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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s642779924
|
p03605
|
u867826040
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 40 |
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
print(["No","Yes"][int("7" in input())])
|
s765784199
|
Accepted
| 17 | 2,940 | 40 |
print(["No","Yes"][int("9" in input())])
|
s662557455
|
p03387
|
u743281086
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 104 |
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.
|
ls = list(map(int, input().split()))
ans = sum(ls) - max(ls)
print(ans//2 if ans%2 == 0 else (ans+3)//2)
|
s885986968
|
Accepted
| 17 | 2,940 | 106 |
ls = list(map(int, input().split()))
ans = max(ls)*3 - sum(ls)
print(ans//2 if ans%2 == 0 else (ans+3)//2)
|
s558476375
|
p03759
|
u147571984
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,188 | 79 |
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful.
|
a, b, c = map(int, input().split())
print('%s' % 'Yes' if b-a == c-b else 'No')
|
s035210360
|
Accepted
| 17 | 2,940 | 79 |
a, b, c = map(int, input().split())
print('%s' % 'YES' if b-a == c-b else 'NO')
|
s995392318
|
p02238
|
u539803218
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,760 | 580 |
Depth-first search (DFS) follows the strategy to search ”deeper” in the graph whenever possible. In DFS, edges are recursively explored out of the most recently discovered vertex $v$ that still has unexplored edges leaving it. When all of $v$'s edges have been explored, the search ”backtracks” to explore edges leaving the vertex from which $v$ was discovered. This process continues until all the vertices that are reachable from the original source vertex have been discovered. If any undiscovered vertices remain, then one of them is selected as a new source and the search is repeated from that source. DFS timestamps each vertex as follows: * $d[v]$ records when $v$ is first discovered. * $f[v]$ records when the search finishes examining $v$’s adjacency list. Write a program which reads a directed graph $G = (V, E)$ and demonstrates DFS on the graph based on the following rules: * $G$ is given in an adjacency-list. Vertices are identified by IDs $1, 2,... n$ respectively. * IDs in the adjacency list are arranged in ascending order. * The program should report the discover time and the finish time for each vertex. * When there are several candidates to visit during DFS, the algorithm should select the vertex with the smallest ID. * The timestamp starts with 1.
|
graph={}
num = int(input())
Adj = [[0 for i in range(num)] for i in range(num)]
for i in range(num):
l = list(map(int,input().split()))
graph[str(l[0])]=l[2:]
color = [0] * num
d = [0] * num
f = [0] * num
t = 0
def dfs(u):
global t
color[u - 1] = 1
t += 1
d[u - 1] = t
for i in range(1, num+1):
if Adj[u - 1][i - 1] == 1 and color[i - 1] == 0:
dfs(i)
color[u - 1] = 2
t += 1
f[u - 1] = t
for i in range(1, num+1):
if color[i-1] == 0:
dfs(i)
for i in range(num):
print("{} {} {}".format(i+1,d[i],f[i]))
|
s735292327
|
Accepted
| 30 | 7,748 | 542 |
# coding: utf-8
def depth(i, adj, Dep, Arr, times):
if Dep[i-1]:
return times
Dep[i-1]=times
times+=1
# t=[depth(j, adj, Dep, f, times) j in adj[i-1]]
for j in adj[i-1]:
times=depth(j, adj, Dep, Arr, times)
Arr[i - 1]=times
times+=1
return times
n=int(input())
adj=[list(map(int, input().split()))[2:] for i in range(n)]
d,f=[0]*n,[0]*n
t=1
dlist=[i for i in range(n) if d[i]==0]
for i in dlist:
t = depth(i + 1, adj, d, f, t)
for i, df in enumerate(zip(d, f)):
print(i + 1, *df)
|
s518305215
|
p03476
|
u201660334
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,106 | 37,632 | 695 |
We say that a odd number N is _similar to 2017_ when both N and (N+1)/2 are prime. You are given Q queries. In the i-th query, given two odd numbers l_i and r_i, find the number of odd numbers x similar to 2017 such that l_i ≤ x ≤ r_i.
|
Q = int(input())
a = []
for i in range(Q):
a.append(list(map(int, input().split())))
number_list = []
for i in range(1,100000):
number_list.append(i+1)
number_list_set = set(number_list)
prime_number_list = []
while True:
try:
prime_number_list.append(min(number_list_set))
synthesis_number = {x for x in number_list if x % min(number_list_set) == 0}
number_list_set = number_list_set - synthesis_number
except:
break
for i in range(Q):
like_2017 = []
for j in range(a[i][0], a[i][1] + 1, 2):
if (j in prime_number_list) and (int((j + 1) / 2) in prime_number_list):
like_2017.append(j)
print(len(like_2017))
|
s021654500
|
Accepted
| 484 | 10,644 | 621 |
def prime(n):
p = []
flag = [True] * (n - 1)
for i in range(n - 1):
if flag[i] == True:
p.append(i + 2)
num = i
while num <= n - 2:
flag[num] = False
num += i + 2
return set(p)
p1 = prime(10 ** 5)
p2 = [0]
for i in range(10 ** 5):
if i % 2 == 1:
p2.append(p2[-1])
continue
else:
if i + 1 in p1 and (i + 2) // 2 in p1:
p2.append(p2[-1] + 1)
else:
p2.append(p2[-1])
q = int(input())
for i in range(q):
l, r = map(int, input().split())
print(p2[r] - p2[l -1])
|
s123042720
|
p03449
|
u772180901
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 297 |
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())
ans = 0
tmp = 0
arr = [ list(map(int,input().split())) for i in range(2)]
for i in range(n):
if i == 0:
tmp += arr[0][0]
else:
tmp += sum(arr[0][:i+1])
tmp += sum(arr[1][i:n])
print(i,n-i)
if ans < tmp:
ans = tmp
tmp = 0
print(ans)
|
s421232994
|
Accepted
| 18 | 3,064 | 280 |
n = int(input())
ans = 0
tmp = 0
arr = [ list(map(int,input().split())) for i in range(2)]
for i in range(n):
if i == 0:
tmp += arr[0][0]
else:
tmp += sum(arr[0][:i+1])
tmp += sum(arr[1][i:n])
if ans < tmp:
ans = tmp
tmp = 0
print(ans)
|
s080368177
|
p03860
|
u646336933
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 28 |
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")
|
s874423371
|
Accepted
| 17 | 2,940 | 49 |
s = input().split()
t = s[1]
print("A"+t[0]+"C")
|
s665604286
|
p02694
|
u083874202
| 2,000 | 1,048,576 |
Wrong Answer
| 22 | 9,168 | 513 |
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
I = int(input())
#N, M = map(int, input().split())
#N = input()
#list = list(N)
#list = list(map(int, input().split()))
X = 100
cnt = 0
while(X <= I):
X = math.floor(X * 1.01)
cnt += 1
print(cnt)
|
s870682470
|
Accepted
| 20 | 9,096 | 542 |
import math
I = int(input())
#N, M = map(int, input().split())
#N = input()
#list = list(N)
#list = list(map(int, input().split()))
X = 100
cnt = 0
while(X < I):
X = math.floor(X * 1.01)
cnt += 1
if I <= 100:
cnt = 0
print(cnt)
|
s906808625
|
p02600
|
u382740487
| 2,000 | 1,048,576 |
Wrong Answer
| 30 | 8,868 | 35 |
M-kun is a competitor in AtCoder, whose highest rating is X. In this site, a competitor is given a _kyu_ (class) according to his/her highest rating. For ratings from 400 through 1999, the following kyus are given: * From 400 through 599: 8-kyu * From 600 through 799: 7-kyu * From 800 through 999: 6-kyu * From 1000 through 1199: 5-kyu * From 1200 through 1399: 4-kyu * From 1400 through 1599: 3-kyu * From 1600 through 1799: 2-kyu * From 1800 through 1999: 1-kyu What kyu does M-kun have?
|
num = input()
print(type(num), num)
|
s015279254
|
Accepted
| 32 | 9,064 | 308 |
if __name__=="__main__":
num = int(input())
if num < 400:
pass
elif num<600:
print(8)
elif num<800:
print(7)
elif num<1000:
print(6)
elif num<1200:
print(5)
elif num<1400:
print(4)
elif num<1600:
print(3)
elif num<1800:
print(2)
elif num<2000:
print(1)
|
s582919070
|
p02393
|
u831971779
| 1,000 | 131,072 |
Wrong Answer
| 40 | 7,560 | 72 |
Write a program which reads three integers, and prints them in ascending order.
|
a,b,c = map(int,input().split())
list = [a,b,c]
list.sort()
print(list)
|
s756426881
|
Accepted
| 20 | 7,716 | 62 |
a,b,c = sorted([int(i) for i in input().split()])
print(a,b,c)
|
s802029223
|
p03998
|
u267300160
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,104 | 3,064 | 500 |
Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.
|
A_cards = list(input())
B_cards = list(input())
C_cards = list(input())
who = "a"
while(True):
if(who == "a"):
if(len(A_cards)==0):
print("A")
exit()
else:
who = A_cards[0]
del A_cards[0]
elif(who == "b"):
if(len(B_cards)==0):
print("B")
exit()
else:
who = B_cards[0]
del B_cards[0]
else:
if(len(C_cards)==0):
print("C")
exit()
|
s488215882
|
Accepted
| 17 | 3,064 | 583 |
A_cards = list(input())
B_cards = list(input())
C_cards = list(input())
who = "a"
while(True):
if(who == "a"):
if(len(A_cards)==0):
print("A")
exit()
else:
who = A_cards[0]
del A_cards[0]
if(who == "b"):
if(len(B_cards)==0):
print("B")
exit()
else:
who = B_cards[0]
del B_cards[0]
if(who == "c"):
if(len(C_cards)==0):
print("C")
exit()
else:
who = C_cards[0]
del C_cards[0]
|
s316516747
|
p03547
|
u330176731
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 98 |
In programming, hexadecimal notation is often used. In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively. In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`. When X and Y are seen as hexadecimal numbers, which is larger?
