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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s964804643
|
p03369
|
u785066634
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 129
|
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
s=list(input())
count=0
for s_ in s:
if s_ =="◯":
count=count+1
else:
count=count
print(700+100*count)
|
s912969681
|
Accepted
| 17
| 2,940
| 100
|
s=list(input())
count=0
for s_ in s:
if s_ =="o":
count=count+1
print(700+100*count)
|
s940770233
|
p03386
|
u696444274
| 2,000
| 262,144
|
Wrong Answer
| 36
| 5,148
| 291
|
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
import math
import itertools
import statistics
#import numpy as np
#import collections
a, b, k = list(map(int, input().split()))
l = []
if k > b-a:
k = b-a
for i in range(k):
l.append(a+i)
l.append(b-i)
l.sort()
ans = list(set(l))
for i in range(len(ans)):
print(ans[i])
|
s130323180
|
Accepted
| 37
| 5,148
| 341
|
import math
import itertools
import statistics
#import numpy as np
#import collections
a, b, k = list(map(int, input().split()))
l = []
for i in range(k):
l.append(a+i)
l.append(b-i)
ans = list(set(l))
ans.sort()
for i in range(len(ans)):
if ans[i] >= a and ans[i] <= b:
print(ans[i])
|
s039974930
|
p02606
|
u090883248
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 9,104
| 156
|
How many multiples of d are there among the integers between L and R (inclusive)?
|
input = input()
print(input)
L, R, d = [int(n) for n in input.split()]
count = 0
for n in range(L, R+1):
if n % d == 0:
count += 1
print(count)
|
s261549158
|
Accepted
| 31
| 9,040
| 143
|
input = input()
L, R, d = [int(n) for n in input.split()]
count = 0
for n in range(L, R+1):
if n % d == 0:
count += 1
print(count)
|
s057270230
|
p03556
|
u004025573
| 2,000
| 262,144
|
Wrong Answer
| 50
| 3,060
| 130
|
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.
|
import math
N=int(input())
for i in range(N,-1,-1):
a=math.ceil(math.sqrt(i))
if a*a==i:
print(a)
break
|
s770734874
|
Accepted
| 46
| 2,940
| 130
|
import math
N=int(input())
for i in range(N,-1,-1):
a=math.ceil(math.sqrt(i))
if a*a==i:
print(i)
break
|
s751525838
|
p04044
|
u745385679
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 45
|
Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.
|
a = input().split()
print(''.join(sorted(a)))
|
s229053431
|
Accepted
| 17
| 3,060
| 82
|
N, L = map(int, input().split())
print(''.join(sorted(input() for _ in range(N))))
|
s651744975
|
p03251
|
u442877951
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,060
| 207
|
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())
xl = list(map(int,input().split()))
yl = list(map(int,input().split()))
for i in range(X+1,Y):
if max(xl) < i and min(yl) >= i:
print("NoWar")
exit()
print("War")
|
s596564945
|
Accepted
| 18
| 3,060
| 208
|
N,M,X,Y = map(int,input().split())
xl = list(map(int,input().split()))
yl = list(map(int,input().split()))
for i in range(X+1,Y):
if max(xl) < i and min(yl) >= i:
print("No War")
exit()
print("War")
|
s240365898
|
p03737
|
u453634575
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 77
|
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
|
s = list(input().split())
print(s[0][0].upper, s[1][0].upper, s[2][0].upper)
|
s952987189
|
Accepted
| 17
| 2,940
| 86
|
s = list(input().split())
print(s[0][0].upper() + s[1][0].upper() + s[2][0].upper())
|
s933131064
|
p03623
|
u397563544
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,316
| 330
|
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|.
|
from collections import Counter
s = list(sorted(str(input())))
l = list(Counter(s))
a = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
if len(l) == 26:
print('None')
else:
for i in range(len(l)):
if l[i]!=a[i]:
print(a[i])
break
|
s161078988
|
Accepted
| 18
| 2,940
| 90
|
x,a,b = map(int,input().split())
if abs(x-a)<abs(x-b):
print('A')
else:
print('B')
|
s289581156
|
p03339
|
u748311048
| 2,000
| 1,048,576
|
Wrong Answer
| 161
| 25,844
| 234
|
There are N people standing in a row from west to east. Each person is facing east or west. The directions of the people is given as a string S of length N. The i-th person from the west is facing east if S_i = `E`, and west if S_i = `W`. You will appoint one of the N people as the leader, then command the rest of them to face in the direction of the leader. Here, we do not care which direction the leader is facing. The people in the row hate to change their directions, so you would like to select the leader so that the number of people who have to change their directions is minimized. Find the minimum number of people who have to change their directions.
|
n = int(input())
s = str(input())
sE = s.count('E')
sW = s.count('W')
l = []
e, w = 0, 0
for ch in s:
if ch == 'E':
e += 1
else:
w += 1
m = (sE-e)+w
l.append(sE-e+w-1 if ch == 'W' else sE-e+w)
print(l)
|
s519957608
|
Accepted
| 147
| 21,432
| 239
|
n = int(input())
s = str(input())
sE = s.count('E')
sW = s.count('W')
l = []
e, w = 0, 0
for ch in s:
if ch == 'E':
e += 1
else:
w += 1
m = (sE-e)+w
l.append(sE-e+w-1 if ch == 'W' else sE-e+w)
print(min(l))
|
s408650588
|
p03456
|
u663089555
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 132
|
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.
|
A=input().split()
B=int(A[0]+A[1])
print(B)
i=0
while i**2<=B:
if i**2==B:
print(i)
break
else:
i+=1
|
s019411256
|
Accepted
| 17
| 2,940
| 76
|
A=input().split()
B=(int(A[0]+A[1]))**.5
print('Yes' if B==int(B) else 'No')
|
s209399407
|
p00041
|
u811733736
| 1,000
| 131,072
|
Wrong Answer
| 40
| 7,812
| 2,338
|
与えられた 4 つの 1 から 9 の整数を使って、答えが 10 になる式をつくります。 4 つの整数 a, b, c, d を入力したとき、下記の条件に従い、答えが 10 になる式を出力するプログラムを作成してください。また、答えが複数ある時は、最初に見つかった答えだけを出力するものとします。答えがない時は、0 と出力してください。 * 演算子として、加算 (+)、減算 (-)、乗算 (*) だけを使います。除算 (/) は使いません。使用できる演算子は３個です。 * 数を4つとも使わなければいけません。 * 4つの数の順番は自由に入れ換えてかまいません。 * カッコを使ってもかまいません。使用できるカッコは３組（６個）以下です。
|
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0041
"""
import sys
from itertools import permutations, product
def has_possibility(digits):
result = False
digits.sort(reverse=True)
total = digits[0]
for i in digits[1:]:
if i < 2:
total += i
else:
total *= i
if total >= 10:
result = True
return result
def make_ten(digits):
for a, b, c, d in permutations(digits, 4):
for op1, op2, op3 in product(['+', '-', '*'], repeat=3):
t1 = do_calc(a, b, op1)
t2 = do_calc(t1, c, op2)
t3 = do_calc(t2, d, op3)
if t3 == 10:
return '(({} {} {}) {} {}) {} {}'.format(a, op1, b, op2, c, op3, d)
t1 = do_calc(a, b, op1)
t2 = do_calc(c, d, op2)
t3 = do_calc(t1, t2, op3)
if t3 == 10:
return '({} {} {}) {} ({} {} {})'.format(a, op1, b, op3, c, op2, d)
return '0'
def do_calc(x, y, op):
result = None
if op == '+':
result = x + y
elif op == '-':
result = x - y
elif op == '*':
result = x * y
return result
Memo = {}
def solve(digits):
digits.sort(reverse=True)
data = ''.join(map(str, digits))
if data in Memo:
pass
else:
if has_possibility(digits):
Memo[data] = make_ten(digits)
else:
Memo[data] = '0'
return Memo[data]
def main(args):
while True:
digits = [int(x) for x in input().strip().split(' ')]
if digits[0] == 0 and digits[1] == 0 and digits[2] == 0 and digits[3] == 0:
break
result = solve(digits)
print(result)
if __name__ == '__main__':
main(sys.argv[1:])
|
s155409515
|
Accepted
| 550
| 7,872
| 2,813
|
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0041
"""
import sys
from itertools import permutations, product
def make_ten(digits):
result = '0'
for a, b, c, d in permutations(digits, 4):
if result != '0':
break
for op1, op2, op3 in product(['+', '-', '*'], repeat=3): # ??????????????????????????????????????????????????????OK
exp = '(({} {} {}) {} {}) {} {}'.format(a, op1, b, op2, c, op3, d)
if eval(exp) == 10:
result = exp
break
# 1, 3, 2
exp= '(({} {} {}) {} ({} {} {}))'.format(a, op1, b, op2, c, op3, d)
if eval(exp) == 10:
result = exp
break
# 2, 1, 3
exp = '(({} {} ({} {} {})) {} {})'.format(a, op1, b, op2, c, op3, d)
if eval(exp) == 10:
result = exp
break
# 3, 1, 2
exp = '({} {} (({} {} {}) {} {}))'.format(a, op1, b, op2, c, op3, d)
if eval(exp) == 10:
result = exp
break
# 3, 2, 1
exp = '({} {} ({} {} ({} {} {})))'.format(a, op1, b, op2, c, op3, d)
if eval(exp) == 10:
result = exp
break
# 2, 3, 1
exp = '(({} {} {}) {} ({} {} {}))'.format(a, op1, b, op2, c, op3, d)
if eval(exp) == 10:
result = exp
break
return result
Memo = {}
def solve(digits):
global Memo
digits.sort(reverse=True)
data = ''.join(map(str, digits))
if not data in Memo:
Memo[data] = make_ten(digits)
return Memo[data]
def main(args):
while True:
digits = [int(x) for x in input().strip().split(' ')]
# digits = [int(x) for x in '{:04d}'.format(count)]
if digits[0] == 0 and digits[1] == 0 and digits[2] == 0 and digits[3] == 0:
break
result = solve(digits)
print(result)
if __name__ == '__main__':
main(sys.argv[1:])
|
s410041463
|
p03556
|
u941884460
| 2,000
| 262,144
|
Wrong Answer
| 22
| 3,064
| 81
|
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())
count = 1
while count*count <N:
count += 1
print(pow(count,2))
|
s193323666
|
Accepted
| 24
| 2,940
| 90
|
N = int(input())
count = 1
while (count+1)*(count+1) <=N:
count += 1
print(pow(count,2))
|
s616348335
|
p03438
|
u060392346
| 2,000
| 262,144
|
Wrong Answer
| 24
| 4,596
| 172
|
You are given two integer sequences of length N: a_1,a_2,..,a_N and b_1,b_2,..,b_N. Determine if we can repeat the following operation zero or more times so that the sequences a and b become equal. Operation: Choose two integers i and j (possibly the same) between 1 and N (inclusive), then perform the following two actions **simultaneously** : * Add 2 to a_i. * Add 1 to b_j.
