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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s655385708
|
p03854
|
u603234915
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,188 | 98 |
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`.
|
l = ['resare', 'esare', 'maerd', 'remaerd']
s = input()[::-1]
for i in l:
s = s.replace(i, '')
|
s094011693
|
Accepted
| 40 | 3,188 | 441 |
S = input()
S = S[::-1]
L = ['dream', 'dreamer', 'erase', 'eraser']
L = [[l[::-1], len(l)] for l in L]
can = True
ind = 0
S_max = len(S)
while can:
if ind == S_max:
break
for l in L:
if ind + l[1] > S_max:
pass
else:
if l[0] == S[ind: ind+l[1]]:
ind += l[1]
break
else:
can = False
if can:
print("YES")
else:
print("NO")
|
s052203182
|
p00016
|
u957021485
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,760 | 336 |
When a boy was cleaning up after his grand father passing, he found an old paper: In addition, other side of the paper says that "go ahead a number of steps equivalent to the first integer, and turn clockwise by degrees equivalent to the second integer". His grand mother says that Sanbonmatsu was standing at the center of town. However, now buildings are crammed side by side and people can not walk along exactly what the paper says in. Your task is to write a program which hunts for the treature on the paper. For simplicity, 1 step is equivalent to 1 meter. Input consists of several pairs of two integers d (the first integer) and t (the second integer) separated by a comma. Input ends with "0, 0". Your program should print the coordinate (x, y) of the end point. There is the treature where x meters to the east and y meters to the north from the center of town. You can assume that d ≤ 100 and -180 ≤ t ≤ 180\.
|
import sys
import itertools
import math
curdeg = 90
xstep = 0
ystep = 0
for line in sys.stdin.readlines():
step, deg = map(int, line.split(","))
xstep += -step * math.cos(math.pi * curdeg / 180)
ystep += step * math.sin(math.pi * curdeg / 180)
curdeg += deg
print("{:.0f}".format(xstep))
print("{:.0f}".format(ystep))
|
s326246766
|
Accepted
| 30 | 7,860 | 310 |
import sys
import itertools
import math
curdeg = 0
xstep = 0
ystep = 0
for line in sys.stdin.readlines():
step, deg = map(int, line.split(","))
xstep += step * math.cos(math.pi * curdeg / 180)
ystep += step * math.sin(math.pi * curdeg / 180)
curdeg += deg
print(int(ystep))
print(int(xstep))
|
s205568726
|
p03338
|
u982594421
| 2,000 | 1,048,576 |
Wrong Answer
| 19 | 3,060 | 212 |
You are given a string S of length N consisting of lowercase English letters. We will cut this string at one position into two strings X and Y. Here, we would like to maximize the number of different letters contained in both X and Y. Find the largest possible number of different letters contained in both X and Y when we cut the string at the optimal position.
|
n = int(input())
s = input().rstrip()
ans = 0
for i in range(1, len(s)):
s1, s2 = s[0:i], s[i:]
c = set(s1).intersection(set(s2))
for ch in c:
ans = max(ans, min(s1.count(ch), s2.count(ch)))
print(ans)
|
s112915180
|
Accepted
| 18 | 2,940 | 172 |
n = int(input())
s = input().rstrip()
ans = 0
for i in range(1, len(s)):
s1, s2 = s[0:i], s[i:]
c = set(s1).intersection(set(s2))
ans = max(ans, len(c))
print(ans)
|
s346013619
|
p03399
|
u396890425
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 48 |
You planned a trip using trains and buses. The train fare will be A yen (the currency of Japan) if you buy ordinary tickets along the way, and B yen if you buy an unlimited ticket. Similarly, the bus fare will be C yen if you buy ordinary tickets along the way, and D yen if you buy an unlimited ticket. Find the minimum total fare when the optimal choices are made for trains and buses.
|
print(min(input(),input())+min(input(),input()))
|
s583962629
|
Accepted
| 17 | 2,940 | 68 |
print(min(int(input()),int(input()))+min(int(input()),int(input())))
|
s978522001
|
p02412
|
u805716376
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,596 | 204 |
Write a program which identifies the number of combinations of three integers which satisfy the following conditions: * You should select three distinct integers from 1 to n. * A total sum of the three integers is x. For example, there are two combinations for n = 5 and x = 9. * 1 + 3 + 5 = 9 * 2 + 3 + 4 = 9
|
n, x = map(int, input().split())
cnt = 0
for i in range(n):
if i >= x-3:break
for j in range(i+1,n):
for k in range(j+1, n):
if i+j+k == x:
cnt += 1
print(cnt)
|
s468074383
|
Accepted
| 450 | 5,592 | 265 |
while 1:
n, x = map(int, input().split())
if n == x == 0:
break
cnt = 0
for i in range(1,n+1):
for j in range(i+1,n+1):
for k in range(j+1, n+1):
if i+j+k == x:
cnt += 1
print(cnt)
|
s844328569
|
p03416
|
u695429668
| 2,000 | 262,144 |
Wrong Answer
| 153 | 6,772 | 327 |
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.
|
# coding : utf-8
A, B = list(map(int, input().split()))
_list = list(range(A, B+1))
count = 0
buff = 0
for elm in _list:
buff = 0
m = str(elm)
_len = len(m) - 1
for i in range(3):
if m[i] == m[4-i]:
buff += 1
if buff >= 3:
print(m)
count += 1
print(count)
|
s409875258
|
Accepted
| 153 | 6,516 | 310 |
# coding : utf-8
A, B = list(map(int, input().split()))
_list = list(range(A, B+1))
count = 0
buff = 0
for elm in _list:
buff = 0
m = str(elm)
_len = len(m) - 1
for i in range(3):
if m[i] == m[4-i]:
buff += 1
if buff >= 3:
count += 1
print(count)
|
s671342340
|
p03095
|
u288430479
| 2,000 | 1,048,576 |
Wrong Answer
| 36 | 10,136 | 140 |
You are given a string S of length N. Among its subsequences, count the ones such that all characters are different, modulo 10^9+7. Two subsequences are considered different if their characters come from different positions in the string, even if they are the same as strings. Here, a subsequence of a string is a concatenation of **one or more** characters from the string without changing the order.
|
from collections import Counter
n = input()
mod = 10**9+7
s = list(Counter(list(input())).values())
an = 1
for i in s:
an *= i
print(an-1)
|
s735063587
|
Accepted
| 36 | 10,048 | 154 |
from collections import Counter
n = input()
mod = 10**9+7
s = list(Counter(list(input())).values())
an = 1
for i in s:
an *= (i+1)%mod
print((an-1)%mod)
|
s074319610
|
p03457
|
u045091221
| 2,000 | 262,144 |
Wrong Answer
| 539 | 18,468 | 463 |
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.
|
import collections
def solve(N, ts, ps):
for i in range(N-1):
dt = ts[i+1] - ts[i]
d = abs(ps[i+1].x - ps[i].x) + abs(ps[i+1].y - ps[i].y)
if d <= dt and dt % 2 == d % 2:
return "YES"
return "NO"
N = int(input())
point = collections.namedtuple('point', ['x', 'y'])
ts = []
ps = []
for i in range(N):
t, x, y = list(map(int, input().split()))
ts.append(t)
ps.append(point(x, y))
print(solve(N, ts, ps))
|
s861009532
|
Accepted
| 568 | 18,560 | 471 |
import collections
def solve(N, ts, ps):
for i in range(N):
dt = ts[i+1] - ts[i]
d = abs(ps[i+1].x - ps[i].x) + abs(ps[i+1].y - ps[i].y)
if d > dt or dt % 2 != d % 2:
return "No"
return "Yes"
N = int(input())
point = collections.namedtuple('point', ['x', 'y'])
ts = [0]
ps = [point(0, 0)]
for i in range(N):
t, x, y = list(map(int, input().split()))
ts.append(t)
ps.append(point(x, y))
print(solve(N, ts, ps))
|
s665872063
|
p02396
|
u987236471
| 1,000 | 131,072 |
Time Limit Exceeded
| 9,990 | 5,596 | 77 |
In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem.
|
i = 0
x = input()
while x != 0:
print("Case {}: {}".format(i,x))
i += i+1
|
s972164768
|
Accepted
| 140 | 5,596 | 99 |
i = 1
while True:
x =int(input())
if x == 0:
break
print("Case {}: {}".format(i,x))
i = i+1
|
s986906445
|
p02614
|
u696444274
| 1,000 | 1,048,576 |
Wrong Answer
| 40 | 10,492 | 2,287 |
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 datetime
from decimal import Decimal, ROUND_DOWN
import heapq
from fractions import gcd
from functools import reduce
from collections import deque
# from math import factorial
import itertools
import collections
import math
import sys
sys.setrecursionlimit(2000000)
# import statistics
# import numpy as np
# n = int(input())
# n, m, p = list(map(int, input().split()))
# a = list(map(int, input().split()))
#
# n = int(input())
h, w, k = list(map(int, input().split()))
# A = list(map(int, input().split()))
# a = list(map(int, input().split()))
# b = list(map(int, input().split()))
# print(min(15*n, 100*(n//10+1), 100*(n//10)+15*(n % 10)))
c = []
kuro = 0
col_sum = []
row_sum = []
for i in range(h):
a = input()
c.append(a)
ac = a.count('#')
kuro += ac
col_sum.append(ac)
for i in range(w):
cc = 0
for j in range(h):
if c[j][i] == "#":
cc += 1
row_sum.append(cc)
# print(col_sum)
# print(row_sum)
# print("----------")
if kuro == k:
print(1)
exit()
ans = 0
lll = row_sum+col_sum
for i in range(1, h+w):
l = list(itertools.combinations(lll, i))
# print(l)
for j in range(len(l)):
_max = max(l[j])
if _max > k:
m_count = 0
tt = _max
for k in range(i):
if l[j][k] != _max or(l[j][k] == _max and m_count == 1):
tt -= l[j][k]
elif l[j][k] == _max and m_count == 0:
m_count += 1
if tt == k:
ans += 1
elif sum(l[j]) == k:
ans += 1
# print(lll)
# print(ll)
print(ans)
|
s197524567
|
Accepted
| 56 | 9,184 | 751 |
h, w, f = list(map(int, input().split()))
c = [input() for i in range(h)]
count = 0
bit_h = []
for k in range(2**h):
hh = []
for l in range(h):
if((k >> l) & 1):
hh.append(1)
else:
hh.append(0)
bit_h.append(hh)
bit_w = []
for i in range(2**w):
ww = []
for j in range(w):
if((i >> j) & 1):
ww.append(1)
else:
ww.append(0)
bit_w.append(ww)
ans = 0
for i in range(len(bit_h)):
for j in range(len(bit_w)):
count = 0
for k in range(h):
for l in range(w):
if bit_h[i][k] == 1 and bit_w[j][l] == 1 and c[k][l] == "#":
count += 1
if count == f:
ans += 1
print(ans)
|
s059449949
|
p02281
|
u938045879
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,604 | 741 |
Binary trees are defined recursively. A binary tree _T_ is a structure defined on a finite set of nodes that either * contains no nodes, or * is composed of three disjoint sets of nodes: \- a root node. \- a binary tree called its left subtree. \- a binary tree called its right subtree. Your task is to write a program which perform tree walks (systematically traverse all nodes in a tree) based on the following algorithms: 1. Print the root, the left subtree and right subtree (preorder). 2. Print the left subtree, the root and right subtree (inorder). 3. Print the left subtree, right subtree and the root (postorder). Here, the given binary tree consists of _n_ nodes and evey node has a unique ID from 0 to _n_ -1.
