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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s814581844
|
p03598
|
u094565093
| 2,000 | 262,144 |
Wrong Answer
| 1,205 | 18,752 | 175 |
There are N balls in the xy-plane. The coordinates of the i-th of them is (x_i, i). Thus, we have one ball on each of the N lines y = 1, y = 2, ..., y = N. In order to collect these balls, Snuke prepared 2N robots, N of type A and N of type B. Then, he placed the i-th type-A robot at coordinates (0, i), and the i-th type-B robot at coordinates (K, i). Thus, now we have one type-A robot and one type-B robot on each of the N lines y = 1, y = 2, ..., y = N. When activated, each type of robot will operate as follows. * When a type-A robot is activated at coordinates (0, a), it will move to the position of the ball on the line y = a, collect the ball, move back to its original position (0, a) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. * When a type-B robot is activated at coordinates (K, b), it will move to the position of the ball on the line y = b, collect the ball, move back to its original position (K, b) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. Snuke will activate some of the 2N robots to collect all of the balls. Find the minimum possible total distance covered by robots.
|
import numpy as np
X=int(input())
Y=int(np.sqrt(X))+1
max=1
for j in range(2,Y+1):
for i in range(1,Y+1):
if i**j<=X and i**j>max:
max=i**j
print(max)
|
s453054315
|
Accepted
| 17 | 3,060 | 190 |
N=int(input())
K=int(input())
S=list(map(int, input().split()))
total=0
for i in range(N):
if abs(S[i]-K)<S[i]:
total+=abs(S[i]-K)*2
else:
total+=2*S[i]
print(total)
|
s833981746
|
p03090
|
u145035045
| 2,000 | 1,048,576 |
Wrong Answer
| 25 | 3,996 | 195 |
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())
lst = []
for i in range(1, n + 1):
for j in range(i, n + 1):
if i != j and i + j != n + 1:
lst.append((i, j))
print(len(lst))
for s in lst:
print(*s)
|
s831262190
|
Accepted
| 25 | 3,996 | 499 |
n = int(input())
if n % 2 == 0:
lst = []
for i in range(1, n + 1):
for j in range(i, n + 1):
if i != j and i + j != n + 1:
lst.append((i, j))
print(len(lst))
for s in lst:
print(*s)
else:
lst = []
for i in range(1, n):
for j in range(i, n):
if i != j and i + j != n:
lst.append((i, j))
for i in range(1, n):
lst.append((i, n))
print(len(lst))
for s in lst:
print(*s)
|
s221255108
|
p03486
|
u995861601
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 66 |
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.
|
print("Yes") if sorted(input()) < sorted(input()) else print("No")
|
s106712651
|
Accepted
| 18 | 2,940 | 80 |
print("Yes") if sorted(input()) < sorted(input(), reverse=True) else print("No")
|
s099238175
|
p02261
|
u128811851
| 1,000 | 131,072 |
Wrong Answer
| 20 | 6,828 | 2,006 |
Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
MARK = 0
NUMBER = 1
def bubble_sort(A):
for i in range(len(A)-1):
for j in reversed(range(i+1, len(A))):
if A[j][NUMBER] < A[j - 1][NUMBER]:
swap(A, j-1, j)
def selection_sort(A):
for i in range(len(A)):
mini = i
for j in range(i, len(A)):
if A[j][NUMBER] < A[mini][NUMBER]:
mini = j
if mini != i:
swap(A, mini, i)
def swap(A, i, j):
tmp = A[j]
A[j] = A[i]
A[i] = tmp
amount = int(input())
cards = input().split() # list of cards (ex:[H2 S8])
same_numbers = []
isStable_bubble = True
isStable_selection = True
# make tuple of mark and number
for i in range(amount):
cards[i] = tuple([cards[i][MARK], int(cards[i][NUMBER])])
# make list of (pairs of) cards which have the same number
for i in range(amount-1):
number = cards[i][NUMBER]
for j in range(i+1, amount):
if number == cards[j][NUMBER]:
same_numbers.append(cards[i])
same_numbers.append(cards[j])
cards_bubble = list(cards)
cards_selection = list(cards)
bubble_sort(cards_bubble)
selection_sort(cards_selection)
# jugle stable or not
for i in range(len(same_numbers)):
for j in range(amount):
if same_numbers[i][NUMBER] == cards_bubble[j][NUMBER]:
if same_numbers[i] == cards_bubble[j]:
isStable_bubble = True
else:
isStable_bubble = False
if same_numbers[i][NUMBER] == cards_selection[j][NUMBER]:
if same_numbers[i] == cards_selection[j]:
isStable_selection = True
else:
isStable_selection = False
print(*[cards_bubble[i][MARK] + str(cards_bubble[i][NUMBER])
for i in range(amount)])
if isStable_bubble:
print("Stable")
else:
print("Not Stable")
print(*[cards_selection[i][MARK] + str(cards_selection[i][NUMBER])
for i in range(amount)])
if isStable_selection:
print("Stable")
else:
print("Not Stable")
|
s571217822
|
Accepted
| 40 | 6,780 | 1,268 |
MARK = 0
NUMBER = 1
def bubble_sort(A):
for i in range(len(A)-1):
for j in reversed(range(i+1, len(A))):
if A[j][NUMBER] < A[j - 1][NUMBER]:
swap(A, j-1, j)
def selection_sort(A):