|
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
for i in range(n):
if a[i] > b[i]:
print("No")
break
else:
print("Yes")
|
s193530308
|
Accepted
| 26
| 4,596
| 241
|
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
u , v = 0, 0
for i in range(n):
if a[i] > b[i]:
u += a[i] - b[i]
else:
v += (b[i] - a[i]) // 2
if u > v:
print("No")
else:
print("Yes")
|
s952779915
|
p03456
|
u777923818
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,828
| 168
|
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.
|
# -*- coding: utf-8 -*-
a, b = map(int, input().split())
x = int(a+b)
ok = [False]*(100100)
for i in range(1, 317):
ok[i*i] = True
print("Yes" if ok[x] else "No")
|
s732387947
|
Accepted
| 19
| 3,828
| 160
|
# -*- coding: utf-8 -*-
a, b = input().split()
x = int(a+b)
ok = [False]*(100101)
for i in range(1, 317):
ok[i*i] = True
print("Yes" if ok[x] else "No")
|
s401326106
|
p03455
|
u694382289
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 93
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = map(int, input().split())
c= a*b
if c%2 == 0:
print('even')
else:
print('odd')
|
s831743745
|
Accepted
| 17
| 2,940
| 93
|
a, b = map(int, input().split())
c= a*b
if c%2 == 0:
print('Even')
else:
print('Odd')
|
s319147403
|
p03455
|
u738622346
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 99
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
sum_input = list(map(int, input().split(" ")))
print("Even" if sum(sum_input) % 2 == 0 else "Odd")
|
s295109630
|
Accepted
| 17
| 2,940
| 80
|
a, b = map(int, input().split(" "))
print("Even" if (a * b) % 2 == 0 else "Odd")
|
s239256518
|
p03044
|
u089376182
| 2,000
| 1,048,576
|
Wrong Answer
| 589
| 43,492
| 284
|
We have a tree with N vertices numbered 1 to N. The i-th edge in the tree connects Vertex u_i and Vertex v_i, and its length is w_i. Your objective is to paint each vertex in the tree white or black (it is fine to paint all vertices the same color) so that the following condition is satisfied: * For any two vertices painted in the same color, the distance between them is an even number. Find a coloring of the vertices that satisfies the condition and print it. It can be proved that at least one such coloring exists under the constraints of this problem.
|
n = int(input())
node_manager = {i:0 for i in range(1, n+1)}
UVW = sorted([list(map(int, input().split())) for _ in range(n-1)])
for uvw in UVW:
u,v,w = uvw
if w%2==0:
node_manager[v] = node_manager[u]
else:
node_manager[v] = 1-node_manager[u]
print(node_manager)
|
s529144822
|
Accepted
| 941
| 68,220
| 495
|
n = int(input())
node = {i:[] for i in range(1, n+1)}
ans = {i:0 for i in range(1, n+1)}
for _ in range(n-1):
u,v,w = map(int, input().split())
node[u].append([v,w])
node[v].append([u,w])
stack = [1]
memory = set()
while stack:
x = stack.pop()
memory.add(x)
for node_i, len_i in node[x]:
if node_i not in memory:
if len_i%2==0:
ans[node_i] = ans[x]
else:
ans[node_i] = 1-ans[x]
stack.append(node_i)
for v in ans.values():
print(v)
|
s502470362
|
p02602
|
u512873531
| 2,000
| 1,048,576
|
Wrong Answer
| 2,206
| 31,360
| 186
|
M-kun is a student in Aoki High School, where a year is divided into N terms. There is an exam at the end of each term. According to the scores in those exams, a student is given a grade for each term, as follows: * For the first through (K-1)-th terms: not given. * For each of the K-th through N-th terms: the multiplication of the scores in the last K exams, including the exam in the graded term. M-kun scored A_i in the exam at the end of the i-th term. For each i such that K+1 \leq i \leq N, determine whether his grade for the i-th term is **strictly** greater than the grade for the (i-1)-th term.
|
n, k = map(int, input().split())
a = list(map(int, input().split()))
g = 1
c = 0
for i in range(n):
g*=a[i]
if i>k:
if g>c: print('Yes')
else: print('No')
c=g
|
s929242844
|
Accepted
| 149
| 31,544
| 160
|
n, k = map(int, input().split())
a = list(map(int, input().split()))
for i in range(n):
if i>=k:
if a[i]>a[i-k]: print('Yes')
else: print('No')
|
s066516208
|
p03795
|
u272377260
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 48
|
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
n = int(input())
print(n * 800 + (n % 15) * 200)
|
s239044319
|
Accepted
| 17
| 2,940
| 49
|
n = int(input())
print(n * 800 - (n // 15) * 200)
|
s379120142
|
p03696
|
u374802266
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 291
|
You are given a string S of length N consisting of `(` and `)`. Your task is to insert some number of `(` and `)` into S to obtain a _correct bracket sequence_. Here, a correct bracket sequence is defined as follows: * `()` is a correct bracket sequence. * If X is a correct bracket sequence, the concatenation of `(`, X and `)` in this order is also a correct bracket sequence. * If X and Y are correct bracket sequences, the concatenation of X and Y in this order is also a correct bracket sequence. * Every correct bracket sequence can be derived from the rules above. Find the shortest correct bracket sequence that can be obtained. If there is more than one such sequence, find the lexicographically smallest one.
|
n=int(input())
s=input()+"1"
a=0
b=[]
for i in range(n):
if s[i]==s[i+1]:
a+=1
else:
b.append(a+1)
a=0
if s[-2]=='(':
b.append(0)
if s[0]==')':
b.insert(0,0)
print(b)
for i in range(len(b)//2):
c=max(b[2*i],b[2*i+1])
print('('*c,end=')'*c)
|
s374273352
|
Accepted
| 18
| 2,940
| 178
|
n=int(input())
s=input()
a=0
b=0
for i in range(n):
if s[i]=='(':
a+=1
else:
if a==0:
b+=1
else:
a-=1
print('('*b+s+')'*a)
|
s948058796
|
p02612
|
u945405878
| 2,000
| 1,048,576
|
Wrong Answer
| 33
| 9,092
| 108
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
a = N // 1000
b = N % 1000
if b == 0:
ans = 0
else:
ans = N - a * 1000
print(ans)
|
s678195221
|
Accepted
| 30
| 9,156
| 114
|
N = int(input())
a = N // 1000
b = N % 1000
if b == 0:
ans = 0
else:
ans = (a + 1) * 1000 - N
print(ans)
|
s194180092
|
p03502
|
u031358594
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 133
|
An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10. Given an integer N, determine whether it is a Harshad number.
|
N=str(input())
print(N)
X=int(N)
fx=0
for i in range(0,len(N)):
fx+=int(N[i])
if X%fx==0:
print("Yes")
else:
print("No")
|
s942621088
|
Accepted
| 17
| 2,940
| 125
|
N=str(input())
X=int(N)
fx=0
for i in range(0,len(N)):
fx+=int(N[i])
if X%fx==0:
print("Yes")
else:
print("No")
|
s055139776
|
p03448
|
u479638406
| 2,000
| 262,144
|
Wrong Answer
| 59
| 3,060
| 296
|
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
A = int(input())
B = int(input())
C = int(input())
X = int(input())
total = 0
count = 0
for a in range(A):
a += 1
for b in range(B):
b += 1
for c in range(C):
c += 1
total = 500*a+100*b+50*c
if total == X:
count += 1
print(count)
|
s668897848
|
Accepted
| 52
| 2,940
| 283
|
A = int(input())
B = int(input())
C = int(input())
X = int(input())
count = 0
for a in range(A+1):
for b in range(B+1):
for c in range(C+1):
total = 500*a+100*b+50*c
if total == X:
count += 1
print(count)
|
s381219875
|
p03759
|
u215018528
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 76
|
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 b - a == c - b else "No")
|
s747533671
|
Accepted
| 17
| 2,940
| 76
|
a, b, c = map(int, input().split())
print("YES" if b - a == c - b else "NO")
|
s332168279
|
p04043
|
u076917070
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 162
|
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
import sys
input=sys.stdin.readline
l = list(map(int, input().split()))
l.sort()
if l.count(5) == 2 and l.count(7) == 1:
print("yes")
else:
print("no")
|
s791753780
|
Accepted
| 17
| 2,940
| 162
|
import sys
input=sys.stdin.readline
l = list(map(int, input().split()))
l.sort()
if l.count(5) == 2 and l.count(7) == 1:
print("YES")
else:
print("NO")
|
s265802700
|
p03337
|
u204260373
| 2,000
| 1,048,576
|
Wrong Answer
| 20
| 3,060
| 165
|
You are given two integers A and B. Find the largest value among A+B, A-B and A \times B.
