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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s384477799
|
p02697
|
u727980193
| 2,000
| 1,048,576
|
Wrong Answer
| 93
| 15,852
| 140
|
You are going to hold a competition of one-to-one game called AtCoder Janken. _(Janken is the Japanese name for Rock-paper-scissors.)_ N players will participate in this competition, and they are given distinct integers from 1 through N. The arena has M playing fields for two players. You need to assign each playing field two distinct integers between 1 and N (inclusive). You cannot assign the same integer to multiple playing fields. The competition consists of N rounds, each of which proceeds as follows: * For each player, if there is a playing field that is assigned the player's integer, the player goes to that field and fight the other player who comes there. * Then, each player adds 1 to its integer. If it becomes N+1, change it to 1. You want to ensure that no player fights the same opponent more than once during the N rounds. Print an assignment of integers to the playing fields satisfying this condition. It can be proved that such an assignment always exists under the constraints given.
|
N,M=list(map(int, input().split()))
src=[i+1 for i in range(M*2)]
for m in range(M):
a=src[m]
b=src[-(m+1)]
print('{} {}'.format(a,b))
|
s253825470
|
Accepted
| 96
| 15,820
| 434
|
N,M=list(map(int, input().split()))
src=[i+1 for i in range(M*2+1)]
if M%2==0:
for m in range(M//2):
a=src[m]
b=src[M-m]
print('{} {}'.format(a,b))
for m in range(M//2):
a=src[M+1+m]
b=src[-(m+1)]
print('{} {}'.format(a,b))
else:
for m in range(M//2):
a=src[m]
b=src[M-1-m]
print('{} {}'.format(a,b))
for m in range(M-(M//2)):
a=src[M+m]
b=src[-(m+1)]
print('{} {}'.format(a,b))
|
s383073029
|
p03434
|
u493491792
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,064
| 221
|
We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.
|
n=int(input())
lista=list(map(int,input().split()))
lista.sort()
alice=list()
bob=list()
for i in range(n):
if i%2==0:
alice.append(lista[i])
else:
bob.append(lista[i])
print(sum(alice)-sum(bob))
|
s658282124
|
Accepted
| 18
| 3,060
| 242
|
n=int(input())
lista=list(map(int,input().split()))
lista.sort(reverse=True)
alice=list()
bob=list()
for i in range(n):
if i%2==0:
alice.append(lista[i])
else:
bob.append(lista[i])
print(sum(alice)-sum(bob))
|
s314026467
|
p03795
|
u290211456
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 58
|
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
n = int(input())
print(n//15)
print(800*n - 200 * (n//15))
|
s510591388
|
Accepted
| 17
| 2,940
| 45
|
n = int(input())
print(800*n - 200 * (n//15))
|
s444974513
|
p03473
|
u414699019
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 22
|
How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December?
|
print(24-int(input()))
|
s637477501
|
Accepted
| 17
| 2,940
| 22
|
print(48-int(input()))
|
s108009883
|
p03129
|
u371467115
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 80
|
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.
|
n,k=map(int,input().split())
if round(n/2)>k:
print("YES")
else:
print("NO")
|
s942385014
|
Accepted
| 17
| 2,940
| 81
|
n,k=map(int,input().split())
if 1+2*(k-1)<=n:
print("YES")
else:
print("NO")
|
s341648497
|
p03386
|
u585290246
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 173
|
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.
|
#!/bin/python
a, b, k = map(int, input().split())
for i in range(a, min(a + k, b)):
print(i)
for i in range(max(a + k, b - k + 1), min(a + k, b + 1)):
print(i)
|
s242464330
|
Accepted
| 17
| 3,060
| 162
|
#!/bin/python
a, b, k = map(int, input().split())
for i in range(a, min(a + k, b + 1)):
print(i)
for i in range(max(a + k, b - k + 1), b + 1):
print(i)
|
s664434652
|
p03730
|
u343437894
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 145
|
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(" "))
for i in range(0, c):
if (a * i) % b == c:
print("YES")
break
else:
print("NO")
|
s979486288
|
Accepted
| 17
| 2,940
| 145
|
a, b, c = map(int, input().split(" "))
for i in range(1, b):
if (a * i) % b == c:
print("YES")
break
else:
print("NO")
|
s148649385
|
p03377
|
u268470352
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 136
|
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 a < x:
print("NO")
else:
if (x - a)>b:
print("NO")
else:
print("YES")
|
s122594656
|
Accepted
| 17
| 2,940
| 136
|
a, b, x = map(int, input().split())
if a > x:
print("NO")
else:
if (x - a)>b:
print("NO")
else:
print("YES")
|
s354278434
|
p03645
|
u547167033
| 2,000
| 262,144
|
Wrong Answer
| 859
| 38,936
| 252
|
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())
g=[[] for i in range(n)]
for _ in range(m):
u,v=map(int,input().split())
g[u-1].append(v-1)
g[v-1].append(u-1)
for x in g[0]:
for y in g[x]:
if y==n-1:
print('POSSIBLE')
else:
print('IMPOSSIBLE')
|
s735278051
|
Accepted
| 897
| 38,320
| 250
|
n,m=map(int,input().split())
g=[[] for i in range(n)]
for _ in range(m):
u,v=map(int,input().split())
g[u-1].append(v-1)
g[v-1].append(u-1)
for x in g[0]:
for y in g[x]:
if y==n-1:
print('POSSIBLE')
exit()
print('IMPOSSIBLE')
|
s874835174
|
p03080
|
u729133443
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 53
|
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.
|
print('YNeos'[int(input())<=input().count('R')*2::2])
|
s297235638
|
Accepted
| 17
| 2,940
| 53
|
print('YNeos'[int(input())>=input().count('R')*2::2])
|
s495607601
|
p03387
|
u934868410
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 129
|
You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order: * Choose two among A, B and C, then increase both by 1. * Choose one among A, B and C, then increase it by 2. It can be proved that we can always make A, B and C all equal by repeatedly performing these operations.
|
import math
a = sorted(list(map(int,input().split())))
print(a[2] - a[1] + math.ceil((a[1] - a[0]) / 2) + (a[1] - a[0]) % 2 == 1)
|
s739577279
|
Accepted
| 23
| 2,940
| 131
|
import math
a = sorted(list(map(int,input().split())))
print(a[2] - a[1] + math.ceil((a[1] - a[0]) / 2) + ((a[1] - a[0]) % 2 == 1))
|
s206896873
|
p03360
|
u050121913
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 87
|
There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations?
|
x = list(map(int,input().split( )))
y = int(input())
M = max(x)
Max = M*2**y
print(Max)
|
s756863609
|
Accepted
| 17
| 2,940
| 106
|
x = list(map(int,input().split( )))
y = int(input())
xx = sorted(x)
xx[2] *= 2**y
ans = sum(xx)
print(ans)
|
s878973871
|
p03494
|
u603234915
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 2,940
| 203
|
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())
l = list(map(int,input().split()))
k = 0
while True:
for i in l :
if i % 2 != 0:
break
elif i == 2:
break
i = i/2
k += 1
print(k)
|
s549921331
|
Accepted
| 19
| 2,940
| 269
|
n = int(input())
l = list(map(int,input().split()))
k = 0
m = 0
for _ in range(100):
for i in range(len(l)) :
if l[i] % 2 != 0 :
m = 10
break
else:
l[i] = l[i]/2
if m != 0 :
break
k += 1
print(k)
|
s928825561
|
p03067
|
u485716382
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 132
|
There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.
|
def main():
a, b, c = map(int, input().split(' '))
if a > c > b:
print('Yes')
else:
print("No")
main()
|
s320170310
|
Accepted
| 17
| 2,940
| 172
|
def main():
a, b, c = map(int, input().split(' '))
if b > c > a:
print('Yes')
elif a > c > b:
print('Yes')
else:
print("No")
main()
|
s648965215
|
p02614
|
u564655959
| 1,000
| 1,048,576
|
Wrong Answer
| 171
| 26,288
| 606
|
We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is `.`, and black if c_{i,j} is `#`. Consider doing the following operation: * Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns. You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices.
|
import itertools
import numpy as np
h, w, k = (int(each) for each in input().split())
c = []
for _ in range(h):
c.append(list(input()))
result = 0
n_c = np.array(c)
for y in itertools.product([0,1], repeat = h):
for x in itertools.product([0,1], repeat = w):
# print(y, x)
n_c2 = np.array(c)
for i, y2 in enumerate(y):
if y2 == 1:
n_c2[i] = 'R'
for i, x2 in enumerate(x):
if x2 == 1:
n_c2[:,i] = 'R'
# print(n_c2)
m = np.count_nonzero(n_c2 == '#')
if k == m:
result += 1
|
s318549962
|
Accepted
| 164
| 27,000
| 622
|
import itertools
import numpy as np
h, w, k = (int(each) for each in input().split())
c = []
for _ in range(h):
c.append(list(input()))
result = 0
n_c = np.array(c)
for y in itertools.product([0,1], repeat = h):
for x in itertools.product([0,1], repeat = w):
# print(y, x)
n_c2 = np.array(c)
for i, y2 in enumerate(y):
if y2 == 1:
n_c2[i] = 'R'
for i, x2 in enumerate(x):
if x2 == 1:
n_c2[:,i] = 'R'
# print(n_c2)
m = np.count_nonzero(n_c2 == '#')
if k == m:
result += 1
print(result)
|
s872699229
|
p03449
|
u372345564
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,060
| 289
|
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j). You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down. The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them. At most how many candies can you collect when you choose the best way to travel?