for i in range(len(A)):
mini = i
for j in range(i, len(A)):
if A[j][NUMBER] < A[mini][NUMBER]:
mini = j
if mini != i:
swap(A, mini, i)
def swap(A, i, j):
tmp = A[j]
A[j] = A[i]
A[i] = tmp
amount = int(input())
cards = input().split() # list of cards (ex:[H2 S8])
isStable_selection = True
# make tuple of mark and number
for i in range(amount):
cards[i] = tuple([cards[i][MARK], int(cards[i][NUMBER])])
cards_bubble = list(cards)
cards_selection = list(cards)
bubble_sort(cards_bubble)
selection_sort(cards_selection)
# judge stable or not (bubble sort is stable)
if cards_bubble == cards_selection:
isStable_selection = True
else:
isStable_selection = False
print(*[cards_bubble[i][MARK] + str(cards_bubble[i][NUMBER])
for i in range(amount)])
print("Stable")
print(*[cards_selection[i][MARK] + str(cards_selection[i][NUMBER])
for i in range(amount)])
if isStable_selection:
print("Stable")
else:
print("Not stable")
|
s517614358
|
p03970
|
u744592787
| 2,000 | 262,144 |
Wrong Answer
| 24 | 3,064 | 99 |
CODE FESTIVAL 2016 is going to be held. For the occasion, Mr. Takahashi decided to make a signboard. He intended to write `CODEFESTIVAL2016` on it, but he mistakenly wrote a different string S. Fortunately, the string he wrote was the correct length. So Mr. Takahashi decided to perform an operation that replaces a certain character with another in the minimum number of iterations, changing the string to `CODEFESTIVAL2016`. Find the minimum number of iterations for the rewrite operation.
|
# -*- coding: utf-8 -*-
s = input()
if s != 'CODEFESTIVAL2016':
s = 'CODEFESTIVAL2016'
print(s)
|
s799080759
|
Accepted
| 22 | 3,064 | 153 |
# -*- coding: utf-8 -*-
ans = 'CODEFESTIVAL2016'
s = input()
count = 0
for i in range(len(ans)):
if ans[i] != s[i]:
count += 1
print(count)
|
s263590022
|
p03729
|
u102126195
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 102 |
You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`.
|
A, B, C = list(input().split())
if A[:1] == B[0] and B[-1] == C[0]:
print("YES")
else:
print("NO")
|
s578360101
|
Accepted
| 17 | 2,940 | 103 |
A, B, C = list(input().split())
if A[-1] == B[0] and B[-1] == C[0]:
print("YES")
else:
print("NO")
|
s182061449
|
p04029
|
u973108807
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 34 |
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
n = int(input())
print(n*(n-1)//2)
|
s091432422
|
Accepted
| 18 | 2,940 | 34 |
n = int(input())
print(n*(n+1)//2)
|
s480991227
|
p03486
|
u089376182
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 57 |
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.
|
print('Yes' if sorted(input())<sorted(input()) else 'No')
|
s657109361
|
Accepted
| 17 | 2,940 | 64 |
print('Yes' if sorted(input())<sorted(input())[::-1] else 'No')
|
s578703114
|
p03760
|
u977661421
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 241 |
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions. You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
|
# -*- coding: utf-8 -*-
o = list(input())
e = list(input())
for i in range(min(len(o), len(e))):
print(o[i], end = '')
print(e[i], end = '')
if len(e) > len(o):
print(e[len(e) - 1])
if len(o) < len(e):
print(o[len(o) - 1])
|
s008659945
|
Accepted
| 17 | 3,064 | 468 |
# -*- coding: utf-8 -*-
o = list(input())
e = list(input())
len_o = len(o)
len_e = len(e)
if len_o == len_e:
for i in range(len_o - 1):
print(o[i], end = '')
print(e[i], end = '')
print(o[len_o - 1], end = '')
print(e[len_e - 1])
else:
for i in range(min(len_o, len_e)):
print(o[i], end = '')
print(e[i], end = '')
if len_e > len_o:
print(e[len(e) - 1])
if len_o > len_e:
print(o[len(o) - 1])
|
s281615562
|
p03657
|
u121732701
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 117 |
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.
|
A, B= map(int, input().split())
if A+B%3==0 or A%3==0 or B%3==0:
print("Possible")
else:
print("Impossible")
|
s073088490
|
Accepted
| 18 | 2,940 | 119 |
A, B= map(int, input().split())
if (A+B)%3==0 or A%3==0 or B%3==0:
print("Possible")
else:
print("Impossible")
|
s041551865
|
p03494
|
u733774002
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,104 | 2,940 | 234 |
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:
i = 0
for i in range(N):
if A[i] % 2 == 0:
A[i] /=2
i += 1
else:
i += 1
break
print(cnt)
|
s286867549
|
Accepted
| 19 | 3,060 | 145 |
N = input()
A = list(map(int, input().split()))
cnt = 0
while all(i % 2 == 0 for i in A):
A = [int(j / 2) for j in A]
cnt += 1
print(cnt)
|
s679724047
|
p03023
|
u902577051
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 150 |
Given an integer N not less than 3, find the sum of the interior angles of a regular polygon with N sides. Print the answer in degrees, but do not print units.