|
A,B=map(int,input().split())
if A*B>A+B>A-B or A*B>A-B>A+B:
print(A*B)
elif A+B>A+B>A-B or A+B>A-B>A+B:
print(A+B)
elif A-B>A+B>A*B or A-B>A*B>A+B:
print(A-B)
|
s060299637
|
Accepted
| 17
| 2,940
| 53
|
a,b=map(int,input().split())
print(max(a+b,a-b,a*b))
|
s713175397
|
p03351
|
u140251125
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 249
|
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.
|
# input
a, b, c, d = map(int, input().split())
if abs(a - b) > abs(a - c):
if abs(a - c) < d:
print('Yes')
else:
print('No')
else:
if abs(a - b) < d and abs(b - c) < d:
print('Yes')
else:
print('No')
|
s769428442
|
Accepted
| 17
| 3,064
| 252
|
# input
a, b, c, d = map(int, input().split())
if abs(a - b) > abs(a - c):
if abs(a - c) <= d:
print('Yes')
else:
print('No')
else:
if abs(a - b) <= d and abs(b - c) <= d:
print('Yes')
else:
print('No')
|
s683715434
|
p02612
|
u516579758
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 9,132
| 29
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n=int(input())
print(n%1000)
|
s678836913
|
Accepted
| 32
| 9,144
| 53
|
n=int(input())
a=0
while a<n:
a+=1000
print(a-n)
|
s765444709
|
p03457
|
u806257533
| 2,000
| 262,144
|
Wrong Answer
| 417
| 27,300
| 355
|
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
N = int(input())
time_coor = []
for n in range(N):
time_coor.append(list(map(int, input().split())))
can = True
for n in range(N-1):
if (abs(time_coor[n+1][1])-abs(time_coor[n][1]) + abs(time_coor[n+1][2])-abs(time_coor[n][2]))!=1:
can = False
break
else:
continue
if can==True:
print("Yes")
else:
print("No")
|
s215589842
|
Accepted
| 450
| 27,324
| 546
|
N = int(input())
time_coor = []
for n in range(N+1):
if n==0:
time_coor.append([0, 0, 0])
else:
time_coor.append(list(map(int, input().split())))
can = True
for n in range(N):
dist = abs(time_coor[n+1][1]-time_coor[n][1]) + \
abs(time_coor[n+1][2]-time_coor[n][2])
time = time_coor[n+1][0]-time_coor[n][0]
if dist>time:
can = False
break
elif dist%2==time%2:
continue
else:
can = False
break
if can==True:
print("Yes")
else:
print("No")
|
s954344858
|
p03416
|
u924467864
| 2,000
| 262,144
|
Wrong Answer
| 45
| 2,940
| 164
|
Find the number of _palindromic numbers_ among the integers between A and B (inclusive). Here, a palindromic number is a positive integer whose string representation in base 10 (without leading zeros) reads the same forward and backward.
|
a, b = map(int, input().split())
flag = True
cnt = 0
for i in range(a, b - a + 2):
s = str(i)
if s[0] == s[4] and s[1] == s[3]:
cnt += 1
print(cnt)
|
s574392273
|
Accepted
| 46
| 2,940
| 148
|
a, b = map(int, input().split())
cnt = 0
for i in range(a, b + 1):
s = str(i)
if s[0] == s[4] and s[1] == s[3]:
cnt += 1
print(cnt)
|
s156133759
|
p03478
|
u059436995
| 2,000
| 262,144
|
Wrong Answer
| 45
| 3,412
| 159
|
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
n, a, b = map(int, input().split())
ans = 0
for i in range(1,n + 1):
if a <= sum([int(j) for j in str(i)]) <=b:
print(i)
ans +=i
print(ans)
|
s727668992
|
Accepted
| 33
| 2,940
| 142
|
n, a, b = map(int, input().split())
ans = 0
for i in range(1,n + 1):
if a <= sum([int(j) for j in str(i)]) <=b:
ans +=i
print(ans)
|
s953719149
|
p03486
|
u228232845
| 2,000
| 262,144
|
Wrong Answer
| 30
| 9,112
| 598
|
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
|
import sys
def input(): return sys.stdin.readline().strip()
def I(): return int(input())
def LI(): return list(map(int, input().split()))
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def S(): return input()
def LS(): return input().split()
s = list(S())
t = list(S())
ns = ''.join(sorted(s))
nt = ''.join(sorted(t)[::-1])
print(ns, nt)
if ns < nt:
print('Yes')
else:
print('No')
|
s749110770
|
Accepted
| 29
| 8,968
| 584
|
import sys
def input(): return sys.stdin.readline().strip()
def I(): return int(input())
def LI(): return list(map(int, input().split()))
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def S(): return input()
def LS(): return input().split()
s = list(S())
t = list(S())
ns = ''.join(sorted(s))
nt = ''.join(sorted(t)[::-1])
if ns < nt:
print('Yes')
else:
print('No')
|
s785033373
|
p03606
|
u823885866
| 2,000
| 262,144
|
Wrong Answer
| 131
| 27,296
| 503
|
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?
|
import sys
import math
import itertools
import collections
import heapq
import re
import numpy as np
from functools import reduce
rr = lambda: sys.stdin.readline().rstrip()
rs = lambda: sys.stdin.readline().split()
ri = lambda: int(sys.stdin.readline())
rm = lambda: map(int, sys.stdin.readline().split())
rl = lambda: list(map(int, sys.stdin.readline().split()))
inf = float('inf')
mod = 10**9 + 7
n = ri()
li = np.zeros(100000)
for _ in range(n):
l, r = rm()
li[l:r+1] = 1
print(sum(li))
|
s126228479
|
Accepted
| 142
| 27,380
| 508
|
import sys
import math
import itertools
import collections
import heapq
import re
import numpy as np
from functools import reduce
rr = lambda: sys.stdin.readline().rstrip()
rs = lambda: sys.stdin.readline().split()
ri = lambda: int(sys.stdin.readline())
rm = lambda: map(int, sys.stdin.readline().split())
rl = lambda: list(map(int, sys.stdin.readline().split()))
inf = float('inf')
mod = 10**9 + 7
n = ri()
li = np.zeros(200000)
for _ in range(n):
l, r = rm()
li[l:r+1] = 1
print(int(sum(li)))
|
s309852921
|
p02841
|
u844789719
| 2,000
| 1,048,576
|
Wrong Answer
| 19
| 3,316
| 142
|
In this problem, a date is written as Y-M-D. For example, 2019-11-30 means November 30, 2019. Integers M_1, D_1, M_2, and D_2 will be given as input. It is known that the date 2019-M_2-D_2 follows 2019-M_1-D_1. Determine whether the date 2019-M_1-D_1 is the last day of a month.
|
m1, d1 = [int(_) for _ in input().split()]
m2, d2 = [int(_) for _ in input().split()]
if d1 + 1 == d2:
print('No')
else:
print('Yes')
|
s259247309
|
Accepted
| 17
| 2,940
| 139
|
m1, d1 = [int(_) for _ in input().split()]
m2, d2 = [int(_) for _ in input().split()]
if d1 + 1 == d2:
print('0')
else:
print('1')
|
s053110056
|
p03847
|
u875291233
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 567
|
You are given a positive integer N. Find the number of the pairs of integers u and v (0≦u,v≦N) such that there exist two non-negative integers a and b satisfying a xor b=u and a+b=v. Here, xor denotes the bitwise exclusive OR. Since it can be extremely large, compute the answer modulo 10^9+7.
|
# coding: utf-8
# Your code here!
odd={0:1}
even={0:1,1:3}
def memo(N):
if N ==0:
return 1
if N==1:
return 2
return memoodd((N-1)//2)+memoeven(N//2)
def memoodd(N):
if N in odd:
return odd[N]
# a=memoodd((N-1)//2)
# b=memoeven(N//2)
odd[N] = memo(N)
return odd[N]
def memoeven(N):
if N in even:
return even[N]
# a=memoodd((N-1)//2)
# b=memoeven(N//2)
# c=memoodd((N-2)//2)
# d=memoeven((N-1)//2)
# return a+b+c+d
even[N] = memo(N)+memo(N-1)
return even[N]
print(memo(32))
|
s237197852
|
Accepted
| 19
| 3,060
| 238
|
# coding: utf-8
# Your code here!
dic_memo={0:1,1:2}
M=10**9+7
def memo(N):
if N in dic_memo:
return dic_memo[N]
dic_memo[N]=(memo((N-1)//2)+memo(N//2)+memo(N//2-1))%M
return dic_memo[N]
n=int(input())
print(memo(n))
|
s033771213
|
p04044
|
u175743386
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 77
|
Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.
|
N, L = map(int, input().split())
''.join(sorted([input() for i in range(N)]))
|
s785759551
|
Accepted
| 20
| 3,060
| 84
|
N, L = map(int, input().split())
print(''.join(sorted([input() for i in range(N)])))
|
s970933737
|
p03523
|
u896741788
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 221
|
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`?