|
n = int(input())
root = set(range(n))
nodes = [0 for i in range(n)]
for i in range(n):
node = list(map(int, input().split()))
children = node[1:3]
root -= set(children)
nodes[node[0]] = children
def preorder(id):
if(id == -1):
return
order.append(id)
preorder(nodes[id][0])
preorder(nodes[id][1])
def inorder(id):
if(id == -1):
return
inorder(nodes[id][0])
order.append(id)
inorder(nodes[id][1])
def postorder(id):
if(id == -1):
return
postorder(nodes[id][0])
postorder(nodes[id][1])
order.append(id)
order = []
preorder(list(root)[0])
print(*order)
order = []
inorder(list(root)[0])
print(*order)
order = []
postorder(list(root)[0])
print(*order)
|
s057732936
|
Accepted
| 20 | 5,616 | 852 |
n = int(input())
root = set(range(n))
nodes = [0 for i in range(n)]
for i in range(n):
node = list(map(int, input().split()))
children = node[1:3]
root -= set(children)
nodes[node[0]] = children
def preorder(id):
if(id == -1):
return
order.append(id)
preorder(nodes[id][0])
preorder(nodes[id][1])
def inorder(id):
if(id == -1):
return
inorder(nodes[id][0])
order.append(id)
inorder(nodes[id][1])
def postorder(id):
if(id == -1):
return
postorder(nodes[id][0])
postorder(nodes[id][1])
order.append(id)
order = []
preorder(list(root)[0])
print('Preorder')
print(' ', end='')
print(*order)
order = []
inorder(list(root)[0])
print('Inorder')
print(' ', end='')
print(*order)
order = []
postorder(list(root)[0])
print('Postorder')
print(' ', end='')
print(*order)
|
s639731965
|
p03698
|
u853900545
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 76 |
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
s = input()
if len(set(s)) < len(s):
print('No')
else:
print('Yes')
|
s207338621
|
Accepted
| 18 | 2,940 | 76 |
s = input()
if len(set(s)) < len(s):
print('no')
else:
print('yes')
|
s601030089
|
p03962
|
u226912938
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 395 |
AtCoDeer the deer recently bought three paint cans. The color of the one he bought two days ago is a, the color of the one he bought yesterday is b, and the color of the one he bought today is c. Here, the color of each paint can is represented by an integer between 1 and 100, inclusive. Since he is forgetful, he might have bought more than one paint can in the same color. Count the number of different kinds of colors of these paint cans and tell him.
|
s = str(input())
g_count = 0
p_count = 0
point = 0
for i in range(len(s)):
if g_count == p_count:
g_count += 1
if s[i] == 'g':
pass
else:
point -= 1
elif s[i] == 'g':
p_count += 1
point += 1
elif s[i] == 'p':
p_count += 1
ans = point
print(ans)
|
s899805610
|
Accepted
| 16 | 2,940 | 66 |
color = set(map(int, input().split()))
ans = len(color)
print(ans)
|
s583058009
|
p03759
|
u717001163
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 65 |
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("YES" if c-b==a-b else "NO")
|
s131155047
|
Accepted
| 17 | 2,940 | 65 |
a,b,c=map(int,input().split())
print("YES" if c-b==b-a else "NO")
|
s690191528
|
p03131
|
u602075385
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 181 |
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())
bis, jpn = 1, 0
if b-a < 2 or k <= a-1:
print(k+1)
else:
bis = a + max(b-a, 2) * (k-a+1)/2 + (k-a+1)%2
print(int(bis))
|
s355208599
|
Accepted
| 18 | 2,940 | 186 |
k, a, b = map(int, input().split())
bis, jpn = 1, 0
if b-a < 2 or k <= a-1:
print(k+1)
else:
bis = a + max(b-a, 2) * int((k-a+1)/2) + (k-a+1)%2
print(int(bis))
|
s972475593
|
p03449
|
u689322583
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 52 |
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?
|
import sys
line = sys.stdin.readlines()
print(line)
|
s070162354
|
Accepted
| 19 | 3,064 | 430 |
#coding: utf-8
N = int(input())
A0 = list(map(int, input().split()))
A1 = list(map(int, input().split()))
sum = []
for n in range(N):
tmp = 0
if(n==0):
tmp += A0[0]
for j in range(N):
tmp += A1[j]
sum.append(tmp)
else:
for i in range(n):
tmp += A0[i]
for j in range(n-1, N):
tmp += A1[j]
sum.append(tmp)
ans = max(sum)
print(ans)
|
s153117604
|
p02865
|
u707808519
| 2,000 | 1,048,576 |
Wrong Answer
| 20 | 3,316 | 73 |
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
N = int(input())
if N%2 == 0:
print(N/2 - 1)
else:
print((N-1)/2)
|
s883387343
|
Accepted
| 17 | 2,940 | 83 |
N = int(input())
if N%2 == 0:
print(int(N/2 - 1))
else:
print(int((N-1)/2))
|
s628407676
|
p03387
|
u677121387
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 223 |
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.
|
a = list(map(int,input().split()))
a.sort()
b = list(map(lambda x:x%2,a))
ans = 0
for i in range(3):
if sum(b) == 1:
a[i] -= b[i]
else:
a[i] += b[i]
ans += (a[2]-a[0])//2+ (a[2]-a[1])//2
print(ans)
|
s892496106
|
Accepted
| 17 | 3,064 | 276 |
a = list(map(int,input().split()))
a.sort()
b = list(map(lambda x:x%2,a))
ans = 0
if sum(b) in (1,2):
ans += 1
for i in range(3):
if sum(b) == 1:
a[i] -= b[i]
else:
a[i] += b[i]
ans += (a[2]-a[0])//2+ (a[2]-a[1])//2
print(ans)
|
s118061811
|
p03503
|
u285443936
| 2,000 | 262,144 |
Wrong Answer
| 166 | 4,364 | 356 |
Joisino is planning to open a shop in a shopping street. Each of the five weekdays is divided into two periods, the morning and the evening. For each of those ten periods, a shop must be either open during the whole period, or closed during the whole period. Naturally, a shop must be open during at least one of those periods. There are already N stores in the street, numbered 1 through N. You are given information of the business hours of those shops, F_{i,j,k}. If F_{i,j,k}=1, Shop i is open during Period k on Day j (this notation is explained below); if F_{i,j,k}=0, Shop i is closed during that period. Here, the days of the week are denoted as follows. Monday: Day 1, Tuesday: Day 2, Wednesday: Day 3, Thursday: Day 4, Friday: Day 5. Also, the morning is denoted as Period 1, and the afternoon is denoted as Period 2. Let c_i be the number of periods during which both Shop i and Joisino's shop are open. Then, the profit of Joisino's shop will be P_{1,c_1}+P_{2,c_2}+...+P_{N,c_N}. Find the maximum possible profit of Joisino's shop when she decides whether her shop is open during each period, making sure that it is open during at least one period.
|
n = int(input())
bits = []
for i in range(n):
bit = int(input().replace(" ", ""), 2)
bits.append(bit)
p = [[int(item) for item in input().split()] for _ in range(n)]
ans = - 10**9
for i in range(1,2**10):
val = 0
for j,mise in enumerate(p):
val += mise[bin(i & bits[j]).count("1")]
print(bin(i & bits[j]))
ans = max(ans, val)
print(ans)
|
s497822078
|
Accepted
| 270 | 3,064 | 368 |
N = int(input())
F = [list(map(int, input().split())) for i in range(N)]
P = [list(map(int, input().split())) for i in range(N)]
ans = -(10**9)
for i in range(1,2**10):
S = [0]*N
tmp = 0
for j in range(10):
for k in range(N):
if (i >> j) & 1 and F[k][j] == 1:
S[k] += 1
for m in range(N):
tmp += P[m][S[m]]
ans = max(ans,tmp)
print(ans)
|
s095191801
|
p04043
|
u150460433
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 114 |
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.
|
L=list(map(int, input().split()))
L.sort()
if L[0]==5 and L[1]==5 and L[2]==7:
print("yes")
else:
print("no")
|
s890626128
|
Accepted
| 17 | 2,940 | 114 |
L=list(map(int, input().split()))
L.sort()
if L[0]==5 and L[1]==5 and L[2]==7:
print("YES")
else:
print("NO")
|
s410302692
|
p03007
|
u097121858
| 2,000 | 1,048,576 |
Wrong Answer
| 2,104 | 17,236 | 698 |
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()))
numbers = {a: 0 for a in set(A)}
for a in A:
numbers[a] += 1
operations = []
for _ in range(N - 2):
numbers_keys = numbers.keys()
x = min(numbers_keys)
y = max(numbers_keys)
operations.append((x, y))
z = x - y
if z in numbers:
numbers[z] += 1
else:
numbers[z] = 1
if numbers[x] == 1:
del numbers[x]
else:
numbers[x] -= 1
if numbers[y] == 1:
del numbers[y]
else:
numbers[y] -= 1
A = []
for k, v in numbers.items():
A.extend([k] * v)
x, y = max(A), min(A)
print(x - y)
operations.append((x, y))
for o in operations:
print(*o, sep=" ")
|
s099531688
|
Accepted
| 221 | 14,144 | 641 |
N = int(input())
A = list(map(int, input().split()))
A.sort()
neg = [n for n in A if n < 0]
pos = [n for n in A if n >= 0]
if len(neg) == 0: # all positive
x = A.pop(0)
print(sum(A) - x)
pivot = A.pop()
for y in A:
print(x, y)
x -= y
print(pivot, x)
elif len(pos) == 0:
x = A.pop()
print(x - sum(A))
for y in A:
print(x, y)
x -= y
else:
print(sum(pos) - sum(neg))
pivot = pos.pop()
x = neg.pop()
for y in pos:
print(x, y)
x -= y
x, pivot = pivot, x
for y in neg:
print(x, y)
x -= y
print(x, pivot)
|
s468343201
|
p02842
|
u871841829
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,064 | 76 |
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
N = int(input())
if N%1.08 != 0:
print(":(")
else:
print(N//1.08)
|
s445966287
|
Accepted
| 32 | 3,064 | 141 |
N = int(input())
import math
import sys
for i in range(1, 50000+1):
if math.floor(i * 1.08) == N:
print(i)
sys.exit()
print(":(")
|
s036379511
|
p02601
|
u019355060
| 2,000 | 1,048,576 |
Wrong Answer
| 28 | 9,176 | 194 |
M-kun has the following three cards: * A red card with the integer A. * A green card with the integer B. * A blue card with the integer C. He is a genius magician who can do the following operation at most K times: * Choose one of the three cards and multiply the written integer by 2. His magic is successful if both of the following conditions are satisfied after the operations: * The integer on the green card is **strictly** greater than the integer on the red card. * The integer on the blue card is **strictly** greater than the integer on the green card. Determine whether the magic can be successful.