|
# -*- coding: utf-8 -*-
s = input()
num_win = s.count('o')
num_rest = 15 - len(s)
if num_win + num_rest > 7:
print('YES')
else:
print('NO')
|
s385245422
|
Accepted
| 17 | 2,940 | 63 |
# -*- coding: utf-8 -*-
n = int(input())
print(180 * (n - 2))
|
s561158056
|
p03457
|
u977349332
| 2,000 | 262,144 |
Wrong Answer
| 857 | 3,316 | 226 |
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 math
n = int(input())
X = 0
Y = 0
T = 0
for i in range(1, n):
t,x,y = [int(i) for i in input().split()]
sum = abs(X-x)+abs(Y-y)
if ( sum )<T-t and sum%2 == T-t:
X = x
Y = y
else:
print('No')
print('Yes')
|
s158057058
|
Accepted
| 198 | 3,060 | 197 |
import math
import sys
n = int(input())
for i in range(n):
t,x,y = map(int, sys.stdin.readline().split())
d = abs(x)+abs(y)
if d > t or (d - t) % 2 != 0:
print('No')
exit()
print('Yes')
|
s255416332
|
p03998
|
u193927973
| 2,000 | 262,144 |
Wrong Answer
| 31 | 9,264 | 369 |
Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.
|
a=input()
b=input()
c=input()
from collections import deque
da=deque(list(a))
db=deque(list(b))
dc=deque(list(c))
now=da.popleft()
while (1):
if now=="a":
if da:
now=da.popleft()
else:
break
elif now=="b":
if db:
now=db.popleft()
else:
break
elif now=="c":
if dc:
now=dc.popleft()
else:
break
print(now)
|
s748322444
|
Accepted
| 32 | 9,360 | 377 |
a=input()
b=input()
c=input()
from collections import deque
da=deque(list(a))
db=deque(list(b))
dc=deque(list(c))
now=da.popleft()
while (1):
if now=="a":
if da:
now=da.popleft()
else:
break
elif now=="b":
if db:
now=db.popleft()
else:
break
elif now=="c":
if dc:
now=dc.popleft()
else:
break
print(now.upper())
|
s106468749
|
p03408
|
u013408661
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 275 |
Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i. Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen. Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.) At most how much can he earn on balance? Note that the same string may be written on multiple cards.
|
n=int(input())
s=[]
for i in range(n):
k=input()
s.append(k)
m=int(input())
t=[]
for i in range(m):
k=input()
t.append(k)
check=[]
ans=0
for i in s:
if i in check:
continue
if s.count(i)>=t.count(i):
ans+=s.count(i)-t.count(i)
check.append(i)
print(ans)
|
s799620290
|
Accepted
| 18 | 3,064 | 258 |
n=int(input())
s=[]
for i in range(n):
k=input()
s.append(k)
m=int(input())
t=[]
for i in range(m):
k=input()
t.append(k)
check=[]
ans=[0]
for i in s:
if i in check:
continue
ans.append(s.count(i)-t.count(i))
check.append(i)
print(max(ans))
|
s798595667
|
p02401
|
u215732964
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,600 | 270 |
Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.
|
while True:
a, op, b = input().split()
if op == '+':
print(int(a) + int(b))
elif op == '-':
print(int(a) - int(b))
elif op == '*':
print(int(a) * int(b))
elif op == '/':
print(int(a) / int(b))
else:
break
|
s753096917
|
Accepted
| 20 | 5,596 | 271 |
while True:
a, op, b = input().split()
if op == '+':
print(int(a) + int(b))
elif op == '-':
print(int(a) - int(b))
elif op == '*':
print(int(a) * int(b))
elif op == '/':
print(int(a) // int(b))
else:
break
|
s140036669
|
p03760
|
u027641915
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 153 |
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions. You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
|
o = input()
e = input()
password = ''
for i in range(len(o)):
password += o[i]
if (i != len(o) -1):
password += e[i]
print(password)
|
s619632584
|
Accepted
| 17 | 3,064 | 188 |
o = input()
e = input()
password = ''
for i in range(len(o)):
password += o[i]
if (len(o) > len(e)) and ((len(o) - 1) == i):
break
password += e[i]
print(password)
|
s841791021
|
p03623
|
u627530854
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 84 |
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) = (int(tok) for tok in input().split())
print(min(abs(x - a), abs(x - b)))
|
s715206629
|
Accepted
| 18 | 2,940 | 143 |
(x, a, b) = (int(tok) for tok in input().split())
dist_a = abs(x - a)
dist_b = abs(x - b)
print("A" if dist_a == min(dist_a, dist_b) else "B")
|
s719561656
|
p03160
|
u426930857
| 2,000 | 1,048,576 |
Wrong Answer
| 116 | 13,980 | 182 |
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())
l=list(map(int,input().split()))
dp=[0 for i in range(n)]
dp[1]=l[1]-l[0]
for i in range(2,n):
dp[i]=min(dp[i-1]+l[i]-l[i-1],dp[i-2]+l[i]-l[i-2])
print(dp[n-1])
|
s840703163
|
Accepted
| 127 | 13,980 | 208 |
n=int(input())
l=list(map(int,input().split()))
dp=[0 for i in range(n)]
dp[1]=abs(l[1]-l[0])
for i in range(2,n):
dp[i]=min(dp[i-1]+abs(l[i]-l[i-1]),dp[i-2]+abs(l[i]-l[i-2]))
print(dp[n-1])
#print(dp)
|
s170335059
|
p03214
|
u154756110
| 2,525 | 1,048,576 |
Wrong Answer
| 17 | 3,064 | 183 |
Niwango-kun is an employee of Dwango Co., Ltd. One day, he is asked to generate a thumbnail from a video a user submitted. To generate a thumbnail, he needs to select a frame of the video according to the following procedure: * Get an integer N and N integers a_0, a_1, ..., a_{N-1} as inputs. N denotes the number of the frames of the video, and each a_i denotes the representation of the i-th frame of the video. * Select t-th frame whose representation a_t is nearest to the average of all frame representations. * If there are multiple such frames, select the frame with the smallest index. Find the index t of the frame he should select to generate a thumbnail.