|
s=input()
l=list("AKIHABARA".split("A"))[1:-1]
print(l)
for ii in range(2**4):
p="A" if ii%2 else ""
i=ii
for j in range(3):
p+=l[j]
p+="A" if i%2 else ""
i//=2
if s==p:print("YES");exit()
print("NO")
|
s011219694
|
Accepted
| 18
| 3,060
| 212
|
s=input()
l=list("AKIHABARA".split("A"))[1:-1]
for ii in range(2**4):
p="A" if ii%2 else ""
i=ii
for j in range(3):
p+=l[j]
i//=2
p+="A" if i%2 else ""
if s==p:print("YES");exit()
print("NO")
|
s012120655
|
p03394
|
u672220554
| 2,000
| 262,144
|
Wrong Answer
| 32
| 5,100
| 560
|
Nagase is a top student in high school. One day, she's analyzing some properties of special sets of positive integers. She thinks that a set S = \\{a_{1}, a_{2}, ..., a_{N}\\} of **distinct** positive integers is called **special** if for all 1 \leq i \leq N, the gcd (greatest common divisor) of a_{i} and the sum of the remaining elements of S is **not** 1. Nagase wants to find a **special** set of size N. However, this task is too easy, so she decided to ramp up the difficulty. Nagase challenges you to find a **special** set of size N such that the gcd of all elements are 1 and the elements of the set does not exceed 30000.
|
n=int(input())
res=[]
count = 0
for i in range(1,50000):
if i % 2 == 0 or i % 3 == 0:
res.append(i)
count += 1
if count == n-1:
break
s=sum(res)
print(s)
t=res[-1]
f1 = s%2
f2 = s%3
if f1==0 and f2 == 0:
flag = 0
elif f1==0 and f2==1:
flag = 2
elif f1==0 and f2==2:
flag = 4
if f1==1 and f2 == 0:
flag = 3
elif f1==1 and f2==1:
flag = 5
elif f1==1 and f2==2:
flag = 1
for i in range(t+1,t+10):
if i % 6 == flag:
res.append(i)
print(s+i)
break
print(" ".join(map(str, res)))
|
s115074275
|
Accepted
| 31
| 5,212
| 1,001
|
n=int(input())
if n==3:
print("2 5 63")
elif n==6:
print("3 4 6 8 9 12")
elif n==7:
print("2 3 4 6 8 10 15")
else:
res=[]
count = 0
for i in range(1,50000):
if i % 2 == 0 or i % 3 == 0:
res.append(i)
count += 1
if count == n-1:
break
s=sum(res)
t=res[-1]
f1 = s%2
f2 = s%3
if f1==0 and f2 == 0:
flag = 0
elif f1==0 and f2==1:
flag = 2
elif f1==0 and f2==2:
flag = 4
if f1==1 and f2 == 0:
flag = 3
elif f1==1 and f2==1:
flag = 5
elif f1==1 and f2==2:
flag = 1
if flag == 5 or flag ==1:
res.pop(1)
f=0
for i in range(t+1,t+10):
if i % 2==0:
res.append(i)
f+=1
if f==2:
break
else:
for i in range(t+1,t+10):
if i % 6 == flag:
res.append(i)
break
print(" ".join(map(str, res)))
|
s273585661
|
p03672
|
u288430479
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 192
|
We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not. You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
|
a=input()
l=len(a)
while l>0:
# print(l)
if l%2!=0:
l -= 1
a = a[:-1]
# print(a)
else:
t =int(l/2)
if a[0:t] == a[t:]:
print(l)
break
else:
l -= 1
|
s904062020
|
Accepted
| 17
| 2,940
| 82
|
s = input()[:-1]
while s[0:len(s)//2] != s[len(s)//2:]:
s = s[:-1]
print(len(s))
|
s193744393
|
p02928
|
u366482170
| 2,000
| 1,048,576
|
Wrong Answer
| 363
| 3,188
| 334
|
We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}. Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2. Find the inversion number of B, modulo 10^9 + 7. Here the inversion number of B is defined as the number of ordered pairs of integers (i,~j)~(0 \leq i < j \leq K \times N - 1) such that B_i > B_j.
|
n,k=map(int,input().split())
a=list(map(int,input().split()))
count=0
countlast=0
for i in range(n-1):
for j in range(i+1,n):
if a[i]>a[j]:
count+=1
# print(i,j)
for i in range(n-1):
if a[n-1]>a[i]:
# print(i)
countlast+=1
#print(count,countlast)
mod=10**9+7
print((k*(k+1)*count+k*(k-1)*countlast)/2%mod)
|
s006952128
|
Accepted
| 871
| 3,188
| 301
|
n,k=map(int,input().split())
a=list(map(int,input().split()))
rcount=0
lcount=0
for i in range(n-1):
for j in range(i+1,n):
if a[i]>a[j]:
rcount+=1
for i in range(n-1):
for j in range(i+1,n):
if a[i]<a[j]:
lcount+=1
mod=10**9+7
print((k*(k+1)*rcount+k*(k-1)*lcount)//2%mod)
|
s023691490
|
p03719
|
u276115223
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 108
|
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
a, b, c = [int(s) for s in input().split()]
print('YES' if a <= c <= b else 'NO')
|
s850984666
|
Accepted
| 17
| 2,940
| 119
|
a, b, c = [int(s) for s in input().split()]
print('Yes' if a <= c <= b else 'No')
|
s471124329
|
p02390
|
u831971779
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,652
| 95
|
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 / (60*60))
m = (S / 60) % 60
s = S % 60
print('{}:{}:{}'.format(h,m,s))
|
s105137495
|
Accepted
| 20
| 7,572
| 101
|
S = int(input())
h = int(S / (60*60))
m = int(S / 60) % 60
s = S % 60
print('{}:{}:{}'.format(h,m,s))
|
s475077740
|
p04031
|
u293523199
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 282
|
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal **integers** by transforming some of them. He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2). Find the minimum total cost to achieve his objective.
|
N = int(input())
A = list(map(int, input().split()))
total_costs = {}
if( len(list(set(A))) ):
print(0)
else:
for i in range(0, 100):
cost = 0
for a in A:
cost += (a - i)*(a - i)
total_costs[i] = cost
print(min(total_costs.values()))
|
s963345499
|
Accepted
| 22
| 3,060
| 284
|
N = int(input())
A = list(map(int, input().split()))
total_costs = {}
if( len(set(A)) == 1 ):
print(0)
else:
for i in range(-100, 101):
cost = 0
for a in A:
cost += (a - i)*(a - i)
total_costs[i] = cost
print(min(total_costs.values()))
|
s970038985
|
p02612
|
u091945878
| 2,000
| 1,048,576
|
Wrong Answer
| 32
| 9,076
| 45
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
mod = N % 1000
print(mod)
|
s160490534
|
Accepted
| 34
| 9,152
| 101
|
N = int(input())
mod = N % 1000
if mod == 0:
result = 0
else:
result = 1000 - mod
print(result)
|
s550898301
|
p03139
|
u623819879
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 63
|
We conducted a survey on newspaper subscriptions. More specifically, we asked each of the N respondents the following two questions: * Question 1: Are you subscribing to Newspaper X? * Question 2: Are you subscribing to Newspaper Y? As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2. What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y? Write a program to answer this question.
|
n, a, b=map(int, input().split())
print(max(a,b),max(0,a+b-n))
|
s498843411
|
Accepted
| 18
| 2,940
| 64
|
n, a, b=map(int, input().split())
print(min(a,b),max(0,a+b-n))
|
s450457263
|
p03160
|
u201387466
| 2,000
| 1,048,576
|
Wrong Answer
| 132
| 13,800
| 219
|
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
N = int(input())
h = list(map(int,input().split()))
H = [0]*N
for i in range(1,N):
if i == 1:
H[i] = abs(h[1]-h[0])
else:
H[i] = max(H[i-1]+abs(h[i]-h[i-1]),H[i-2]+abs(h[i]-h[i-2]))
print(H[N-1])
|
s597679175
|
Accepted
| 148
| 13,980
| 277
|
N = int(input())
h = list(map(int,input().split()))
H = [0]*N
for i in range(1,N):
if i == 1:
H[i] = abs(h[1]-h[0])
else:
H[i] = H[i-1]+abs(h[i]-h[i-1])
if H[i] > H[i-2]+abs(h[i]-h[i-2]):
H[i] = H[i-2]+abs(h[i]-h[i-2])
print(H[N-1])
|
s791045803
|
p02390
|
u403901064
| 1,000
| 131,072
|
Wrong Answer
| 30
| 8,084
| 177
|
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.
|
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
from datetime import timedelta
def main():
sec = timedelta(seconds= int(input()))
print(sec)
if __name__ == '__main__':
main()
|
s525866338
|
Accepted
| 40
| 7,708
| 198
|
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
def main():
total= int(input())
ti = divmod(divmod(total,60)[0],60)
print("%d:%d:%d" % (ti[0],ti[1],total % 60))
if __name__ == '__main__':
main()
|
s996724367
|
p03501
|
u954693495
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 57
|
You are parking at a parking lot. You can choose from the following two fee plans: * Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours. * Plan 2: The fee will be B yen, regardless of the duration. Find the minimum fee when you park for N hours.
|
n,a,b=[int(x) for x in input().split()]
print(max(n*a,b))
|
s239276061
|
Accepted
| 18
| 2,940
| 57
|
n,a,b=[int(x) for x in input().split()]
print(min(n*a,b))
|
s110989925
|
p04011
|
u842388336
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 92
|
There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.
