=1
print(l)
|
s609442877
|
Accepted
| 472
| 39,852
| 491
|
from collections import deque
N=int(input())
visited=[-1]*N
l=[[] for i in range(N)]
for i in range(N-1):
a,b,c=map(int,input().split())
a-=1;b-=1
l[a].append((b,c%2))
l[b].append((a,c%2))
def ans(node,dis):
visited[node]=0
que=deque(l[node])
while que:
node,dis=que.popleft()
visited[node]=dis
for i,j in l[node]:
if visited[i]==-1:
visited[i]=(dis+j)%2
que.append((i,(dis+j)%2))
ans(0,0)
for i in visited:
print(i)
|
s378476269
|
p02796
|
u272336707
| 2,000
| 1,048,576
|
Wrong Answer
| 2,105
| 30,244
| 379
|
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.
|
n = int(input())
li=[]
lis=[]
count_lis=[0]*n
for i in range(n):
li = list(map(int, input().split()))
lis.append(li)
lis.sort()
for i in range(n):
for j in range(i+1, n):
if lis[i][0] + lis[i][1] > lis[j][0] - lis[j][1]:
count_lis[i] += 1
count_lis[j] += 1
else:
break
num = n- int(sum(count_lis)/2)
print(str(num))
|
s231696329
|
Accepted
| 521
| 20,096
| 309
|
n = int(input())
lis=[]
count=0
for i in range(n):
li=[None]*2
x, l = map(int, input().split())
li[0] = x+l
li[1] = x-l
lis.append(li)
lis.sort()
kari_r = -(10 ** 18)
for r, l in lis:
if kari_r > l:
continue
count+=1
kari_r = r
print(count)
|
s675774474
|
p03370
|
u706159977
| 2,000
| 262,144
|
Wrong Answer
| 354
| 3,064
| 200
|
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.
|
m,x = [int(i) for i in input().split()]
s=[]
for i in range(m):
list.append(s, int(input()))
x = x-sum(s)
print(sum(s))
print(x)
ct = 0
while x>min(s):
x = x - min(s)
ct = ct+1
print(ct+m)
|
s031939020
|
Accepted
| 355
| 3,060
| 178
|
m,x = [int(i) for i in input().split()]
s=[]
for i in range(m):
list.append(s, int(input()))
x = x-sum(s)
ct = 0
while x>=min(s):
x = x - min(s)
ct = ct+1
print(ct+m)
|
s921308107
|
p03836
|
u143976525
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 233
|
Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him.
|
sx,sy,tx,ty=map(int,input().split())
ans=""
y=ty-sy
x=tx-sx
ans +="U"*y
ans +="R"*x
ans +="D"*y
ans +="L"*x
ans +="L"
ans +="U"*y
ans +="R"*x
ans +="D"
ans +="R"
ans +="D"*y
ans +="L"*x
ans +="U"
print(ans)
|
s225235510
|
Accepted
| 17
| 3,064
| 321
|
sx,sy,tx,ty=map(int,input().split())
ans=""
y=ty-sy
x=tx-sx
ans +="U"*y
ans +="R"*x
ans +="D"*y
ans +="L"*x
ans +="L"
ans +="U"*(y+1)
ans +="R"*(x+1)
ans +="D"
ans +="R"
ans +="D"*(y+1)
ans +="L"*(x+1)
ans +="U"
print(ans)
|
s714891429
|
p00085
|
u546285759
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,564
| 203
|
昔、ヨセフのおイモというゲームがありました。n 人が参加しているとしましょう。参加者は中心を向いて円陣を組み、1 から順番に番号が振られます。アツアツのおイモがひとつ、参加者 n (左の図内側の大きい数字の 30 )に渡されます。おイモを渡された参加者は右隣の参加者にそのおイモを渡します。 m 回目に渡された人は右隣の人に渡して円陣から抜けます(左の図では m = 9 の場合を表しています) 。 回渡す毎に一人ずつぬけ、最後に残った人が勝者となり、おイモをいただきます。 n ,m が決まってから、実際におイモを渡し始める前にどこにいたら勝てるかわかるといいですよね。上の図は 30 人の参加者で 9 人ごとに抜けるというルールでこのゲームをした場合を書き表しています。内側の大きい数字が参加者に振られた番号、外側の小さい数字が抜ける順番です。それによると、9,18,27,6,16,26 という順番で円陣から抜け出し、最後には 21 が残ることになります。すなわち 21 が勝者となります(小さい数字が 30 になっています)。 ゲーム参加者数 n と円陣から抜け出す参加者の間隔 m を入力し、勝者の番号を出力するプログラムを作成してください。ただし、m, n < 1000 とします。
|
n, m = map(int, input().split())
p, i = [i+1 for i in range(n)], m
while True:
p.pop(m-1)
if len(p)==1:
print(p[0])
break
tmp = m+i-1
m = tmp%len(p) if tmp>len(p) else tmp
|
s685147294
|
Accepted
| 30
| 7,720
| 219
|
while True:
n, m = map(int, input().split())
if n == 0:
break
p = [i for i in range(1, n+1)]
idx = m-1
while len(p) != 1:
v = p.pop(idx)
idx = (idx+m-1) % len(p)
print(*p)
|
s192931388
|
p02608
|
u343744945
| 2,000
| 1,048,576
|
Wrong Answer
| 2,206
| 9,312
| 928
|
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).
|
import math
N=int(input())
sqN = int(math.sqrt(N))+1
sq=list()
for i in range(1,N):
sq.append(i*i)
def isSquare(n):
if n in sq:
return True
else:
return False
ans=list()
for i in range(min(N,6)):
ans.append(0)
if N>5:
for m in range(6,N+1):
cnt = 0
for x in range(1,sqN):
for y in range(1,sqN):
b = x+y
D = -3*(x*x+y*y)-2*x*y+4*m
print(m,x,y,D,isSquare(D))
if D > 0 and isSquare(D):
sqD=sq.index(D)+1
if (sqD-b) > 0 and (sqD-b) % 2 == 0:
# print(x,y,z)
# print(x*x+y*y+z*z+x*y+y*z-m)
cnt+=1
if D < 0:
break
ans.append(cnt)
for a in ans:
print(a)
|
s163197746
|
Accepted
| 58
| 9,264
| 659
|
N=int(input())
cnts=[0 for i in range(N+1)]
for x in range(1,100):
x2=x*x
for y in range(x,100):
y2=y*y
xy=x*y
if x2+y*y+x*y+x+y > N:
break
for z in range(y,100):
z2 = z*z
M = x2+y2+z2+xy+y*z+z*x
if M > N:
break
if x!=y:
if y!=z:
cnts[M]+=6
else:
cnts[M]+=3
else:
if y!=z:
cnts[M]+=3
else:
cnts[M]+=1
for a in cnts[1:N+1]:
print(a)
|
s147818109
|
p02412
|
u017435045
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,592
| 258
|
Write a program which identifies the number of combinations of three integers which satisfy the following conditions: * You should select three distinct integers from 1 to n. * A total sum of the three integers is x. For example, there are two combinations for n = 5 and x = 9. * 1 + 3 + 5 = 9 * 2 + 3 + 4 = 9
|
while 1:
n, x = map(int,input().split())
if n ==0 and x ==0:
break
ans = 0
for i in range(n-1):
for j in range(i,n):
for k in range(j,n):
if i + j +k ==x:
ans += 1
print(ans)
|
s134780393
|
Accepted
| 510
| 5,596
| 264
|
while 1:
n, x = map(int,input().split())
if n == x ==0:
break
ans = 0
for i in range(1, n+1):
for j in range(i+1,n+1):
for k in range(j+1,n+1):
if i + j +k ==x:
ans += 1
print(ans)
|
s705689954
|
p03672
|
u827202523
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 275
|
We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not. You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
|
def judge(S):
half = int(len(S)/2)
if S[:half] == S[half:]:
return True
else:
return False
S= input()
l = len(S)
if l%2 == 1:
S = S[:-1]
else:
S = S[:-2]
print(S)
while S != "":
if judge(S):
break
S = S[:-2]
print(len(S))
|
s174907134
|
Accepted
| 17
| 3,060
| 266
|
def judge(S):
half = int(len(S)/2)
if S[:half] == S[half:]:
return True
else:
return False
S= input()
l = len(S)
if l%2 == 1:
S = S[:-1]
else:
S = S[:-2]
while S != "":
if judge(S):
break
S = S[:-2]
print(len(S))
|
s171990764
|
p03380
|
u794173881
| 2,000
| 262,144
|
Wrong Answer
| 154
| 14,428
| 245
|
Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order. From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized. If there are multiple pairs that maximize the value, any of them is accepted.