|
A,B,C=map(int,input().split())
K=int(input())
for i in range(K):
if A>=B:
B=B*2
print(A,B,C)
elif B>=C:
C=C*2
print(A,B,C)
if (A<B)&(B<C):
print("yes")
else:
print("No")
|
s530070522
|
Accepted
| 28 | 9,168 | 162 |
A,B,C=map(int,input().split())
K=int(input())
for i in range(K):
if A>=B:
B=B*2
elif B>=C:
C=C*2
if ((A<B)&(B<C)):
print("Yes")
else:
print("No")
|
s350888205
|
p03795
|
u359007262
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 74 |
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())
pay = 800 * n
get = n // 15
ans = pay - get
print(ans)
|
s160262508
|
Accepted
| 17 | 2,940 | 82 |
n = int(input())
pay = 800 * n
get = (n // 15) * 200
ans = pay - get
print(ans)
|
s094318246
|
p03485
|
u821588465
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 66 |
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
import math
a, b = map(int, input().split())
print(math.ceil(b/a))
|
s580634322
|
Accepted
| 18 | 2,940 | 70 |
import math
a, b = map(int, input().split())
print(math.ceil((a+b)/2))
|
s610835945
|
p03576
|
u391731808
| 2,000 | 262,144 |
Wrong Answer
| 120 | 3,064 | 675 |
We have N points in a two-dimensional plane. The coordinates of the i-th point (1 \leq i \leq N) are (x_i,y_i). Let us consider a rectangle whose sides are parallel to the coordinate axes that contains K or more of the N points in its interior. Here, points on the sides of the rectangle are considered to be in the interior. Find the minimum possible area of such a rectangle.
|
N,K = map(int,input().split())
XY = [list(map(int,input().split())) for _ in [0]*N]
iX = sorted(x for x,y in XY)
iY = sorted(y for x,y in XY)
X = {x:i for i,x in enumerate(iX)}
Y = {y:i for i,y in enumerate(iY)}
c = [[0]*(N+1) for i in [0]*(N+1)]
for x,y in XY:
c[Y[y]+1][X[x]+1] = 1
for i in range(N):
for j in range(N):
c[i+1][j+1] += c[i+1][j] + c[i][j+1]
ans = 10**20
for l in range(N):
for r in range(l+1,N):
u = 0
d = 1
dX = iX[r] - iX[l]
while d<N:
if c[d+1][r+1]+c[u][l]-c[u][r+1]-c[d+1][l] >=K:
ans = min(ans, dX*(iY[d]-iY[u]))
u+=1
else:d+=1
print(ans)
|
s191911583
|
Accepted
| 60 | 3,188 | 990 |
def main():
N,K = map(int,input().split())
XY = [list(map(int,input().split())) for _ in [0]*N]
iX = [0]*N
iY = [0]*N
for i,xy in enumerate(XY):
x,y = xy
iX[i] = x
iY[i] = y
iX.sort()
iY.sort()
X = {x:i for i,x in enumerate(iX)}
Y = {y:i for i,y in enumerate(iY)}
c = [[0]*(N+1) for i in [0]*(N+1)]
for x,y in XY:
c[Y[y]+1][X[x]+1] = 1
for i in range(N):
ci1 = c[i+1]
for j in range(N):
ci1[j+1] += ci1[j]
for i in range(N):
c[i+1] = [ci1j+cij for ci1j,cij in zip(c[i+1],c[i])]
ans = 10**20
for u in range(N):
for d in range(u+K-1,N):
l = 0
r = K-1
dY = iY[d]-iY[u]
cd = c[d+1]
cu = c[u]
while r<N:
if cd[r+1]+cu[l]-cu[r+1]-cd[l] >=K:
ans = min(ans, (iX[r] - iX[l])*dY)
l+=1
else:r+=1
print(ans)
main()
|
s406338409
|
p03351
|
u440129511
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 78 |
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a,b,c,d= map(int,input().split())
if abs(c-a)<=d:print('Yes')
else:print('No')
|
s107604357
|
Accepted
| 17 | 2,940 | 117 |
a,b,c,d= map(int,input().split())
if abs(c-a)<=d or abs(a - b) <= d and abs(b - c) <= d:print('Yes')
else:print('No')
|
s071613182
|
p03455
|
u907676137
| 2,000 | 262,144 |
Wrong Answer
| 26 | 8,976 | 94 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
b, c = map(int, input().split())
if (b * c //2) == 0:
print('Even')
else:
print('Odd')
|
s237497960
|
Accepted
| 29 | 9,040 | 96 |
b, c = map(int, input().split())
if ((b * c) % 2) == 0:
print('Even')
else:
print('Odd')
|
s152167530
|
p03493
|
u873839198
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 90 |
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.
|
number = input()
count = 0
for n in number:
if n == 1:
count += 1
print(count)
|
s336672042
|
Accepted
| 17 | 2,940 | 93 |
number = str(input())
count = 0
for n in number:
if n == "1":
count += 1
print(count)
|
s786961366
|
p03637
|
u731322489
| 2,000 | 262,144 |
Wrong Answer
| 51 | 14,252 | 158 |
We have a sequence of length N, a = (a_1, a_2, ..., a_N). Each a_i is a positive integer. Snuke's objective is to permute the element in a so that the following condition is satisfied: * For each 1 ≤ i ≤ N - 1, the product of a_i and a_{i + 1} is a multiple of 4. Determine whether Snuke can achieve his objective.
|
n = int(input())
A = list(map(int, input().split()))
count = len([a for a in A if a % 4 == 0])
print(count, n // 2)
print("Yes" if count >= n // 2 else "No")
|
s897410396
|
Accepted
| 58 | 14,224 | 326 |
n = int(input())
A = list(map(int, input().split()))
count_4 = len([a for a in A if a % 4 == 0])
count_2 = len([a for a in A if a % 2 == 0]) - count_4
others = n - (count_4 + count_2)
ans = ""
if count_2 == 0:
ans = "Yes" if others <= count_4 + 1 else "No"
else:
ans = "Yes" if others <= count_4 else "No"
print(ans)
|
s258514210
|
p03761
|
u105302073
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,188 | 280 |
Snuke loves "paper cutting": he cuts out characters from a newspaper headline and rearranges them to form another string. He will receive a headline which contains one of the strings S_1,...,S_n tomorrow. He is excited and already thinking of what string he will create. Since he does not know the string on the headline yet, he is interested in strings that can be created regardless of which string the headline contains. Find the longest string that can be created regardless of which string among S_1,...,S_n the headline contains. If there are multiple such strings, find the lexicographically smallest one among them.
|
n = int(input())
s = [input() for _ in range(n)]
a = "abcdefghijklmnopqrstuvwxyz"
cnt = []
for i in a:
mi = 50
for j in s:
print(i, j, j.count(i))
mi = min(mi, j.count(i))
cnt.append(mi)
ans = ""
for i in range(26):
ans += a[i] * cnt[i]
print(ans)
|
s113745094
|
Accepted
| 18 | 3,060 | 248 |
n = int(input())
s = [input() for _ in range(n)]
a = "abcdefghijklmnopqrstuvwxyz"
cnt = []
for i in a:
mi = 50
for j in s:
mi = min(mi, j.count(i))
cnt.append(mi)
ans = ""
for i in range(26):
ans += a[i] * cnt[i]
print(ans)
|
s066755486
|
p02613
|
u554590385
| 2,000 | 1,048,576 |
Wrong Answer
| 156 | 9,172 | 303 |
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())
c1=0
c2=0
c3=0
c4=0
for I in range(1, n):
s = str(input())
if(s == "AC"):
c1+= 1
elif(s == "WA"):
c2+=1
elif(s == "TLE"):
c3+=1
elif(s == "RE"):
c4 +=1
print(f"AC x {c1}")
print(f"WA x {c2}")
print(f"TLE x {c3}")
print(f"RE x {c4}")
|
s326109701
|
Accepted
| 158 | 9,152 | 305 |
n = int(input())
c1=0
c2=0
c3=0
c4=0
for I in range(1, n+1):
s = str(input())
if(s == "AC"):
c1+= 1
elif(s == "WA"):
c2+=1
elif(s == "TLE"):
c3+=1
elif(s == "RE"):
c4 +=1
print(f"AC x {c1}")
print(f"WA x {c2}")
print(f"TLE x {c3}")
print(f"RE x {c4}")
|
s309467406
|
p03377
|
u556477263
| 2,000 | 262,144 |
Wrong Answer
| 29 | 9,072 | 84 |
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,c = map(int,input().split())
if a+b <= c:
print('YES')
else:
print('NO')
|
s541959071
|
Accepted
| 25 | 9,092 | 99 |
a,b,c = map(int,input().split())
if a <= c and a+b >= c:
print('YES')
else:
print('NO')
|
s399343818
|
p02260
|
u749243807
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,600 | 484 |
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.
|
count = int(input());
data = [int(n) for n in input().split(" ")];
def selection_sort(data):
count = len(data);
o = 0;
for i in range(count):
minI = i;
for j in range(i + 1, count):
if data[j] < data[minI]:
minI = j;
temp = data[i];
data[i] = data[minI];
data[minI] = temp;
o += 1;
show(data);
print(o);
def show(data):
print(" ".join(str(n) for n in data));
selection_sort(data);
|
s929455939
|
Accepted
| 20 | 5,600 | 522 |
count = int(input());
data = [int(n) for n in input().split(" ")];
def selection_sort(data):
count = len(data);
o = 0;
for i in range(count):
minI = i;
for j in range(i + 1, count):
if data[j] < data[minI]:
minI = j;
if minI != i:
temp = data[i];
data[i] = data[minI];
data[minI] = temp;
o += 1;
show(data);
print(o);
def show(data):
print(" ".join(str(n) for n in data));
selection_sort(data);
|
s029578761
|
p02613
|
u756311765
| 2,000 | 1,048,576 |
Wrong Answer
| 145 | 16,268 | 221 |
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 = []
for _ in range(N):
S.append(input())
AC = S.count('AC')
WA = S.count('WA')
TLE = S.count('TLE')
RE = S.count('RE')
print('AC ×', AC)
print('WA ×', WA)
print('TLE ×', TLE)
print('RE ×', RE)
|
s920109176
|
Accepted
| 151 | 16,284 | 217 |
N = int(input())
S = []
for _ in range(N):
S.append(input())
AC = S.count('AC')
WA = S.count('WA')
TLE = S.count('TLE')
RE = S.count('RE')
print('AC x', AC)
print('WA x', WA)
print('TLE x', TLE)
print('RE x', RE)
|
s155796765
|
p03778
|
u278670845
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 132 |
AtCoDeer the deer found two rectangles lying on the table, each with height 1 and width W. If we consider the surface of the desk as a two-dimensional plane, the first rectangle covers the vertical range of AtCoDeer will move the second rectangle horizontally so that it connects with the first rectangle. Find the minimum distance it needs to be moved.
|
import sys
w,a,b = map(int, input().split())
if a+w>b or b+w>a:
print(0)
elif a+w<b:
print(b-a-w+1)
elif b+w<a:
print(a-b-w+1)
|
s962546596
|
Accepted
| 17 | 3,060 | 136 |
import sys
w,a,b = map(int, input().split())
if a+w>=b>=a or b+w>=a>=b:
print(0)
elif a+w<b:
print(b-a-w)
elif b+w<a:
print(a-b-w)
|
s775954496
|
p02613
|
u060927350
| 2,000 | 1,048,576 |
Wrong Answer
| 150 | 9,212 | 273 |
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.