|
x,y = input().split()
if min(x,y) == x: print('>')
elif min(x,y) == y: print('<')
else: print('=')
|
s941369726
|
Accepted
| 17 | 3,060 | 105 |
x,y = input().split()
if x == y: print('=')
elif min(x,y) == x: print('<')
elif min(x,y) == y: print('>')
|
s548467243
|
p02389
|
u069789944
| 1,000 | 131,072 |
Wrong Answer
| 40 | 6,720 | 72 |
Write a program which calculates the area and perimeter of a given rectangle.
|
a_b = input().split(' ')
a = int(a_b[0])
b = int(a_b[1])
print(a * b)
|
s003412291
|
Accepted
| 30 | 6,720 | 64 |
s,n=map(int,input().split())
print("{} {}".format(s*n, (s+n)*2))
|
s950091922
|
p03434
|
u143903328
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 215 |
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())
a = list(map(int,input().split()))
alice = 0
bob = 0
a.sort()
a.reverse()
for i in range (0,n):
if i % 2:
bob += a[i]
else:
alice += a[i]
print(alice - bob)
print(a)
|
s983819240
|
Accepted
| 17 | 3,060 | 203 |
n = int(input())
a = list(map(int,input().split()))
alice = 0
bob = 0
a.sort()
a.reverse()
for i in range (0,n):
if i % 2:
bob += a[i]
else:
alice += a[i]
print(alice - bob)
|
s570689533
|
p03090
|
u844902298
| 2,000 | 1,048,576 |
Wrong Answer
| 25 | 3,700 | 251 |
You are given an integer N. Build an undirected graph with N vertices with indices 1 to N that satisfies the following two conditions: * The graph is simple and connected. * There exists an integer S such that, for every vertex, the sum of the indices of the vertices adjacent to that vertex is S. It can be proved that at least one such graph exists under the constraints of this problem.
|
n = int(input())
a = [[0]*(n+1) for i in range(n+1)]
np = (n//2)*2
for i in range(0,np//2):
a[i+1][np-i] = -1
for i in range(1,n+1):
for j in range(i+1,n+1):
if a[i][j] == -1:
continue
else:
print(i,j)
|
s563037062
|
Accepted
| 24 | 3,700 | 281 |
n = int(input())
a = [[0]*(n+1) for i in range(n+1)]
np = (n//2)*2
for i in range(0,np//2):
a[i+1][np-i] = -1
print(int(n*(n-1)/2-(n//2)))
for i in range(1,n+1):
for j in range(i+1,n+1):
if a[i][j] == -1:
continue
else:
print(i,j)
|
s588294133
|
p02409
|
u058433718
| 1,000 | 131,072 |
Wrong Answer
| 40 | 7,712 | 925 |
You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room. For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth floor at building b. If v is negative, it means that −v persons left. Assume that initially no person lives in the building.
|
import sys
def make_room_data():
room_data = []
for i in range(4):
one_house = []
for j in range(3):
one_floor = []
for k in range(10):
one_floor.append(0)
one_house.append(one_floor)
room_data.append(one_house)
return room_data
def main():
room_data = make_room_data()
n = int(sys.stdin.readline().strip())
for _ in range(n):
data = sys.stdin.readline().strip().split(' ')
b, f, r, v = [int(i) for i in data]
room_data[b-1][f-1][r-1] += v
for i in range(4):
for j in range(3):
for k in range(10):
print(' %d' % (room_data[i][j][k]), end='')
print('')
if i != 3:
print('*' * 20)
if __name__ == '__main__':
main()
|
s555194189
|
Accepted
| 30 | 7,748 | 925 |
import sys
def make_room_data():
room_data = []
for i in range(4):
one_house = []
for j in range(3):
one_floor = []
for k in range(10):
one_floor.append(0)
one_house.append(one_floor)
room_data.append(one_house)
return room_data
def main():
room_data = make_room_data()
n = int(sys.stdin.readline().strip())
for _ in range(n):
data = sys.stdin.readline().strip().split(' ')
b, f, r, v = [int(i) for i in data]
room_data[b-1][f-1][r-1] += v
for i in range(4):
for j in range(3):
for k in range(10):
print(' %d' % (room_data[i][j][k]), end='')
print('')
if i != 3:
print('#' * 20)
if __name__ == '__main__':
main()
|
s019364953
|
p03501
|
u504715104
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 65 |
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.
|
moji=[]
moji=input().split()
max_1=min(moji)
print(max_1)
|
s919024606
|
Accepted
| 17 | 2,940 | 236 |
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Wed Oct 18 17:04:29 2017
@author: goto
"""
int1,int2,int3=map(int, input().split())
t=int1*int2
if t<int3:
minmoji=t
else:
minmoji=int3
print(minmoji)
|
s449018688
|
p03434
|
u556657484
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 91 |
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())
A = sorted(map(int, input().split()))
print(sum(A[0::2]) - sum(A[1::2]))
|
s178679544
|
Accepted
| 17 | 2,940 | 105 |
N = int(input())
A = sorted(map(int, input().split()), reverse=True)
print(sum(A[0::2]) - sum(A[1::2]))
|
s113354916
|
p03455
|
u178317679
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 93 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
def calc(a, b):
if (a * b) % 2 == 0:
print('Even')
else:
print('Odd')
|
s704513422
|
Accepted
| 17 | 2,940 | 90 |
a, b = map(int, input().split())
if a*b % 2 == 0:
print("Even")
else:
print("Odd")
|
s302359765
|
p02613
|
u690419532
| 2,000 | 1,048,576 |
Wrong Answer
| 150 | 16,280 | 260 |
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())
l = []
for i in range(N):
l.append(input())
AC = l.count('AC')
WA = l.count('WA')
TLE = l.count('TLE')
RE = l.count('RE')
print("AC × {}".format(AC))
print("WA × {}".format(WA))
print("TLE × {}".format(TLE))
print("RE × {}".format(RE))
|
s134733814
|
Accepted
| 148 | 9,116 | 340 |
N = int(input())
AC = 0
WA = 0
TLE = 0
RE = 0
for i in range(N):
test = input()
if test == 'AC':
AC += 1
elif test == 'WA':
WA += 1
elif test == 'TLE':
TLE += 1
else:
RE += 1
print('AC x {}'.format(AC))
print('WA x {}'.format(WA))
print('TLE x {}'.format(TLE))
print('RE x {}'.format(RE))
|
s450518910
|
p02842
|
u285443936
| 2,000 | 1,048,576 |
Wrong Answer
| 34 | 3,060 | 142 |
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
import math
N = int(input())
for i in range(50001):
if math.floor(i*1.08) == N:
print(math.floor(i*1.08))
exit()
else:
print(":(")
|
s503721283
|
Accepted
| 33 | 3,060 | 127 |
import math
N = int(input())
for i in range(1,50001):
if math.floor(i*1.08) == N:
print(i)
exit()
else:
print(":(")
|
s384623115
|
p03448
|
u256106029
| 2,000 | 262,144 |
Wrong Answer
| 52 | 3,060 | 266 |
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())
count = 0
for i in range(a):
for j in range(b):
for k in range(c):
total = 500 * i + 100 * j + 50 * k
if total == x:
count += 1
print(count)
|
s099063138
|
Accepted
| 55 | 3,064 | 278 |
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):
total = 500 * i + 100 * j + 50 * k
if total == x:
count += 1
print(count)
|
s076000868
|
p03386
|
u849229491
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 185 |
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
a,b,k = map(int,input().split())
ans = []
for i in range(k):
t1 = a+i
t2 = b-i
ans.append(t1)
ans.append(t2)
A = list(set(ans))
for i in range(len(A)):
print(A[i])
|
s431352586
|
Accepted
| 17 | 3,064 | 265 |
import sys
a,b,k = map(int,input().split())
ans = []
if k > b-a:
for i in range(a,b+1):
print(i)
sys.exit()
for i in range(k):
t1 = a+i
t2 = b-i
ans.append(t1)
ans.append(t2)
A = list(set(ans))
A.sort()
for i in A:
print(i)
|
s983613509
|
p03643
|
u366959492
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 29 |
This contest, _AtCoder Beginner Contest_ , is abbreviated as _ABC_. When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example, ABC680 is the 680th round of ABC. What is the abbreviation for the N-th round of ABC? Write a program to output the answer.
|
n=str(input())
print("ABC",n)
|
s034829564
|
Accepted
| 17 | 2,940 | 25 |
n=input()
print("ABC"+n)
|
s953420471
|
p02422
|
u130834228
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,700 | 389 |
Write a program which performs a sequence of commands to a given string $str$. The command is one of: * print a b: print from the a-th character to the b-th character of $str$ * reverse a b: reverse from the a-th character to the b-th character of $str$ * replace a b p: replace from the a-th character to the b-th character of $str$ with p Note that the indices of $str$ start with 0.