|
N=int(input())
A=list(map(int,input().split()))
sum=0.0
for i in range(N):
sum+=A[i]
sum/=N
X=0
ans=0
for i in range(N):
A[i]=abs(A[i]-sum)
if(A[i]<A[X]):
ans=i
print(ans+1)
|
s405935814
|
Accepted
| 18 | 3,064 | 176 |
N=int(input())
A=list(map(int,input().split()))
sum=0
for i in range(N):
sum+=A[i]
sum
ans=0
for i in range(N):
A[i]=abs(A[i]*N-sum)
if(A[i]<A[ans]):
ans=i
print(ans)
|
s355524288
|
p04011
|
u008357982
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 50 |
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,k,x,y=map(int,open(0));print([n*k,(n-k)*y][n<k])
|
s488775257
|
Accepted
| 17 | 2,940 | 54 |
n,k,x,y=map(int,open(0));print([n*x,k*x+(n-k)*y][n>k])
|
s598411799
|
p03251
|
u422272120
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 175 |
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()))
if max(x) + 1 < min(y):
print ("No War")
else:
print ("War")
|
s380483493
|
Accepted
| 17 | 2,940 | 200 |
N,M,X,Y = map(int,input().split())
x = list(map(int,input().split()))
y = list(map(int,input().split()))
print ("No War") if max(x) < min(y) and (X < max(x) < Y or X < min(y) <= Y) else print ("War")
|
s981552279
|
p03477
|
u074220993
| 2,000 | 262,144 |
Wrong Answer
| 26 | 9,016 | 168 |
A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R. Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan. Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right.
|
S = list(input())
lenS = len(S)
S = S + ['.']
K = lenS
i = 0
ps = S[0]
for s in S:
if s != ps:
K = min(K, max(i, lenS-i))
ps = s
i += 1
print(K)
|
s239928415
|
Accepted
| 29 | 9,080 | 134 |
A, B, C, D = map(int, input().split())
if A+B == C+D:
print('Balanced')
elif A+B > C+D:
print('Left')
else:
print('Right')
|
s870094019
|
p03407
|
u662449766
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,060 | 174 |
An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.
|
import sys
input = sys.stdin.readline
def main():
a, b, c = map(int, input().split())
print("YES" if c <= a + b else "NO")
if __name__ == "__main__":
main()
|
s025690802
|
Accepted
| 17 | 2,940 | 174 |
import sys
input = sys.stdin.readline
def main():
a, b, c = map(int, input().split())
print("Yes" if c <= a + b else "No")
if __name__ == "__main__":
main()
|
s373276518
|
p03998
|
u654558363
| 2,000 | 262,144 |
Wrong Answer
| 21 | 3,316 | 421 |
Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.
|
from collections import deque
if __name__ == "__main__":
stacks = []
for i in range(3):
stacks.append(deque([x for x in input()]))
winner = False
turn = 0
while not winner:
turn = ord(stacks[turn].popleft()) - ord('a')
for i in range(3):
if len(stacks[i]) == 0:
print(i)
winner = chr(97 + i)
break
print(winner)
|
s038603971
|
Accepted
| 21 | 3,316 | 345 |
from collections import deque
if __name__ == "__main__":
stacks = []
for i in range(3):
stacks.append(deque([x for x in input()]))
winner = False
turn = 0
while not winner:
turn = ord(stacks[turn].popleft()) - ord('a')
if len(stacks[turn]) == 0:
winner = chr(65 + turn)
print(winner)
|
s690400989
|
p03079
|
u920204936
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 137 |
You are given three integers A, B and C. Determine if there exists an equilateral triangle whose sides have lengths A, B and C.