|
arr=[0]*4
for _ in range(int(input())):
s=input()
if s=='AC':
arr[0]+=1
elif s=='WA':
arr[1]+=1
elif s=='TLE':
arr[2]+=1
else:
arr[3]+=1
print("AC *",arr[0])
print("WA *",arr[1])
print("TLE *",arr[2])
print("RE *",arr[3])
|
s077371239
|
Accepted
| 151 | 9,092 | 273 |
arr=[0]*4
for _ in range(int(input())):
s=input()
if s=='AC':
arr[0]+=1
elif s=='WA':
arr[1]+=1
elif s=='TLE':
arr[2]+=1
else:
arr[3]+=1
print("AC x",arr[0])
print("WA x",arr[1])
print("TLE x",arr[2])
print("RE x",arr[3])
|
s318047891
|
p02694
|
u332793228
| 2,000 | 1,048,576 |
Wrong Answer
| 25 | 9,096 | 66 |
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())
i=1
while X <= 100*(1.01**i):
print(i)
i+=1
|
s428864416
|
Accepted
| 25 | 9,168 | 139 |
import math
x=int(input())
i=1
a=100
while i>0:
a=math.floor(a*1.01)
if a>=x:
print(i)
break
else:
i+=1
|
s950091038
|
p03555
|
u749491107
| 2,000 | 262,144 |
Wrong Answer
| 29 | 8,980 | 109 |
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
a = input()
b = input()
if a[0] == b[2] and a[1] == b[1] and a[2] == b[0]:
print("Yes")
else:
print("No")
|
s936389832
|
Accepted
| 31 | 9,044 | 109 |
a = input()
b = input()
if a[0] == b[2] and a[1] == b[1] and a[2] == b[0]:
print("YES")
else:
print("NO")
|
s345625650
|
p03485
|
u064563749
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 86 |
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())
x=(a+b)/2
if x%1==0:
print(x)
else:
print(int(x))
|
s529750408
|
Accepted
| 17 | 2,940 | 54 |
a,b=map(int,input().split())
x=int((a+b+1)/2)
print(x)
|
s769426131
|
p03251
|
u992910889
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,060 | 225 |
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
N,M,X,Y=map(int,input().split())
x=list(map(int,input().split()))
y=list(map(int,input().split()))
for i in range(X+1,Y+1):
if max(x)<i and min(y)<=i:
print('No War')
elif i==Y:
print('War')
|
s623451698
|
Accepted
| 17 | 2,940 | 293 |
# coding: utf-8
# Your code here!
N,M,X,Y=map(int,input().split())
x=list(map(int,input().split()))
y=list(map(int,input().split()))
for i in range(X+1,Y+1):
if max(x)<i and min(y)>=i:
print('No War')
#print(i)
break
elif i==Y:
print('War')
|
s334858660
|
p02243
|
u487861672
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,660 | 1,500 |
For a given weighted graph $G = (V, E)$, find the shortest path from a source to each vertex. For each vertex $u$, print the total weight of edges on the shortest path from vertex $0$ to $u$.
|
from sys import maxsize
from heapq import heappop, heappush
class Node:
WHITE = 0
GRAY = 1
BLACK = 2
def __init__(self):
self.adjs = []
self.color = Node.WHITE
self.dist = maxsize
def __lt__(self, other):
return self.dist < other.dist
def __repr__(self):
return "Node({},{},{})".format(self.adjs, self.color, self.dist)
def read_adj():
n = int(input())
nodes = [Node() for _ in range(n)]
for i in range(n):
u, _, *rest = [int(x) for x in input().split()]
for i in range(0, len(rest) - 1, 2):
nodes[u].adjs.append((rest[i], rest[i+1]))
return nodes
def shortest_path(nodes):
h = []
nodes[0].dist = 0
nodes[0].color = Node.GRAY
heappush(h, (0, 0))
while len(h) > 0:
min_node, min_dist = heappop(h)
if nodes[min_node].dist < min_dist:
continue
nodes[min_node].color = Node.BLACK
for adj, c in nodes[min_node].adjs:
if nodes[min_node].color == Node.BLACK:
continue
if nodes[min_node].dist + c < nodes[adj].dist:
nodes[adj].dist = nodes[min_node].dist + c
nodes[adj].color = Node.GRAY
heappush(h, (adj, nodes[adj].dist))
def print_dist(nodes):
for u, n in enumerate(nodes):
print(u, n.dist)
def main():
nodes = read_adj()
shortest_path(nodes)
print_dist(nodes)
if __name__ == '__main__':
main()
|
s958503111
|
Accepted
| 580 | 68,620 | 1,495 |
from sys import maxsize
from heapq import heappop, heappush
class Node:
WHITE = 0
GRAY = 1
BLACK = 2
def __init__(self):
self.adjs = []
self.color = Node.WHITE
self.dist = maxsize
def __lt__(self, other):
return self.dist < other.dist
def __repr__(self):
return "Node({},{},{})".format(self.adjs, self.color, self.dist)
def read_adj():
n = int(input())
nodes = [Node() for _ in range(n)]
for i in range(n):
u, _, *rest = [int(x) for x in input().split()]
for i in range(0, len(rest) - 1, 2):
nodes[u].adjs.append((rest[i], rest[i+1]))
return nodes
def shortest_path(nodes):
h = []
nodes[0].dist = 0
nodes[0].color = Node.GRAY
heappush(h, (0, 0))
while len(h) > 0:
min_dist, min_node = heappop(h)
nodes[min_node].color = Node.BLACK
if nodes[min_node].dist < min_dist:
continue
for adj, c in nodes[min_node].adjs:
if nodes[adj].color == Node.BLACK:
continue
if nodes[min_node].dist + c < nodes[adj].dist:
nodes[adj].dist = nodes[min_node].dist + c
nodes[adj].color = Node.GRAY
heappush(h, (nodes[adj].dist, adj))
def print_dist(nodes):
for u, n in enumerate(nodes):
print(u, n.dist)
def main():
nodes = read_adj()
shortest_path(nodes)
print_dist(nodes)
if __name__ == '__main__':
main()
|
s841355100
|
p03006
|
u334712262
| 2,000 | 1,048,576 |
Wrong Answer
| 1,837 | 14,716 | 1,802 |
There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i). We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation: * Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1. Find the minimum total cost required to collect all the balls when we optimally choose p and q.
|
# -*- coding: utf-8 -*-
import bisect
import heapq
import math
import random
import sys
from collections import Counter, defaultdict, deque
from decimal import ROUND_CEILING, ROUND_HALF_UP, Decimal
from functools import lru_cache, reduce
from itertools import combinations, combinations_with_replacement, product, permutations
from operator import add, mul, sub
sys.setrecursionlimit(100000)
input = sys.stdin.readline
def read_int():
return int(input())
def read_int_n():
return list(map(int, input().split()))
def read_float():
return float(input())
def read_float_n():
return list(map(float, input().split()))
def read_str():
return input().strip()
def read_str_n():
return list(map(str, input().split()))
def error_print(*args):
print(*args, file=sys.stderr)
def mt(f):
import time
def wrap(*args, **kwargs):
s = time.time()
ret = f(*args, **kwargs)
e = time.time()
error_print(e - s, 'sec')
return ret
return wrap
@mt
def slv(N, XY):
XY = [tuple(v) for v in XY]
XY.sort()
@lru_cache(maxsize=None)
def f(i, v, x, y):
ans = 0
for j, u in enumerate(XY[i+1:]):
if v[0] - u[0] == x and v[1] - u[1] == y:
ans = max(ans, f(i+j+1, u, x, y) + 1)
else:
ans = max(ans, f(i+j+1, u, x, y))
return ans
ans = 0
for i in range(N):
print(i)
for j in range(i+1, N):
if XY[i][0]-XY[j][0] == 0 and XY[i][1]-XY[j][1] == 0:
continue
ans = max(ans, f(i, XY[i], XY[i][0]-XY[j][0], XY[i][1]-XY[j][1]))
return N - ans
def main():
N = read_int()
XY = [read_int_n() for _ in range(N)]
print(slv(N, XY))
if __name__ == '__main__':
main()
|
s564732131
|
Accepted
| 53 | 6,468 | 1,442 |
# -*- coding: utf-8 -*-
import bisect
import heapq
import math
import random
import sys
from collections import Counter, defaultdict, deque
from decimal import ROUND_CEILING, ROUND_HALF_UP, Decimal
from functools import lru_cache, reduce
from itertools import combinations, combinations_with_replacement, product, permutations
from operator import add, mul, sub
sys.setrecursionlimit(100000)
input = sys.stdin.readline
def read_int():
return int(input())
def read_int_n():
return tuple(map(int, input().split()))
def read_float():
return float(input())
def read_float_n():
return list(map(float, input().split()))
def read_str():
return input().strip()
def read_str_n():
return list(map(str, input().split()))
def error_print(*args):
print(*args, file=sys.stderr)
def mt(f):
import time
def wrap(*args, **kwargs):
s = time.time()
ret = f(*args, **kwargs)
e = time.time()
error_print(e - s, 'sec')
return ret
return wrap
@mt
def slv(N, XY):
c = Counter()
for i in range(N):
x, y = XY[i]
for j in range(N):
if i == j:
continue
a, b = XY[j]
c[(x-a, y-b)] += 1
if not c:
return N
return N - c.most_common(1)[0][1]
def main():
N = read_int()
XY = [read_int_n() for _ in range(N)]
print(slv(N, XY))
if __name__ == '__main__':
main()
|
s043000268
|
p02608
|
u325660636
| 2,000 | 1,048,576 |
Wrong Answer
| 2,208 | 66,612 | 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).
|
import numpy as np
N = int(input())
l = []
for x in range(1,101):
for y in range(1,101):
for z in range(1,101):
fn = x**2 + y**2 + z**2 + x*y + y*z + z*x
l.append(fn)
for i in range(N+1):
print(l.count(i))
|
s910971975
|
Accepted
| 885 | 9,516 | 380 |
import itertools
N = int(input())
num = int(N**0.5)+1
ans = [0]*(N+1)
# for x in range(1,num+1):
# for y in range(1,num+1):
for x,y,z in itertools.product(range(1, num+1), repeat=3):
fn = x**2 + y**2 + z**2 + x*y + y*z + z*x
if fn > N:
continue
else:
ans[fn] += 1
for i in range(1,len(ans)):
print(ans[i])
|
s948703292
|
p03193
|
u735211927
| 2,000 | 1,048,576 |
Wrong Answer
| 20 | 3,060 | 174 |
There are N rectangular plate materials made of special metal called AtCoder Alloy. The dimensions of the i-th material are A_i \times B_i (A_i vertically and B_i horizontally). Takahashi wants a rectangular plate made of AtCoder Alloy whose dimensions are exactly H \times W. He is trying to obtain such a plate by choosing one of the N materials and cutting it if necessary. When cutting a material, the cuts must be parallel to one of the sides of the material. Also, the materials have fixed directions and cannot be rotated. For example, a 5 \times 3 material cannot be used as a 3 \times 5 plate. Out of the N materials, how many can produce an H \times W plate if properly cut?