|
n = int(input())
a = []
for i in range(2):
a.append([int(num) for num in input().split()])
#print(a)
result = 0
for i in range(0, n):
point = sum(a[0][0:i+1]) + sum(a[1][i:])
print(a[0][0:i+1])
print(a[1][i:])
if(result < point):
result = point
print(result)
|
s152607243
|
Accepted
| 19
| 3,060
| 291
|
n = int(input())
a = []
for i in range(2):
a.append([int(num) for num in input().split()])
#print(a)
result = 0
for i in range(0, n):
point = sum(a[0][0:i+1]) + sum(a[1][i:])
# print(a[0][0:i+1])
# print(a[1][i:])
if(result < point):
result = point
print(result)
|
s150698734
|
p02694
|
u744695362
| 2,000
| 1,048,576
|
Time Limit Exceeded
| 2,205
| 9,072
| 91
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
n = int(input())
b = 100
ans = 0
while b<=n :
b *= int(b*0.01)
ans += 1
print(ans)
|
s935962088
|
Accepted
| 22
| 9,160
| 89
|
n = int(input())
b = 100
ans = 0
while b<n :
b = int(b*1.01)
ans += 1
print(ans)
|
s034763473
|
p02972
|
u226912938
| 2,000
| 1,048,576
|
Wrong Answer
| 49
| 7,148
| 63
|
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()))
print(-1)
|
s533314073
|
Accepted
| 264
| 17,228
| 366
|
n = int(input())
A = list(map(int, input().split()))
Bo = [0 for _ in range(n)]
Ans = []
for i in range(n-1, -1, -1):
tot = sum(Bo[i::i+1])
if (tot % 2 == A[i]):
pass
elif (tot % 2 != A[i]):
Ans.append(str(i+1))
Bo[i] += 1
else:
pass
if len(Ans) ==0:
print(0)
else:
print(len(Ans))
print(' '.join(Ans))
|
s008235894
|
p03433
|
u807948625
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 100
|
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
N=int(input())
A=int(input())
amari = N % 500
if amari > A:
print("no")
else:
print("yes")
|
s678346712
|
Accepted
| 17
| 2,940
| 100
|
N=int(input())
A=int(input())
amari = N % 500
if amari > A:
print("No")
else:
print("Yes")
|
s534749345
|
p03697
|
u271044469
| 2,000
| 262,144
|
Wrong Answer
| 20
| 2,940
| 82
|
You are given two integers A and B as the input. Output the value of A + B. However, if A + B is 10 or greater, output `error` instead.
|
a,b = map(int, input().split())
if a+b<10:
print('error')
else:
print(a+b)
|
s949752713
|
Accepted
| 17
| 2,940
| 84
|
a,b = map(int, input().split())
if a+b>=10:
print('error')
else:
print(a+b)
|
s416710939
|
p03730
|
u063073794
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 115
|
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="No"
for i in range(1,100):
if a*i%b==c:
flag="Yes"
break
print(flag)
|
s960293173
|
Accepted
| 17
| 2,940
| 115
|
a,b,c=map(int,input().split())
flag="NO"
for i in range(1,100):
if a*i%b==c:
flag="YES"
break
print(flag)
|
s103062337
|
p03550
|
u207707177
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,188
| 164
|
We have a deck consisting of N cards. Each card has an integer written on it. The integer on the i-th card from the top is a_i. Two people X and Y will play a game using this deck. Initially, X has a card with Z written on it in his hand, and Y has a card with W written on it in his hand. Then, starting from X, they will alternately perform the following action: * Draw some number of cards from the top of the deck. Then, discard the card in his hand and keep the last drawn card instead. Here, at least one card must be drawn. The game ends when there is no more card in the deck. The score of the game is the absolute difference of the integers written on the cards in the two players' hand. X will play the game so that the score will be maximized, and Y will play the game so that the score will be minimized. What will be the score of the game?
|
n,z,w = [int(i) for i in input().split()]
a = list(int(i) for i in input().split())
print(a)
abs1 = abs(a[n-1] - w)
abs2 = abs(a[n-2]-a[n-1])
print(max(abs1,abs2))
|
s533176958
|
Accepted
| 18
| 3,188
| 165
|
n,z,w = [int(i) for i in input().split()]
a = list(int(i) for i in input().split())
#print(a)
abs1 = abs(a[n-1] - w)
abs2 = abs(a[n-2]-a[n-1])
print(max(abs1,abs2))
|
s615894886
|
p03813
|
u317044805
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 250
|
Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise.
|
# -*- coding: utf-8 -*-
x = int(input())
quot, rem = divmod(x, 11)
quot *= 2
if rem>6:
quot += 2
elif rem>0:
quot += 1
print(quot)
|
s181784968
|
Accepted
| 17
| 2,940
| 67
|
x = int(input())
if x<1200:
print("ABC")
else:
print("ARC")
|
s225895932
|
p03141
|
u191635495
| 2,000
| 1,048,576
|
Wrong Answer
| 659
| 28,964
| 281
|
There are N dishes of cuisine placed in front of Takahashi and Aoki. For convenience, we call these dishes Dish 1, Dish 2, ..., Dish N. When Takahashi eats Dish i, he earns A_i points of _happiness_ ; when Aoki eats Dish i, she earns B_i points of happiness. Starting from Takahashi, they alternately choose one dish and eat it, until there is no more dish to eat. Here, both of them choose dishes so that the following value is maximized: "the sum of the happiness he/she will earn in the end" minus "the sum of the happiness the other person will earn in the end". Find the value: "the sum of the happiness Takahashi earns in the end" minus "the sum of the happiness Aoki earns in the end".
|
# input
n = int(input())
ab = sorted([list(map(int, input().split())) for _ in range(n)])
res = 0
for i in range(n):
if i % 2 == 0:
abi = ab.pop(len(ab)-1)
res += abi[0]
else:
abi = ab.pop(len(ab)-1)
res -= abi[1]
print(abi)
print(res)
|
s470202792
|
Accepted
| 1,880
| 8,284
| 248
|
n = int(input())
dd = []
sb = 0
for _ in range(n):
a, b = map(int, input().split())
dd.append(a+b)
sb -= b
abd = sorted(dd, reverse=True)
res = 0
for i in range(n):
d = abd.pop(0)
if i % 2 == 0:
res += d
print(res+sb)
|
s003454246
|
p03400
|
u245870380
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,316
| 192
|
Some number of chocolate pieces were prepared for a training camp. The camp had N participants and lasted for D days. The i-th participant (1 \leq i \leq N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on. As a result, there were X chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces. Find the number of chocolate pieces prepared at the beginning of the camp.
|
N = int(input())
D, X = map(int, input().split())
A = [int(input()) for i in range(N)]
choco = 0
for i in A:
for j in range(1,D+1,i):
print(j)
choco += 1
print(choco + X)
|
s586137930
|
Accepted
| 18
| 3,064
| 175
|
N = int(input())
D, X = map(int, input().split())
A = [int(input()) for i in range(N)]
choco = 0
for i in A:
for j in range(1,D+1,i):
choco += 1
print(choco + X)
|
s289295494
|
p02615
|
u284120954
| 2,000
| 1,048,576
|
Wrong Answer
| 2,208
| 31,504
| 276
|
Quickly after finishing the tutorial of the online game _ATChat_ , you have decided to visit a particular place with N-1 players who happen to be there. These N players, including you, are numbered 1 through N, and the **friendliness** of Player i is A_i. The N players will arrive at the place one by one in some order. To make sure nobody gets lost, you have set the following rule: players who have already arrived there should form a circle, and a player who has just arrived there should cut into the circle somewhere. When each player, except the first one to arrive, arrives at the place, the player gets **comfort** equal to the smaller of the friendliness of the clockwise adjacent player and that of the counter-clockwise adjacent player. The first player to arrive there gets the comfort of 0. What is the maximum total comfort the N players can get by optimally choosing the order of arrivals and the positions in the circle to cut into?
|
N = int(input())
A = []
a, *an = map(int, input().split())
A.append(a)
A += an
A.sort()
Node = []
score = 0
Node.append(A.pop())
print(A)
for n in range(N - 1):
score += Node.pop(0)
num = A.pop()
Node.append(num)
Node.append(num)
#print(Node)
print(score)
|
s458668900
|
Accepted
| 210
| 31,560
| 328
|
N = int(input())
A = []
a, *an = map(int, input().split())
A.append(a)
A += an
A.sort()
Node = []
score = 0
Node.append(A.pop())
#print(A)
start = 0
for n in range(N - 1):
score += Node[start]
num = A.pop()
start += 1
#Node = Node[1:]
Node.append(num)
Node.append(num)
#print(Node)
print(score)
|
s944739409
|
p03386
|
u277802731
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 127
|
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.
|
#93b
a,b,k=map(int,input().split())
for i in range(k):
print(a+i)
for i in range(k):
print(b-k+i)
|
s537657782
|
Accepted
| 17
| 3,060
| 120
|
#93b
a,b,k=map(int,input().split())
c=min(a+k,b)
for i in range(a,c):print(i)
for i in range(max(c,b-k+1),b+1):print(i)
|
s103741683
|
p03729
|
u694422786
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 90
|
You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`.
|
a,b,c = map(str,input().split())
print("Yes" if a[-1] == b[0] and b[-1] == c[0] else "No")
|
s737738585
|
Accepted
| 17
| 2,940
| 92
|
a,b,c = map(str,input().split())
print("YES" if (a[-1] == b[0] and b[-1] == c[0]) else "NO")
|
s198877826
|
p03080
|
u517447467
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 121
|
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.
|
hat = input()
num = 0
for i in hat:
if i == "R":
num += 1
if num > (len(hat)/2):
print("Yes")
else:
print("No")
|
s924854221
|
Accepted
| 18
| 2,940
| 129
|
input()
hat = input()
num = 0
for i in hat:
if i == "R":
num += 1
if num > (len(hat)/2):
print("Yes")
else:
print("No")
|
s133941464
|
p02853
|
u747602774
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 104
|
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())
if X*Y == 1:
print(10^6)
else:
print(10^5*(max(0,4-X)+max(0,4-X)))
|
s624801513
|
Accepted
| 17
| 2,940
| 106
|
X,Y = map(int,input().split())
if X*Y == 1:
print(10**6)
else:
print(10**5*(max(0,4-X)+max(0,4-Y)))
|
s479804196
|
p03719
|
u772649753
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 89
|
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
A,B,C = map(int,input().split())
if A <= C and C <= B:
print("YES")
else:
print("NO")
|
s727703895
|
Accepted
| 17
| 2,940
| 84
|
a,b,c = map(int, input().split())
if a <= c <= b:
print("Yes")
else:
print("No")
|
s693353997
|
p03501
|
u752552310
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 78
|
You are parking at a parking lot. You can choose from the following two fee plans: * Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours. * Plan 2: The fee will be B yen, regardless of the duration. Find the minimum fee when you park for N hours.