|
length = list(map(int, input().split()))
if length[0] == length[1] and length[1] == length[2]:
print(True)
else:
print(False)
|
s018558844
|
Accepted
| 18 | 2,940 | 137 |
length = list(map(int, input().split()))
if length[0] == length[1] and length[1] == length[2]:
print("Yes")
else:
print("No")
|
s969816211
|
p02258
|
u500396695
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,532 | 136 |
You can obtain profits from foreign exchange margin transactions. For example, if you buy 1000 dollar at a rate of 100 yen per dollar, and sell them at a rate of 108 yen per dollar, you can obtain (108 - 100) × 1000 = 8000 yen. Write a program which reads values of a currency $R_t$ at a certain time $t$ ($t = 0, 1, 2, ... n-1$), and reports the maximum value of $R_j - R_i$ where $j > i$ .
|
n = int(input())
R = []
for i in range(n):
R.append(int(input()))
r = sorted(R)
maximum_profit = r[n-1] - r[0]
print(maximum_profit)
|
s034708120
|
Accepted
| 510 | 7,664 | 186 |
n = int(input())
# R[0]
minv = int(input())
maxv = - 10 ** 10
# R[1..N-1]
for j in range(1, n):
r = int(input())
maxv = max(maxv, r - minv)
minv = min(minv, r)
print(maxv)
|
s338644676
|
p04029
|
u439392790
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 33 |
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
x=int(input())
print((x/2)*(x+1))
|
s791976061
|
Accepted
| 17 | 2,940 | 37 |
n=int(input())
print(int (n*(n+1)/2))
|
s707711957
|
p02601
|
u517389396
| 2,000 | 1,048,576 |
Wrong Answer
| 34 | 9,224 | 317 |
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.
|
import sys
def check(r,g,b):
return r<g and g<b
r,g,b=input().split()
r,g,b=int(r),int(g),int(b)
K=int(input())
for i in range(K):
for j in range(K):
for k in range(K):
if check(r*2**(i+1),g*2**(j+1),b*2**(k+1)):
print("True")
sys.exit(0)
print("False")
|
s978479855
|
Accepted
| 34 | 9,160 | 332 |
import sys
def check(r,g,b):
return r<g and g<b
r,g,b=input().split()
r,g,b=int(r),int(g),int(b)
K=int(input())+1
for i in range(K):
for j in range(K-i):
for k in range(K-i-j):
if check(r*2**(i),g*2**(j),b*2**(k)):
print("Yes")
sys.exit(0)
print("No")
|
s584486600
|
p03007
|
u952467214
| 2,000 | 1,048,576 |
Wrong Answer
| 235 | 14,228 | 830 |
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.
|
import sys
sys.setrecursionlimit(10 ** 7)
input = sys.stdin.readline
n = int(input())
a = list( map(int, input().split()))
a.sort()
from bisect import bisect_left
from bisect import bisect_right
pos = n - bisect_left(a, 1)
neg = bisect_right(a, -1)
if pos>=neg:
aa = a[neg-1:n-neg+1]
for i in range(neg-1):
print(a[-1-i], a[i])
aa.append(a[-1-i] - a[i])
aa.sort()
tmp = aa[0]
for i in range(1,len(aa)-1):
print(tmp, a[i])
tmp = tmp-a[i]
print(a[-1], tmp)
exit()
if pos<=neg:
aa = a[pos-1:n-pos+1]
for i in range(pos-1):
print(a[i], a[-1-i])
aa.append(a[i] - a[-1-i])
aa.sort()
tmp = aa[-1]
print(aa[-1], aa[0])
tmp = aa[-1]-aa[0]
for i in range(1,len(aa)-1):
print(tmp, a[i])
tmp = tmp-a[i]
exit()
|
s954581493
|
Accepted
| 273 | 19,988 | 1,001 |
import sys
sys.setrecursionlimit(10 ** 7)
input = sys.stdin.readline
n = int(input())
a = list( map(int, input().split()))
a.sort()
from bisect import bisect_left
from bisect import bisect_right
pos = n - bisect_left(a, 1)
neg = bisect_right(a, -1)
ans_exec = []
if pos>=neg:
aa = a[max(0,neg-1):n-neg+1]
for i in range(neg-1):
ans_exec.append((a[-1-i], a[i]))
aa.append(a[-1-i] - a[i])
aa.sort()
tmp = aa[0]
for i in range(1,len(aa)-1):
ans_exec.append((tmp, aa[i]))
tmp = tmp-aa[i]
ans_exec.append((aa[-1], tmp))
ans = aa[-1] - tmp
elif pos<=neg:
aa = a[max(0,pos-1):n-pos+1]
for i in range(pos-1):
ans_exec.append((a[i], a[-1-i]))
aa.append(a[i] - a[-1-i])
aa.sort()
tmp = aa[-1]
ans_exec.append((aa[-1], aa[0]))
tmp = aa[-1]-aa[0]
for i in range(1,len(aa)-1):
ans_exec.append((tmp, aa[i]))
tmp = tmp-aa[i]
ans = tmp
print(ans)
for a,b in ans_exec:
print(a,b)
|
s645370860
|
p03385
|
u544050502
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 24 |
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
print(len(set(input())))
|
s644560292
|
Accepted
| 17 | 2,940 | 46 |
print("Yes" if len(set(input()))==3 else "No")
|
s709540678
|
p03351
|
u341855122
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 158 |
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.