|
N = int(input())
K = int(input())
X = int(input())
Y = int(input())
print(N*X+min(0,N-K)*Y)
|
s714521245
|
Accepted
| 17
| 2,940
| 99
|
N = int(input())
K = int(input())
X = int(input())
Y = int(input())
print(min(N,K)*X+max(0,N-K)*Y)
|
s735348569
|
p03860
|
u839873388
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,068
| 43
|
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
s = input()
x = s[0]
print("A" + x + "C")
|
s033093513
|
Accepted
| 18
| 2,940
| 46
|
a,b,c = input().split()
print("A" +b[0] +"C")
|
s406364707
|
p03494
|
u732870425
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 3,064
| 214
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
N = int(input())
A = list(map(int, input().split()))
cnt = 0
while True:
for i in range(len(A)):
if A[i] % 2 != 0:
break
else:
A = [a/2 for a in A]
cnt += 1
print(cnt)
|
s953837564
|
Accepted
| 19
| 3,060
| 282
|
N = int(input())
A = list(map(int, input().split()))
cnt = 0
while True:
for i in range(len(A)):
if A[i] % 2 != 0:
break
else:
A = [a/2 for a in A]
cnt += 1
continue
break
print(cnt)
|
s487018277
|
p03024
|
u758973277
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 78
|
Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.
|
S = input()
if S.count('o')+(15-len(S))>=8:
print('Yes')
else:
print('No')
|
s866586858
|
Accepted
| 17
| 2,940
| 78
|
S = input()
if S.count('o')+(15-len(S))>=8:
print('YES')
else:
print('NO')
|
s064250458
|
p03470
|
u253422591
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 775
|
An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?
|
n, total = 9, 45000
min = n * 1000
max = n * 10000
if (total > max) or (total < min):
print('-1 -1 -1')
else:
min_man = (total - (5000 * n)) // 10000
n -= min_man
total -= min_man*10000
min_sen = (total % 5000) // 1000
total -= min_sen*1000
n -= min_sen
success = False
for i in range(int(total // 10000)+1):
if success:
break
ii = total - i*10000
for j in range(int(ii // 5000)+1):
jj = ii - j*5000
k = jj//1000
if i + j + k + min_sen + min_man == n:
print(str(i+min_man) + ' ' + str(j) + ' ' + str(k+min_sen))
success = True
break
if not success:
print('-1 -1 -1')
|
s277439199
|
Accepted
| 20
| 2,940
| 101
|
n = int(input())
mochis = []
for i in range(n):
mochis.append(int(input()))
print(len(set(mochis)))
|
s714020126
|
p04043
|
u252828980
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 72
|
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 == 35 else "NO")
|
s592349750
|
Accepted
| 17
| 2,940
| 72
|
a,b,c = map(int,input().split())
print("YES" if a*b*c ==175 else "NO")
|
s082983478
|
p03493
|
u901687869
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 140
|
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.
|
a = input()
b = a[0:0]
c = a[1:0]
d = a[2:0]
ball = 0
if(b == 1):
ball += 1
if(c == 1):
ball += 1
if(d == 1):
ball += 1
print(1)
|
s110385931
|
Accepted
| 18
| 2,940
| 132
|
a = input()
count = 0
if a[0:1] == "1":
count += 1
if a[1:2] == "1":
count += 1
if a[2:3] == "1":
count += 1
print(count)
|
s614019364
|
p03719
|
u950376354
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 81
|
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
A,B,C = map(int, input().split())
if A<=C<=B:
print('YES')
else:
print('NO')
|
s102588937
|
Accepted
| 17
| 2,940
| 81
|
A,B,C = map(int, input().split())
if A<=C<=B:
print('Yes')
else:
print('No')
|
s710274402
|
p03192
|
u636311816
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 8
|
You are given an integer N that has exactly four digits in base ten. How many times does `2` occur in the base-ten representation of N?
|
print(1)
|
s385707372
|
Accepted
| 17
| 2,940
| 72
|
s = input()
cnt=0
for c in s:
if c =="2":
cnt+=1
print(cnt)
|
s796976234
|
p03379
|
u625963200
| 2,000
| 262,144
|
Wrong Answer
| 305
| 25,556
| 177
|
When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N.
|
n=int(input())
X=list(map(int,input().split()))
X.sort()
median1=X[n//2-1]
median2=X[n//2]
for i in range(n):
if X[i]<=median1:
print(median2)
else:
print(median1)
|
s548355865
|
Accepted
| 301
| 25,620
| 172
|
n=int(input())
X=list(map(int,input().split()))
sortedX=sorted(X)
m1=sortedX[n//2-1]
m2=sortedX[n//2]
for i in range(n):
if X[i]<=m1:
print(m2)
else:
print(m1)
|
s792380491
|
p03338
|
u562015767
| 2,000
| 1,048,576
|
Wrong Answer
| 34
| 9,116
| 238
|
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())
l = list(map(str,input()))
ans = 0
for i in range(n):
a = l[0:i]
b = l[i:n-1]
tmp = []
for j in a:
if j in b:
tmp.append(j)
tmp = set(tmp)
ans = (max(ans,len(tmp)))
print(ans)
|
s027743013
|
Accepted
| 32
| 9,120
| 234
|
n = int(input())
l = list(map(str,input()))
ans = 0
for i in range(n):
a = l[:i]
b = l[i:]
tmp = []
for j in a:
if j in b:
tmp.append(j)
tmp = set(tmp)
ans = (max(ans,len(tmp)))
print(ans)
|
s653630906
|
p03644
|
u440975163
| 2,000
| 262,144
|
Wrong Answer
| 29
| 9,416
| 529
|
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
n = int(input())
ans = 0
def factorization(n):
arr = []
temp = n
for i in range(2, int(-(-n**0.5//1))+1):
if temp%i==0:
cnt=0
while temp%i==0:
cnt+=1
temp //= i
arr.append([i, cnt])
if temp!=1:
arr.append([temp, 1])
if arr==[]:
arr.append([n, 1])
return arr
for i in range(n + 1):
k = factorization(i)
for j in k:
if j[0] == 2:
if ans < j[1]:
ans = j[1]
print(ans)
|
s499388865
|
Accepted
| 33
| 9,128
| 121
|
import sys
n = int(input())
ln = [64, 32, 16, 8, 4, 2, 1]
for i in ln:
if i <= n:
print(i)
sys.exit()
|
s771547751
|
p03352
|
u617037231
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 152
|
You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.
|
import sys
X = int(input())
L = []
for i in range(1,32):
for j in range(2,10):
if X <= i**j:
L.append(i**(j-1))
break
print(max(L))
|
s960380692
|
Accepted
| 18
| 3,060
| 265
|
import sys
X = int(input())
L = []
for i in range(1,32):
for j in range(1,10):
if X < i**j:
if j != 2:
L.append(i**(j-1))
break
break
elif X == i**j and j != 1:
print(i**j)
sys.exit(0)
print(max(L))
|
s489982268
|
p04011
|
u052499405
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 123
|
There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.
|
n = int(input())
k = int(input())
x = int(input())
y = int(input())
ans = min(k, n) * x
ans += min(0, n - k) * y
print(ans)
|
s783737697
|
Accepted
| 17
| 3,064
| 214
|
#!/usr/bin/env python3
import sys
sys.setrecursionlimit(10**8)
input = sys.stdin.readline
n = int(input())
k = int(input())
x = int(input())
y = int(input())
ans = min(k, n) * x
ans += max(0, n - k) * y
print(ans)
|
s194756607
|
p03069
|
u591295155
| 2,000
| 1,048,576
|
Wrong Answer
| 352
| 6,012
| 265
|
There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is `.`, and the stone is black if the character is `#`. Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored.
|
n = int(input())
s = input()
black = 0
min_val = 200001
tot_white = s.count('.')
for i in range(len(s)):
if s[i] == '#':
black += 1
else:
tot_white -=1
print(tot_white, black)
min_val = min(min_val, tot_white+black)
print(min_val)
|
s043669630
|
Accepted
| 76
| 11,264
| 322
|
n = int(input())
s = input()
black_right, min_val = 0, 2000001
white_left = s.count('.')
ans = [white_left+black_right]
for moji in s:
if moji == '.':
white_left -=1
else:
black_right += 1
ans.append(white_left+black_right)
#min_val = min(min_val, white_left+black_right)
print(min(ans))
|
s706671775
|
p03997
|
u129978636
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 49
|
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
print((int(input())+int(input()))*int(input())/2)
|
s621953426
|
Accepted
| 18
| 2,940
| 50
|
print((int(input())+int(input()))*int(input())//2)
|
s463274106
|
p03759
|
u442948527
| 2,000
| 262,144
|
Wrong Answer
| 25
| 9,156
| 62
|
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(["NO","YES"][2*b-a-c!=0])
|
s794311063
|
Accepted
| 29
| 9,060
| 60
|
a,b,c=map(int,input().split())
print(["NO","YES"][2*b==a+c])
|
s028078680
|
p03069
|
u200346982
| 2,000
| 1,048,576
|
Wrong Answer
| 2,104
| 18,848
| 558
|
There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is `.`, and the stone is black if the character is `#`. Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored.
|
N = int(input())
S = input()
cnt=0
l = []
b = 1 if S[0] == "." else 0
for i in range(1, N):
if S[i]==S[i-1]:
cnt += 1
if i == N-1:
l.append(cnt+1)
else:
l.append(cnt+1)
if i == N-1:
l.append(1)
cnt = 0
if len(l)<=1:
print("0")
exit()
s = 0
m = 1000000000
for i in range(b, len(l), 2):
for j in range(b, i, 2):
print(i, j, l[j])
s += l[j]
for j in range(i+1, len(l), 2):
print(i, j, l[j])
s += l[j]
m = min(s, m)
print(l)
print(m)
|
s102248970
|
Accepted
| 196
| 5,220
| 692
|
N = int(input())
S = input()
cnt=0
l = []
b = 1 if S[0] == "." else 0
for i in range(1, N):
if S[i]==S[i-1]:
cnt += 1
if i == N-1:
l.append(cnt+1)
else:
l.append(cnt+1)
if i == N-1:
l.append(1)
cnt = 0
if len(l)<=1:
print("0")
exit()
s = 0
q = 0
q2 = 0
for c in S:
if c=="#":
q += 1
else:
q2 += 1
for j in range(abs(b-1), len(l), 2):
#print(j, l[j])
s += l[j]
m = min(q,q2,s)
#print(s)
for j in range(b, len(l), 2):
#print(j-2,l[j-2], j-1, l[j-1])
if j-2 >= 0:
s += l[j-2]-l[j-1]
elif j-1 >= 0:
s += -l[j-1]
m = min(s, m)
#print(l)
print(m)
|
s177188028
|
p03658
|
u399721252
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 133
|
Snuke has N sticks. The length of the i-th stick is l_i. Snuke is making a snake toy by joining K of the sticks together. The length of the toy is represented by the sum of the individual sticks that compose it. Find the maximum possible length of the toy.