|
str1 = list(input())
q = int(input())
for i in range(q):
order = [str(x) for x in input().split()]
if order[0] == 'replace':
str1[int(order[1]):int(order[2])+1] = list(order[3])
print(str1)
elif order[0] == 'reverse':
str2 = str1[int(order[1]):int(order[2])+1]
str1[int(order[1]):int(order[2])+1] = reversed(str2)
else: #print
print(str1[int(order[1]):int(order[2])+1])
|
s704152050
|
Accepted
| 20 | 7,792 | 399 |
str1 = list(input())
q = int(input())
for i in range(q):
order = [str(x) for x in input().split()]
if order[0] == 'replace':
str1[int(order[1]):int(order[2])+1] = list(order[3])
#print(str1)
elif order[0] == 'reverse':
str2 = str1[int(order[1]):int(order[2])+1]
str1[int(order[1]):int(order[2])+1] = reversed(str2)
else: #print
print(*str1[int(order[1]):int(order[2])+1], sep='')
|
s862838134
|
p02608
|
u503111914
| 2,000 | 1,048,576 |
Time Limit Exceeded
| 2,206 | 8,960 | 217 |
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
N =int(input())
for l in range(1,N+1):
ans = 0
for i in range(1,101):
for j in range(1,101):
for k in range(1,101):
f = i*i+j*j+k*k+i*j+j*k+k*i
if f == l:
ans += 1
print(ans)
|
s549271072
|
Accepted
| 503 | 9,308 | 231 |
N = int(input())
ans = [0 for _ in range(10050)]
for i in range(1,105):
for j in range(1,105):
for k in range(1,105):
v = i*i+j*j+k*k+i*j+j*k+k*i;
if v<10050:
ans[v]+=1
for l in range(N):
print(ans[l+1])
|
s216612864
|
p02842
|
u721425712
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 102 |
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
n = int(input())
import math
x = math.ceil(n/1.08)
if x == n*1.08:
print(x)
else:
print(':(')
|
s115281695
|
Accepted
| 17 | 3,060 | 108 |
n = int(input())
import math
x = math.ceil(n/1.08)
if n == math.floor(x*1.08):
print(x)
else:
print(':(')
|
s796684936
|
p03385
|
u121732701
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 93 |
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
S = list(input())
A = ["a","b","c"]
if set(S) == A:
print("Yes")
else:
print("No")
|
s943309008
|
Accepted
| 17 | 2,940 | 90 |
S = input()
if "a" in S and "b" in S and "c" in S:
print("Yes")
else:
print("No")
|
s158370212
|
p03693
|
u669382434
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 96 |
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
r,g,b=[int(i) for i in input().split()]
if 100*r+10*g+b%4==0:
print("YES")
else:
print("NO")
|
s782446478
|
Accepted
| 17 | 2,940 | 98 |
r,g,b=[int(i) for i in input().split()]
if (100*r+10*g+b)%4==0:
print("YES")
else:
print("NO")
|
s780127279
|
p03476
|
u001769145
| 2,000 | 262,144 |
Wrong Answer
| 229 | 27,060 | 425 |
We say that a odd number N is _similar to 2017_ when both N and (N+1)/2 are prime. You are given Q queries. In the i-th query, given two odd numbers l_i and r_i, find the number of odd numbers x similar to 2017 such that l_i ≤ x ≤ r_i.
|
# 69
import itertools
q = int(input())
lr_l = [[int(x) for x in input().split()] for y in range(q)]
n_max = max(itertools.chain.from_iterable(lr_l))
isp_l = [True] * (n_max+1)
isp_l[0] = False
isp_l[1] = False
for i in range(2, int(n_max ** 0.5 + 2)):
if not isp_l[i]:
continue
for j in range(i * 2, n_max + 1, i):
isp_l[j] = False
p_l = [int(x) for x in range(n_max+1) if isp_l[x]]
print(p_l)
|
s792042672
|
Accepted
| 1,205 | 27,268 | 651 |
# 69
import itertools
import bisect
q = int(input())
lr_l = [[int(x) for x in input().split()] for y in range(q)]
n_max = max(itertools.chain.from_iterable(lr_l))
isp_l = [True] * (n_max+1)
isp_l[0] = False
isp_l[1] = False
for i in range(2, int(n_max ** 0.5 + 2)):
if not isp_l[i]:
continue
for j in range(i * 2, n_max + 1, i):
isp_l[j] = False
p_l = [int(x) for x in range(n_max+1) if isp_l[x]]
l_l = []
for i in range(len(p_l)):
if (p_l[i]+1)//2 in p_l:
l_l.append(p_l[i])
for i in range(q):
_l,_r = lr_l[i]
_s = bisect.bisect_left(l_l, _l)
_e = bisect.bisect_right(l_l, _r)
print(_e-_s)
|
s480457504
|
p04011
|
u370429695
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 136 |
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.
|
total = int(input())
limit = int(input())
first = int(input())
second = int(input())
print(limit * first + (total - limit + 1) * second)
|
s991408434
|
Accepted
| 17 | 2,940 | 181 |
total = int(input())
limit = int(input())
first = int(input())
second = int(input())
if total > limit:
print(limit * first + (total - limit) * second)
else:
print(total * first)
|
s835040128
|
p03623
|
u869265610
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 71 |
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
a,b,c=map(int,input().split())
print("A" if abs(a-b)>abs(a-c) else "B")
|
s136055409
|
Accepted
| 18 | 2,940 | 72 |
a,b,c=map(int,input().split())
print("A" if abs(a-b)<abs(a-c) else "B")
|
s202941085
|
p03853
|
u011062360
| 2,000 | 262,144 |
Wrong Answer
| 24 | 3,828 | 155 |
There is an image with a height of H pixels and a width of W pixels. Each of the pixels is represented by either `.` or `*`. The character representing the pixel at the i-th row from the top and the j-th column from the left, is denoted by C_{i,j}. Extend this image vertically so that its height is doubled. That is, print a image with a height of 2H pixels and a width of W pixels where the pixel at the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division is rounded down).
|
h, w = map(int, input().split())
list_score = [ list(input()) for _ in range(h) ]
for l in list_score:
print(*l)
for l in list_score:
print(*l)
|
s711000156
|
Accepted
| 18 | 3,060 | 143 |
h, w = map(int, input().split())
list_score = [ input() for _ in range(h) ]
for l in list_score:
print("".join(l))
print("".join(l))
|
s214795889
|
p04043
|
u886902015
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,068 | 116 |
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.
|
li=a,b,c=list(map(int,input().split()))
if li.count(7)==2 and li.count(5)==1:
print("YES")
else:
print("NO")
|
s414424552
|
Accepted
| 25 | 8,980 | 117 |
li=a,b,c=list(map(int,input().split()))
if li.count(7)==1 and li.count(5)==2:
print("YES")
else:
print("NO")
|
s970006215
|
p03911
|
u952165951
| 2,000 | 262,144 |
Wrong Answer
| 491 | 61,116 | 253 |
On a planet far, far away, M languages are spoken. They are conveniently numbered 1 through M. For _CODE FESTIVAL 20XX_ held on this planet, N participants gathered from all over the planet. The i-th (1≦i≦N) participant can speak K_i languages numbered L_{i,1}, L_{i,2}, ..., L_{i,{}K_i}. Two participants A and B can _communicate_ with each other if and only if one of the following conditions is satisfied: * There exists a language that both A and B can speak. * There exists a participant X that both A and B can communicate with. Determine whether all N participants can communicate with all other participants.
|
from numpy import*
from scipy.sparse import*
(n,m),*p=[map(int,o.split())for o in open(0)]
e,*f=array([(i,j+n-1,1)for i,q in enumerate(p)for j in list(q)[1:]]).T[:3]
print("YNEOS"[1in csgraph.connected_components(csr_matrix(((e,f)),[n+m]*2))[1][:n]::2])
|
s927475137
|
Accepted
| 249 | 29,476 | 1,198 |
#!/usr/bin/env python3
class UnionFind():
def __init__(self, n):
self.n = n
self.parents = [-1] * n
def find(self, x):
if self.parents[x] < 0:
return x
else:
self.parents[x] = self.find(self.parents[x])
return self.parents[x]
def union(self, x, y):
x = self.find(x)
y = self.find(y)
if x == y:
return
if self.parents[x] > self.parents[y]:
x, y = y, x
self.parents[x] += self.parents[y]
self.parents[y] = x
def size(self, x):
return -self.parents[self.find(x)]
def same(self, x, y):
return self.find(x) == self.find(y)
def members(self, x):
root = self.find(x)
return [i for i in range(self.n) if self.find(i) == root]
def roots(self):
return [i for i, x in enumerate(self.parents) if x < 0]
def group_count(self):
return len(self.roots())
(n, m), *LL = [[*map(int,o.split())] for o in open(0)]
UF = UnionFind(n + m)
for e, (k, *L) in enumerate(LL):
for l in L:
UF.union(e, l + n - 1)
print("YES" if len([*filter(lambda i: i < n, UF.roots())]) == 1 else "NO")
|
s039480809
|
p03192
|
u255499778
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 48 |
You are given an integer N that has exactly four digits in base ten. How many times does `2` occur in the base-ten representation of N?
|
n = list(map(int, input()))
print(n.count("2"))
|
s346737033
|
Accepted
| 18 | 2,940 | 37 |
n = list(input())
print(n.count("2"))
|
s659222114
|
p02409
|
u669360983
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,716 | 338 |
You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room. For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth floor at building b. If v is negative, it means that −v persons left. Assume that initially no person lives in the building.