|
n, a, b = map(int, input().split())
if n*a > b:
print(n*a)
else:
print(b)
|
s321035363
|
Accepted
| 17
| 2,940
| 80
|
n, a, b = map(int, input().split())
if n*a >= b:
print(b)
else:
print(n*a)
|
s533775667
|
p02972
|
u821284362
| 2,000
| 1,048,576
|
Wrong Answer
| 240
| 10,560
| 356
|
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_ls = list(map(int,input().split(" ")))
a_ls_rev = a_ls[::-1]
N_rev = [0]*N
l = len(a_ls)
for i,a in enumerate(a_ls_rev):
s_2 = sum(N_rev[::(i+1)])%2
if s_2 != a:
N_rev[i]+=1
s = ""
for i,n in enumerate(N_rev[::-1]):
if n == 1:
s += "{} ".format(i+1)
if len(s) == 0:
print(0)
else:
print(s[:-1])
|
s020034564
|
Accepted
| 301
| 9,912
| 421
|
N = int(input())
a_ls = list(map(int,input().split(" ")))
N = [0]*N
l = len(a_ls)
for i in range(l):
a = a_ls[-(i+1)]
s = sum(N[(l-i)-1::(l-i)])%2
# print(s)
# print(a)
if a != s:
N[-(i+1)]=1
sen = ""
for i,n in enumerate(N):
if n == 1:
sen += "{} ".format(i+1)
if len(sen) == 0:
print(0)
else:
print(sum(N))
print(sen[:-1])
|
s976543731
|
p03861
|
u209619667
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 87
|
You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x?
|
A = list(map(int,input().split()))
a=A[0]
b=A[1]
x=A[2]
s = 0
n = 0
s=(b-a)//x
print(s)
|
s771406015
|
Accepted
| 17
| 2,940
| 98
|
A = list(map(int,input().split()))
a=A[0]
b=A[1]
x=A[2]
a = (a-1) // x
b = b // x
s = b-a
print(s)
|
s872687241
|
p02608
|
u457802431
| 2,000
| 1,048,576
|
Wrong Answer
| 2,207
| 47,312
| 250
|
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).
|
def func(x,y,z):
return x**2+y**2+z**2+x*y+y*z+z*x
n = int(input())
ans = 0
arr = []
for x in range(1,100):
for y in range(1,100):
for z in range(1,100):
arr.append(func(x,y,z))
for i in range(n):
print(arr.count(i))
|
s024088619
|
Accepted
| 291
| 19,824
| 395
|
import itertools
import math
import collections
def func(x,y,z):
return x**2+y**2+z**2+x*y+y*z+z*x
n = int(input())
ans = 0
arr = []
for x in range(1,100):
for y in range(1,100):
for z in range(1,100):
ans = func(x,y,z)
arr.append(ans)
if ans > n:
break
c = collections.Counter(arr)
for i in range(n):
print(c[i+1])
|
s213168194
|
p03228
|
u009414205
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,060
| 256
|
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().split())
for i in range(k):
if i % 2 == 0:
if a % 2 == 1:
a -= 1
b += a / 2
a /= 2
else:
if b % 2 == 1:
b -= 1
a += b / 2
b /= 2
print(a, b)
|
s507420614
|
Accepted
| 17
| 2,940
| 260
|
a, b, k = map(int, input().split())
for i in range(k):
if i % 2 == 0:
if a % 2 == 1:
a -= 1
b += a // 2
a //= 2
else:
if b % 2 == 1:
b -= 1
a += b // 2
b //= 2
print(a, b)
|
s760348727
|
p03110
|
u712082626
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 165
|
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?
|
N = int(input())
list = []
for i in range(N):
a, b = input().split()
list.append(float(a) if b == "JPY" else float(a) * 380000)
print(list)
print(sum(list))
|
s623379775
|
Accepted
| 17
| 2,940
| 153
|
N = int(input())
list = []
for i in range(N):
a, b = input().split()
list.append(float(a) if b == "JPY" else float(a) * 380000)
print(sum(list))
|
s863434145
|
p00008
|
u742178809
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,528
| 215
|
Write a program which reads an integer n and identifies the number of combinations of a, b, c and d (0 ≤ a, b, c, d ≤ 9) which meet the following equality: a + b + c + d = n For example, for n = 35, we have 4 different combinations of (a, b, c, d): (8, 9, 9, 9), (9, 8, 9, 9), (9, 9, 8, 9), and (9, 9, 9, 8).
|
s = int(input())
results=set()
for a in range(10):
for b in range(10):
for c in range(10):
d = s-a-b-c
if d<0:break
if d<=9: results.add((a,b,c,d))
print(len(results))
|
s095801391
|
Accepted
| 40
| 7,712
| 379
|
import sys
#from me.io import dup_file_stdin
def solve():
for s in map(int,sys.stdin):
results=set()
for a in range(10):
for b in range(10):
for c in range(10):
d = s-a-b-c
if d<0:break
if d<=9: results.add((a,b,c,d))
print(len(results))
solve()
|
s904112890
|
p04011
|
u241190800
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 152
|
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.
|
NKXY = [int(input()) for i in range(4)]
total=0
for i in range(NKXY[0]):
if i<=NKXY[1]:
total += NKXY[2]
else:
total += NKXY[3]
print(total)
|
s528791439
|
Accepted
| 20
| 2,940
| 151
|
NKXY = [int(input()) for i in range(4)]
total=0
for i in range(NKXY[0]):
if i<NKXY[1]:
total += NKXY[2]
else:
total += NKXY[3]
print(total)
|
s511039289
|
p02613
|
u840570107
| 2,000
| 1,048,576
|
Wrong Answer
| 147
| 16,112
| 166
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
n = int(input())
lis = [input() for _ in range(n)]
dic = {"AC":0, "WA":0, "TLE":0, "RE":0}
for x in lis:
dic[x] += 1
for y in dic.keys():
print(y, "×", dic[y])
|
s305680114
|
Accepted
| 144
| 16,152
| 165
|
n = int(input())
lis = [input() for _ in range(n)]
dic = {"AC":0, "WA":0, "TLE":0, "RE":0}
for x in lis:
dic[x] += 1
for y in dic.keys():
print(y, "x", dic[y])
|
s165499554
|
p02612
|
u810066979
| 2,000
| 1,048,576
|
Wrong Answer
| 25
| 9,136
| 30
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n = int(input())
print(n%1000)
|
s238653723
|
Accepted
| 29
| 9,152
| 74
|
n = int(input())
ans = 1000 - n%1000
if ans ==1000:
ans -=1000
print(ans)
|
s113756270
|
p03485
|
u141410514
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 50
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
a,b = map(int,input().split())
print(round(a+b)/2)
|
s395665705
|
Accepted
| 17
| 2,940
| 69
|
import math
a,b = map(int,input().split())
print(math.ceil((a+b)/2))
|
s515225215
|
p02603
|
u924594299
| 2,000
| 1,048,576
|
Wrong Answer
| 34
| 9,252
| 699
|
To become a millionaire, M-kun has decided to make money by trading in the next N days. Currently, he has 1000 yen and no stocks - only one kind of stock is issued in the country where he lives. He is famous across the country for his ability to foresee the future. He already knows that the price of one stock in the next N days will be as follows: * A_1 yen on the 1-st day, A_2 yen on the 2-nd day, ..., A_N yen on the N-th day. In the i-th day, M-kun can make the following trade **any number of times** (possibly zero), **within the amount of money and stocks that he has at the time**. * Buy stock: Pay A_i yen and receive one stock. * Sell stock: Sell one stock for A_i yen. What is the maximum possible amount of money that M-kun can have in the end by trading optimally?
|
N = int(input())
A = list(map(int, input().split()))
# dp[i]:
dp = [0] * N
dp[0] = 1000
for i in range(1, N):
for j in range(0, i):
v = int(dp[j] / A[j])
w = dp[j] - v*A[j] + v*A[i]
print(v, w)
dp[i] = max(dp[i-1], w)
print(dp)
|
s523659706
|
Accepted
| 29
| 9,136
| 683
|
N = int(input())
A = list(map(int, input().split()))
# dp[i]:
dp = [0] * N
dp[0] = 1000
for i in range(1, N):
for j in range(0, i):
v = int(dp[j] / A[j])
w = dp[j] - v*A[j] + v*A[i]
dp[i] = max(dp[i-1], w)
print(max(dp))
|
s705661947
|
p03457
|
u637289184
| 2,000
| 262,144
|
Wrong Answer
| 370
| 30,672
| 465
|
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
N = int(input())
t=[[0,0,0]]
for i in range(N):
A=list(map(int, input().strip().split()))
t.append(A)
flag3=0
for i in range(N):
flag=0
flag2=0
time=t[i+1][0]-t[i][0]
x=t[i+1][1]-t[i][1]
y=t[i+1][2]-t[i][2]
time=abs(time)
x=abs(x)
y=abs(y)
if time==(x+y):
flag=1
elif (time-(x+y))%2==0 and time>x+y:
flag=1
if flag!=1:
flag3=1
if flag3==1:
print("No")
else:
print("Yes")
print(t)
|
s783808345
|
Accepted
| 323
| 27,120
| 429
|
N = int(input())
t=[[0,0,0]]
for i in range(N):
A=list(map(int, input().strip().split()))
t.append(A)
flag=0
flag2=0
for i in range(N):
flag=0
time=t[i+1][0]-t[i][0]
x=t[i+1][1]-t[i][1]
y=t[i+1][2]-t[i][2]
time=abs(time)
x=abs(x)
y=abs(y)
if (time-(x+y))%2==0 and time>=(x+y):
flag=1
if flag!=1:
flag2=1
if flag2==1:
print("No")
else:
print("Yes")
|
s417681071
|
p03555
|
u238510421
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 126
|
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.