|
n = list(map(int,input().split()))
if ((n[3] > abs(n[0]-n[1])) and( n[3] > abs(n[1]-n[2]))) or( n[3] > abs(n[0]-n[2])):
print('Yes')
else:
print('No')
|
s640273188
|
Accepted
| 17 | 3,064 | 161 |
n = list(map(int,input().split()))
if ((n[3] >= abs(n[0]-n[1])) and( n[3] >= abs(n[1]-n[2]))) or( n[3] >= abs(n[0]-n[2])):
print('Yes')
else:
print('No')
|
s948834815
|
p02742
|
u870518235
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 112 |
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
h, w = map(int, input().split())
if h % 2 != 0 and w % 2 != 0:
print((h*w-1)/2 + 1)
else:
print((h*w)/2)
|
s850020381
|
Accepted
| 18 | 2,940 | 179 |
h, w = map(int, input().split())
if h == 1 or w == 1:
print(1)
else:
if h % 2 != 0 and w % 2 != 0:
print(int((h*w-1)/2 + 1))
else:
print(int((h*w)/2))
|
s074529987
|
p03251
|
u672475305
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,064 | 297 |
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())
lstx = list(map(int,input().split()))
lsty = list(map(int,input().split()))
lstx.append(x)
lsty.append(y)
maxx = max(lstx)
miny = min(lsty)
for z in range(x,y+1):
if (maxx<z) and (miny>=z):
print(z)
print('No War')
exit()
print('War')
|
s997432322
|
Accepted
| 17 | 3,064 | 256 |
n,m,x,y = map(int,input().split())
lstx = list(map(int,input().split()))
lsty = list(map(int,input().split()))
max_x = max(lstx)
min_y = min(lsty)
for z in range(x+1,y+1):
if (max_x<z) and (min_y>=z):
print('No War')
exit()
print('War')
|
s722541466
|
p03910
|
u463655976
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,060 | 114 |
The problem set at _CODE FESTIVAL 20XX Finals_ consists of N problems. The score allocated to the i-th (1≦i≦N) problem is i points. Takahashi, a contestant, is trying to score exactly N points. For that, he is deciding which problems to solve. As problems with higher scores are harder, he wants to minimize the highest score of a problem among the ones solved by him. Determine the set of problems that should be solved.
|
import math
N = int(input()) * 2
i = math.floor(math.sqrt(N))
while N - 2 * i * (i+1) > 2 * i:
i += 1
print(i)
|
s884798956
|
Accepted
| 21 | 3,408 | 174 |
import math
N = int(input())
i = math.floor(math.sqrt(2*N))
while i * (i+1) // 2 - N < 0:
i += 1
x = i * (i+1) // 2 - N
for j in range(1, i+1):
if j != x:
print(j)
|
s535239610
|
p02409
|
u636711749
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,720 | 360 |
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.
|
data = [[[0 for r in range(10)] for f in range(3)] for b in range(4)]
n = int(input())
for nc in range(n):
(b, f, r, v) = [int(i) for i in input().split()]
data[b - 1][f - 1][r - 1] += v
for b in range(4):
for f in range(3):
for r in range(10):
print(' {0}'.format(data[b][f][r]), end='')
print()
print('#' * 20)
|
s898129215
|
Accepted
| 30 | 6,724 | 391 |
data = [[[0 for r in range(10)] for f in range(3)] for b in range(4)]
n = int(input())
for nc in range(n):
(b, f, r, v) = [int(i) for i in input().split()]
data[b - 1][f - 1][r - 1] += v
for b in range(4):
for f in range(3):
for r in range(10):
print(' {0}'.format(data[b][f][r]), end='')
print()
print('#' * 20) if b < 4 -1 else print(end='')
|
s144691920
|
p03162
|
u050698451
| 2,000 | 1,048,576 |
Wrong Answer
| 2,106 | 90,080 | 504 |
Taro's summer vacation starts tomorrow, and he has decided to make plans for it now. The vacation consists of N days. For each i (1 \leq i \leq N), Taro will choose one of the following activities and do it on the i-th day: * A: Swim in the sea. Gain a_i points of happiness. * B: Catch bugs in the mountains. Gain b_i points of happiness. * C: Do homework at home. Gain c_i points of happiness. As Taro gets bored easily, he cannot do the same activities for two or more consecutive days. Find the maximum possible total points of happiness that Taro gains.