|
import sys
n=int(input())
data={}
for i in range(1,5):
for j in range(1,4):
for k in range(1,11):
data[(i,j,k)]=0
for i in range(1,n+1):
b,r,f,v=map(int,input().split())
data[(b,r,f)]+=v
for i in range(1,5):
for j in range(1,4):
for k in range(1,11):
sys.stdout.write(str(data[(i,j,k)])+' ')
print('')
print('#'*20)
|
s737060818
|
Accepted
| 30 | 6,724 | 348 |
import sys
n=int(input())
data={}
for i in range(1,5):
for j in range(1,4):
for k in range(1,11):
data[(i,j,k)]=0
for i in range(1,n+1):
b,r,f,v=map(int,input().split())
data[(b,r,f)]+=v
for i in range(1,5):
for j in range(1,4):
for k in range(1,11):
sys.stdout.write(' '+str(data[(i,j,k)]))
print('')
if i<4:
print('#'*20)
|
s220748387
|
p02854
|
u469953228
| 2,000 | 1,048,576 |
Wrong Answer
| 101 | 26,764 | 177 |
Takahashi, who works at DISCO, is standing before an iron bar. The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters. Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length. However, this may not be possible as is, so he will do the following operations some number of times **before** he does the cut: * Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan). * Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen. Find the minimum amount of money needed before cutting the bar into two parts with the same length.
|
n=int(input())
A=list(map(int,input().split()))
L=sum(A)
l=0
for i in range(n):
l+=A[i]
if l >= L/2:
r=l-A[i]
break
print(l,r)
print(min(abs((L-r)-r),abs((L-l)-l)))
|
s714174082
|
Accepted
| 102 | 26,764 | 166 |
n=int(input())
A=list(map(int,input().split()))
L=sum(A)
l=0
for i in range(n):
l+=A[i]
if l >= L/2:
r=l-A[i]
break
print(min(abs((L-r)-r),abs((L-l)-l)))
|
s949451000
|
p02842
|
u353548710
| 2,000 | 1,048,576 |
Wrong Answer
| 26 | 9,168 | 208 |
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
import math
n = int(input())
x = n/1.08
x1, x2 = math.floor( x ), math.ceil( x )
print(x1, x2)
if n == math.floor( x1 * 1.08 ) :
print(x1)
elif n == math.floor( x2 * 1.08 ):
print(x2)
else:
print(':(')
|
s643922755
|
Accepted
| 26 | 9,036 | 194 |
import math
n = int(input())
x = n/1.08
x1, x2 = math.floor( x ), math.ceil( x )
if n == math.floor( x1 * 1.08 ) :
print(x1)
elif n == math.floor( x2 * 1.08 ):
print(x2)
else:
print(':(')
|
s879327354
|
p03416
|
u587213169
| 2,000 | 262,144 |
Wrong Answer
| 76 | 2,940 | 157 |
Find the number of _palindromic numbers_ among the integers between A and B (inclusive). Here, a palindromic number is a positive integer whose string representation in base 10 (without leading zeros) reads the same forward and backward.
|
a, b =map(int, input().split())
count=0
for i in range(a,b):
m = list(str(i))
if m[0]==m[3]:
if m[1]==m[2]:
count+=1
print(count)
|
s434924208
|
Accepted
| 77 | 2,940 | 159 |
a, b =map(int, input().split())
count=0
for i in range(a,b+1):
m = list(str(i))
if m[0]==m[4]:
if m[1]==m[3]:
count+=1
print(count)
|
s563417314
|
p02261
|
u935329231
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,832 | 1,416 |
Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
# -*- coding: utf-8 -*-
def bubble_sort(cards, num):
for left in range(num-1):
for right in range(left+1, num):
if cards[right][1] < cards[left][1]:
cards[left], cards[right] = cards[right], cards[left]
print(list_to_str(cards))
def selection_sort(cards, num):
for head in range(num):
min_i = head
for target in range(head+1, num):
if cards[target][1] < cards[min_i][1]:
min_i = target
cards[head], cards[min_i] = cards[min_i], cards[head]
print(list_to_str(cards))
def is_stable(before, after, num):
for i in range(num):
for j in range(i+1, num):
for a in range(num):
for b in range(a+1, num):
if before[i][1] == before[j][1]\
and before[i] == after[b]\
and before[j] == after[a]:
print('Not stable')
print('Stable')
def list_to_str(l, delimiter=' '):
# The default delimiter is one space.
return delimiter.join([str(v) for v in l])
if __name__ == '__main__':
num = int(input())
cards = input().split()
cards_for_bubble = cards.copy()
cards_for_selection = cards.copy()
bubble_sort(cards_for_bubble, num)
is_stable(cards, cards_for_bubble, num)
selection_sort(cards_for_selection, num)
is_stable(cards, cards_for_selection, num)
|
s938324727
|
Accepted
| 120 | 7,856 | 1,462 |
# -*- coding: utf-8 -*-
def bubble_sort(cards, num):
for tail in range(num-1, 0, -1):
for right in range(0, tail):
left = right + 1
if cards[left][1] < cards[right][1]:
cards[right], cards[left] = cards[left], cards[right]
print(list_to_str(cards))
def selection_sort(cards, num):
for head in range(num):
min_i = head
for target in range(head+1, num):
if cards[target][1] < cards[min_i][1]:
min_i = target
cards[head], cards[min_i] = cards[min_i], cards[head]
print(list_to_str(cards))
def is_stable(before, after, num):
for i in range(num):
for j in range(i+1, num):
for a in range(num):
for b in range(a+1, num):
if before[i][1] == before[j][1]\
and before[i] == after[b]\
and before[j] == after[a]:
return 'Not stable'
return 'Stable'
def list_to_str(l, delimiter=' '):
# The default delimiter is one space.
return delimiter.join([str(v) for v in l])
if __name__ == '__main__':
num = int(input())
cards = input().split()
cards_for_bubble = cards.copy()
cards_for_selection = cards.copy()
bubble_sort(cards_for_bubble, num)
print(is_stable(cards, cards_for_bubble, num))
selection_sort(cards_for_selection, num)
print(is_stable(cards, cards_for_selection, num))
|
s854833401
|
p03997
|
u295961023
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 135 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
def main():
a, b, h = [int(input()) for _ in range(3)]
s = (a + b) * h / 2
print(s)
if __name__ == "__main__":
main()
|
s225332016
|
Accepted
| 17 | 2,940 | 140 |
def main():
a, b, h = [int(input()) for _ in range(3)]
s = (a + b) * h / 2
print(int(s))
if __name__ == "__main__":
main()
|
s530008718
|
p03385
|
u243159381
| 2,000 | 262,144 |
Wrong Answer
| 28 | 9,008 | 69 |
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
s=input()
if sorted(s)=='abc':
print('Yes')
else:
print('No')
|
s350065729
|
Accepted
| 26 | 9,064 | 78 |
s=input()
if ''.join(sorted(s))=='abc':
print('Yes')
else:
print('No')
|
s092572033
|
p03457
|
u381068238
| 2,000 | 262,144 |
Wrong Answer
| 337 | 3,064 | 669 |
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.
|
_ = int(input())
bef_time = 0
bef_x = 0
bef_y = 0
ans = True
for i in range(_):
next = input().split()
next_time = int(next[0])
next_x = int(next[1])
next_y = int(next[2])
range = abs((next_x - bef_x) + (next_y - bef_y))
if (next_time - bef_time) >= range:
if (next_time - bef_time) % 2 == 0:
if range % 2 != 0:
ans = False
break
else:
if range % 2 == 0:
ans = False
break
else:
ans = False
break
bef_time = next_time
bef_y = next_y
bef_x = next_x
if ans == True:
print('YES')
else:
print('NO')
|
s777548413
|
Accepted
| 352 | 3,064 | 645 |
_ = int(input())
bef_time = 0
bef_x = 0
bef_y = 0
ans = True
for i in range(_):
next = input().split()
next_time = int(next[0])
next_x = int(next[1])
next_y = int(next[2])
time = next_time - bef_time
range = abs(next_x - bef_x) + abs(next_y - bef_y)
if (next_time - bef_time) < range:
ans = False
break
if time % 2 == 0:
if range % 2 != 0:
ans = False
break
else:
if range % 2 == 0 or range == 0:
ans = False
break
bef_time = next_time
bef_y = next_y
bef_x = next_x
if ans == True:
print('Yes')
else:
print('No')
|
s192414133
|
p03227
|
u310466455
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 83 |
You are given a string S of length 2 or 3 consisting of lowercase English letters. If the length of the string is 2, print it as is; if the length is 3, print the string after reversing it.
|
a = input()
if len(a) == 2:
print(a)
else:
new_str = ''.join(list(reversed(a)))
|
s277583418
|
Accepted
| 17 | 2,940 | 100 |
a = input()
if len(a) == 2:
print(a)
else:
new_str = ''.join(list(reversed(a)))
print(new_str)
|
s390509534
|
p03719
|
u732491892
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 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(' '))
print("YES") if c >= a and c <= b else print("NO")
|
s977337334
|
Accepted
| 17 | 2,940 | 90 |
a, b, c = map(int, input().split(' '))
print("Yes") if c >= a and c <= b else print("No")
|
s328310560
|
p02393
|
u302561071
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,328 | 94 |
Write a program which reads three integers, and prints them in ascending order.
|
data = input().split()
data.sort
print(str(data[0]) + " " + str(data[1]) + " " + str(data[2]))
|
s514526649
|
Accepted
| 20 | 7,452 | 96 |
data = input().split()
data.sort()
print(str(data[0]) + " " + str(data[1]) + " " + str(data[2]))
|
s512105099
|
p03795
|
u821251381
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 36 |
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
N = int(input())
print(N*800-N//15)
|
s611977038
|
Accepted
| 17 | 2,940 | 41 |
N = int(input())
print(N*800-N//15 *200)
|
s588873035
|
p02694
|
u740267532
| 2,000 | 1,048,576 |
Wrong Answer
| 25 | 9,180 | 127 |
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, b = 100, 1
while 1:
a += a * 0.01
if a >= X:
print(b)
break
b += 1
print(a)
|
s775754891
|
Accepted
| 22 | 9,168 | 119 |
X = int(input())
a, b = 100, 1
while 1:
a += int(a * 0.01)
if a >= X:
print(b)
break
b += 1
|
s827406079
|
p03494
|
u288786530
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 199 |
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.