|
n = int(input())
a = list(map(int,input().split()))
sort_a = sorted(a,reverse=True)
max_a = sort_a[0]
ans=0
for i in range(n):
if min(abs(max_a//2-sort_a[i]),abs(max_a//2-ans)) == abs(max_a//2-sort_a[i]):
ans = sort_a[i]
print(max_a,ans)
|
s029248815
|
Accepted
| 166
| 14,428
| 242
|
n = int(input())
a = list(map(int,input().split()))
sort_a = sorted(a,reverse=True)
max_a = sort_a[0]
ans=0
for i in range(n):
if min(abs(max_a/2-sort_a[i]),abs(max_a/2-ans)) == abs(max_a/2-sort_a[i]):
ans = sort_a[i]
print(max_a,ans)
|
s472034381
|
p03486
|
u311379832
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 148
|
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
|
s = list(input())
t = list(input())
s.sort()
t.sort(reverse=True)
ss = ''.join(s)
ts = ''.join(t)
if s < t:
print('YES')
else:
print('NO')
|
s956109656
|
Accepted
| 17
| 3,060
| 148
|
s = list(input())
t = list(input())
s.sort()
t.sort(reverse=True)
ss = ''.join(s)
ts = ''.join(t)
if s < t:
print('Yes')
else:
print('No')
|
s472826336
|
p03068
|
u243572357
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 143
|
You are given a string S of length N consisting of lowercase English letters, and an integer K. Print the string obtained by replacing every character in S that differs from the K-th character of S, with `*`.
|
n = int(input())
a = input()
l = int(input())
for i in range(n):
if a[l-1] != a[i]:
print('*', sep='')
else:
print(a[l-1], sep='')
|
s361658932
|
Accepted
| 17
| 2,940
| 89
|
input()
s = input()
n = int(input())
print(*[i if i==s[n-1] else "*" for i in s], sep='')
|
s093519404
|
p02261
|
u159356473
| 1,000
| 131,072
|
Wrong Answer
| 50
| 7,680
| 852
|
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).