|
N = int(input())
abc = []
for _ in range(N):
a, b, c = map(int, input().split())
abc.append([a,b,c])
dp = [[0, 0, 0] for _ in range(N+1)]
print(abc)
for i in range(1,N+1):
dp[i][0] = max(dp[i][0], dp[i-1][1]+abc[i-1][0])
dp[i][0] = max(dp[i][0], dp[i-1][2]+abc[i-1][0])
dp[i][1] = max(dp[i][1], dp[i-1][0]+abc[i-1][1])
dp[i][1] = max(dp[i][1], dp[i-1][2]+abc[i-1][1])
dp[i][2] = max(dp[i][2], dp[i-1][0]+abc[i-1][2])
dp[i][2] = max(dp[i][2], dp[i-1][1]+abc[i-1][2])
print(dp)
print(max(dp[N]))
|
s130301814
|
Accepted
| 724 | 42,532 | 508 |
N = int(input())
abc = []
for _ in range(N):
a, b, c = map(int, input().split())
abc.append([a,b,c])
dp = [[0, 0, 0] for _ in range(N+1)]
# print(abc)
for i in range(1,N+1):
dp[i][0] = max(dp[i][0], dp[i-1][1]+abc[i-1][0])
dp[i][0] = max(dp[i][0], dp[i-1][2]+abc[i-1][0])
dp[i][1] = max(dp[i][1], dp[i-1][0]+abc[i-1][1])
dp[i][1] = max(dp[i][1], dp[i-1][2]+abc[i-1][1])
dp[i][2] = max(dp[i][2], dp[i-1][0]+abc[i-1][2])
dp[i][2] = max(dp[i][2], dp[i-1][1]+abc[i-1][2])
# print(dp)
print(max(dp[N]))
|
s199706062
|
p02417
|
u879471116
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,560 | 341 |
Write a program which counts and reports the number of each alphabetical letter. Ignore the case of characters.
|
alphabet = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k',
'l', 'm', 'n', 'o', 'p', 'r', 's', 't', 'u', 'v', 'w',
'x', 'y', 'z']
dic = {}
for c in alphabet:
dic[c] = 0
string = list(input().rstrip().lower())
for c in string:
if c in alphabet:
dic[c] = dic[c] + 1
for key in dic:
print('{} : {}'.format(key, dic[key]))
|
s252183724
|
Accepted
| 20 | 5,560 | 394 |
alphabet = ['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']
dic = {}
for c in alphabet:
dic[c] = 0
while True:
try:
string = list(input().rstrip().lower())
except EOFError:
break
for c in string:
if c in alphabet:
dic[c] = dic[c] + 1
for key in dic:
print('{} : {}'.format(key, dic[key]))
|
s781226185
|
p03469
|
u427984570
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 34 |
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
s = input()
print("2018" + s[:4])
|
s267345239
|
Accepted
| 18 | 2,940 | 35 |
s = input()
print("2018" + s[4:])
|
s215953341
|
p03110
|
u681110193
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 150 |
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?
|
n=int(input())
for i in range(n):
ans=0
a,b=map(str,input().split())
if b=='JPY':
ans+=int(a)
else:
ans+=float(a)*380000
print(ans)
|
s402235580
|
Accepted
| 17 | 2,940 | 148 |
n=int(input())
ans=0
for i in range(n):
a,b=map(str,input().split())
if b=='JPY':
ans+=int(a)
else:
ans+=float(a)*380000
print(ans)
|
s177166091
|
p03150
|
u137724047
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 166 |
A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string.
|
#!/usr/bin/python3
s = input().strip()
t = "keyence"
for x in range(0,len(t)+1):
if s.startswith(t[x:]) and s.endswith(t[:x]):
print("YES")
exit(0)
print("NO")
|
s229964997
|
Accepted
| 17 | 2,940 | 166 |
#!/usr/bin/python3
s = input().strip()
t = "keyence"
for x in range(0,len(t)+1):
if s.startswith(t[:x]) and s.endswith(t[x:]):
print("YES")
exit(0)
print("NO")
|
s806489107
|
p03377
|
u417835834
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 95 |
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
A,B,X = map(int,input().split())
if X >= A and X <= A+B:
print('Yes')
else:
print('No')
|
s103990943
|
Accepted
| 17 | 2,940 | 95 |
A,B,X = map(int,input().split())
if X >= A and X <= A+B:
print('YES')
else:
print('NO')
|
s971196520
|
p03486
|
u288430479
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 114 |
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
|
s = sorted(input())
t = sorted(input())
s = ''.join(s)
t = ''.join(t)
if s < t:
print('Yes')
else:
print('No')
|
s189237873
|
Accepted
| 23 | 2,940 | 120 |
s = sorted(input())
t = sorted(input())[::-1]
s = ''.join(s)
t = ''.join(t)
if s < t:
print('Yes')
else:
print('No')
|
s188878459
|
p03455
|
u052499405
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 117 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = [int(item) for item in input().split()]
if a % 2 == 1 and b % 2 == 1:
print("Even")
else:
print("Odd")
|
s060806784
|
Accepted
| 18 | 2,940 | 117 |
a, b = [int(item) for item in input().split()]
if a % 2 == 1 and b % 2 == 1:
print("Odd")
else:
print("Even")
|
s381688614
|
p03160
|
u345966487
| 2,000 | 1,048,576 |
Wrong Answer
| 2,109 | 22,684 | 293 |
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 numpy as np
import sys
f = sys.stdin
n = int(f.readline())
h = list(map(int, f.readline().split()))
h = np.array(h)
dp = np.zeros(n, int)
for i in range(n - 1):
np.minimum(
dp[1:-1] + np.abs(h[2:] - h[1:-1]), dp[:-2] + np.abs(h[2:] - h[:-2]), out=dp[2:]
)
print(dp[-1])
|
s855591084
|
Accepted
| 115 | 13,980 | 201 |
n = int(input())
h = list(map(int, input().split()))
p0, p1 = 0, abs(h[1] - h[0])
for i in range(2, n):
p2 = min(p1 + abs(h[i] - h[i - 1]), p0 + abs(h[i] - h[i - 2]))
p0, p1 = p1, p2
print(p1)
|
s082326875
|
p03448
|
u881116515
| 2,000 | 262,144 |
Wrong Answer
| 50 | 3,060 | 211 |
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
a = int(input())
b = int(input())
c = int(input())
x = int(input())
ans = 0
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
if i*500+j*100*k*50 == x:
ans += 1
print(ans)
|
s380544629
|
Accepted
| 49 | 3,060 | 212 |
a = int(input())
b = int(input())
c = int(input())
x = int(input())
ans = 0
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
if i*500+j*100+k*50 == x:
ans += 1
print(ans)
|
s711028269
|
p03697
|
u622011073
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 48 |
You are given two integers A and B as the input. Output the value of A + B. However, if A + B is 10 or greater, output `error` instead.