|
l = list()
n = int(input())
original_list = list(range(1,n))
for x in original_list:
ans = 0
while x % 2 == 0:
x = x / 2
ans = ans + 1
l.append(ans)
print(max(l))
|
s554088204
|
Accepted
| 19 | 2,940 | 147 |
n = int(input())
a = list(map(int, input().split()))
l = []
for i in a:
s = 0
while i % 2 == 0:
i = i/2
s += 1
l.append(s)
print(min(l))
|
s349849200
|
p03694
|
u744898490
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 89 |
It is only six months until Christmas, and AtCoDeer the reindeer is now planning his travel to deliver gifts. There are N houses along _TopCoDeer street_. The i-th house is located at coordinate a_i. He has decided to deliver gifts to all these houses. Find the minimum distance to be traveled when AtCoDeer can start and end his travel at any positions.
|
a= input().split(' ')
a = [int(x) for x in a]
mi = min(a)
ma = max(a)
print(int(ma- mi))
|
s604729391
|
Accepted
| 17 | 2,940 | 101 |
w = input()
a= input().split(' ')
a = [int(x) for x in a]
mi = min(a)
ma = max(a)
print(int(ma- mi))
|
s784227664
|
p03658
|
u518064858
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 118 |
Snuke has N sticks. The length of the i-th stick is l_i. Snuke is making a snake toy by joining K of the sticks together. The length of the toy is represented by the sum of the individual sticks that compose it. Find the maximum possible length of the toy.
|
n,k=map(int,input().split())
l=sorted(list(map(int,input().split())),reverse=True)
s=0
for i in range(k):
s+=l[i]
|
s798487151
|
Accepted
| 18 | 2,940 | 129 |
n,k=map(int,input().split())
l=sorted(list(map(int,input().split())),reverse=True)
s=0
for i in range(k):
s+=l[i]
print(s)
|
s974045464
|
p03449
|
u918601425
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 184 |
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())
lsa=[int(s) for s in input().split()]
lsb=[int(s) for s in input().split()]
ls=[lsa[0]+max(lsb)]
for i in range(1,N):
ls.append(ls[-1]+lsa[i]-lsb[i-1])
print(max(ls))
|
s839427238
|
Accepted
| 17 | 2,940 | 184 |
N=int(input())
lsa=[int(s) for s in input().split()]
lsb=[int(s) for s in input().split()]
ls=[lsa[0]+sum(lsb)]
for i in range(1,N):
ls.append(ls[-1]+lsa[i]-lsb[i-1])
print(max(ls))
|
s979605742
|
p03644
|
u212328220
| 2,000 | 262,144 |
Wrong Answer
| 26 | 9,028 | 82 |
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
n = int(input())
x = 1
cnt = 1
while x < n:
x = 2 ** cnt
cnt += 1
print(x)
|
s754855450
|
Accepted
| 28 | 9,028 | 83 |
n = int(input())
x = 1
cnt = 1
while x <= n:
x = 2 ** cnt
cnt += 1
print(x//2)
|
s291570959
|
p03777
|
u045408189
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 100 |
Two deer, AtCoDeer and TopCoDeer, are playing a game called _Honest or Dishonest_. In this game, an honest player always tells the truth, and an dishonest player always tell lies. You are given two characters a and b as the input. Each of them is either `H` or `D`, and carries the following information: If a=`H`, AtCoDeer is honest; if a=`D`, AtCoDeer is dishonest. If b=`H`, AtCoDeer is saying that TopCoDeer is honest; if b=`D`, AtCoDeer is saying that TopCoDeer is dishonest. Given this information, determine whether TopCoDeer is honest.
|
a,b=input().split()
if a=='H':
print('b')
else:
if b=='H':
print('D')
else:
print('H')
|
s973665733
|
Accepted
| 17 | 2,940 | 99 |
a,b=input().split()
if a=='H':
print(b)
else:
if b=='H':
print('D')
else:
print('H')
|
s717366316
|
p03494
|
u879870653
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 399 |
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
N = int(input())
A = list(map(int,input().split()))
B = []
answer = 0
for i in range(N) :
if A[i] % 2 == 0 :
B.append(A[i])
if len(B) == 0 :
print(1)
else :
for i in range(N) :
ans = 0
for j in range(1,10) :
if A[i] % 2**j == 0 :
ans += 1
else :
answer = max(answer,ans)
break
print(answer)
|
s671382675
|
Accepted
| 18 | 3,060 | 261 |
N = int(input())
A = list(map(int,input().split()))
P = [2**i for i in range(30)]
P = sorted(P,reverse = True)
ans = 100
for i in range(N) :
for j in range(len(P)) :
if A[i] % P[j] == 0 :
ans = min(ans,29-j)
break
print(ans)
|
s721074057
|
p02612
|
u598684283
| 2,000 | 1,048,576 |
Wrong Answer
| 30 | 9,040 | 74 |
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.
|
input1 = int(input())
while input1 > 999:
input1 -= 1000
print(input1)
|
s622389053
|
Accepted
| 30 | 9,044 | 82 |
input1 = int(input())
while input1 > 1000:
input1 -= 1000
print(1000 - input1)
|
s015956762
|
p03434
|
u698919163
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 215 |
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())
a = list(map(int,input().split()))
a.sort(reverse=True)
print(a)
Alice = 0
Bob = 0
for i in range(N):
if i%2 == 0:
Alice += a[i]
else:
Bob += a[i]
print(Alice-Bob)
|
s048811078
|
Accepted
| 18 | 3,060 | 206 |
N = int(input())
a = list(map(int,input().split()))
a.sort(reverse=True)
Alice = 0
Bob = 0
for i in range(N):
if i%2 == 0:
Alice += a[i]
else:
Bob += a[i]
print(Alice-Bob)
|
s979224619
|
p02694
|
u021849254
| 2,000 | 1,048,576 |
Wrong Answer
| 30 | 9,008 | 67 |
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?
|
a=int(input())
c=100
x=0
while c<=a:
c=c+c//100
x=x+1
print(x)
|
s204679178
|
Accepted
| 31 | 8,996 | 66 |
a=int(input())
c=100
x=0
while c<a:
c=c+c//100
x=x+1
print(x)
|
s771717192
|
p03401
|
u988402778
| 2,000 | 262,144 |
Wrong Answer
| 243 | 13,920 | 618 |
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.
|
N = int(input())
a = list(int(i) for i in input().split())
sum = 0
for k in range(N):
if k == 0:
sum += abs(a[k])
elif k == (N-1):
sum += abs(a[k])
print(k)
print(sum)
else:
sum += abs(a[k] - a[k-1])
print(sum)
for j in range(N):
if j == 0:
if (a[j+1] - a[j]) * (a[j] -0) > 0:
print(sum)
else:
print(sum - 2 * abs(a[j]))
elif j == N-1:
if (a[j] - a[j-1]) * (0 - a[j]) > 0:
print(sum)
else:
print(sum - 2 * abs(a[j]))
else:
if (a[j] - a[j-1] ) * (a[j+1] - a[j]) > 0:
print(sum)
else:
print(sum - 2 * abs(a[j]))
|
s867939389
|
Accepted
| 279 | 13,928 | 670 |
N = int(input())
a = list(int(i) for i in input().split())
sum = 0
for k in range(N):
if k == 0:
sum += abs(a[k])
elif k == (N-1):
sum += abs(a[k] - a[k-1]) + abs(a[k])
else:
sum += abs(a[k] - a[k-1])
for j in range(N):
if j == 0:
if (a[j+1] - a[j]) * (a[j] -0) >= 0:
print(sum)
else:
print(sum - 2 * min(abs(a[j]),abs(a[j]-a[j+1])))
elif j == N-1:
if (a[j] - a[j-1]) * (0 - a[j]) >= 0:
print(sum)
else:
print(sum - 2 * min(abs(a[j]-a[j-1]),abs(a[j])))
else:
if (a[j] - a[j-1] ) * (a[j+1] - a[j]) >= 0:
print(sum)
else:
print(sum - 2 * min(abs(a[j]-a[j-1]),abs(a[j+1]-a[j])))
|
s184093003
|
p02399
|
u606989659
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,504 | 89 |
Write a program which reads two integers a and b, and calculates the following values: * a ÷ b: d (in integer) * remainder of a ÷ b: r (in integer) * a ÷ b: f (in real number)
|
a,b = map(int,input().split())
d = a / b
r = a % b
f = float(a) / float(b)
print(d, r, f)
|
s328653823
|
Accepted
| 20 | 7,588 | 100 |
a,b = map(int,input().split())
d = a // b
r = a % b
f = a / b
print('{0} {1} {2:.5f}'.format(d,r,f))
|
s078224962
|
p00077
|
u075836834
| 1,000 | 131,072 |
Wrong Answer
| 50 | 7,548 | 276 |
文字列が連続した場合、ある規則で文字を置き換え文字列を短くすることができます。たとえば、AAAA という文字列の場合、@4A と表現すれば 1 文字分圧縮されます。この規則で圧縮された文字列を入力してもとの文字列に復元するプログラムを作成してください。ただし、復元した文字列に＠文字は出現しないものとします。 また、原文の文字列は英大文字、英小文字、数字、記号であり 100 文字以内、連続する文字は 9 文字以内です。
|
while True:
try:
line=list(map(str,input()))
string=""
for i in range(len(line)):
if (line[i]!='@' and line[i-1]!='@'):
string+=line[i]
elif line[i]=='@':
string+=line[i+2]*int(line[i+1])
else:
pass
print(string)
except EOFError:
break
|
s980554386
|
Accepted
| 30 | 7,592 | 295 |
while True:
try:
line=list(map(str,input()))
string=""
for i in range(len(line)):
if (line[i]!='@' and line[i-1]!='@' and line[i-2]!='@'):
string+=line[i]
elif line[i]=='@':
string+=line[i+2]*int(line[i+1])
else:
pass
print(string)
except EOFError:
break
|
s058770699
|
p03759
|
u187516587
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 106 |
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful.