|
N, H, W = map(int, input().split())
A = []
B = []
counter = 0
for i in range(0,N):
a,b = map(int, input().split())
if a>H and b>W:
counter +=1
print(counter)
|
s655772634
|
Accepted
| 20 | 3,060 | 176 |
N, H, W = map(int, input().split())
A = []
B = []
counter = 0
for i in range(0,N):
a,b = map(int, input().split())
if a>=H and b>=W:
counter +=1
print(counter)
|
s524591688
|
p03712
|
u805011545
| 2,000 | 262,144 |
Wrong Answer
| 32 | 9,304 | 242 |
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 = [int(x) for x in input().split()]
pic = [list(input()) for _ in range(H)]
ans = [['#' for i in range(W+2)] for j in range(H+2)]
for i in range(W):
for j in range(H):
ans[j+1][i+1] = pic[j][i]
for i in range(H+2):
print(*ans[i])
|
s302083622
|
Accepted
| 31 | 9,104 | 302 |
H, W = [int(x) for x in input().split()]
pic = [list(input()) for _ in range(H)]
ans = [['#' for i in range(W+2)] for j in range(H+2)]
for i in range(H):
for j in range(W):
ans[i+1][j+1] = pic[i][j]
low = ''
for i in range(H+2):
for j in range(W+2):
low += ans[i][j]
print(low)
low = ''
|
s404702706
|
p02669
|
u860002137
| 2,000 | 1,048,576 |
Wrong Answer
| 23 | 9,280 | 1,367 |
You start with the number 0 and you want to reach the number N. You can change the number, paying a certain amount of coins, with the following operations: * Multiply the number by 2, paying A coins. * Multiply the number by 3, paying B coins. * Multiply the number by 5, paying C coins. * Increase or decrease the number by 1, paying D coins. You can perform these operations in arbitrary order and an arbitrary number of times. What is the minimum number of coins you need to reach N? **You have to solve T testcases.**
|
def solve(n, a, b, c, d, bi=1):
curr = n
coin = d
while curr != 1:
ca = curr / 2 / a
cb = curr * 2 / 3 / b
cc = curr * 4 / 5 / c
cd = 1 / d
cs = [(ca, 2, a), (cb, 3, b), (cc, 5, c), (cd, 1, d)]
cs.sort(reverse=True)
if cs[0][1] == 1:
curr = 1
coin += (curr - 1) * d
break
if curr % 2 != 0 and curr % 3 != 0 and curr % 5 != 0:
curr += bi
coin += d
continue
if curr % cs[0][1] == 0 and cs[0][1] != 1 and curr >= cs[0][1]:
curr //= cs[0][1]
coin += cs[0][2]
continue
if curr % cs[1][1] == 0 and cs[1][1] != 1 and curr >= cs[1][1]:
curr //= cs[1][1]
coin += cs[1][2]
continue
if curr % cs[2][1] == 0 and cs[2][1] != 1 and curr >= cs[2][1]:
curr //= cs[2][1]
coin += cs[2][2]
continue
if curr % cs[3][1] == 0 and cs[3][1] != 1 and curr >= cs[3][1]:
curr //= cs[3][1]
coin += cs[3][2]
continue
coin = float("inf")
break
return coin
t = int(input())
ans = []
for i in range(t):
n, a, b, c, d = map(int, input().split())
ans.append(min(solve(n, a, b, c, d, -1), solve(n, a, b, c, d, 1)))
print(*ans, sep="\n")
|
s706737137
|
Accepted
| 218 | 20,764 | 1,663 |
from functools import lru_cache
def solve(n, a, b, c, d):
@lru_cache(maxsize=None)
def func(n):
if n == 0:
return 0
if n == 1:
return d
else:
coin = d * n
div, mod = divmod(n, 2)
if mod == 0:
coin = min(coin, func(div) + a)
else:
coin = min(coin, func(div) + a + d, func(div + 1) + a + d)
div, mod = divmod(n, 3)
if mod == 0:
coin = min(coin, func(div) + b)
elif mod == 1:
coin = min(coin, func(div) + b + d, func(div + 1) + b + 2 * d)
else:
coin = min(coin, func(div) + b + 2 * d, func(div + 1) + b + d)
div, mod = divmod(n, 5)
if mod == 0:
coin = min(coin, func(div) + c)
elif mod == 1:
coin = min(coin, func(div) + c + d, func(div + 1) + c + 4 * d)
elif mod == 2:
coin = min(coin, func(div) + c + 2 * d, func(div + 1) + c + 3 * d)
elif mod == 3:
coin = min(coin, func(div) + c + 3 * d, func(div + 1) + c + 2 * d)
else:
coin = min(coin, func(div) + c + 4 * d, func(div + 1) + c + d)
return coin
return func(n)
t = int(input())
ans = []
for i in range(t):
n, a, b, c, d = map(int, input().split())
ans.append(solve(n, a, b, c, d))
print(*ans, sep="\n")
|
s395987195
|
p02275
|
u599130514
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,604 | 439 |
Counting sort can be used for sorting elements in an array which each of the n input elements is an integer in the range 0 to k. The idea of counting sort is to determine, for each input element x, the number of elements less than x as C[x]. This information can be used to place element x directly into its position in the output array B. This scheme must be modified to handle the situation in which several elements have the same value. Please see the following pseudocode for the detail: Counting-Sort(A, B, k) 1 for i = 0 to k 2 do C[i] = 0 3 for j = 1 to length[A] 4 do C[A[j]] = C[A[j]]+1 5 /* C[i] now contains the number of elements equal to i */ 6 for i = 1 to k 7 do C[i] = C[i] + C[i-1] 8 /* C[i] now contains the number of elements less than or equal to i */ 9 for j = length[A] downto 1 10 do B[C[A[j]]] = A[j] 11 C[A[j]] = C[A[j]]-1 Write a program which sorts elements of given array ascending order based on the counting sort.
|
def counting_sort(A, B, k):
len_A = len(A)
C = [0 for _ in range(k + 1)]
for j in range(len_A):
C[A[j]] += 1
for i in range(1, k + 1):
C[i] = C[i] + C[i - 1]
for j in range(len_A - 1, -1, -1):
B[C[A[j]] - 1] = A[j]
C[A[j]] -= 1
n = int(input())
input_list = list(map(int, input().split(' ')))
B = [0 for _ in range(n)]
counting_sort(input_list, B, max(input_list) + 1)
print(B)
|
s080222712
|
Accepted
| 2,530 | 224,388 | 440 |
def counting_sort(A, B, k):
len_A = len(A)
C = [0 for _ in range(k + 1)]
for j in range(len_A):
C[A[j]] += 1
for i in range(1, k + 1):
C[i] = C[i] + C[i - 1]
for j in range(len_A - 1, -1, -1):
B[C[A[j]] - 1] = A[j]
C[A[j]] -= 1
n = int(input())
input_list = list(map(int, input().split(' ')))
B = [0 for _ in range(n)]
counting_sort(input_list, B, max(input_list) + 1)
print(*B)
|
s483760158
|
p02390
|
u535719732
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,588 | 97 |
Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively.
|
s = int(input())
h = s / 3600
s = s % 3600
m = s / 360
s = s % 360
print("%d:%d:%d" % (h,m,s))
|
s489115005
|
Accepted
| 20 | 5,588 | 97 |
s = int(input())
h = s // 3600
s = s % 3600
m = s // 60
s = s % 60
print("%d:%d:%d" % (h,m,s))
|
s042814161
|
p03377
|
u532502139
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 131 |
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.
|
def catOrDog(A, B, X):
if X < A:
print('NO')
elif A + B > X + 1:
print('YES')
else:
print('NO')
|
s814788478
|
Accepted
| 17 | 2,940 | 124 |
a,b,x=[int(x) for x in input().split()]
if x < a:
print('NO')
elif a + b > x + 1:
print('YES')
else:
print('NO')
|
s707313277
|
p03168
|
u501842214
| 2,000 | 1,048,576 |
Wrong Answer
| 2,119 | 179,956 | 750 |
Let N be a positive odd number. There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i. Taro has tossed all the N coins. Find the probability of having more heads than tails.
|
def solve(N, probs):
# dp[i][j]: probability of having j heads in i coin tosses
dp = [[0 for _ in range(N + 1)] for _ in range(N + 1)]
dp[0][0] = 1
for toss_id in range(1, N + 1):
head_prob = probs[toss_id - 1]
tail_prob = 1 - probs[toss_id - 1]
# only until i needed
for j in range(toss_id + 1):
if j == 0:
dp[toss_id][j] += dp[toss_id - 1][j] * tail_prob
else:
dp[toss_id][j] += (dp[toss_id - 1][j] * tail_prob) + \
(dp[toss_id - 1][j - 1] * head_prob)
ans = 0
for j in range(1, N + 1):
if j >= (N + 1) // 2:
ans += dp[N][j]
def iin(): return int(input())
def fln(): return list(map(float, input().split()))
N = iin()
probs = fln()
solve(N, probs)
|
s184689212
|
Accepted
| 952 | 3,316 | 710 |
def solve(N, probs):
# dp[i][j]: probability of having j heads in i coin tosses
# shorter version: dp[j]: probability of j heads
dp = [0 for _ in range(N + 1)]
dp[0] = 1
for toss_id in range(1, N + 1):
head_prob = probs[toss_id - 1]
tail_prob = 1.0 - probs[toss_id - 1]
# only until i needed
for j in range(toss_id, -1, -1):
if j == 0:
dp[j] = dp[j] * tail_prob
else:
dp[j] = (dp[j] * tail_prob) + (dp[j - 1] * head_prob)
# sum all probabilies where heads > tails
ans = sum(dp[N // 2 + 1:])
print(ans)
def iin(): return int(input())
def fln(): return list(map(float, input().split()))
N = iin()
probs = fln()
solve(N, probs)
|
s540484025
|
p03155
|
u428132025
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 73 |
It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board. The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns. How many ways are there to choose where to put the notice so that it completely covers exactly HW squares?
|
n = int(input())
h = int(input())
w = int(input())
print((n-h+1)*(h-w+1))
|
s607968652
|
Accepted
| 17 | 2,940 | 73 |
n = int(input())
h = int(input())
w = int(input())
print((n-h+1)*(n-w+1))
|
s778682704
|
p03556
|
u037221289
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 92 |
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,1,-1):
if (i*0.5).is_integer():
print(i)
exit()
|
s043319713
|
Accepted
| 37 | 3,060 | 99 |
N = int(input())
for i in range(N,0,-1):
if (i**0.5).is_integer():
print(int(i))
exit()
|
s983649918
|
p03090
|
u913965975
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 3,828 | 341 |
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())
matrix = [[1 for i in range(N)] for i in range(N)]
for i in range(N):
matrix[i][i] = 0
if N%2==0:
for i in range(N):
matrix[i][N-i-1] = 0
else:
for i in range(N-1):
matrix[i][N-i-2] = 0
for i in range(N):
for j in range(i+1,N):
if matrix[i][j] == 1:
print(i+1,"",j+1)
|
s842528151
|
Accepted
| 24 | 3,700 | 371 |
N = int(input())
matrix = [[1 for i in range(N)] for i in range(N)]
for i in range(N):
matrix[i][i] = 0
if N%2==0:
for i in range(N):
matrix[i][N-i-1] = 0
else:
for i in range(N-1):
matrix[i][N-i-2] = 0
print(int((N*(N-1)/2)-int(N/2)))
for i in range(N):
for j in range(i+1,N):
if matrix[i][j] == 1:
print(i+1,j+1)
|
s485675936
|
p03712
|
u589087860
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 351 |
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())
a = []
for i in range(h):
a.append(input())
print(a)
ans = []
for i in range(h + 2):
if i == 0 or i == h + 1:
s = ""
for j in range(w + 2):
s = s + "#"
ans.append(s)
else:
s = "#" + a[i - 1] + "#"
ans.append(s)
#print(ans)
print('\n'.join(ans))
|
s207879914
|
Accepted
| 18 | 3,060 | 352 |
h, w = map(int, input().split())
a = []
for i in range(h):
a.append(input())
#print(a)
ans = []
for i in range(h + 2):
if i == 0 or i == h + 1:
s = ""
for j in range(w + 2):
s = s + "#"
ans.append(s)
else:
s = "#" + a[i - 1] + "#"
ans.append(s)
#print(ans)
print('\n'.join(ans))
|
s205751402
|
p04012
|
u094191970
| 2,000 | 262,144 |
Wrong Answer
| 21 | 3,316 | 245 |
Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful.