|
#coding:UTF-8
def BS(N,A):
for i in range(N):
for j in reversed(range(i+1,N)):
if int(A[j][1:])<int(A[j-1][1:]):
swap=A[j]
A[j]=A[j-1]
A[j-1]=swap
return A
def SS(N,A):
for i in range(N):
minj=i
for j in range(i,N):
if int(A[j][1:])<int(A[minj][1:]):
minj=j
swap=A[i]
A[i]=A[minj]
A[minj]=swap
return A
def Stable(A1,A2):
ans="Stable"
for i in range(len(A1)):
if A1[i][1:]==A2[i][1:] and A1[i][:1]!=A2[i][:1]:
ans="Not stable"
return ans
def SS2(N,A):
A1=BS(N,A)
print(" ".join(A1))
print("Stable")
A2=SS(N,A)
print(" ".join(A2))
print(Stable(A1,A2))
if __name__=="__main__":
N=int(input())
a=input()
A=a.split(" ")
SS2(N,A)
|
s658822519
|
Accepted
| 50
| 7,796
| 886
|
#coding:UTF-8
def BS(N,A):
for i in range(N):
for j in reversed(range(i+1,N)):
if int(A[j][1:])<int(A[j-1][1:]):
swap=A[j]
A[j]=A[j-1]
A[j-1]=swap
return A
def SS(N,A):
for i in range(N):
minj=i
for j in range(i,N):
if int(A[j][1:]) < int(A[minj][1:]):
minj=j
swap=A[i]
A[i]=A[minj]
A[minj]=swap
return A
def Stable(A1,A2):
ans="Stable"
for i in range(len(A1)):
if A1[i][1:]==A2[i][1:] and A1[i][:1]!=A2[i][:1]:
ans="Not stable"
return ans
def SS2(N,A,B):
A1=BS(N,A)
print(" ".join(A1))
print("Stable")
A2=SS(N,B)
print(" ".join(A2))
print(Stable(A1,A2))
if __name__=="__main__":
N=int(input())
a=input()
A=a.split(" ")
B=a.split(" ")
SS2(N,A,B)
|
s175040961
|
p03386
|
u672494157
| 2,000
| 262,144
|
Wrong Answer
| 25
| 3,812
| 823
|
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.
|
import sys
from functools import reduce
import copy
import math
from pprint import pprint
sys.setrecursionlimit(4100000)
def inputs(num_of_input):
ins = [input() for i in range(num_of_input)]
return ins
def int_inputs(num_of_input):
ins = [int(input()) for i in range(num_of_input)]
return ins
def solve(inputs):
[A, B, K] = string_to_int(inputs[0])
ret = set()
count = 0
for v in range(A, B+1):
if count >= K:
break
ret.add(v)
count += 1
count = 0
for v in reversed(range(A, B+1)):
if count >= K:
break
ret.add(v)
count += 1
return ret
def string_to_int(string):
return list(map(int, string.split()))
if __name__ == "__main__":
ret = solve(inputs(1))
for r in ret:
print(r)
|
s651745766
|
Accepted
| 40
| 4,068
| 837
|
import sys
from functools import reduce
import copy
import math
from pprint import pprint
sys.setrecursionlimit(4100000)
def inputs(num_of_input):
ins = [input() for i in range(num_of_input)]
return ins
def int_inputs(num_of_input):
ins = [int(input()) for i in range(num_of_input)]
return ins
def solve(inputs):
[A, B, K] = string_to_int(inputs[0])
ret = set()
count = 0
for v in range(A, B+1):
if count >= K:
break
ret.add(v)
count += 1
count = 0
for v in reversed(range(A, B+1)):
if count >= K:
break
ret.add(v)
count += 1
return sorted(list(ret))
def string_to_int(string):
return list(map(int, string.split()))
if __name__ == "__main__":
ret = solve(inputs(1))
for r in ret:
print(r)
|
s166658261
|
p03369
|
u292810930
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 140
|
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
S = input()
100
t = 0
for i in range(len(S)):
if S[i] == 'o':
t =+ 1
print(700 + t*100)
|
s122639953
|
Accepted
| 17
| 2,940
| 160
|
S = input()
t = 0
for i in range(len(S)):
if S[i] == 'o':
t += 1
print(700 + t*100)
|
s250078492
|
p03044
|
u227082700
| 2,000
| 1,048,576
|
Wrong Answer
| 2,105
| 109,564
| 185
|
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())
a=[0]*n
b=[]
for i in range(n-1):b.append(list(map(int,input().split())))
b.sort()
for i,j,k in b:
if k%2==0:a[j-1]=a[i-1]
else:a[j-1]=-(a[i-1]-1)
for i in a:print(a)
|
s139357246
|
Accepted
| 693
| 61,716
| 509
|
from sys import stdin
input = stdin.readline
n=int(input())
b=[]
for i in range(n-1):b.append(list(map(int,input().split())))
def BFS(K,edges,N):
roots=[ [] for i in range(N)]
for a,b,c in edges:
roots[a-1]+=[(b-1,c)]
roots[b-1]+=[(a-1,c)]
dist=[-1]*N
stack=[]
stack.append(K)
dist[K]=0
while stack:
label=stack.pop(-1)
for i,c in roots[label]:
if dist[i]==-1:
dist[i]=dist[label]+c
stack+=[i]
return dist
a=BFS(0,b,n)
for i in a:print(0 if i%2==0 else 1)
|
s077305925
|
p02612
|
u827627481
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 9,148
| 106
|
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())
for i in range(0,11):
if i*1000 <= N and N < (i+1)*1000:
break
print(N-1000*i)
|
s640314608
|
Accepted
| 28
| 9,156
| 112
|
N = int(input())
for i in range(0,11):
if i*1000 < N and N <= (i+1)*1000:
break
print(1000*(i+1) - N)
|
s351400302
|
p03699
|
u149249972
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,176
| 270
|
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?