|
n, k = [ int(v) for v in input().split() ]
bar_list = sorted([ int(v) for v in input().split() ], reverse = True)
print(bar_list[:k])
|
s386114683
|
Accepted
| 17
| 2,940
| 138
|
n, k = [ int(v) for v in input().split() ]
bar_list = sorted([ int(v) for v in input().split() ], reverse = True)
print(sum(bar_list[:k]))
|
s790872203
|
p02388
|
u998185318
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,584
| 83
|
Write a program which calculates the cube of a given integer x.
|
print('整数値を入力してください')
x = input()
x = int(x)
print(x ** 3)
|
s945972453
|
Accepted
| 20
| 5,580
| 130
|
x = input()
x = int(x)
if 1 <= x and x <= 100:
print(x ** 3)
else:
print('number must larger equal 1 and less equal 100')
|
s558601312
|
p03501
|
u556589653
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 97
|
You are parking at a parking lot. You can choose from the following two fee plans: * Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours. * Plan 2: The fee will be B yen, regardless of the duration. Find the minimum fee when you park for N hours.
|
N,A,B = map(int,input().split())
K = N*A
if K<B:
print(A)
elif B<K:
print(B)
else:
print(A)
|
s166061835
|
Accepted
| 17
| 3,060
| 97
|
N,A,B = map(int,input().split())
K = N*A
if K<B:
print(K)
elif B<K:
print(B)
else:
print(K)
|
s825949461
|
p04029
|
u379716238
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 67
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
N = int(input())
ans = 0
for i in range(N):
ans += i
print(ans)
|
s149883446
|
Accepted
| 17
| 2,940
| 71
|
N = int(input())
ans = 0
for i in range(N):
ans += i + 1
print(ans)
|
s781867961
|
p03476
|
u284854859
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 15,384
| 607
|
We say that a odd number N is _similar to 2017_ when both N and (N+1)/2 are prime. You are given Q queries. In the i-th query, given two odd numbers l_i and r_i, find the number of odd numbers x similar to 2017 such that l_i ≤ x ≤ r_i.
|
import bisect
def prime(n):
seq = list(range(2, n))
while len(seq) > 0:
prime = seq.pop(0)
yield prime
seq = [i for i in seq if not i % prime == 0]
q=int(input())
l = []
r = []
for i in range(q):
x,y = map(int,input().split())
l.append(x)
r.append(y)
a = list(prime(max(r)+1))
print('a',a)
c = []
for i in range(len(a)):
if (a[i]+1)/2 in a:
c.append(a[i])
c.sort()
print('c',c)
for i in range(q):
h = bisect.bisect_left(c,l[i])
j = bisect.bisect_right(c,r[i])
print(j-h)
|
s386052839
|
Accepted
| 1,030
| 7,156
| 536
|
import math
Q = int(input())
n = 100000
P = [False] * n
P[2] = True
for i in range(3, n, 2):
k = True
for j in range(3, int(math.sqrt(i)) + 1, 2):
if i % j == 0:
k = False
break
if k:
P[i] = True
a = [0] * n
for i in range(1, n, 2):
j = (i + 1) // 2
if P[i] and P[j]:
a[i] = 1
for i in range(n - 1):
a[i + 1] += a[i]
for _ in range(Q):
l, r = map(int, input().split())
print(a[r] - a[l - 1])
|
s493117951
|
p03555
|
u325264482
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 147
|
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
S = input()
T = input()
ans = "Yes"
if S[0] != T[2]:
ans = "No"
if S[1] != T[1]:
ans = "No"
if S[2] != T[0]:
ans = "No"
print(ans)
|
s205164676
|
Accepted
| 17
| 3,060
| 147
|
S = input()
T = input()
ans = "YES"
if S[0] != T[2]:
ans = "NO"
if S[1] != T[1]:
ans = "NO"
if S[2] != T[0]:
ans = "NO"
print(ans)
|
s965370941
|
p03369
|
u690536347
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 26
|
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
700+input().count("o")*100
|
s310707231
|
Accepted
| 17
| 2,940
| 33
|
print(700+input().count("o")*100)
|
s740358830
|
p03089
|
u977193988
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,064
| 638
|
Snuke has an empty sequence a. He will perform N operations on this sequence. In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1). You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
|
import sys
def input():
return sys.stdin.readline().strip()
N = int(input())
B = list(map(int, input().split()))
hand = []
def check(B, cnt):
n = len(B)
for i in range(n):
if B[n - i - 1] == (n - i):
hand.append(B[n - i - 1])
B = B[: n - i - 1] + B[n - i :]
break
if len(B) != n and len(B) > 0:
return B
else:
return False
cnt = 1
while B:
B = check(B, cnt)
if len(hand) < N:
print(-1)
sys.exit()
for i in range(N):
if hand[i] > i + 1:
print(-1)
sys.exit()
answer = hand[::-1]
print(*answer, sep="\n")
|
s745914178
|
Accepted
| 18
| 3,064
| 640
|
import sys
def input():
return sys.stdin.readline().strip()
N = int(input())
B = list(map(int, input().split()))
hand = []
def check(B, cnt):
n = len(B)
for i in range(n):
if B[n - i - 1] == (n - i):
hand.append(B[n - i - 1])
B = B[: n - i - 1] + B[n - i :]
break
if len(B) != n and len(B) > 0:
return B
else:
return False
cnt = 1
while B:
B = check(B, cnt)
if len(hand) < N:
print(-1)
sys.exit()
answer = hand[::-1]
for i in range(N):
if answer[i] > i + 1:
print(-1)
sys.exit()
print(*answer, sep="\n")
|
s095897253
|
p02263
|
u091533407
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,620
| 526
|
An expression is given in a line. Two consequtive symbols (operand or operator) are separated by a space character. You can assume that +, - and * are given as the operator and an operand is a positive integer less than 106
|
if __name__=="__main__":
a = list(input().split())
print(a)
b = []
for i in range(len(a)):
if a[i] == "+":
y = b.pop()
x = b.pop()
z = x + y
b.append(z)
elif a[i] == "-":
y = b.pop()
x = b.pop()
z = x - y
b.append(z)
elif a[i] == "*":
y = b.pop()
x = b.pop()
z = x * y
b.append(z)
else:
b.append(int(a[i]))
print(b[0])
|
s464273150
|
Accepted
| 20
| 7,692
| 513
|
if __name__=="__main__":
a = list(input().split())
b = []
for i in range(len(a)):
if a[i] == "+":
y = b.pop()
x = b.pop()
z = x + y
b.append(z)
elif a[i] == "-":
y = b.pop()
x = b.pop()
z = x - y
b.append(z)
elif a[i] == "*":
y = b.pop()
x = b.pop()
z = x * y
b.append(z)
else:
b.append(int(a[i]))
print(b[0])
|
s692114722
|
p03149
|
u206570055
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 133
|
You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974".
|
li = list(map(int, input().split()))
print(li)
if 1 in li and 9 in li and 7 in li and 4 in li:
print('YES')
else:
print('NO')
|
s833345658
|
Accepted
| 17
| 2,940
| 123
|
li = list(map(int, input().split()))
if 1 in li and 9 in li and 7 in li and 4 in li:
print('YES')
else:
print('NO')
|
s845710011
|
p03501
|
u947327691
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 49
|
You are parking at a parking lot. You can choose from the following two fee plans: * Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours. * Plan 2: The fee will be B yen, regardless of the duration. Find the minimum fee when you park for N hours.
|
n,a,b=map(int,input().split())
print(max(a*n,b))
|
s130289718
|
Accepted
| 17
| 2,940
| 49
|
n,a,b=map(int,input().split())
print(min(a*n,b))
|
s263545895
|
p03024
|
u497952650
| 2,000
| 1,048,576
|
Wrong Answer
| 19
| 2,940
| 70
|
Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.