|
W = input()
import collections
switch = 0
W_list = list(W)
W_dic = collections.Counter(W_list)
for num in W_dic.values():
if num%2 != 0:
switch = 1
break
print(switch)
if switch == 0:
print('Yes')
else:
print('No')
|
s399636107
|
Accepted
| 21 | 3,316 | 231 |
W = input()
import collections
switch = 0
W_list = list(W)
W_dic = collections.Counter(W_list)
for num in W_dic.values():
if num%2 != 0:
switch = 1
break
if switch == 0:
print('Yes')
else:
print('No')
|
s664474787
|
p03623
|
u813450984
| 2,000 | 262,144 |
Wrong Answer
| 21 | 2,940 | 168 |
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 = input()
alphabet = "abcdefghijklmnopqrstuvwxyz"
found = [i for i in alphabet if i not in S]
found.sort()
if len(found) == 0:
print('None')
else:
print(found[0])
|
s653825323
|
Accepted
| 19 | 2,940 | 146 |
place = map(int, input().split())
X, A, B = [i for i in place]
X_A = abs(X - A)
X_B = abs(X - B)
if X_A <= X_B:
print('A')
else:
print('B')
|
s265284893
|
p03623
|
u003501233
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 70 |
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())
A = x - a
B = x - b
print(min(A,B))
|
s880143392
|
Accepted
| 17 | 2,940 | 88 |
x,a,b = map(int,input().split())
if abs(x-a) < abs(x-b):
print("A")
else:
print("B")
|
s699353885
|
p03494
|
u404561212
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,316 | 304 |
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
def devide_counter(num):
devide_count = 0
while num%2==0:
num /= 2
devide_count += 1
return devide_count
#N = int(input())
A = list(map(int, input().split()))
devide_count_list=[]
for a in A:
devide_count_list.append(devide_counter(a))
print(min(devide_count_list))
|
s628435539
|
Accepted
| 19 | 3,316 | 303 |
def devide_counter(num):
devide_count = 0
while num%2==0:
num /= 2
devide_count += 1
return devide_count
N = int(input())
A = list(map(int, input().split()))
devide_count_list=[]
for a in A:
devide_count_list.append(devide_counter(a))
print(min(devide_count_list))
|
s878974975
|
p03713
|
u663101675
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 752 |
There is a bar of chocolate with a height of H blocks and a width of W blocks. Snuke is dividing this bar into exactly three pieces. He can only cut the bar along borders of blocks, and the shape of each piece must be a rectangle. Snuke is trying to divide the bar as evenly as possible. More specifically, he is trying to minimize S_{max} \- S_{min}, where S_{max} is the area (the number of blocks contained) of the largest piece, and S_{min} is the area of the smallest piece. Find the minimum possible value of S_{max} - S_{min}.
|
H, W = list(map(int, input().split()))
def Diff(H, W):
H_min = int(H / 3)
H_max = int(H / 3) + 1
W_short = int(W / 2)
W_long = W - W_short
S_1_1 = [H_min * W, (H - H_min) * W_short, (H - H_min) * W_long]
S_1_2 = [H_max * W, (H - H_max) * W_short, (H - H_max) * W_long]
D_1_1 = max(S_1_1) - min(S_1_1)
D_1_2 = max(S_1_2) - min(S_1_2)
return min(D_1_1, D_1_2)
R = []
R.append(H)
R.append(W)
R.append(Diff(H, W))
R.append(Diff(W, H))
print(min(R))
|
s146185463
|
Accepted
| 17 | 3,064 | 805 |
H, W = list(map(int, input().split()))
def Diff(H, W):
H_min = int(H / 3)
H_max = int(H / 3) + 1
W_short = int(W / 2)
W_long = W - W_short
S_1_1 = [H_min * W, (H - H_min) * W_short, (H - H_min) * W_long]
S_1_2 = [H_max * W, (H - H_max) * W_short, (H - H_max) * W_long]
D_1_1 = max(S_1_1) - min(S_1_1)
D_1_2 = max(S_1_2) - min(S_1_2)
return min(D_1_1, D_1_2)
R = []
R.append(H)
R.append(W)
R.append(Diff(H, W))
R.append(Diff(W, H))
if H % 3 == 0 or W % 3 == 0:
print(0)
else:
print(min(R))
|
s950530329
|
p03556
|
u018679195
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 67 |
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.
|
if __name__=="main":
i=int(input())
print((int(i**0.5))**2)
|
s172346272
|
Accepted
| 21 | 3,316 | 566 |
import sys
def fastio():
from io import StringIO
from atexit import register
global input
sys.stdin = StringIO(sys.stdin.read())
input = lambda : sys.stdin.readline().rstrip('\r\n')
sys.stdout = StringIO()
register(lambda : sys.__stdout__.write(sys.stdout.getvalue()))
fastio()
INF = 10**20
MOD = 10**9 + 7
I = lambda:list(map(int,input().split()))
# from math import gcd
from math import ceil
from collections import defaultdict as dd, Counter
from bisect import bisect_left as bl, bisect_right as br
n, = I()
print(int(n ** 0.5) ** 2)
|
s969454646
|
p02601
|
u342801789
| 2,000 | 1,048,576 |
Wrong Answer
| 36 | 9,212 | 437 |
M-kun has the following three cards: * A red card with the integer A. * A green card with the integer B. * A blue card with the integer C. He is a genius magician who can do the following operation at most K times: * Choose one of the three cards and multiply the written integer by 2. His magic is successful if both of the following conditions are satisfied after the operations: * The integer on the green card is **strictly** greater than the integer on the red card. * The integer on the blue card is **strictly** greater than the integer on the green card. Determine whether the magic can be successful.
|
numbers = [int(s) for s in input().split()]
K = int(input())
for i in range(K):
if(numbers[2] < numbers[0]):
numbers[2] *= 2
elif(numbers[1] < numbers[0]):
numbers[1] *= 2
elif(numbers[2] < numbers[1]):
numbers[2] *= 2
else:
break
print(numbers)
if(numbers[1] > numbers[0] and numbers[2] > numbers[1]):
print("Yes")
else:
print("No")
|
s659435844
|
Accepted
| 31 | 9,204 | 398 |
numbers = [int(s) for s in input().split()]
K = int(input())
for i in range(K):
if(numbers[2] <= numbers[0]):
numbers[2] *= 2
elif(numbers[1] <= numbers[0]):
numbers[1] *= 2
elif(numbers[2] <= numbers[1]):
numbers[2] *= 2
else:
break
if(numbers[1] > numbers[0] and numbers[2] > numbers[1]):
print("Yes")
else:
print("No")
|
s326606951
|
p04012
|
u218838821
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 358 |
Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful.
|
w = list(map(str,input()))
w_ord = []
for i in range(len(w)):
w_ord.append(ord(w[i]))
w_ord.sort()
j = 1
if len(w) % 2 == 1:
print("NO")
else:
for i in range((len(w)+1)//2):
if not w_ord[2*i] == w_ord[2*i + 1]:
print("NO")
break
else:
if i == (len(w)+1)//2 -1:
print("YES")
|
s621887563
|
Accepted
| 27 | 9,040 | 226 |
w = input()
W = []
for i in range(len(w)):
W.append(w[i])
W.sort()
if len(w) % 2 == 1:
print("No")
exit()
for i in range(len(w)//2):
if not W[2*i] == W[2*i+1]:
print("No")
exit()
print("Yes")
|
s125277438
|
p02260
|
u362520072
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,592 | 325 |
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.
|
def selectionSort(a, n):
count = 0
for i in range(0, n):
minj = i
for j in range(i, n):
if a[j] < a[minj]:
minj = j
w = a[i]
a[i] = a[minj]
a[minj] = w
count += 1
print(' '.join(map(str, a)))
print(count)
n = int(input())
a = list(map(int, input().split()))
selectionSort(a, n)
|
s756476032
|
Accepted
| 20 | 5,600 | 351 |
def selectionSort(a, n):
count = 0
for i in range(0, n):
minj = i
for j in range(i, n):
if a[j] < a[minj]:
minj = j
if i != minj:
w = a[i]
a[i] = a[minj]
a[minj] = w
count += 1
print(' '.join(map(str, a)))
print(count)
n = int(input())
a = list(map(int, input().split()))
selectionSort(a, n)
|
s209890526
|
p03543
|
u268516119
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 75 |
We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**?
|
A=input()
print('NYOE S'[(A[1]==A[2]) and
(A[0]==A[1] or A[1]==A[3])
::2])
|
s394734070
|
Accepted
| 17 | 2,940 | 75 |
A=input()
print('NYoe s'[(A[1]==A[2]) and
(A[0]==A[1] or A[1]==A[3])
::2])
|
s986004725
|
p03377
|
u642823003
| 2,000 | 262,144 |
Wrong Answer
| 20 | 2,940 | 144 |
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
a, b, x = map(int, input().split())
if x - a < 0:
print("No")
else:
if b < x - a:
print("No")
else:
print("Yes")
|
s204004828
|
Accepted
| 17 | 2,940 | 144 |
a, b, x = map(int, input().split())
if x - a < 0:
print("NO")
else:
if b < x - a:
print("NO")
else:
print("YES")
|
s722506308
|
p03110
|
u832871520
| 2,000 | 1,048,576 |
Wrong Answer
| 39 | 5,296 | 1,183 |
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?