|
g = int(input())
grades = []
for i in range(g):
grade = int(input())
grades.append(grade)
total = sum(grades)
grades.sort()
grades = [0] + grades
for grade in grades:
if (total - grade) % 10 != 0:
print(total - grade)
break
print(0)
|
s581332691
|
Accepted
| 25
| 9,096
| 333
|
g = int(input())
grades = []
for i in range(g):
grade = int(input())
grades.append(grade)
total = sum(grades)
grades.sort()
grades = [0] + grades
count = 0
for grade in grades:
if (total - grade) % 10 != 0:
print(total - grade)
break
else:
count += 1
if count == len(grades):
print(0)
|
s360039838
|
p02612
|
u303039933
| 2,000
| 1,048,576
|
Wrong Answer
| 48
| 10,712
| 2,709
|
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.
|
# -*- coding: utf-8 -*-
import sys
import math
import os
import itertools
import string
import heapq
import _collections
from collections import Counter
from collections import defaultdict
from collections import deque
from functools import lru_cache
import bisect
import re
import queue
import decimal
class Scanner():
@staticmethod
def int():
return int(sys.stdin.readline().rstrip())
@staticmethod
def string():
return sys.stdin.readline().rstrip()
@staticmethod
def map_int():
return [int(x) for x in Scanner.string().split()]
@staticmethod
def string_list(n):
return [Scanner.string() for i in range(n)]
@staticmethod
def int_list_list(n):
return [Scanner.map_int() for i in range(n)]
@staticmethod
def int_cols_list(n):
return [Scanner.int() for i in range(n)]
class Math():
@staticmethod
def gcd(a, b):
if b == 0:
return a
return Math.gcd(b, a % b)
@staticmethod
def lcm(a, b):
return (a * b) // Math.gcd(a, b)
@staticmethod
def divisor(n):
lower, upper = [], []
i = 1
while i * i <= n:
if n % i == 0:
lower.append(i)
if i != n // i:
upper.append(n // i)
i += 1
return lower + upper[::-1]
@staticmethod
def round_up(a, b):
return -(-a // b)
@staticmethod
def is_prime(n):
if n < 2:
return False
if n == 2:
return True
if n % 2 == 0:
return False
d = int(n ** 0.5) + 1
for i in range(3, d + 1, 2):
if n % i == 0:
return False
return True
@staticmethod
def fact(N):
res = {}
tmp = N
for i in range(2, int(N ** 0.5 + 1) + 1):
cnt = 0
while tmp % i == 0:
cnt += 1
tmp //= i
if cnt > 0:
res[i] = cnt
if tmp != 1:
res[tmp] = 1
if res == {}:
res[N] = 1
return res
def pop_count(x):
x = x - ((x >> 1) & 0x5555555555555555)
x = (x & 0x3333333333333333) + ((x >> 2) & 0x3333333333333333)
x = (x + (x >> 4)) & 0x0f0f0f0f0f0f0f0f
x = x + (x >> 8)
x = x + (x >> 16)
x = x + (x >> 32)
return x & 0x0000007f
def solve():
N = Scanner.int()
print(N % 1000)
def main():
# T = Scanner.int()
# for _ in range(T):
# solve()
# print('YNeos'[not solve()::2])
solve()
if __name__ == "__main__":
main()
|
s597029114