|
l=input().split()
a,b,c=int(l[0]),int(l[1]),int(l[2])
if a-b==b-c:
print("Yes")
else:
print("No")
|
s204273192
|
Accepted
| 17 | 2,940 | 106 |
l=input().split()
a,b,c=int(l[0]),int(l[1]),int(l[2])
if a-b==b-c:
print("YES")
else:
print("NO")
|
s795198105
|
p03564
|
u319818856
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 537 |
Square1001 has seen an electric bulletin board displaying the integer 1. He can perform the following operations A and B to change this value: * Operation A: The displayed value is doubled. * Operation B: The displayed value increases by K. Square1001 needs to perform these operations N times in total. Find the minimum possible value displayed in the board after N operations.
|
def addition_and_multiplication(N: int, K: int)->int:
min_d = float('inf')
for i in range(1 << N):
d = 1
for _ in range(N):
if i & 1:
d = d * 2
else:
d = d + K
min_d = min(d, min_d)
return min_d
if __name__ == "__main__":
N = int(input())
K = int(input())
ans = addition_and_multiplication(N, K)
print(ans)
|
s793328508
|
Accepted
| 19 | 3,060 | 557 |
def addition_and_multiplication(N: int, K: int)->int:
min_d = float('inf')
for i in range(1 << N):
d = 1
for _ in range(N):
if i & 1:
d = d * 2
else:
d = d + K
i >>= 1
min_d = min(d, min_d)
return min_d
if __name__ == "__main__":
N = int(input())
K = int(input())
ans = addition_and_multiplication(N, K)
print(ans)
|
s847284050
|
p02613
|
u486074261
| 2,000 | 1,048,576 |
Wrong Answer
| 59 | 23,072 | 238 |
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.
|
from sys import stdin
n, *s = (x.rstrip() for x in stdin.readlines())
ac = s.count('AC')
wa = s.count('WA')
tle = s.count('TLE')
re = s.count('RE')
print(f'AC × {ac}')
print(f'WA × {wa}')
print(f'TLE × {tle}')
print(f'RE × {re}')
|
s006625136
|
Accepted
| 57 | 23,128 | 234 |
from sys import stdin
n, *s = (x.rstrip() for x in stdin.readlines())
ac = s.count('AC')
wa = s.count('WA')
tle = s.count('TLE')
re = s.count('RE')
print(f'AC x {ac}')
print(f'WA x {wa}')
print(f'TLE x {tle}')
print(f'RE x {re}')
|
s253399842
|
p03379
|
u253952966
| 2,000 | 262,144 |
Wrong Answer
| 310 | 25,624 | 212 |
When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N.
|
n = int(input())
an = list(map(int, input().split()))
an.sort()
a, b = an[n//2-1:n//2+1]
if a == b:
for _ in range(n):
print(n)
else:
for ai in an:
if ai <= a:
print(b)
else:
print(a)
|
s422575252
|
Accepted
| 304 | 26,180 | 222 |
n = int(input())
xn = list(map(int, input().split()))
xsort = sorted(xn)
a, b = xsort[n//2-1:n//2+1]
if a == b:
for _ in range(n):
print(a)
else:
for x in xn:
if x <= a:
print(b)
else:
print(a)
|
s506426211
|
p03698
|
u644907318
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 78 |
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
if len(set(list(input().strip())))==26:
print("yes")
else:
print("no")
|
s726489720
|
Accepted
| 17 | 2,940 | 94 |
S = input().strip()
n = len(S)
if len(set(list(S)))==n:
print("yes")
else:
print("no")
|
s966249965
|
p02747
|
u277920185
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 76 |
A Hitachi string is a concatenation of one or more copies of the string `hi`. For example, `hi` and `hihi` are Hitachi strings, while `ha` and `hii` are not. Given a string S, determine whether S is a Hitachi string.
|
s=input()
s.replace('hi', '')
if s == '':
print('Yes')
else:
print('No')
|
s896228912
|
Accepted
| 17 | 2,940 | 81 |
s=input()
x=len(s) // 2
a=x*'hi'
if a == s:
print('Yes')
else:
print('No')
|
s388561476
|
p03693
|
u737321654
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 121 |
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
x = list(map(int, input().split()))
num = x[0] + x[1] + x[2]
if int(num) % 4 == 0:
print('YES')
else:
print('No')
|
s819960501
|
Accepted
| 17 | 2,940 | 106 |
x = input().split()
num = x[0] + x[1] + x[2]
if int(num) % 4 == 0:
print('YES')
else:
print('NO')
|
s112205933
|
p03194
|
u636311816
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 68 |
There are N integers a_1, a_2, ..., a_N not less than 1. The values of a_1, a_2, ..., a_N are not known, but it is known that a_1 \times a_2 \times ... \times a_N = P. Find the maximum possible greatest common divisor of a_1, a_2, ..., a_N.
|
n,p= map(int,input().split())
if n==1:
print(p)
else :print(1)
|
s137870419
|
Accepted
| 312 | 2,940 | 119 |
n,p= map(int,input().split())
for i in range(round(p**(1/n)),0,-1):
if p%(i**n)==0:
print(i)
exit()
|
s680109881
|
p03623
|
u266171694
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 96 |
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x, a, b = map(int, input().split())
if abs(x - a) > abs(x - b):
print("A")
else:
print("B")
|
s024651712
|
Accepted
| 18 | 2,940 | 96 |
x, a, b = map(int, input().split())
if abs(x - a) > abs(x - b):
print("B")
else:
print("A")
|
s923668605
|
p02409
|
u201482750
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,628 | 386 |
You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room. For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth floor at building b. If v is negative, it means that −v persons left. Assume that initially no person lives in the building.
|
n=int(input())
room=[[[ 0 for i in range(10)] for j in range(3)] for k in range(4)]
for i in range(n) :
b,f,r,v=map(int,input().split())
room[b-1][f-1][r-1]+=v
for i in range(4) :
for j in range(3) :
for k in range(10) :
print(" {}".format(room[i][j][k]),end="")
print("")
if i!=3 :
for l in range(20) :
print("#",end="")
|
s477002314
|
Accepted
| 20 | 5,628 | 409 |
room=[ [ [ 0 for k in range(10) ] for j in range(3) ] for i in range(4)]
n=int(input())
for i in range(n) :
b,f,r,v=map(int,input().split())
room[b-1][f-1][r-1]+=v
for i in range(4) :
for j in range(3) :
for k in range(10) :
print(" {}".format(room[i][j][k]),end="")
print("")
if i!=3 :
for l in range(20) :
print("#",end="")
print("")
|
s676701095
|
p03386
|
u370331385
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 26,704 | 280 |
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
A,B,K = map(int,input().split())
ans = []
if(K <= A-B+1):
for i in range(A,A+K):
ans.append(i)
for j in range(B-K+1,B+1):
ans.append(j)
ans = set(ans)
ans = list(ans)
ans.sort()
for i in range(len(ans)):
print(ans[i])
else:
for i in range(A,B+1):
print(i)
|
s664205655
|
Accepted
| 18 | 3,060 | 250 |
A,B,K = map(int,input().split())
ans = []
for i in range(A,A+K):
if(A<= i<= B):
ans.append(i)
for j in range(B-K+1,B+1):
if(A<= j<= B):
ans .append(j)
ans = set(ans)
ans = list(ans)
ans.sort()
for i in range(len(ans)):
print(ans[i])
|
s979781744
|
p04025
|
u703890795
| 2,000 | 262,144 |
Wrong Answer
| 25 | 3,316 | 165 |
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal **integers** by transforming some of them. He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2). Find the minimum total cost to achieve his objective.
|
N = int(input())
A = list(map(int, input().split()))
s = 0
m = 100000000
for j in range(-100,101):
for i in range(N):
s += (A[i]-j)**2
m = min(m, s)
print(m)
|
s549332313
|
Accepted
| 25 | 2,940 | 173 |
N = int(input())
A = list(map(int, input().split()))
s = 0
m = 100000000
for j in range(-100,101):
for i in range(N):
s += (A[i]-j)**2
m = min(m, s)
s = 0
print(m)
|
s649120913
|
p02647
|
u470560351
| 2,000 | 1,048,576 |
Wrong Answer
| 2,206 | 51,588 | 335 |
We have N bulbs arranged on a number line, numbered 1 to N from left to right. Bulb i is at coordinate i. Each bulb has a non-negative integer parameter called intensity. When there is a bulb of intensity d at coordinate x, the bulb illuminates the segment from coordinate x-d-0.5 to x+d+0.5. Initially, the intensity of Bulb i is A_i. We will now do the following operation K times in a row: * For each integer i between 1 and N (inclusive), let B_i be the number of bulbs illuminating coordinate i. Then, change the intensity of each bulb i to B_i. Find the intensity of each bulb after the K operations.