|
C1 = list(input())
C2 = list(input())
if (C1[0]==C2[2])&(C1[1]==C2[1])&(C1[2]==C2[0]):
print("Yes")
else:
print("No")
|
s363718861
|
Accepted
| 17
| 2,940
| 126
|
C1 = list(input())
C2 = list(input())
if (C1[0]==C2[2])&(C1[1]==C2[1])&(C1[2]==C2[0]):
print("YES")
else:
print("NO")
|
s355434779
|
p03455
|
u151625340
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 90
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a,b = map(int,input().split())
if (a*b)% 2 == 0:
print('Odd')
else:
print('Even')
|
s780283632
|
Accepted
| 17
| 2,940
| 91
|
a,b = map(int,input().split())
if (a*b)% 2 == 1:
print('Odd')
else:
print('Even')
|
s110111866
|
p03836
|
u218838821
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 178
|
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())
print("U"*(ty-sy) + "R"*(tx-sx) + "U" + "L"*(tx-sx+1) + "D"*(ty-sy+1) + "R"*(tx-sx)+ "U"*(ty-sy) + "R" + "D"*(ty-sy+1) + "L"*(tx-sx+1))
|
s638474820
|
Accepted
| 17
| 3,064
| 191
|
sx, sy, tx, ty = map(int,input().split())
print("U"*(ty-sy) + "R"*(tx-sx) + "D"*(ty-sy) + "L"*(tx-sx) + "L" + "U"*(ty-sy+1) + "R"*(tx-sx+1) + "D" + "R" + "D"*(ty-sy+1) + "L"*(tx-sx+1) + "U")
|
s355981839
|
p03752
|
u351480677
| 1,000
| 262,144
|
Wrong Answer
| 176
| 3,636
| 636
|
There are N buildings along the line. The i-th building from the left is colored in color i, and its height is currently a_i meters. Chokudai is a mayor of the city, and he loves colorful thigs. And now he wants to see at least K buildings from the left. You can increase height of buildings, but it costs 1 yens to increase 1 meters. It means you cannot make building that height is not integer. You cannot decrease height of buildings. Calculate the minimum cost of satisfying Chokudai's objective. Note: "Building i can see from the left" means there are no j exists that (height of building j) ≥ (height of building i) and j < i.
|
n, k = map(int,input().split())
a = list(map(int,input().split()))
res = float("inf")
for i in range(2**n):
if(i%2 == 0):
continue
useidx = [0 for _ in range(n)]
usecnt = 0
for j in range(n):
if (i>>j)&1:
useidx[j] = 1
usecnt = usecnt+1
if (usecnt < k):
continue
sm = 0
mx = a[0]
for j in range(1,n):
if useidx[j]==1:
if a[j]<=mx:
print("a[j]="+str(a[j]))
sm = sm + (mx+1)-a[j]
print("max="+str(mx+1))
mx = mx +1
mx = max(mx, a[j])
res = min(res,sm)
print(sm)
|
s577544404
|
Accepted
| 115
| 3,064
| 1,723
|
n, k = map(int, input().split())
a = list(map(int, input(). split()))
res = 1 << 60
for i in range(1<<n):
if(i%2 == 0): continue
use_idx = [0 for _ in range(n)]
use_cnt = 0
for j in range(n):
if(i & (1<<j)):
use_idx[j] = 1
use_cnt += 1
if(use_cnt != k): continue
sm = 0
mx = a[0]
for j in range(1, n):
if(use_idx[j]):
if(a[j] <= mx):
sm += mx+1 - a[j]
mx += 1
mx = max(mx, a[j])
res = min(res, sm)
print(res)
|
s159060055
|
p03129
|
u191635495
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 108
|
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.
|
import math
n, k = map(int, input().split())
if k <= math.ceil(n/2):
print("Yes")
else:
print("No")
|
s120043529
|
Accepted
| 17
| 3,064
| 108
|
import math
n, k = map(int, input().split())
if k <= math.ceil(n/2):
print("YES")
else:
print("NO")
|
s868020712
|
p02694
|
u579746769
| 2,000
| 1,048,576
|
Wrong Answer
| 22
| 9,172
| 91
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
X=int(input())
a=100
count=0
while a<=X:
a=int(a+(a*0.01))
count=count+1
print(count)
|
s305501944
|
Accepted
| 25
| 9,164
| 90
|
X=int(input())
a=100
count=0
while a<X:
a=int(a+(a*0.01))
count=count+1
print(count)
|
s846399056
|
p03371
|
u620480037
| 2,000
| 262,144
|
Wrong Answer
| 108
| 7,048
| 503
|
"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively. Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.
|
input_line =input()
input_line = input_line.rstrip().split(" ")
a = int(input_line[0])
b = int(input_line[1])
c = int(input_line[2])
x = int(input_line[3])
y = int(input_line[4])
list1=[]
for i in range(max(x,y)+1):
if x-i<=0:
cost = b*(y-i)+2*c*i
list1.append(cost)
i +=1
elif y-i<=0:
cost = a*(x-i)+2*c*i
list1.append(cost)
i +=1
else:
cost = a*(x-i)+b*(y-i)+2*c*i
list1.append(cost)
i +=1
print(min(list1))
|
s019434257
|
Accepted
| 120
| 3,064
| 202
|
A,B,C,X,Y=map(int,input().split())
ans=10**10
for i in range(max(X,Y)+1):
cnt=0
cnt+=C*2*i
x=X-i
y=Y-i
cnt+=max(0,x*A)
cnt+=max(0,y*B)
if ans>cnt:
ans=cnt
print(ans)
|
s749837905
|
p03090
|
u690536347
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 3,740
| 290
|
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.
|
from itertools import combinations as comb
N = int(input())
n = N-1 if N%2 else N
s = set()
for i in range(1, N+1):
if N%2:
a, b = i, N-i
else:
a, b = i, N-i+1
if a<b:
s.add((a, b))
for i in comb(range(1, N+1), 2):
if not i in s:
print(*i)
|
s439297057
|
Accepted
| 24
| 3,996
| 216
|
from itertools import combinations as comb
N = int(input())
n = N-N%2
s = {(i, n+1-i) for i in range(1, N+1) if i<n+1-i}
l = [i for i in comb(range(1, N+1), 2) if not i in s]
print(len(l))
for i in l:
print(*i)
|
s802855115
|
p02602
|
u458925609
| 2,000
| 1,048,576
|
Wrong Answer
| 274
| 31,612
| 340
|
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 = list(map(int, input().split()))
A = list(map(int, input().split()))
results = list()
for first, second in zip(A[0:N - K], A[K:N]):
print(first, second)
if second > first:
results.append(True)
else:
results.append(False)
for res in results:
if res:
print('Yes')
else:
print('No')
|
s723507313
|
Accepted
| 155
| 31,760
| 315
|
N, K = list(map(int, input().split()))
A = list(map(int, input().split()))
results = list()
for first, second in zip(A[0:N - K], A[K:N]):
if second > first:
results.append(True)
else:
results.append(False)
for res in results:
if res:
print('Yes')
else:
print('No')
|
s677711004
|
p03544
|
u101627912
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 126
|
It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers. You are given an integer N. Find the N-th Lucas number. Here, the i-th Lucas number L_i is defined as follows: * L_0=2 * L_1=1 * L_i=L_{i-1}+L_{i-2} (i≥2)
|
# coding: utf-8
N=int(input())
lst=[2,1]
for i in range(N):
lst.append(lst[i]+lst[i+1])
print(lst[i+2])
print(lst[N])
|
s132489601
|
Accepted
| 17
| 2,940
| 106
|
# coding: utf-8
N=int(input())
lst=[2,1]
for i in range(N):
lst.append(lst[i]+lst[i+1])
print(lst[N])
|
s341427372
|
p03448
|
u189397279
| 2,000
| 262,144
|
Wrong Answer
| 49
| 3,064
| 263
|
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.
|
# -*- coding: utf-8 -*-
A = int(input())
B = int(input())
C = int(input())
X = int(input())
count = 0
for i in range(1, A + 1):
for j in range(1, B + 1):
for k in range(1, C + 1):
if 500 * i + 100 * j + 50 * k == X:
count += 1
print(count)
|
s474259751
|
Accepted
| 52
| 3,060
| 256
|
# -*- coding: utf-8 -*-
A = int(input())
B = int(input())
C = int(input())
X = int(input())
count = 0
for i in range(A + 1):
for j in range(B + 1):
for k in range(C + 1):
if 500 * i + 100 * j + 50 * k == X:
count += 1
print(count)
|
s024261579
|
p02273
|
u782850499
| 2,000
| 131,072
|
Wrong Answer
| 20
| 7,712
| 793
|
Write a program which reads an integer _n_ and draws a Koch curve based on recursive calles of depth _n_. The Koch curve is well known as a kind of You should start (0, 0), (100, 0) as the first segment.