|
Accepted
| 42
| 10,708
| 2,769
|
# -*- coding: utf-8 -*-
import sys
import math
import os
import itertools
import string
import heapq
import _collections
from collections import Counter
from collections import defaultdict
from collections import deque
from functools import lru_cache
import bisect
import re
import queue
import decimal
class Scanner():
@staticmethod
def int():
return int(sys.stdin.readline().rstrip())
@staticmethod
def string():
return sys.stdin.readline().rstrip()
@staticmethod
def map_int():
return [int(x) for x in Scanner.string().split()]
@staticmethod
def string_list(n):
return [Scanner.string() for i in range(n)]
@staticmethod
def int_list_list(n):
return [Scanner.map_int() for i in range(n)]
@staticmethod
def int_cols_list(n):
return [Scanner.int() for i in range(n)]
class Math():
@staticmethod
def gcd(a, b):
if b == 0:
return a
return Math.gcd(b, a % b)
@staticmethod
def lcm(a, b):
return (a * b) // Math.gcd(a, b)
@staticmethod
def divisor(n):
lower, upper = [], []
i = 1
while i * i <= n:
if n % i == 0:
lower.append(i)
if i != n // i:
upper.append(n // i)
i += 1
return lower + upper[::-1]
@staticmethod
def round_up(a, b):
return -(-a // b)
@staticmethod
def is_prime(n):
if n < 2:
return False
if n == 2:
return True
if n % 2 == 0:
return False
d = int(n ** 0.5) + 1
for i in range(3, d + 1, 2):
if n % i == 0:
return False
return True
@staticmethod
def fact(N):
res = {}
tmp = N
for i in range(2, int(N ** 0.5 + 1) + 1):
cnt = 0
while tmp % i == 0:
cnt += 1
tmp //= i
if cnt > 0:
res[i] = cnt
if tmp != 1:
res[tmp] = 1
if res == {}:
res[N] = 1
return res
def pop_count(x):
x = x - ((x >> 1) & 0x5555555555555555)
x = (x & 0x3333333333333333) + ((x >> 2) & 0x3333333333333333)
x = (x + (x >> 4)) & 0x0f0f0f0f0f0f0f0f
x = x + (x >> 8)
x = x + (x >> 16)
x = x + (x >> 32)
return x & 0x0000007f
def solve():
N = Scanner.int()
if N % 1000 == 0:
print(0)
else:
print(1000 - N % 1000)
def main():
# T = Scanner.int()
# for _ in range(T):
# solve()
# print('YNeos'[not solve()::2])
solve()
if __name__ == "__main__":
main()
|
s522004811
|
p02602
|
u501451051
| 2,000
| 1,048,576
|
Wrong Answer
| 2,206
| 31,632
| 315
|
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.
|
n ,k = map(int, input().split())
lis = list(map(int, input().split()))
t = 1
for i in range(k):
t *= lis[i]
print(t)
for i in range(k, n):
tmp = lis[i-(k-1):i+1]
a = 1
for j in tmp:
a *= j
if a > t:
print('Yes')
else:
print('No')
t = a
a = 1
|
s133356482
|
Accepted
| 162
| 31,440
| 192
|
n ,k = map(int, input().split())
lis = list(map(int, input().split()))
for i in range(k, n):
l = lis[i-k]
r = lis[i]
if r > l:
print('Yes')
else:
print('No')
|
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