|
import sys
import math
from collections import Counter
import itertools
import fractions
import datetime
from decimal import Decimal
#from functools import reduce
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SR(): return sys.stdin.readline().rstrip()
ascii_lowercase = 'abcdefghijklmnopqrstuvwxyz'
ascii_uppercase = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
ascii_uppercase2 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZABCDEFGHIJKLMNOPQRSTUVWXYZ'
p2D = lambda x: print(*x, sep="\n")
p2E = lambda x: print(''.join(x))
p2S = lambda x: print(*x, sep=" ")
# ###########################################
N=II()
XU=[list(SR().split()) for _ in range(N)]
print(XU)
ans = 0
for xu in XU:
if xu[1] == 'JPY':
ans += int(xu[0])
else:
ans += float(xu[0])*380000
print(ans)
|
s113242577
|
Accepted
| 37 | 5,296 | 1,173 |
import sys
import math
from collections import Counter
import itertools
import fractions
import datetime
from decimal import Decimal
#from functools import reduce
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SR(): return sys.stdin.readline().rstrip()
ascii_lowercase = 'abcdefghijklmnopqrstuvwxyz'
ascii_uppercase = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
ascii_uppercase2 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZABCDEFGHIJKLMNOPQRSTUVWXYZ'
p2D = lambda x: print(*x, sep="\n")
p2E = lambda x: print(''.join(x))
p2S = lambda x: print(*x, sep=" ")
# ###########################################
N=II()
XU=[list(SR().split()) for _ in range(N)]
ans = 0
for xu in XU:
if xu[1] == 'JPY':
ans += int(xu[0])
else:
ans += float(xu[0])*380000
print(ans)
|
s507183757
|
p03302
|
u340947941
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 10 |
You are given two integers a and b. Determine if a+b=15 or a\times b=15 or neither holds. Note that a+b=15 and a\times b=15 do not hold at the same time.
|
print("a")
|
s223284992
|
Accepted
| 18 | 2,940 | 119 |
a,b = list(map(int, input().split()))
if a+b == 15:
print("+")
elif a*b == 15:
print("*")
else:
print("x")
|
s370523653
|
p03854
|
u120810144
| 2,000 | 262,144 |
Wrong Answer
| 210 | 14,800 | 541 |
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`.
|
import sys
import math
import numpy as np
import copy
def main():
s = input()
a, b, c, d = "dream", "dreamer", "erase", "eraser"
ra, rb, rc, rd = a[::-1], b[::-1], c[::-1], d[::-1]
rs = s[::-1]
while rs:
flag = False
for ele in [ra, rb, rc, rd]:
if rs.startswith(ele):
rs = rs.replace(ele, "", 1)
flag = True
if not flag:
print("No")
return 0
print("Yes")
return 0
if __name__ == '__main__':
sys.exit(main())
|
s858868761
|
Accepted
| 208 | 13,216 | 541 |
import sys
import math
import numpy as np
import copy
def main():
s = input()
a, b, c, d = "dream", "dreamer", "erase", "eraser"
ra, rb, rc, rd = a[::-1], b[::-1], c[::-1], d[::-1]
rs = s[::-1]
while rs:
flag = False
for ele in [ra, rb, rc, rd]:
if rs.startswith(ele):
rs = rs.replace(ele, "", 1)
flag = True
if not flag:
print("NO")
return 0
print("YES")
return 0
if __name__ == '__main__':
sys.exit(main())
|
s141190747
|
p03657
|
u797674884
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 196 |
Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.
|
l = list(map(int, input().split()))
if l[0] % 3 == 0:
print("Possible")
elif l[1] % 3 == 0:
print("Possible")
elif l[0]+l[1] % 3 == 0:
print("Possible")
else:
print("Impossible")
|
s056732088
|
Accepted
| 17 | 2,940 | 202 |
l = list(map(int, input().split()))
a = l[0]+l[1]
if l[0] % 3 == 0:
print("Possible")
elif l[1] % 3 == 0:
print("Possible")
elif a % 3 == 0:
print("Possible")
else:
print("Impossible")
|
s897646900
|
p02408
|
u811841526
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,896 | 366 |
Taro is going to play a card game. However, now he has only n cards, even though there should be 52 cards (he has no Jokers). The 52 cards include 13 ranks of each of the four suits: spade, heart, club and diamond.
|
from collections import OrderedDict
cards = OrderedDict()
cards['S'] = set()
cards['H'] = set()
cards['D'] = set()
cards['C'] = set()
n = int(input())
for _ in range(n):
suit, num = input().split()
num = int(num)
cards[suit].add(num)
for suit, nums in cards.items():
for num in range(1,14):
if num not in nums:
print(suit, num)
|
s203379100
|
Accepted
| 20 | 5,604 | 202 |
n = int(input())
deck = []
for i in range(n):
deck.append(input())
for suit in 'SHCD':
for i in range(1, 14):
card = f'{suit} {i}'
if card not in deck:
print(card)
|
s711109460
|
p02600
|
u514229192
| 2,000 | 1,048,576 |
Wrong Answer
| 23 | 9,088 | 87 |
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?
|
x=float(input("m?"))
for i in range(8):
if 400+200*i<=x<400+200*(i+1):
print(8-i)
|
s276963014
|
Accepted
| 32 | 9,056 | 83 |
x=float(input())
for i in range(8):
if 400+200*i<=x<400+200*(i+1):
print(8-i)
|
s083594437
|
p02409
|
u286589639
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,620 | 375 |
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())
S = [[[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())
S[b-1][f-1][r-1] += v
for i in range(4):
for j in range(3):
for k in range(10):
print(" " + str(S[i][j][k]), end = "")
print("")
if i == 4:
break
print("####################")
|
s294442556
|
Accepted
| 20 | 7,756 | 375 |
n =int(input())
S = [[[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())
S[b-1][f-1][r-1] += v
for i in range(4):
for j in range(3):
for k in range(10):
print(" " + str(S[i][j][k]), end = "")
print("")
if i == 3:
break
print("####################")
|
s903630385
|
p02612
|
u726823037
| 2,000 | 1,048,576 |
Wrong Answer
| 28 | 9,160 | 204 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
import sys
def Ii():return int(sys.stdin.readline())
def Mi():return map(int,sys.stdin.buffer.readline().split())
def Li():return list(map(int,sys.stdin.buffer.readline().split()))
n = Ii()
print(n%1000)
|
s774715013
|
Accepted
| 26 | 9,180 | 244 |
import sys
def Ii():return int(sys.stdin.readline())
def Mi():return map(int,sys.stdin.buffer.readline().split())
def Li():return list(map(int,sys.stdin.buffer.readline().split()))
n = Ii()
if n%1000 == 0:
print(0)
else:
print(1000-n%1000)
|
s903704301
|
p02612
|
u486536494
| 2,000 | 1,048,576 |
Wrong Answer
| 36 | 10,020 | 110 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
import datetime
import string
import re
import math
N = int(input())
while N > 1000:
N -= 1000
print(N)
|
s578999802
|
Accepted
| 38 | 10,020 | 117 |
import datetime
import string
import re
import math
N = int(input())
while N > 1000:
N -= 1000
print(1000 - N)
|
s086879457
|
p03448
|
u735468069
| 2,000 | 262,144 |
Wrong Answer
| 43 | 3,316 | 452 |
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.
|
from collections import defaultdict, Counter
from heapq import heapify, heappop, heappush
from sys import stdin
def main():
A = int(input())
B = int(input())
C = int(input())
x = int(input())
ans = 0
for a in range(1, A+1):
for b in range(1, B+1):
for c in range(1, C+1):
if 500*a+50*b+10*c == x:
ans += 1
print(ans)
input = lambda: stdin.readline().rstrip()
main()
|
s075035520
|
Accepted
| 47 | 3,444 | 453 |
from collections import defaultdict, Counter
from heapq import heapify, heappop, heappush
from sys import stdin
def main():
A = int(input())
B = int(input())
C = int(input())
x = int(input())
ans = 0
for a in range(0, A+1):
for b in range(0, B+1):
for c in range(0, C+1):
if 500*a+100*b+50*c == x:
ans += 1
print(ans)
input = lambda: stdin.readline().rstrip()
main()
|
s338608236
|
p02613
|
u730807152
| 2,000 | 1,048,576 |
Wrong Answer
| 148 | 9,204 | 365 |
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())
countAC=0
countWA=0
countTLE=0
countRE=0
for i in range(n):
s=input()
if s=="AC":
countAC += 1
elif s=="WA":
countWA += 1
elif s=="TLE":
countTLE += 1
else:
countRE += 1
print("AC × " + str(countAC))
print("WA × " + str(countWA))
print("TLE × " + str(countTLE))
print("RE × " + str(countRE))
|
s795808474
|
Accepted
| 141 | 9,200 | 361 |
n=int(input())
countAC=0
countWA=0
countTLE=0
countRE=0
for i in range(n):
s=input()
if s=="AC":
countAC += 1
elif s=="WA":
countWA += 1
elif s=="TLE":
countTLE += 1
else:
countRE += 1
print("AC x " + str(countAC))
print("WA x " + str(countWA))
print("TLE x " + str(countTLE))
print("RE x " + str(countRE))
|
s229960700
|
p03730
|
u629540524
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 101 |
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())
print('No' if all((a*i)%b != c for i in range(1,b+1)) else 'Yes')
|
s066060342
|
Accepted
| 17 | 2,940 | 101 |
a, b, c = map(int, input().split())
print('NO' if all((a*i)%b != c for i in range(1,b+1)) else 'YES')
|
s438003536
|
p02613
|
u193927973
| 2,000 | 1,048,576 |
Wrong Answer
| 148 | 9,192 | 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())
a=[]
d=dict(AC=0, WA=0, TLE=0, RE=0)
for _ in range(N):
a=input()
d[a]+=1
for k, v in d.items():
ans=k+" × "+str(v)
print(ans)
|
s217620629
|
Accepted
| 148 | 9,192 | 149 |
N=int(input())
a=[]
d=dict(AC=0, WA=0, TLE=0, RE=0)
for _ in range(N):
a=input()
d[a]+=1
for k, v in d.items():
ans=k+" x "+str(v)
print(ans)
|
s973555942
|
p02833
|
u518042385
| 2,000 | 1,048,576 |
Wrong Answer
| 2,104 | 3,060 | 163 |
For an integer n not less than 0, let us define f(n) as follows: * f(n) = 1 (if n < 2) * f(n) = n f(n-2) (if n \geq 2) Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N).
|
n=int(input())
if n%2==1:
print(0)
else:
now=1
ans=0
while 1:
if n%(10**now)!=0:
break
else:
ans+=n%(10**now)
now+=1
print(ans)
|
s998192716
|
Accepted
| 17 | 2,940 | 176 |
n=int(input())
if n%2==1:
print(0)
else:
now=1
ans=0
while 1:
if (n//(5**now))//2==0:
break
else:
ans+=(n//(5**now))//2
now+=1
print(ans)
|
s917275184
|
p03564
|
u453642820
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 103 |
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.
|
N=int(input())
K=int(input())
ans=[]
for i in range(N):
ans.append((i+1)*2+(N-i-1))
print(min(ans))
|
s284964147
|
Accepted
| 17 | 2,940 | 90 |
N=int(input())
K=int(input())
ans=1
for i in range(N):
ans=min(ans*2,ans+K)
print(ans)
|
s504757039
|
p02396
|
u042882066
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,524 | 48 |
In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem.
|
x=int(input())
if x != 0 :
print("Case i: ", x)
|
s990639317
|
Accepted
| 140 | 7,492 | 99 |
i = 0
while True:
n = int(input())
i += 1
if n == 0:
break
print("Case {}: {}".format(i, n))
|
s986607129
|
p02833
|
u821432765
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 209 |
For an integer n not less than 0, let us define f(n) as follows: * f(n) = 1 (if n < 2) * f(n) = n f(n-2) (if n \geq 2) Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N).