|
import math as mt
def kock(n, p1, p2):
if n == 0:
return
s =[(2 * p1[0] + p2[0]) / 3 , (2 * p1[1] + p2[1]) / 3]
t = [(p1[0] + 2 * p2[0]) / 3, (p1[1] + 2 * p2[1]) / 3]
u = [(t[0]-s[0])*mt.cos(mt.radians(60))
-(t[1]-s[1])*mt.sin(mt.radians(60)) + s[0],
(t[0]-s[0])*mt.sin(mt.radians(60))
+(t[1]-s[1])*mt.cos(mt.radians(60)) + s[1]]
kock(n-1, p1, s)
print("{0:.8f} {1:.8f}".format(s[1],s[0]))
kock(n-1, s, u)
print("{0:.8f} {1:.8f}".format(u[1],u[0]))
kock(n-1, u, t)
print("{0:.8f} {1:.8f}".format(t[1],t[0]))
kock(n-1, t, p2)
if __name__ == '__main__':
n = 2
p1 = (0,0)
p2 = (0,100)
print("{0:.8f} {1:.8f}".format(p1[0],p1[1]))
kock(n, p1, p2)
print("{0:.8f} {1:.8f}".format(p2[0],p2[1]))
|
s338306445
|
Accepted
| 30
| 7,992
| 804
|
import math as mt
def kock(n, p1, p2):
if n == 0:
return
s =[(2 * p1[0] + p2[0]) / 3 , (2 * p1[1] + p2[1]) / 3]
t = [(p1[0] + 2 * p2[0]) / 3, (p1[1] + 2 * p2[1]) / 3]
u = [(t[0]-s[0])*mt.cos(mt.radians(60))
-(t[1]-s[1])*mt.sin(mt.radians(60)) + s[0],
(t[0]-s[0])*mt.sin(mt.radians(60))
+(t[1]-s[1])*mt.cos(mt.radians(60)) + s[1]]
kock(n-1, p1, s)
print("{0:.8f} {1:.8f}".format(s[0],s[1]))
kock(n-1, s, u)
print("{0:.8f} {1:.8f}".format(u[0],u[1]))
kock(n-1, u, t)
print("{0:.8f} {1:.8f}".format(t[0],t[1]))
kock(n-1, t, p2)
if __name__ == '__main__':
n = int(input())
p1 = (0,0)
p2 = (100,0)
print("{0:.8f} {1:.8f}".format(p1[0],p1[1]))
kock(n, p1, p2)
print("{0:.8f} {1:.8f}".format(p2[0],p2[1]))
|
s829864557
|
p03599
|
u092650292
| 3,000
| 262,144
|
Wrong Answer
| 25
| 3,064
| 505
|
Snuke is making sugar water in a beaker. Initially, the beaker is empty. Snuke can perform the following four types of operations any number of times. He may choose not to perform some types of operations. * Operation 1: Pour 100A grams of water into the beaker. * Operation 2: Pour 100B grams of water into the beaker. * Operation 3: Put C grams of sugar into the beaker. * Operation 4: Put D grams of sugar into the beaker. In our experimental environment, E grams of sugar can dissolve into 100 grams of water. Snuke will make sugar water with the highest possible density. The beaker can contain at most F grams of substances (water and sugar combined), and there must not be any undissolved sugar in the beaker. Find the mass of the sugar water Snuke will make, and the mass of sugar dissolved in it. If there is more than one candidate, any of them will be accepted. We remind you that the sugar water that contains a grams of water and b grams of sugar is \frac{100b}{a + b} percent. Also, in this problem, pure water that does not contain any sugar is regarded as 0 percent density sugar water.
|
a,b,c,d,e,f = map(int, input().split(' '))
w = 0
s = 0
ratio = 0
for i in range((f // (b*100))+1):
for j in range(((f-(i*b*100))//a)+1):
if i == 0 and j == 0:
continue
smax = min(f-(j*a+i*b)*100,(j*a+i*b)*e)
sm = 0
for k in range((smax//d)+1):
sm = max(sm, k*d+c*((smax-(k*d))//c))
ratio = sm/((j*a+i*b)*100+sm)
w = (j*a+i*b)*100
s = sm
print("%i %i" %(w+s,s))
|
s449513463
|
Accepted
| 19
| 3,064
| 510
|
a,b,c,d,e,f = map(int, input().split(' '))
w = 0
s = 0
ratio = 0
for i in range((f // (b*100))+1):
for j in range(((f-(i*b*100))//(a*100))+1):
if i == 0 and j == 0:
continue
smax = min(f-(j*a+i*b)*100,(j*a+i*b)*e)
sm = 0
for k in range((smax//d)+1):
sm = max(sm, k*d+c*((smax-(k*d))//c))
if ratio <= sm/((j*a+i*b)*100+sm):
ratio = sm/((j*a+i*b)*100+sm)
w = (j*a+i*b)*100
s = sm
print("%i %i" %(w+s,s))
|
s336737585
|
p03433
|
u684330841
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,188
| 121
|
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
N=int(input())
A=int(input())
for i in range(A):
if (N-i)%500==0:
print("Yes")
else:
print("No")
|
s418196095
|
Accepted
| 17
| 2,940
| 86
|
N=int(input())
A=int(input())
x=N%500
if x<=A:
print("Yes")
else:
print("No")
|
s894853951
|
p03131
|
u995062424
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 9,176
| 144
|
Snuke has one biscuit and zero Japanese yen (the currency) in his pocket. He will perform the following operations exactly K times in total, in the order he likes: * Hit his pocket, which magically increases the number of biscuits by one. * Exchange A biscuits to 1 yen. * Exchange 1 yen to B biscuits. Find the maximum possible number of biscuits in Snuke's pocket after K operations.
|
K, A, B = map(int, input().split())
if(B-A <= 2):
print(K+1)
else:
m = max(0, K-(A-1)//2)*(B-A)+(K+1)-2*max(0, K-(A-1)//2)
print(m)
|
s906553675
|
Accepted
| 26
| 9,036
| 148
|
K, A, B = map(int, input().split())
if(B-A <= 2):
print(K+1)
else:
m = max(0, (K-(A-1))//2)*(B-A)+(K+1)-2*max(0, (K-(A-1))//2)
print(m)
|
s315546320
|
p04013
|
u789205676
| 2,000
| 262,144
|
Wrong Answer
| 2,108
| 21,832
| 269
|
Tak has N cards. On the i-th (1 \leq i \leq N) card is written an integer x_i. He is selecting one or more cards from these N cards, so that the average of the integers written on the selected cards is exactly A. In how many ways can he make his selection?
|
import numpy as np
n, a = map(int, input().split())
x = np.array(list(map(int, input().split())))
dp = {0:1}
for y in x-a:
for k, v in dp.copy().items():
print("before",dp)
dp[y+k] = dp.get(y+k, 0) + v
print("after",dp)
print(dp[0]-1)
|
s291375065
|
Accepted
| 330
| 21,772
| 215
|
import numpy as np
n, a = map(int, input().split())
x = np.array(list(map(int, input().split())))
dp = {0:1}
for y in x-a:
for k, v in dp.copy().items():
dp[y+k] = dp.get(y+k, 0) + v
print(dp[0]-1)
|
s934686963
|
p02936
|
u501724534
| 2,000
| 1,048,576
|
Wrong Answer
| 2,109
| 91,976
| 953
|
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations.
|
# -*- coding: utf-8 -*-
"""
Created on Mon Sep 16 21:29:27 2019
@author: lamducanhndgv
"""
class Node:
def __init__(self):
self.value = 0
self.next = []
def __add__(self, value):
self.value += value
return self
def add_next(self, node):
self.next.append(node)
def solution(t):
q = []
q.append((0, 0))
while q:
addValue, tNode = q.pop(0)
t[tNode] += addValue
for i in t[tNode].next:
q.append((t[tNode].value, i))
if __name__ == "__main__":
N, Q = list(map(int, input().split()))
tree = [Node() for i in range(N)]
for _ in range(N - 1):
a, b = list(map(int, input().split()))
tree[a - 1].add_next(b - 1)
for _ in range(Q):
p, x = list(map(int, input().split()))
tree[p - 1].value += x
solution(tree)
print(*(i.value for i in tree), end = "")
|
s605820547
|
Accepted
| 1,863
| 56,160
| 895
|
import collections
n,q=map(int,input().split())
cnt=[0]*(n+1)
g=[[] for _ in range(n+1)]
for _ in range(n-1):
a,b=map(int,input().split())
g[a].append(b)
g[b].append(a)
for _ in range(q):
v,val=map(int,input().split())
cnt[v]+=val
q=collections.deque()
q.append(1)
checked=[0]*(n+1)
while q:
v=q.pop()
checked[v]=1
for u in g[v]:
if checked[u]==1:
continue
cnt[u]+=cnt[v]
q.append(u)
print(*cnt[1:])
|
s169600701
|
p03712
|
u934868410
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 198
|
You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1.
|
h,w = map(int, input().split())
for i in range(w):
print('#', end='')
print()
for i in range(h):
print('#', end='')
print(input(), end='')
print('#')
for i in range(w):
print('#', end='')
|
s187269831
|
Accepted
| 18
| 3,060
| 209
|
h,w = map(int, input().split())
for i in range(w+2):
print('#', end='')
print()
for i in range(h):
print('#', end='')
print(input(), end='')
print('#')
for i in range(w+2):
print('#', end='')
print()
|
s630049861
|
p03007
|
u625729943
| 2,000
| 1,048,576
|
Wrong Answer
| 244
| 14,024
| 860
|
There are N integers, A_1, A_2, ..., A_N, written on a blackboard. We will repeat the following operation N-1 times so that we have only one integer on the blackboard. * Choose two integers x and y on the blackboard and erase these two integers. Then, write a new integer x-y. Find the maximum possible value of the final integer on the blackboard and a sequence of operations that maximizes the final integer.
|
N = int(input())
A = list(map(int, input().split()))
A.sort()
if N==2:
print(A[-1] - A[0])
print(A[-1], A[0])
elif A[0]<0 and A[-1]<0:
tmpA = A.copy()
print(-sum(A[:-2]) + (A[-1]-A[-2]))
for i in range(N-1):
if i==0:
print(A[-1], A[-2])
tmp = A[-1] - A[-2]
elif i==N-2:
print(tmp, A[-i-1])
else:
print(tmp, A[-i-1])
tmp = tmp - A[i-1]
elif A[0]>0 and A[-1]>0:
tmpA = A.copy()
print(sum(A[2:]) + (A[1] - A[0]))
for i in range(N-1):
if i==0:
print(A[0], A[1])
tmp = A[0] - A[1]
elif i==N-2:
print(A[i+1], tmp)
else:
print(tmp, A[i+1])
tmp = tmp - A[i+1]
else:
tmpA = A.copy()
len_of_Apos = len([a for a in A if a>=0])
len_of_Aneg = len([a for a in A if a<0])
zero_in = 0 in A
minv = min(len_of_Aneg, len_of_Apos)
print(sum([abs(a) for a in A]))
|
s081118959
|
Accepted
| 129
| 18,132
| 490
|
def c(N, A):
A.sort()
maximum = A.pop()
minimum = A.pop(0)
operation = []
for a in A:
if a >= 0:
operation.append('{} {}'.format(minimum, a))
minimum -= a
else:
operation.append('{} {}'.format(maximum, a))
maximum -= a
operation.append('{} {}'.format(maximum, minimum))
return str(maximum - minimum) + '\n' + '\n'.join(operation)
N = int(input())
A = list(map(int, input().split()))
print(c(N, A))
|
s513964958
|
p03385
|
u126823513
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 134
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
str_i = input()
if str_i.count('a') == 1 and str_i.count('b') == 1 and str_i.count('c') == 1:
print('YES')
else:
print('No')
|
s367801148
|
Accepted
| 17
| 2,940
| 134
|
str_i = input()
if str_i.count('a') == 1 and str_i.count('b') == 1 and str_i.count('c') == 1:
print('Yes')
else:
print('No')
|
s855995061
|
p03493
|
u416758623
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 118
|
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.