|
import numpy as np
N,K = map(int, input().split())
A =list(map(int, input().split()))
ans = np.array(A, np.int8)
for i in range(K):
k = np.array([0]*N, np.int8)
for n in range(N):
k[max(0, n - ans[n]):min(N, n + ans[n])+1] += 1
ans = k
for i in ans:
print(min(i, N))
|
s044906316
|
Accepted
| 892 | 125,032 | 378 |
import numpy as np
from numba import njit
N,K = map(int, input().split())
A = np.array(list(map(int, input().split())))
@njit
def main(N, K, A):
for _ in range(min(K, 42)):
A_ = np.zeros_like(A)
for n in range(N):
A_[max(0, n - A[n])] += 1
if n + A[n] + 1 < N:
A_[n + A[n] + 1] -= 1
A = A_.cumsum()
return A
print(*main(N, K, A))
|
s736528728
|
p00050
|
u990228206
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,552 | 329 |
福島県は果物の産地としても有名で、その中でも特に桃とりんごは全国でも指折りの生産量を誇っています。ところで、ある販売用の英文パンフレットの印刷原稿を作ったところ、手違いでりんごに関する記述と桃に関する記述を逆に書いてしまいました。 あなたは、apple と peach を修正する仕事を任されましたが、なにぶん面倒です。１行の英文を入力して、そのなかの apple という文字列を全て peach に、peach という文字列を全てapple に交換した英文を出力するプログラムを作成してください。
|
w=str(input())
print(len(w))
for i in range(len(w)):
if i>len(w):break
if w[i]=="a" and w[i+1]=="p" and w[i+2]=="p" and w[i+3]=="l" and w[i+4]=="e":
w=w[:i]+"peach"+w[i+5:]
continue
if w[i]=="p" and w[i+1]=="e" and w[i+2]=="a" and w[i+3]=="c" and w[i+4]=="h":
w=w[:i]+"apple"+w[i+5:]
print(w)
|
s361441134
|
Accepted
| 20 | 5,556 | 315 |
w=str(input())
for i in range(len(w)):
if i>len(w):break
if w[i]=="a" and w[i+1]=="p" and w[i+2]=="p" and w[i+3]=="l" and w[i+4]=="e":
w=w[:i]+"peach"+w[i+5:]
continue
if w[i]=="p" and w[i+1]=="e" and w[i+2]=="a" and w[i+3]=="c" and w[i+4]=="h":
w=w[:i]+"apple"+w[i+5:]
print(w)
|
s307937579
|
p03485
|
u274907892
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 59 |
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(float, input().split())
print(round((a+b)/2, 0))
|
s895201203
|
Accepted
| 17 | 2,940 | 77 |
import math
a, b = map(float, input().split())
print(int(math.ceil((a+b)/2)))
|
s407101120
|
p03623
|
u357751375
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 257 |
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
s = list(input())
s = set(s)
s = sorted(s)
m = 'abcdefghijklmnopqrstuvwxyz'
for i in range(len(m)):
if s[i] != m[i]:
print(m[i])
exit(0)
if i + 1 != len(m) and i + 1 == len(s):
print(m[i + 1])
exit(0)
print('None')
|
s606295237
|
Accepted
| 17 | 2,940 | 97 |
x,a,b = map(int,input().split())
if abs(x - a) > abs(x - b):
print('B')
else:
print('A')
|
s172777091
|
p03854
|
u441253420
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,188 | 162 |
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
S = input()
Mozi = ["dreamer","eraser","dream","erase"]
for i in Mozi:
T = S.replace(i,"X")
S = T
if S not in "X":
print("No")
else:
print("Yes")
|
s113525391
|
Accepted
| 19 | 3,188 | 156 |
S = input()
Mozi = ["resare","esare","remaerd","maerd"]
S = S[::-1]
for i in Mozi:
S = S.replace(i,"")
if not S:
print("YES")
else:
print("NO")
|
s733894968
|
p03659
|
u026102659
| 2,000 | 262,144 |
Wrong Answer
| 145 | 24,952 | 322 |
Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer a_i written on it. They will share these cards. First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards. Here, both Snuke and Raccoon have to take at least one card. Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively. They would like to minimize |x-y|. Find the minimum possible value of |x-y|.
|
import sys
input = sys.stdin.readline
N = int(input())
nums = list(map(int, input().split()))
s = sum(nums)
x = 0
y = 0
for i in range(N):
a = nums[i]
if x + a <= s//2:
x += a
else:
y = x + a
break
print(x)
if x == 0:
x = nums[0]
if y == s:
y = 0
print(min(s-2*x, 2*y-2))
|
s246800830
|
Accepted
| 208 | 23,780 | 260 |
import sys
input = sys.stdin.readline
n = int(input())
A = list(map(int, input().split()))
x = 0
s = sum(A)
ab = -1
for i in range(n-1):
a = A[i]
x += a
if ab == -1:
ab = abs(s - 2*x)
else:
ab = min(ab, abs(s - 2*x))
print(ab)
|
s016782334
|
p03024
|
u691018832
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 153 |
Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.
|
# sys.stdin.readline()
import sys
input = sys.stdin.readline
s = input()
if 15-len(s)+s.count('o') >= 8:
ans = 'YES'
else:
ans = 'NO'
print(ans)
|
s330967918
|
Accepted
| 17 | 2,940 | 91 |
s = input()
if 15-len(s)+s.count('o') >= 8:
ans = 'YES'
else:
ans = 'NO'
print(ans)
|
s476085695
|
p03251
|
u190406011
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,064 | 312 |
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 = [int(i) for i in input().split()]
x_ls = [int(i) for i in input().split()]
y_ls = [int(i) for i in input().split()]
max_x = max(x_ls)
min_y = min(y_ls)
flg = 0
for z in range(int(X + 0.00001), int(Y)):
if max_x < z <= min_y:
flg = 1
if flg > 0:
print("War")
else:
print("No War")
|
s132030479
|
Accepted
| 18 | 3,064 | 335 |
import math
N, M, X, Y = [int(i) for i in input().split()]
x_ls = [int(i) for i in input().split()]
y_ls = [int(i) for i in input().split()]
max_x = max(x_ls)
min_y = min(y_ls)
flg = 0
for z in range(int(math.ceil(X + 0.00001)), int(Y)):
if max_x < z <= min_y:
flg = 1
if flg > 0:
print("No War")
else:
print("War")
|
s647697359
|
p03860
|
u294385082
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 28 |
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
s = input()
print("A"+s+"C")
|
s824089841
|
Accepted
| 17 | 2,940 | 33 |
st = input()
print("A"+st[8]+"C")
|
s409028822
|
p03860
|
u865826312
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 41 |
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
a,b,c=input().split()
print("a"+b[0]+"c")
|
s298326898
|
Accepted
| 17 | 2,940 | 43 |
a,b,c=input().split()
print(a[0]+b[0]+c[0])
|
s643170399
|
p03860
|
u665598835
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 47 |
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
S = input().split(" ")
print([i[0] for i in S])
|
s450074413
|
Accepted
| 17 | 2,940 | 56 |
S = input().split(" ")
print("".join([i[0] for i in S]))
|
s582209181
|
p02608
|
u185948224
| 2,000 | 1,048,576 |
Wrong Answer
| 106 | 9,452 | 262 |
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
N = int(input())
ans = [0]*(N+1)
for i in range(1, int(N**0.5)+1):
for j in range(1, 101 - i):
for k in range(1, 101 - i - j):
f = i*i + j*j + k*k +i*j + j*k + k*i
if f <= N:
ans[f] += 1
print(*ans, sep='\n')
|
s418496941
|
Accepted
| 453 | 9,528 | 274 |
N = int(input())
ans = [0]*(N+1)
for i in range(1, int(N**0.5)+1):
for j in range(1, int(N**0.5)+2):
for k in range(1, int(N**0.5)+2):
f = i*i + j*j + k*k +i*j + j*k + k*i
if f <= N:
ans[f] += 1
print(*ans[1:], sep='\n')
|
s485491323
|
p03457
|
u858523893
| 2,000 | 262,144 |
Wrong Answer
| 388 | 11,840 | 509 |
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, x, y = [], [], []
for i in range(N) :
_t, _x, _y = map(int, input().split())
t.append(_t)
x.append(_x)
y.append(_y)
cur_x, cur_y = 0, 0
cur_t = 0
flg_impossible = "YES"
for i in range(N) :
dloc = (x[i] - cur_x) + (y[i] - cur_y)
dt = t[i] - cur_t
if ((dloc % 2 == 0 and dt % 2 == 0) or (dloc % 2 != 0 and dt % 2 != 0)) and dloc <= dt :
cur_x, cur_y, cur_t = x[i], y[i], t[i]
else :
flg_impossible = "NO"
print(flg_impossible)
|
s408934656
|
Accepted
| 410 | 11,728 | 551 |
N = int(input())
t, x, y = [], [], []
for i in range(N) :
_t, _x, _y = map(int, input().split())
t.append(_t)
x.append(_x)
y.append(_y)
cur_x = 0
cur_y = 0
cur_time = 0
flg_possible = True
for i in range(N) :
min_dist = abs(x[i] - cur_x) + abs(y[i] - cur_y)
if min_dist > t[i] or not((t[i] - cur_time) % 2 == min_dist % 2):
print("No")
flg_possible = False
break
else :
cur_x, cur_y, cur_time = x[i], y[i], t[i]
if flg_possible == True :
print("Yes")
|
s724934407
|
p03456
|
u551109821
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 125 |
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.