|
N = int(input())
if N%2==1:
print(0)
quit()
x = N
fives = 0
i = 0
while x:
if i % 2 == 1:
i +=1
continue
i += 1
x //= 5
fives += x
print(x, fives)
print(fives)
|
s756331232
|
Accepted
| 17 | 2,940 | 150 |
N = int(input())
if N%2==1:
print(0)
quit()
x = N
fives = 0
while x:
x //= 5
fives += (x//2)
# print(x, fives)
print(fives)
|
s128081691
|
p03338
|
u331036636
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 4 |
You are given a string S of length N consisting of lowercase English letters. We will cut this string at one position into two strings X and Y. Here, we would like to maximize the number of different letters contained in both X and Y. Find the largest possible number of different letters contained in both X and Y when we cut the string at the optimal position.
|
s195572359
|
Accepted
| 19 | 3,064 | 256 |
n = int(input())
s = input()
c = 0
list_s = [sorted(list(set(s[:i]))) for i in range(1,n)]
list_l = [sorted(list(set(s[i:]))) for i in range(1,n)]
print(max([sum([1 if list_s[j][i] in list_l[j] else 0 for i in range(len(list_s[j]))]) for j in range(n-1)]))
|
|
s369167552
|
p03778
|
u079022116
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 121 |
AtCoDeer the deer found two rectangles lying on the table, each with height 1 and width W. If we consider the surface of the desk as a two-dimensional plane, the first rectangle covers the vertical range of AtCoDeer will move the second rectangle horizontally so that it connects with the first rectangle. Find the minimum distance it needs to be moved.
|
w,a,b=map(int,input().split())
if min(a,b) + w >= max(a,b):
print(0)
else:
ans = max(a,b) - min(a,b) + w
print(ans)
|
s600363905
|
Accepted
| 17 | 2,940 | 122 |
w,a,b=map(int,input().split())
if min(a,b) + w >= max(a,b):
print(0)
else:
ans = max(a,b) - min(a,b) - w
print(ans)
|
s419704104
|
p03543
|
u414558682
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 148 |
We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**?
|
N = input()
print(N)
if N[0] == N[1] and N[1] == N[2]:
print('Yes')
elif N[1] == N[2] and N[2] == N[3]:
print('Yes')
else:
print('No')
|
s832628882
|
Accepted
| 17 | 3,060 | 185 |
# coding: utf-8
# Your code here!
N = input()
# print(N)
if N[0] == N[1] and N[1] == N[2]:
print('Yes')
elif N[1] == N[2] and N[2] == N[3]:
print('Yes')
else:
print('No')
|
s410309666
|
p03523
|
u409306788
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 331 |
You are given a string S. Takahashi can insert the character `A` at any position in this string any number of times. Can he change S into `AKIHABARA`?
|
import sys
input = sys.stdin.readline
# A - AKIBA
substring = ['', 'KIH', 'B', 'R']
akihabara = []
for i in range(16):
b = format(i, 'b')
bin_str = b.zfill(4)
s = ''
for j in range(4):
s += substring[j]
if bin_str[j] == '1':
s += 'A'
akihabara.append(s)
if input() in akihabara:
print('YES')
else:
print('NO')
|
s925474528
|
Accepted
| 17 | 3,060 | 292 |
# A - AKIBA
substring = ['', 'KIH', 'B', 'R']
akihabara = []
for i in range(16):
b = format(i, 'b')
bin_str = b.zfill(4)
s = ''
for j in range(4):
s += substring[j]
if bin_str[j] == '1':
s += 'A'
akihabara.append(s)
if input() in akihabara:
print('YES')
else:
print('NO')
|
s759581607
|
p03228
|
u734876600
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 247 |
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 == 1:
if a % 2 == 1:
a = (a - 1) // 2
b += a
else:
if b % 2 == 1:
b = (b - 1) // 2
a += b
print(a,b)
|
s793866193
|
Accepted
| 18 | 3,060 | 260 |
a,b,k = map(int,input().split())
for i in range(k):
if i % 2 == 0:
if a % 2 == 1:
a -= 1
a = a // 2
b += a
else:
if b % 2 == 1:
b = b - 1
b = b // 2
a += b
print(a,b)
|
s351477532
|
p03962
|
u439392790
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 102 |
AtCoDeer the deer recently bought three paint cans. The color of the one he bought two days ago is a, the color of the one he bought yesterday is b, and the color of the one he bought today is c. Here, the color of each paint can is represented by an integer between 1 and 100, inclusive. Since he is forgetful, he might have bought more than one paint can in the same color. Count the number of different kinds of colors of these paint cans and tell him.
|
a,b,c=map(int, input().split())
if(a+b==c)or(b+c==a)or(a+c==b):
print("Yes")
else:
print("No")
|
s808649794
|
Accepted
| 18 | 2,940 | 142 |
a,b,c=map(int,input().split())
if a==b==c:
print(1)
elif a==b or a==c or b==c:
print(2)
elif a!=b and a!=c and b!=c:
print(3)
|
s290679320
|
p03457
|
u325282913
| 2,000 | 262,144 |
Wrong Answer
| 326 | 3,060 | 234 |
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.
|
# 10
import sys
N = int(input())
for _ in range(N):
t, x, y = map(int, input().split())
if ((x + y) % 2) != (t % 2):
print('NO')
sys.exit()
if t < x + y:
print('NO')
sys.exit()
print('YES')
|
s194620522
|
Accepted
| 323 | 3,060 | 234 |
# 10
import sys
N = int(input())
for _ in range(N):
t, x, y = map(int, input().split())
if ((x + y) % 2) != (t % 2):
print('No')
sys.exit()
if t < x + y:
print('No')
sys.exit()
print('Yes')
|
s240870998
|
p02255
|
u646461156
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,592 | 151 |
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.
|
n=int(input())
a=list(map(int,input().split()))
for i in range(1,n):
j=i-1
while j>=0 and a[j]>a[i]:
a[j+1]=a[j]
j-=1
a[j+1]=a[i]
print(*a)
|
s757557347
|
Accepted
| 20 | 5,980 | 151 |
n=int(input())
a=list(map(int,input().split()))
for i in range(n):
v=a[i]
j=i-1
while j>=0 and a[j]>v:
a[j+1]=a[j]
j-=1
a[j+1]=v
print(*a)
|
s109055771
|
p02613
|
u110311725
| 2,000 | 1,048,576 |
Wrong Answer
| 170 | 16,188 | 323 |
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 = []
c1 = 0
c2 = 0
c3 = 0
c4 = 0
for i in range(n):
s.append(input())
if s[i] == 'AC':
c1 += 1
elif s[i] == 'WA':
c2 += 1
elif s[i] == 'TLE':
c3 += 1
elif s[i] == 'RE':
c4 += 1
print('AC ×',c1)
print('WA ×',c2)
print('TLE ×',c3)
print('RE ×',c4)
|
s117522950
|
Accepted
| 162 | 16,332 | 319 |
n = int(input())
s = []
c1 = 0
c2 = 0
c3 = 0
c4 = 0
for i in range(n):
s.append(input())
if s[i] == 'AC':
c1 += 1
elif s[i] == 'WA':
c2 += 1
elif s[i] == 'TLE':
c3 += 1
elif s[i] == 'RE':
c4 += 1
print('AC x',c1)
print('WA x',c2)
print('TLE x',c3)
print('RE x',c4)
|
s324084946
|
p03556
|
u460245024
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 36 |
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())
print(int(N**0.5))
|
s942553225
|
Accepted
| 17 | 2,940 | 39 |
N = int(input())
print(int(N**0.5)**2)
|
s266593745
|
p03494
|
u256833330
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,060 | 209 |
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()))
c=0
while True:
for i in range(n):
if l[i]%2 !=0:
print(c)
exit(0)
else:
l[i] = l[i]//2
c+=1
|
s581515549
|
Accepted
| 19 | 2,940 | 201 |
n=int(input())
l=list(map(int,input().split()))
c=0
while True:
for i in range(n):
if l[i]%2 !=0:
print(c)
exit(0)
else:
l[i] = l[i]//2
c+=1
|
s404169096
|
p03433
|
u894348341
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 93 |
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
N = int(input())
A = int(input())
if (N - A) // 500 == 0:
print('Yes')
else:
print('No')
|
s469675863
|
Accepted
| 18 | 2,940 | 85 |
n = int(input())
a = int(input())
n%=500
if n<=a:
print('Yes')
else:
print('No')
|
s813263493
|
p02407
|
u350064373
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,456 | 146 |
Write a program which reads a sequence and prints it in the reverse order.
|
input()
list1 = list(map(int, input().split()))
list1.sort()
result=""
for i in range(0,len(list1)):
result += str(list1[i])+" "
print(result)
|
s632336955
|
Accepted
| 20 | 5,560 | 38 |
input()
print(*input().split()[::-1])
|
s249923495
|
p03380
|
u504562455
| 2,000 | 262,144 |
Wrong Answer
| 82 | 14,056 | 220 |
Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order. From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized. If there are multiple pairs that maximize the value, any of them is accepted.
|
import bisect
n = int(input())
a = [int(i) for i in input().split()]
a.sort()
ai = a[-1]
idx = bisect.bisect_left(a, ai//2)
if abs(ai//2-a[idx]) >= abs(ai//2-a[idx-1]):
print(ai, a[idx-1])
else:
print(ai, a[idx])
|
s940745510
|
Accepted
| 82 | 14,052 | 369 |
import bisect
n = int(input())
a = [int(i) for i in input().split()]
a.sort()
ai = a[-1]
idx = bisect.bisect_left(a, ai//2)
if abs(ai//2-a[idx]) == abs(ai//2-a[idx-1]):
if ai % 2 == 0 or idx == len(a)-1:
print(ai, a[idx-1])
else:
print(ai, a[idx])
elif abs(ai//2-a[idx]) > abs(ai//2-a[idx-1]):
print(ai, a[idx-1])
else:
print(ai, a[idx])
|
s277747236
|
p04043
|
u396495667
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 70 |
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,b,c = map(int, input().split())
print('Yes' if a+b+c ==17 else 'No')
|
s357706912
|
Accepted
| 18 | 3,060 | 98 |
a = [int(_) for _ in input().split()]
a.sort()
print('YES' if a[0]==a[1]==5 and a[2]==7 else 'NO')
|
s368067579
|
p03386
|
u773981351
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 145 |
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())
for i in range(a, min(a + k, b + 1)):
print(i)
for j in range(max(i + 1, b - k + 1), b + 1):
print(i)
|
s437353328
|
Accepted
| 17 | 3,060 | 145 |
a, b, k = map(int, input().split())
for i in range(a, min(a + k, b + 1)):
print(i)
for j in range(max(i + 1, b - k + 1), b + 1):
print(j)
|
s816242126
|
p03606
|
u969848070
| 2,000 | 262,144 |
Wrong Answer
| 20 | 2,940 | 101 |
Joisino is working as a receptionist at a theater. The theater has 100000 seats, numbered from 1 to 100000. According to her memo, N groups of audiences have come so far, and the i-th group occupies the consecutive seats from Seat l_i to Seat r_i (inclusive). How many people are sitting at the theater now?
|
n = int(input())
x = 0
for i in range(n):
a, b = map(int, input().split())
x += a - b +1
print(x)
|
s448178868
|
Accepted
| 20 | 2,940 | 99 |
n = int(input())
x = 0
for i in range(n):
a, b = map(int, input().split())
x += b-a +1
print(x)
|
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