|
sentence = input().split()
count = 0
for i in range(len(sentence)):
if sentence[i] == 1:
count+=1
print(count)
|
s478994599
|
Accepted
| 17
| 2,940
| 91
|
s = input()
ans = 0
for i in range(len(s)):
if s[i] == "1":
ans += 1
print(ans)
|
s053386187
|
p03110
|
u261891508
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 140
|
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?
|
n=int(input())
ans=0
for i in range(n):
x,u=input().split()
x=float(x)
if u=="JPY":
ans+=x
else:
ans+=3800*x
print(int(ans))
|
s254647977
|
Accepted
| 18
| 2,940
| 137
|
n=int(input())
ans=0
for i in range(n):
x,u=input().split()
x=float(x)
if u=="JPY":
ans+=x
else:
ans+=380000*x
print(ans)
|
s558858965
|
p03574
|
u846634344
| 2,000
| 262,144
|
Wrong Answer
| 29
| 3,064
| 690
|
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.
|
import sys
if sys.platform =='ios':
sys.stdin=open('input.txt')
h, w = [int(x) for x in input().split()]
board = [input() for _ in range(h)]
mines=['' for _ in range(h)]
print(board)
for row in range(h):
for col in range(w):
if board[row][col] != '#':
counts = 0
dx = [x for x in range(-1, 2) if (col+x>=0 and col+x<w)]
dy = [y for y in range(-1, 2) if (row+y>=0 and row+y<h)]
#print('row:{}, col:{}, dx:{}, dy:{}'.format(row,col,dx,dy))
for x in dx:
for y in dy:
#print('row+y:{}, col+x:{}'.format(row+y, col+x))
if board[row+y][col+x] == '#':
counts += 1
mines[row] += str(counts)
else:
mines[row] += '#'
for mine in mines:
print(mine)
|
s900589186
|
Accepted
| 29
| 3,064
| 765
|
h, w = [int(x) for x in input().split()]
board = [input() for _ in range(h)]
mines=['' for _ in range(h)]
#print(board)
for row in range(h):
for col in range(w):
if board[row][col] != '#':
counts = 0
dx = [x for x in range(-1, 2) if (col+x>=0 and col+x<w)]
dy = [y for y in range(-1, 2) if (row+y>=0 and row+y<h)]
#print('row:{}, col:{}, dx:{}, dy:{}'.format(row,col,dx,dy))
for x in dx:
for y in dy:
#print('row+y:{}, col+x:{}'.format(row+y, col+x))
if board[row+y][col+x] == '#':
counts += 1
mines[row] += str(counts)
else:
mines[row] += '#'
for mine in mines:
print(mine)
|
s996614369
|
p04043
|
u342106216
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 235
|
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
data = input().split()
def main(data):
five_num = data.count(5)
seven_num = data.count(7)
if five_num == 2 and seven_num == 1:
print("YES")
else:
print("NO")
if __name__ == "__main__":
main(data)
|
s753267174
|
Accepted
| 17
| 2,940
| 239
|
data = input().split()
def main(data):
five_num = data.count("5")
seven_num = data.count("7")
if five_num == 2 and seven_num == 1:
print("YES")
else:
print("NO")
if __name__ == "__main__":
main(data)
|
s392386253
|
p03699
|
u606174760
| 2,000
| 262,144
|
Wrong Answer
| 2,125
| 339,052
| 454
|
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?
|
# -*- coding:utf-8 -*-
import sys, itertools
N = int(input())
Sn = sorted([int(input()) for i in range(N)])
total = sum(Sn)
print(total)
if total % 10 != 0:
print(total)
sys.exit()
sub_set = set(Sn)
for i in range(2, N):
comb = [sum(l) for l in itertools.combinations(Sn, i)]
for c in comb:
sub_set.add(c)
for sub in sorted(sub_set):
if (total - sub) % 10 != 0:
print(total - sub)
break
else:
print(0)
|
s040493080
|
Accepted
| 18
| 3,060
| 353
|
# -*- coding:utf-8 -*-
import sys
N = int(input())
Sn = sorted([int(input()) for i in range(N)])
total = sum(Sn)
if total % 10 != 0:
print(total)
sys.exit()
else:
i = 0
while True:
if (total - Sn[i]) % 10 != 0:
print(total - Sn[i])
sys.exit()
i += 1
if i == N:
break
print(0)
|
s438547710
|
p01102
|
u481571686
| 8,000
| 262,144
|
Wrong Answer
| 20
| 5,584
| 442
|
The programming contest named _Concours de Programmation Comtemporaine Interuniversitaire_ (CPCI) has a judging system similar to that of ICPC; contestants have to submit correct outputs for two different inputs to be accepted as a correct solution. Each of the submissions should include the program that generated the output. A pair of submissions is judged to be a correct solution when, in addition to the correctness of the outputs, they include an identical program. Many contestants, however, do not stop including a different version of their programs in their second submissions, after modifying a single string literal in their programs representing the input file name, attempting to process different input. The organizers of CPCI are exploring the possibility of showing a special error message for such _close_ submissions, indicating contestants what's wrong with such submissions. Your task is to detect such close submissions.
|
A=[]
while True:
a=input().split('"')
if a[0]==".":
break
b=input().split('"')
cnt=0
if len(a)!=len(b):
A.append(-1)
else:
for i in range(len(a)):
if(a[i]==b[i]):
pass
else:
cnt+=1
if cnt==2:
A.append(-1)
if cnt==1:
A.append(0)
if cnt==0:
A.append(1)
A.reverse()
for i in range(len(A)):
tmp=A.pop()
if tmp==-1:
print("DIFFERENT")
elif tmp==0:
print("CLOSE")
else:
print("IDENTICAL")
|
s510258139
|
Accepted
| 20
| 5,592
| 487
|
A=[]
while True:
a=input().split('"')
if a[0]==".":
break
b=input().split('"')
cnt=0
if len(a)!=len(b):
A.append(-1)
else:
for i in range(len(a)):
if(a[i]==b[i]):
pass
else:
if i%2==1:
cnt+=1
else:
cnt+=2
if cnt>=2:
A.append(-1)
if cnt==1:
A.append(0)
if cnt==0:
A.append(1)
A.reverse()
#print(A)
for i in range(len(A)):
tmp=A.pop()
if tmp==-1:
print("DIFFERENT")
elif tmp==0:
print("CLOSE")
else:
print("IDENTICAL")
|
s663595754
|
p03400
|
u331997680
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,188
| 218
|
Some number of chocolate pieces were prepared for a training camp. The camp had N participants and lasted for D days. The i-th participant (1 \leq i \leq N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on. As a result, there were X chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces. Find the number of chocolate pieces prepared at the beginning of the camp.
|
N = int(input())
D, X = map(int, input().split())
A = [int(input()) for i in range(N)]
B = []
for i in A:
if D//i > 0:
B.append(D//i+1)
elif D//i == 0:
B.append(D//i)
else:
B.append(1)
print(sum(B)+X)
|
s220813485
|
Accepted
| 19
| 3,060
| 152
|
N = int(input())
D, X = map(int, input().split())
A = [int(input()) for i in range(N)]
B = []
for i in A:
n = (D-1)//i+1
B.append(n)
print(sum(B)+X)
|
s718347695
|
p00013
|
u776559258
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,492
| 174
|
This figure shows railway tracks for reshuffling cars. The rail tracks end in the bottom and the top-left rail track is used for the entrace and the top- right rail track is used for the exit. Ten cars, which have numbers from 1 to 10 respectively, use the rail tracks. We can simulate the movement (comings and goings) of the cars as follow: * An entry of a car is represented by its number. * An exit of a car is represented by 0 For example, a sequence 1 6 0 8 10 demonstrates that car 1 and car 6 enter to the rail tracks in this order, car 6 exits from the rail tracks, and then car 8 and car 10 enter. Write a program which simulates comings and goings of the cars which are represented by the sequence of car numbers. The program should read the sequence of car numbers and 0, and print numbers of cars which exit from the rail tracks in order. At the first, there are no cars on the rail tracks. You can assume that 0 will not be given when there is no car on the rail tracks.
|
# coding: utf-8
L=[]
while True:
try:
n=int(input())
if n==0:
L.pop()
else:
L.append(n)
except EOFError:
break
|
s222976081
|
Accepted
| 20
| 7,528
| 199
|
# coding: utf-8
L=[]
while True:
try:
n=int(input())
if n==0:
print(L[-1])
L.pop()
else:
L.append(n)
except EOFError:
break
|
s232836489
|
p02612
|
u092614249
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 9,152
| 98
|
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())
if n%1000 == 0:
print(0)
else:
while n > 1000:
n = n - 1000
print(n)
|
s211391658
|
Accepted
| 29
| 9,116
| 105
|
n = int(input())
if n%1000 == 0:
print(0)
else:
while n > 1000:
n = n - 1000
print(1000 - n)
|
s054499080
|
p03378
|
u532502139
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 243
|
There are N + 1 squares arranged in a row, numbered 0, 1, ..., N from left to right. Initially, you are in Square X. You can freely travel between adjacent squares. Your goal is to reach Square 0 or Square N. However, for each i = 1, 2, ..., M, there is a toll gate in Square A_i, and traveling to Square A_i incurs a cost of 1. It is guaranteed that there is no toll gate in Square 0, Square X and Square N. Find the minimum cost incurred before reaching the goal.