|
a,b = map(int,input().split())
for i in range(100):
if int(a+b) == i**2:
print('Yes')
exit()
print('No')
|
s434713932
|
Accepted
| 82 | 2,940 | 130 |
a,b = map(str,input().split())
for i in range(1,100100):
if int(a+b) == i**2:
print('Yes')
exit()
print('No')
|
s007265677
|
p02975
|
u326744360
| 2,000 | 1,048,576 |
Wrong Answer
| 2,108 | 64,436 | 393 |
Snuke has N hats. The i-th hat has an integer a_i written on it. There are N camels standing in a circle. Snuke will put one of his hats on each of these camels. If there exists a way to distribute the hats to the camels such that the following condition is satisfied for every camel, print `Yes`; otherwise, print `No`. * The bitwise XOR of the numbers written on the hats on both adjacent camels is equal to the number on the hat on itself. What is XOR? The bitwise XOR x_1 \oplus x_2 \oplus \ldots \oplus x_n of n non- negative integers x_1, x_2, \ldots, x_n is defined as follows: - When x_1 \oplus x_2 \oplus \ldots \oplus x_n is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if the number of integers among x_1, x_2, \ldots, x_n whose binary representations have 1 in the 2^k's place is odd, and 0 if that count is even. For example, 3 \oplus 5 = 6.
|
a = [input() for i in range(2)]
N = int(a[0])
data = a[1].split(' ')
data = [int(i) for i in data]
result = []
for i in range(N):
for j in range(N):
if i <= j:
continue
result.append(data[i] ^ data[j])
sorted_data = data.sort()
sorted_result = result.sort()
if sorted_data == sorted_result:
print('Yes')
print('No')
|
s915231055
|
Accepted
| 61 | 14,980 | 223 |
a = [input() for i in range(2)]
N = int(a[0])
data = a[1].split(' ')
data = [int(i) for i in data]
result = []
a = 0
for i in range(N):
a = a ^ data[i]
if a == 0:
print('Yes')
exit()
print('No')
|
s769496278
|
p03400
|
u172823566
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,316 | 308 |
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_list = list(map(int, [input() for i in range(N)]))
counter = 0
for a in a_list:
counter += 1
base = 1
for i in range(1, D):
base = (i * a + 1)
print(base)
if base <= D:
counter += 1
else:
break
counter += X
print(counter)
|
s192800369
|
Accepted
| 18 | 3,064 | 292 |
N = int(input())
D, X = map(int, input().split())
a_list = list(map(int, [input() for i in range(N)]))
counter = 0
for a in a_list:
counter += 1
base = 1
for i in range(1, D):
base = (i * a + 1)
if base <= D:
counter += 1
else:
break
counter += X
print(counter)
|
s688084366
|
p03130
|
u930480241
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 202 |
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads. Determine if we can visit all the towns by traversing each of the roads exactly once.
|
array = [0] * 4
for i in range(3):
thisData = list(map(int, input().split()))
for j in thisData:
array[j - 1] += 1
print(array)
if max(array) == 3:
print("NO")
else:
print("YES")
|
s197116669
|
Accepted
| 17 | 2,940 | 190 |
array = [0] * 4
for i in range(3):
thisData = list(map(int, input().split()))
for j in thisData:
array[j - 1] += 1
if max(array) == 3:
print("NO")
else:
print("YES")
|
s182171460
|
p03456
|
u940516059
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 160 |
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.
|
a,b = [int(x) for x in input().split()]
s = str(a) + str(b)
for n in range(1, 101):
if int(s) % n*n ==0:
print("Yes")
else:
print("No")
|
s290228612
|
Accepted
| 18 | 2,940 | 154 |
import math
a,b = [int(x) for x in input().split()]
s = str(a) + str(b)
if math.sqrt(int(s)).is_integer() == True:
print("Yes")
else:
print("No")
|
s344886511
|
p02260
|
u148477094
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,596 | 220 |
Write a program of the Selection Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: SelectionSort(A) 1 for i = 0 to A.length-1 2 mini = i 3 for j = i to A.length-1 4 if A[j] < A[mini] 5 mini = j 6 swap A[i] and A[mini] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 6 of the pseudocode in the case where i ≠ mini.
|
n=int(input())
a=list(map(int,input().split()))
b=0
for i in a:
minj=i
for j in range(i,n-1):
if a[minj]>a[j]:
a[minj],a[j]=a[j],a[minj]
minj=j
b+=1
print(*a)
print(b)
|
s490647033
|
Accepted
| 20 | 5,604 | 233 |
n=int(input())
a=list(map(int,input().split()))
b=0
for i in range(n):
minj=i
for j in range(i,n):
if a[minj]>a[j]:
minj=j
if i!=minj:
a[i],a[minj]=a[minj],a[i]
b+=1
print(*a)
print(b)
|
s975443200
|
p03612
|
u343675824
| 2,000 | 262,144 |
Wrong Answer
| 68 | 10,612 | 148 |
You are given a permutation p_1,p_2,...,p_N consisting of 1,2,..,N. You can perform the following operation any number of times (possibly zero): Operation: Swap two **adjacent** elements in the permutation. You want to have p_i ≠ i for all 1≤i≤N. Find the minimum required number of operations to achieve this.
|
N = int(input())
A = map(int, input().split())
ans = -1
for i, a in enumerate(A):
if a == i+1:
ans += 1
ans = max((ans, 0))
print(ans)
|
s199611765
|
Accepted
| 80 | 14,008 | 237 |
N = int(input())
a = list(map(int, input().split()))
ans = 0
for i in range(N-1):
if i < N-1 and a[i] == i+1:
ans += 1
a[i], a[i+1] = a[i+1], a[i]
if a[-1] == N:
ans += 1
a[-1], a[-2] = a[-2], a[-1]
print(ans)
|
s966178447
|
p04043
|
u697386253
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 115 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
a = list(map(int, input().split()))
if a.count(5) == 2 and a.count(7) == 1:
print('Yes')
else:
print('No')
|
s616484020
|
Accepted
| 17 | 2,940 | 115 |
a = list(map(int, input().split()))
if a.count(5) == 2 and a.count(7) == 1:
print('YES')
else:
print('NO')
|
s403870273
|
p03386
|
u701318346
| 2,000 | 262,144 |
Wrong Answer
| 2,190 | 1,464,484 | 135 |
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
A, B, K = map(int, input().split())
S = list(i for i in range(A, B + 1))
S = S[:K] + S[-K:]
ans = list(set(S))
for i in ans:
print(i)
|
s540562601
|
Accepted
| 17 | 3,060 | 202 |
A, B, K = map(int, input().split())
S1 = [i for i in range(A, min(A + K, B + 1))]
S2 = [i for i in range(max(A, B - K + 1),B + 1)]
S = list(set(S1 + S2))
S.sort()
for i in range(len(S)):
print(S[i])
|
s104305521
|
p03592
|
u170183831
| 2,000 | 262,144 |
Wrong Answer
| 407 | 3,060 | 237 |
We have a grid with N rows and M columns of squares. Initially, all the squares are white. There is a button attached to each row and each column. When a button attached to a row is pressed, the colors of all the squares in that row are inverted; that is, white squares become black and vice versa. When a button attached to a column is pressed, the colors of all the squares in that column are inverted. Takahashi can freely press the buttons any number of times. Determine whether he can have exactly K black squares in the grid.
|
n, m, k = map(int, input().split())
exists = False
for i in range(n):
for j in range(m):
if (i + 1) * m + (j + 1) * n - (i + 1) * (j + 1) == k:
exists = True
break
if exists:
break
print('Yes' if exists else 'No')
|
s053214948
|
Accepted
| 342 | 3,060 | 225 |
n, m, k = map(int, input().split())
exists = False
for i in range(n + 1):
for j in range(m + 1):
if i * m + j * n - 2 * i * j == k:
exists = True
break
if exists:
break
print('Yes' if exists else 'No')
|
s007682528
|
p02742
|
u992622610
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 109 |
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
H,W = map(int, input().split())
if H * W % 2 == 0:
print( (H * W) /2 )
else:
print((H * W) / 2 + 1)
|
s607753378
|
Accepted
| 18 | 3,188 | 181 |
import math
H,W = map(int, input().split())
if H == 1 or W == 1:
print(1)
elif H * W % 2 == 0:
print( math.floor((H * W) /2 ))
else:
print(math.floor((H * W) / 2) + 1)
|
s731231225
|
p03719
|
u591287669
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 86 |
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 <= b:
print("YES")
else:
print("NO")
|
s138371740
|
Accepted
| 17 | 2,940 | 86 |
a,b,c=map(int,input().split())
if a <= c <= b:
print("Yes")
else:
print("No")
|
s725086407
|
p04011
|
u357751375
| 2,000 | 262,144 |
Wrong Answer
| 31 | 3,316 | 332 |
There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.
|
N = input()
K = input()
X = input()
Y = input()
S = 1
while S <= int(K):
print(str(S)+'泊目は'+X+'円')
S = S + 1
while S <= int(N):
print(str(S)+'泊目は'+Y+'円')
S = S + 1
|
s039302921
|
Accepted
| 24 | 3,060 | 573 |
N = input()
K = input()
X = input()
Y = input()
S = 1
P = 0
if int(N) >= int(K):
while S <= int(K):
P = P + int(X)
S = S + 1
while S <= int(N):
P = P + int(Y)
S = S + 1
else :
while S <= int(N):
P = P + int(X)
S = S + 1
print(P)
|
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