|
S = input()
if S.count("x")>=8:
print("No")
else:
print("Yes")
|
s278691259
|
Accepted
| 17
| 2,940
| 70
|
S = input()
if S.count("x")>=8:
print("NO")
else:
print("YES")
|
s535323656
|
p03854
|
u107267797
| 2,000
| 262,144
|
Wrong Answer
| 62
| 9,060
| 376
|
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
S = input()[::-1]
dreams = ['dream', 'dreamer', 'erase', 'eraser']
dreams = [dream[::-1] for dream in dreams]
i = 0
while i < len(S):
flag = False
for j in dreams:
if S[i:i+len(j)] == j:
print(S[i:i+len(j)], j)
print(i)
i = i + len(j)
flag = True
if not flag:
break
print("YES" if flag else "NO")
|
s076843883
|
Accepted
| 42
| 9,256
| 319
|
S = input()[::-1]
dreams = ['dream', 'dreamer', 'erase', 'eraser']
dreams = [dream[::-1] for dream in dreams]
i = 0
while i < len(S):
flag = False
for j in dreams:
if S[i:i+len(j)] == j:
i = i + len(j)
flag = True
if not flag:
break
print("YES" if flag else "NO")
|
s785448253
|
p03545
|
u867005447
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 505
|
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
a = [int(x) for x in list(input())]
op = 3
ops = [0] * op
for i in range(2 ** op):
result = a[0]
bry = list(format(i, "0{}b".format(op)))
for index, value in enumerate(bry):
if value == "1":
result += a[index + 1]
ops[index] = "+"
else:
result -= a[index + 1]
ops[index] = "-"
if result == 7:
ans = str(a[0])
for j in range(3):
ans += "{}{}".format(ops[j], a[j+1])
print(ans)
break
|
s257177654
|
Accepted
| 18
| 2,940
| 171
|
a = input()
for i in range(2 ** 3):
s = a[0]
for j in range(3):
s += "+-"[i >> j & 1] + a[j+1]
if eval(s) == 7:
print(s + "=7")
break
|
s368057603
|
p03712
|
u016323272
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 138
|
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.
|
#ABC062.B
H,W = map(int, input().split())
for i in range(H):
a = input()
n = len(a)
print('#'*(n+2))
print('#'+a+'#')
print('#'*(n+2))
|
s957646350
|
Accepted
| 18
| 3,060
| 120
|
H,W = map(int,input().split())
print('#'*(W+2))
for i in range(H):
a = input()
print('#'+a+'#')
print('#'*(W+2))
|
s886780175
|
p02603
|
u099300899
| 2,000
| 1,048,576
|
Wrong Answer
| 2,211
| 119,076
| 521
|
To become a millionaire, M-kun has decided to make money by trading in the next N days. Currently, he has 1000 yen and no stocks - only one kind of stock is issued in the country where he lives. He is famous across the country for his ability to foresee the future. He already knows that the price of one stock in the next N days will be as follows: * A_1 yen on the 1-st day, A_2 yen on the 2-nd day, ..., A_N yen on the N-th day. In the i-th day, M-kun can make the following trade **any number of times** (possibly zero), **within the amount of money and stocks that he has at the time**. * Buy stock: Pay A_i yen and receive one stock. * Sell stock: Sell one stock for A_i yen. What is the maximum possible amount of money that M-kun can have in the end by trading optimally?
|
import sys
sys.setrecursionlimit(10**8)
n = int(input())
a = [int(i) for i in input().split()]
def rec(k, ans, money, n):
x = a[k]
for i in range(k+1, len(a)):
if (x < a[i]):
n = money // a[i-1]
money -= a[i-1] * n
money += a[i] * n
rec(i, ans, money, n)
return ans.append(money)
def main():
ans = [0]
money = 1000
n = 0
rec(0, ans, money, n)
print(ans)
print(max(ans))
if __name__ == '__main__':
main()
|
s782405515
|
Accepted
| 32
| 8,828
| 300
|
n = int(input())
a = [int(i) for i in input().split()]
def main():
money = 1000
n = 0
for i in range(1, len(a)):
if (a[i-1] < a[i]):
n = money // a[i-1]
money += (a[i] - a[i-1]) * n
print(money)
if __name__ == '__main__':
main()
|
s036972226
|
p03050
|
u745514010
| 2,000
| 1,048,576
|
Wrong Answer
| 145
| 9,204
| 130
|
Snuke received a positive integer N from Takahashi. A positive integer m is called a _favorite number_ when the following condition is satisfied: * The quotient and remainder of N divided by m are equal, that is, \lfloor \frac{N}{m} \rfloor = N \bmod m holds. Find all favorite numbers and print the sum of those.
|
n = int(input())
ans = 0
for i in range(1, int(n ** 0.5)):
if n % i == 0:
k = (n // i) - 1
ans += k
print(ans)
|
s650649902
|
Accepted
| 147
| 9,328
| 156
|
n = int(input())
ans = 0
for i in range(1, int(n ** 0.5) + 1):
if n % i == 0:
k = (n // i) - 1
if k > i:
ans += k
print(ans)
|
s886628639
|
p03693
|
u709799578
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 88
|
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
a = int(''.join(input().split()))
if a % 4 == 0:
print('Yes')
else:
print('No')
|
s005740470
|
Accepted
| 17
| 2,940
| 88
|
a = int(''.join(input().split()))
if a % 4 == 0:
print('YES')
else:
print('NO')
|
s266382040
|
p03160
|
u713492631
| 2,000
| 1,048,576
|
Wrong Answer
| 124
| 14,692
| 540
|
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
import sys
input = lambda :sys.stdin.readline()
def main():
n = int(input())
h = list(map(int, input().split()))
dp = [float('inf')] * n
dp[0] = 0
dp[1] = abs(h[1] - h[0])
for i in range(2, n):
chmin(dp, i, dp[i-1] + abs(h[i] - h[i-1]))
chmin(dp, i, dp[i-2] + abs(h[i] - h[i-2]))
print(dp)
print(dp[n-1])
def chmin(array, idx, value):
if array[idx] > value:
array[idx] = value
def chmax(array, idx, value):
if array[idx] < value:
array[idx] = value
main()
|
s551427255
|
Accepted
| 115
| 13,716
| 514
|
import sys
stdin = lambda : sys.stdin.readline()
stdout = lambda *args: sys.stdout.write(' '.join(map(str, args))+'\n')
def main():
n = int(stdin())
h = list(map(int, stdin().split()))
dp = [float('inf')] * n
dp[0] = 0
dp[1] = abs(h[1] - h[0])
for i in range(2, n):
chmin(dp, i, dp[i-1] + abs(h[i] - h[i-1]))
chmin(dp, i, dp[i-2] + abs(h[i] - h[i-2]))
stdout(dp[n-1])
def chmin(array, idx, value):
if array[idx] > value:
array[idx] = value
main()
|
s036256091
|
p02841
|
u982591663
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 118
|
In this problem, a date is written as Y-M-D. For example, 2019-11-30 means November 30, 2019. Integers M_1, D_1, M_2, and D_2 will be given as input. It is known that the date 2019-M_2-D_2 follows 2019-M_1-D_1. Determine whether the date 2019-M_1-D_1 is the last day of a month.
|
#A
M1, D1 = map(int, input().split())
M2, D2 = map(int, input().split())
if M1 == M2:
print(1)
else:
print(0)
|
s486817448
|
Accepted
| 17
| 2,940
| 118
|
#A
M1, D1 = map(int, input().split())
M2, D2 = map(int, input().split())
if M1 == M2:
print(0)
else:
print(1)
|
s253467688
|
p03545
|
u830054172
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 305
|
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
A = list(input())
for i in range(2**3):
num = int(A[0])
ans = A[0]
for j in range(3):
if ((i >> j)&1):
num += int(A[j+1])
ans += "+"+A[j+1]
else:
num -= int(A[j+1])
ans += "-"+A[j+1]
if num == 7:
print(ans)
break
|
s142946929
|
Accepted
| 17
| 2,940
| 231
|
A = input()
for i in range(2**3):
ans = A[0]
for j in range(3):
if ((i >> j)&1):
ans += "+"+A[j+1]
else:
ans += "-"+A[j+1]
if eval(ans) == 7:
print(ans+"=7")
break
|
s114695862
|
p02842
|
u811436126
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 152
|
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
import math
n = int(input())
ans = n // 1.08 + 1
val = math.floor(ans * 1.08)
print(ans, val)
if val == n:
print(int(ans))
else:
print(':(')
|
s194745553
|
Accepted
| 20
| 2,940
| 198
|
n = int(input())
ans = 0
for i in range(100):
val = 100 * n + i
if val % 108 == 0:
ans = val // 108
break
else:
pass
print(':(') if ans == 0 else print(ans)
|
s809548537
|
p03944
|
u588081069
| 2,000
| 262,144
|
Wrong Answer
| 155
| 3,064
| 1,028
|
There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white. Snuke plotted N points into the rectangle. The coordinate of the i-th (1 ≦ i ≦ N) point was (x_i, y_i). Then, he created an integer sequence a of length N, and for each 1 ≦ i ≦ N, he painted some region within the rectangle black, as follows: * If a_i = 1, he painted the region satisfying x < x_i within the rectangle. * If a_i = 2, he painted the region satisfying x > x_i within the rectangle. * If a_i = 3, he painted the region satisfying y < y_i within the rectangle. * If a_i = 4, he painted the region satisfying y > y_i within the rectangle. Find the area of the white region within the rectangle after he finished painting.