|
a=[int(x) for x in input().split()]
n = a[0]
m = a[1]
x = a[2]
arr = a[3:3+m]
goLeft = 0
goRight = 0
for i in arr:
if i < x:
goLeft += 1
else:
goRight += 1
if goLeft > goRight:
print(goRight)
else:
print(goLeft)
|
s600285908
|
Accepted
| 17
| 3,060
| 206
|
n,m,x=[int(x) for x in input().split()]
arr=[int(x) for x in input().split()]
goLeft = 0
goRight = 0
for i in arr:
if i < x:
goLeft += 1
else:
goRight += 1
print(min(goLeft,goRight))
|
s627735543
|
p02393
|
u294922877
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,544
| 56
|
Write a program which reads three integers, and prints them in ascending order.
|
nums=list(map(int, input().split()))
print(sorted(nums))
|
s895144397
|
Accepted
| 20
| 5,584
| 66
|
print(" ".join(map(str, sorted(list(map(int,input().split()))))))
|
s105217162
|
p04029
|
u881472317
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 39
|
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 + 1) * n / 2)
|
s630150067
|
Accepted
| 19
| 2,940
| 45
|
n = int(input())
print(int((n + 1) * n / 2))
|
s348858239
|
p03719
|
u013756322
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 94
|
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
a, b, c = map(int, input().split())
if (a <= c and c <= b):
print('YES')
else:
print('NO')
|
s411753106
|
Accepted
| 17
| 2,940
| 88
|
a, b, c = map(int, input().split())
if (a <= c <= b):
print('Yes')
else:
print('No')
|
s999661727
|
p04043
|
u686253980
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 202
|
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
s = [int(i) for i in input().split(' ')]
n = [0] * 5
for i in s:
if i == 5:
n[0] += 1
elif i == 7:
n[1] += 1
if n[0] == 2 and n[1] == 1:
print('Yes')
else:
print('No')
|
s890795783
|
Accepted
| 17
| 3,060
| 205
|
s = [int(i) for i in input().split(' ')]
n = [0] * 2
for i in s:
if i == 5:
n[0] += 1
elif i == 7:
n[1] += 1
if n[0] == 2 and n[1] == 1:
print('YES')
else:
print('NO')
|
s326159088
|
p03739
|
u760802228
| 2,000
| 262,144
|
Wrong Answer
| 256
| 14,468
| 588
|
You are given an integer sequence of length N. The i-th term in the sequence is a_i. In one operation, you can select a term and either increment or decrement it by one. At least how many operations are necessary to satisfy the following conditions? * For every i (1≤i≤n), the sum of the terms from the 1-st through i-th term is not zero. * For every i (1≤i≤n-1), the sign of the sum of the terms from the 1-st through i-th term, is different from the sign of the sum of the terms from the 1-st through (i+1)-th term.
|
n = int(input())
a = list(map(int, input().split()))
def check(f):
flag = f
t = 0
c = 0
for i in range(n):
if t + a[i] <= 0 and flag == 1:
c += 1 - t - a[i]
t = 1
elif t + a[i] >= 0 and flag == -1:
c += 1 + t + a[i]
t = -1
else:
t += a[i]
flag *= -1
print(t)
return c
total_p = check(1)
total_m = check(-1)
print(min(total_p, total_m))
|
s116832632
|
Accepted
| 110
| 14,468
| 563
|
n = int(input())
a = list(map(int, input().split()))
def check(f):
flag = f
t = 0
c = 0
for i in range(n):
if t + a[i] <= 0 and flag == 1:
c += 1 - t - a[i]
t = 1
elif t + a[i] >= 0 and flag == -1:
c += 1 + t + a[i]
t = -1
else:
t += a[i]
flag *= -1
return c
total_p = check(1)
total_m = check(-1)
print(min(total_p, total_m))
|
s898659067
|
p03044
|
u839857256
| 2,000
| 1,048,576
|
Wrong Answer
| 2,225
| 1,993,844
| 796
|
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.
|
from collections import deque
def bfs(s):
flag_list[s] = 1
color_list[s] = 0
q.append(s)
while len(q)>0:
u = q.popleft()
for v in range(n):
if m_list[u][v] > 0 and flag_list[v] == 0:
flag_list[v] = 1
if color_list[u] == 0:
color_list[v] = 1
else:
color_list[v] = 0
q.append(v)
flag_list[u] = 2
if __name__ == "__main__":
n = int(input())
m_list = [[0]*n for _ in range(n)]
color_list = [0]*n
flag_list = [0]*n
q = deque([])
for _ in range(n-1):
u, v, w = map(int, input().split())
m_list[u-1][v-1] = w
m_list[v-1][u-1] = w
bfs(0)
for color in color_list:
print(color)
|
s297908651
|
Accepted
| 774
| 46,140
| 854
|
import sys
from collections import deque
def bfs(s):
color_list[s] = 0
flag_list[s] = 1
q.append(s)
while len(q) > 0:
u = q.popleft()
for v, w in m_list[u]:
if flag_list[v] == 0:
flag_list[v] = 1
if w%2 == 0:
color_list[v] = color_list[u]
else:
color_list[v] = (color_list[u]+1)%2
q.append(v)
flag_list[u] = 2
if __name__ == "__main__":
input = sys.stdin.readline
n = int(input())
m_list = [[]*n for _ in range(n)]
color_list = [0]*n
flag_list = [0]*n
q = deque([])
for _ in range(n-1):
u, v, w = map(int, input().split())
m_list[u-1].append([v-1, w])
m_list[v-1].append([u-1, w])
bfs(0)
for color in color_list:
print(color)
|
s793857291
|
p03251
|
u996434204
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 211
|
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
n,m,X,Y=map(int,input().split())
p_x=[int(j) for j in input().split()]
p_y=[int(i) for i in input().split()]
max_x=max(p_x)
Z=min(p_y)
if max_x<Z and (X<Z and X<=Y):
print('No war')
else:
print('War')
|
s365706867
|
Accepted
| 17
| 3,060
| 211
|
n,m,X,Y=map(int,input().split())
p_x=[int(j) for j in input().split()]
p_y=[int(i) for i in input().split()]
max_x=max(p_x)
Z=min(p_y)
if max_x<Z and (X<Z and Z<=Y):
print('No War')
else:
print('War')
|
s428605935
|
p02865
|
u921773161
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 12
|
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
print('Yes')
|
s442159135
|
Accepted
| 18
| 2,940
| 46
|
#%%
n = int(input())
ans = (n-1)//2
print(ans)
|
s036470580
|
p03469
|
u584153341
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 30
|
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
print('2017'+str(input())[4:])
|
s483108474
|
Accepted
| 17
| 2,940
| 31
|
print('2018'+str(input())[4:])
|
s930282939
|
p03401
|
u360116509
| 2,000
| 262,144
|
Wrong Answer
| 188
| 14,048
| 326
|
There are N sightseeing spots on the x-axis, numbered 1, 2, ..., N. Spot i is at the point with coordinate A_i. It costs |a - b| yen (the currency of Japan) to travel from a point with coordinate a to another point with coordinate b along the axis. You planned a trip along the axis. In this plan, you first depart from the point with coordinate 0, then visit the N spots in the order they are numbered, and finally return to the point with coordinate 0. However, something came up just before the trip, and you no longer have enough time to visit all the N spots, so you decided to choose some i and cancel the visit to Spot i. You will visit the remaining spots as planned in the order they are numbered. You will also depart from and return to the point with coordinate 0 at the beginning and the end, as planned. For each i = 1, 2, ..., N, find the total cost of travel during the trip when the visit to Spot i is canceled.
|
def main():
N = int(input()) + 2
A = list(map(int, input().split()))
A.insert(0, 0)
A.append(0)
D = sum([abs(A[i + 1] - A[i]) for i in range(N - 1)])
print(D)
for i in range(1, N - 1):
print(D - abs(A[i - 1] - A[i]) -
abs(A[i] - A[i + 1]) + abs(A[i - 1] - A[i + 1]))
main()
|
s695930082
|
Accepted
| 195
| 14,044
| 313
|
def main():
N = int(input()) + 2
A = list(map(int, input().split()))
A.insert(0, 0)
A.append(0)
D = sum([abs(A[i + 1] - A[i]) for i in range(N - 1)])
for i in range(1, N - 1):
print(D - abs(A[i - 1] - A[i]) -
abs(A[i] - A[i + 1]) + abs(A[i - 1] - A[i + 1]))
main()
|
s490484156
|
p02694
|
u423282882
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 9,168
| 168
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
import math
X = int(input())
money = 100
counts=0
while True:
counts+=1
money=money+math.floor((money*0.01))
if X<money:
print(counts)
break
|
s966319298
|
Accepted
| 24
| 9,164
| 169
|
import math
X = int(input())
money = 100
counts=0
while True:
money=money+math.floor((money*0.01))
counts+=1
if X<=money:
print(counts)
break
|
s156563838
|
p03162
|
u480138356
| 2,000
| 1,048,576
|
Wrong Answer
| 498
| 30,580
| 375
|
Taro's summer vacation starts tomorrow, and he has decided to make plans for it now. The vacation consists of N days. For each i (1 \leq i \leq N), Taro will choose one of the following activities and do it on the i-th day: * A: Swim in the sea. Gain a_i points of happiness. * B: Catch bugs in the mountains. Gain b_i points of happiness. * C: Do homework at home. Gain c_i points of happiness. As Taro gets bored easily, he cannot do the same activities for two or more consecutive days. Find the maximum possible total points of happiness that Taro gains.
|
N = int(input())
h = []
for i in range(N):
h.append(list(map(int, input().split(" "))))
cost = h[-1]
today_cost = [0 for i in range(3)]
for day in range(N-2, -1, -1):
today_cost[0] = h[day][0] + max(cost[1], cost[2])
today_cost[1] = h[day][1] + max(cost[0], cost[2])
today_cost[2] = h[day][2] + max(cost[0], cost[1])
cost = today_cost
print(max(cost))
|
s664825357
|
Accepted
| 513
| 30,580
| 369
|
N = int(input())
h = []
for i in range(N):
h.append(list(map(int, input().split(" "))))
cost = h[-1]
for day in range(N-2, -1, -1):
today_cost = []
today_cost.append(h[day][0] + max(cost[1], cost[2]))
today_cost.append(h[day][1] + max(cost[0], cost[2]))
today_cost.append(h[day][2] + max(cost[0], cost[1]))
cost = today_cost
print(max(cost))
|
s339102795
|
p03377
|
u768896740
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 98
|
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 or x > a + b:
print('No')
else:
print('Yes')
|
s305210212
|
Accepted
| 17
| 2,940
| 98
|
a, b, x = map(int, input().split())
if x < a or x > a + b:
print('NO')
else:
print('YES')
|
s979732917
|
p02694
|
u189604332
| 2,000
| 1,048,576
|
Wrong Answer
| 25
| 9,248
| 153
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
import math
X = int(input())
res = 100
count = 0
while X > res:
res = math.floor(res * 1.01)
print(str(res))
count +=1
print(str(count))
|
s842501287
|
Accepted
| 22
| 9,180
| 133
|
import math
X = int(input())
res = 100
count = 0
while X > res:
res = math.floor(res * 1.01)
count +=1
print(str(count))
|
s626550680
|
p02602
|
u033839917
| 2,000
| 1,048,576
|
Wrong Answer
| 2,206
| 33,984
| 259
|
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 = [int(i)for i in input().split()]
A = [int(i) for i in input().split()]
S = []
for i in range(K-1,N):
s = 1
for j in range(i,i-K,-1):
s *= A[j]
S.append(s)
print(S)
for i in range(1,len(S)):
print("Yes" if S[i-1]<S[i] else "No")
|
s789691648
|
Accepted
| 152
| 31,628
| 178
|
N,K = [int(i)for i in input().split()]
A = [int(i) for i in input().split()]
for i in range(K,N):
if A[i] / A[i-K] > 1 :
print("Yes")
else :
print("No")
|
s773438760
|
p04044
|
u842689614
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 91
|
Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.
|
N,L=map(int,input().split())
s=[]
for n in range(N):
s.append(input())
"".join(sorted(s))
|
s364637151
|
Accepted
| 18
| 3,060
| 98
|
N,L=map(int,input().split())
s=[]
for n in range(N):
s.append(input())
print("".join(sorted(s)))
|
s121340536
|
p02612
|
u114365796
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 9,032
| 125
|
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.
|
amount = int(input())
if(amount % 1000 == 0):
print(0)
else:
change = (amount / 1000 + 1) * 1000 - amount
print(change)
|
s005451991
|
Accepted
| 29
| 9,036
| 126
|
amount = int(input())
if(amount % 1000 == 0):
print(0)
else:
change = (amount // 1000 + 1) * 1000 - amount
print(change)
|
s907280064
|
p04044
|
u006167882
| 2,000
| 262,144
|
Wrong Answer
| 16
| 3,060
| 121
|
Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.
|
N, L = map(int,input().split())
l = []
for i in range(N):
l.append(input())
l.sort
result = ''.join(l)
print(result)
|
s139660160
|
Accepted
| 17
| 3,060
| 123
|
N, L = map(int,input().split())
l = []
for i in range(N):
l.append(input())
l.sort()
result = ''.join(l)
print(result)
|
s760055483
|
p03719
|
u720417458
| 2,000
| 262,144
|
Wrong Answer
| 20
| 3,316
| 90
|
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
A, B, C = map(int, input().split())
if C < B:
if C > A:
print('YES')
else:
print('NO')
|
s660736346
|
Accepted
| 17
| 2,940
| 84
|
A, B, C = map(int, input().split())
if A <= C <= B:
print('Yes')
else:
print('No')
|
s289025938
|
p02613
|
u030819239
| 2,000
| 1,048,576
|
Wrong Answer
| 150
| 16,048
| 150
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
N = int(input())
S = list()
for i in range(N):
S.append(input())
for k in ['AC', 'TLE', 'WA', 'RE']:
print('{0} x {1}'.format(k, S.count(k)))
|
s909039523
|
Accepted
| 147
| 16,240
| 150
|
N = int(input())
S = list()
for i in range(N):
S.append(input())
for k in ['AC', 'WA', 'TLE', 'RE']:
print('{0} x {1}'.format(k, S.count(k)))
|
s629474975
|
p03860
|
u171014488
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 25
|
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.
|
print("A"+input()[0]+"C")
|
s994617165
|
Accepted
| 17
| 2,940
| 25
|
print("A"+input()[8]+"C")
|
s900130341
|
p03545
|
u634079249
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 936
|
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.
|
import sys
import os
ii = lambda: int(sys.stdin.buffer.readline().rstrip())
il = lambda: list(map(int, sys.stdin.buffer.readline().split()))
iln = lambda n: [int(sys.stdin.buffer.readline().rstrip()) for _ in range(n)]
iss = lambda: sys.stdin.buffer.readline().decode().rstrip()
isn = lambda n: [sys.stdin.buffer.readline().decode().rstrip() for _ in range(n)]
def main():
if os.getenv("LOCAL"):
sys.stdin = open("input.txt", "r")
*S, = iss()
print(S)
E = ['+', '-']
tmp = [0] * 7
for e in E:
tmp[0] = S[0]
tmp[1] = e
for e in E:
tmp[2] = S[1]
tmp[3] = e
for e in E:
tmp[4] = S[2]
tmp[5] = e
tmp[6] = S[3]
ans = ''.join(map(str, tmp))
if eval(ans) == 7:
print(ans + '=7')
exit()
if __name__ == '__main__':
main()
|
s451117400
|
Accepted
| 18
| 3,064
| 923
|
import sys
import os
ii = lambda: int(sys.stdin.buffer.readline().rstrip())
il = lambda: list(map(int, sys.stdin.buffer.readline().split()))
iln = lambda n: [int(sys.stdin.buffer.readline().rstrip()) for _ in range(n)]
iss = lambda: sys.stdin.buffer.readline().decode().rstrip()
isn = lambda n: [sys.stdin.buffer.readline().decode().rstrip() for _ in range(n)]
def main():
if os.getenv("LOCAL"):
sys.stdin = open("input.txt", "r")
*S, = iss()
E = ['+', '-']
tmp = [0] * 7
for e in E:
tmp[0] = S[0]
tmp[1] = e
for e in E:
tmp[2] = S[1]
tmp[3] = e
for e in E:
tmp[4] = S[2]
tmp[5] = e
tmp[6] = S[3]
ans = ''.join(map(str, tmp))
if eval(ans) == 7:
print(ans + '=7')
exit()
if __name__ == '__main__':
main()
|
s051952495
|
p03556
|
u507116804
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 19,464
| 117
|
Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer.
|
n=int(input())
for i in range(n):
if (float.is_integer(n**0.5)):
print(n)
else:
n=n-1
|
s174669253
|
Accepted
| 50
| 3,444
| 113
|
n=int(input())
a=n
for i in range(n):
if (float.is_integer(a**0.5)):
print(a)
break
else:
a=a-1
|
s954019779
|
p03472
|
u771365068
| 2,000
| 262,144
|
Wrong Answer
| 436
| 36,820
| 430
|
You are going out for a walk, when you suddenly encounter a monster. Fortunately, you have N katana (swords), Katana 1, Katana 2, …, Katana N, and can perform the following two kinds of attacks in any order: * Wield one of the katana you have. When you wield Katana i (1 ≤ i ≤ N), the monster receives a_i points of damage. The same katana can be wielded any number of times. * Throw one of the katana you have. When you throw Katana i (1 ≤ i ≤ N) at the monster, it receives b_i points of damage, and you lose the katana. That is, you can no longer wield or throw that katana. The monster will vanish when the total damage it has received is H points or more. At least how many attacks do you need in order to vanish it in total?
|
import math
N, H = map(int, input().split())
swords = [list(map(int, input().split())) for _ in range(N)]
swords_T = [list(x) for x in zip(*swords)]
#print(swords_T)
swing_max = max(swords_T[0])
order = sorted(swords_T[1])
print(order)
count = 0
while H > 0 and len(order) != 0:
tmp = order.pop()
if tmp > swing_max:
H -= tmp
count += 1
#print(f'{H=}')
count += math.ceil(H/swing_max)
print(count)
|
s938405191
|
Accepted
| 427
| 36,820
| 444
|
import math
N, H = map(int, input().split())
swords = [list(map(int, input().split())) for _ in range(N)]
swords_T = [list(x) for x in zip(*swords)]
#print(swords_T)
swing_max = max(swords_T[0])
order = sorted(swords_T[1])
#print(order)
count = 0
while H > 0 and len(order) != 0:
tmp = order.pop()
if tmp > swing_max:
H -= tmp
count += 1
#print(f'{H=}')
if H > 0:
count += math.ceil(H/swing_max)
print(count)
|
s358054946
|
p03448
|
u145035045
| 2,000
| 262,144
|
Wrong Answer
| 50
| 3,316
| 212
|
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
a = int(input())
b = int(input())
c = int(input())
x = int(input())
ans = 0
for i in range(a):
for j in range(b):
for k in range(c):
if (a * 500 + b* 100 + c * 50 == x):
ans += 1
print(ans)
|
s016229622
|
Accepted
| 48
| 3,060
| 221
|
a = int(input())
b = int(input())
c = int(input())
x = int(input())
ans = 0
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
if ((i * 500 + j * 100 + k * 50) == x):
ans += 1
print(ans)
|
s826960194
|
p03493
|
u188272222
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 18
|
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.
|
input().count('1')
|
s667554464
|
Accepted
| 17
| 2,940
| 25
|
print(input().count('1'))
|
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