|
W, H, N = list(map(int, input().split()))
input_list = []
for i in range(N):
input_list.append(list(map(int, input().split())))
input_list[i][1] = H - input_list[i][1]
C = [[0] * W] * H
for col in input_list:
if col[2] == 1:
for w in range(len(C)):
for h in range(len(C[w])):
if w <= col[0]:
C[w][h] = 1
elif col[2] == 2:
for w in range(len(C)):
for h in range(len(C[w])):
if w >= col[0]:
C[w][h] = 1
elif col[2] == 3:
for w in range(len(C)):
for h in range(len(C[w])):
if h <= col[1]:
C[w][h] = 1
elif col[2] == 4:
for w in range(len(C)):
for h in range(len(C[w])):
if h >= col[1]:
C[w][h] = 1
cnt = 0
for col in C:
cnt += col.count(1)
print(cnt)
|
s911192226
|
Accepted
| 17
| 3,064
| 531
|
W, H, N = list(map(int, input().split()))
L, R = 0, W
D, T = 0, H
for i in range(N):
x, y, a = list(map(int, input().split()))
if a == 1:
if L < x:
L = x
elif a == 2:
if x < R:
R = x
elif a == 3:
if D < y:
D = y
elif a == 4:
if y < T:
T = y
if (R - L) > 0 and (T - D) > 0:
print((R - L) * (T - D))
elif (R - L) > 0 and (T - D) < 0:
print((R - L))
elif (R - L) < 0 and (T - D) > 0 > 0:
print(T - D)
else:
print(0)
|
s439743891
|
p02842
|
u326609687
| 2,000
| 1,048,576
|
Wrong Answer
| 36
| 3,060
| 142
|
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
N = int(input())
x = int(N / 1.08)
for y in range(x, x + 3):
p = int(1.08 * y)
if p == N:
print(p)
exit(0)
print(':(')
|
s886500059
|
Accepted
| 17
| 2,940
| 143
|
N = int(input())
x = int(N / 1.08)
for y in range(x, x + 3):
p = int(1.08 * y)
if p == N:
print(y)
exit(0)
print(':(')
|
s419954102
|
p02741
|
u695655590
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 324
|
Print the K-th element of the following sequence of length 32: 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51
|
K = int(input())
Ans = 0
if K == 4 or K == 6 or K == 9 or K == 10 or K == 14 or K == 21 or K == 22 or K == 25 or K == 26:
Ans = 2
elif K == 28 or K == 30:
Ans = 4
elif K == 8 or K == 12 or K == 18 or K == 20 or K == 27:
Ans = 5
elif K == 16:
Ans = 14
elif K == 24:
Ans = 15
elif K == 32:
Ans = 51
else:
Ans = 1
|
s617639597
|
Accepted
| 19
| 3,064
| 338
|
K = int(input())
Ans = 0
if K == 4 or K == 6 or K == 9 or K == 10 or K == 14 or K == 21 or K == 22 or K == 25 or K == 26:
Ans = 2
elif K == 28 or K == 30:
Ans = 4
elif K == 8 or K == 12 or K == 18 or K == 20 or K == 27:
Ans = 5
elif K == 16:
Ans = 14
elif K == 24:
Ans = 15
elif K == 32:
Ans = 51
else:
Ans = 1
print(Ans)
|
s514936310
|
p02406
|
u037441960
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,584
| 140
|
In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; }
|
n = int(input())
for i in range(1, n) :
x = i
if(x % 3 == 0 or x % 10 == 3) :
print(" ", i, end = "")
i += 1
else :
i += 1
print()
|
s878866953
|
Accepted
| 20
| 6,164
| 147
|
n = int(input())
for i in range(3, n + 1) :
if(i % 3 == 0 or "3" in str(i)) :
print(" ", i, sep = "", end = "")
i += 1
else :
pass
print()
|
s483543075
|
p02414
|
u914146430
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,732
| 282
|
Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula: \\[ c_{ij} = \sum_{k=1}^m a_{ik}b_{kj} \\] where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively.
|
n,m,l=list(map(int,input().split()))
m_A=[list(map(int,input().split())) for i in range(n)]
m_B=[list(map(int,input().split())) for i in range(m)]
for i in range(n):
for k in range(l):
s=0
for j in range(m):
s+=m_A[i][j]*m_B[j][k]
print(s)
|
s256474216
|
Accepted
| 420
| 8,628
| 317
|
n,m,l=list(map(int,input().split()))
m_A=[list(map(int,input().split())) for i in range(n)]
m_B=[list(map(int,input().split())) for i in range(m)]
for i in range(n):
t=[]
for k in range(l):
s=0
for j in range(m):
s+=m_A[i][j]*m_B[j][k]
t.append(s)
print(*t)
|
s486718124
|
p02396
|
u489071923
| 1,000
| 131,072
|
Wrong Answer
| 50
| 7,968
| 94
|
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.
|
import sys
x = 0
for i in sys.stdin.readlines():
x += 1
print("Case "+ str(x)+":"+ i )
|
s569046704
|
Accepted
| 60
| 7,952
| 144
|
import sys
i = 0
for x in sys.stdin.readlines():
x = x.rstrip()
if x == "0":
break
i += 1
print("Case "+ str(i)+": "+ x)
|
s060496223
|
p03487
|
u760771686
| 2,000
| 262,144
|
Wrong Answer
| 98
| 22,448
| 209
|
You are given a sequence of positive integers of length N, a = (a_1, a_2, ..., a_N). Your objective is to remove some of the elements in a so that a will be a **good sequence**. Here, an sequence b is a **good sequence** when the following condition holds true: * For each element x in b, the value x occurs exactly x times in b. For example, (3, 3, 3), (4, 2, 4, 1, 4, 2, 4) and () (an empty sequence) are good sequences, while (3, 3, 3, 3) and (2, 4, 1, 4, 2) are not. Find the minimum number of elements that needs to be removed so that a will be a good sequence.
|
from collections import defaultdict
N = int(input())
A = list(map(int,input().split()))
hashmap = defaultdict(int)
for a in A:
hashmap[a]+=1
ans = 0
for k, v in hashmap.items():
ans+=min(v,k-v)
print(ans)
|
s652749573
|
Accepted
| 93
| 22,488
| 240
|
from collections import defaultdict
N = int(input())
A = list(map(int,input().split()))
hashmap = defaultdict(int)
for a in A:
hashmap[a]+=1
ans = 0
for k, v in hashmap.items():
if k < v:
ans+=v-k
elif k > v:
ans+=v
print(ans)
|
s380790319
|
p02259
|
u599130514
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,600
| 426
|
Write a program of the Bubble Sort algorithm which sorts a sequence _A_ in ascending order. The algorithm should be based on the following pseudocode: BubbleSort(A) 1 for i = 0 to A.length-1 2 for j = A.length-1 downto i+1 3 if A[j] < A[j-1] 4 swap A[j] and A[j-1] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 4 of the pseudocode.
|
def bubbleSort(A, N):
flag = True
count = 0
while flag:
flag = 0
for j in range(N-1, 0, -1):
if A[j] < A[j-1]:
tmp = A[j]
A[j] = A[j-1]
A[j-1] = tmp
flag = 1
count += 1
print(A)
print(count)
arr_length = int(input())
arr_num = [int(i) for i in input().split(" ")]
bubbleSort(arr_num, arr_length)
|
s255429630
|
Accepted
| 30
| 5,608
| 427
|
def bubbleSort(A, N):
flag = True
count = 0
while flag:
flag = 0
for j in range(N-1, 0, -1):
if A[j] < A[j-1]:
tmp = A[j]
A[j] = A[j-1]
A[j-1] = tmp
flag = 1
count += 1
print(*A)
print(count)
arr_length = int(input())
arr_num = [int(i) for i in input().split(" ")]
bubbleSort(arr_num, arr_length)
|
s462139069
|
p03251
|
u477650749
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 252
|
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())
list_x = [int(i) for i in input().split()]
list_y = [int(j) for j in input().split()]
if X >= Y:
print("War")
else:
if max(list_x) +1 >= min(list_y):
print("War")
else:
print("No War")
|
s180159618
|
Accepted
| 17
| 2,940
| 267
|
N, M, X, Y = map(int, input().split())
list_x = [int(i) for i in input().split()]
list_y = [int(j) for j in input().split()]
if max(list_x) >= min(list_y):
print('War')
else:
if X < max(list_x) + 1 <= Y:
print('No War')
else:
print('War')
|
s553352797
|
p03578
|
u500297289
| 2,000
| 262,144
|
Wrong Answer
| 277
| 41,816
| 384
|
Rng is preparing a problem set for a qualification round of CODEFESTIVAL. He has N candidates of problems. The difficulty of the i-th candidate is D_i. There must be M problems in the problem set, and the difficulty of the i-th problem must be T_i. Here, one candidate of a problem cannot be used as multiple problems. Determine whether Rng can complete the problem set without creating new candidates of problems.
|
N = int(input())
D = list(map(int, input().split()))
M = int(input())
T = list(map(int, input().split()))
dic = {}
for d in D:
if d in dic.keys():
dic[d] += 1
else:
dic[d] = 1
for t in T:
if t in dic.keys():
if dic[t] == 0:
print('No')
exit()
dic[t] -= 1
else:
print('No')
exit()
print('Yes')
|
s467873877
|
Accepted
| 267
| 41,316
| 384
|
N = int(input())
D = list(map(int, input().split()))
M = int(input())
T = list(map(int, input().split()))
dic = {}
for d in D:
if d in dic.keys():
dic[d] += 1
else:
dic[d] = 1
for t in T:
if t in dic.keys():
if dic[t] == 0:
print('NO')
exit()
dic[t] -= 1
else:
print('NO')
exit()
print('YES')
|
s551329003
|
p02578
|
u589734885
| 2,000
| 1,048,576
|
Wrong Answer
| 82
| 32,176
| 232
|
N persons are standing in a row. The height of the i-th person from the front is A_i. We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person: Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool. Find the minimum total height of the stools needed to meet this goal.
|
def main():
n = int(input())
a = list(map(int, input().split(" ")))
top = a[0]
ans = 0
for height in a:
if top < height:
ans += height - top
top = height
print(ans)
main()
|
s499169650
|
Accepted
| 92
| 32,256
| 242
|
def main():
n = int(input())
a = list(map(int, input().split(" ")))
top = a[0]
ans = 0
for i in range(n):
if a[i] < top:
ans += top - a[i]
else:
top = a[i]
print(ans)
main()
|
s019231152
|
p02865
|
u660245210
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 71
|
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
N = int(input())
if N % 2 == 1:
print((N-1) / 2)
else: print(N/2 - 1)
|
s536570298
|
Accepted
| 17
| 2,940
| 34
|
N = int(input())
print((N-1) // 2)
|
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