|
a=input();print('no'*len(a)==len(set(a))or'yes')
|
s809504684
|
Accepted
| 17 | 2,940 | 56 |
a,b=map(int,input().split());print([a+b,'error'][a+b>9])
|
s550057627
|
p02612
|
u617225232
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 9,028 | 65 |
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.
|
a = int(input())
print(0) if a%1000==0 else print(1000-(a//1000))
|
s613169332
|
Accepted
| 29 | 9,064 | 65 |
a = int(input())
print(0) if a%1000==0 else print(1000-(a%1000))
|
s995128097
|
p03049
|
u448354193
| 2,000 | 1,048,576 |
Wrong Answer
| 43 | 3,188 | 468 |
Snuke has N strings. The i-th string is s_i. Let us concatenate these strings into one string after arranging them in some order. Find the maximum possible number of occurrences of `AB` in the resulting string.
|
import re
n = int(input())
s = []
compiled = re.compile(r"AB")
pre = 0
xa=0
bx=0
ba=0
for i in range(n):
one = input()
pre+=len(compiled.findall(one))
f=one[0]
l=one[-1]
if f=="B" and l=="A":
ba+=1
elif f=="B" and l!="A":
bx+=1
elif f!="B" and l=="A":
xa+=1
print(xa,bx,ba,pre)
cnt=ba-1
if xa>=1:
xa-=1
cnt+1
if bx>=1:
bx-=1
cnt+1
cnt += xa if xa>bx else bx
print(cnt+pre)
|
s405170297
|
Accepted
| 43 | 3,316 | 500 |
import re
n = int(input())
s = []
compiled = re.compile(r"AB")
pre = 0
xa=0
bx=0
ba=0
for i in range(n):
one = input()
pre+=len(compiled.findall(one))
f=one[0]
l=one[-1]
if f=="B" and l=="A":
ba+=1
elif f=="B" and l!="A":
bx+=1
elif f!="B" and l=="A":
xa+=1
#print(ba,bx,xa,pre)
cnt=0
if ba>=1:
cnt+=ba-1
if xa>=1:
xa-=1
cnt+=1
if bx>=1:
bx-=1
cnt+=1
cnt += xa if xa<bx else bx
print(cnt+pre)
|
s116664281
|
p02613
|
u302701160
| 2,000 | 1,048,576 |
Wrong Answer
| 153 | 16,156 | 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())
L =[]
for i in range(N):
S = input()
L.append(S)
print("AC * " + str(L.count("AC")))
print("WA * " + str(L.count("WA")))
print("TLE * " + str(L.count("TLE")))
print("RE * " + str(L.count("RE")))
|
s777973860
|
Accepted
| 154 | 16,072 | 222 |
L =[]
N = int(input())
for i in range(N):
S = input()
L.append(S)
print("AC x " + str(L.count("AC")))
print("WA x " + str(L.count("WA")))
print("TLE x " + str(L.count("TLE")))
print("RE x " + str(L.count("RE")))
|
s615575980
|
p03379
|
u513434790
| 2,000 | 262,144 |
Wrong Answer
| 268 | 25,556 | 164 |
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 = [0] + list(map(int, input().split()))
for i in range(N):
if i+1 <= N//2:
print(X[(N // 2 + 1)])
else:
print(X[N // 2])
|
s708301344
|
Accepted
| 355 | 26,772 | 186 |
N = int(input())
a = [0] + list(map(int, input().split()))
X = sorted(a)
for i in range(1,N+1):
if a[i] <= X[N//2]:
print(X[(N // 2 + 1)])
else:
print(X[N // 2])
|
s511663127
|
p04044
|
u521323621
| 2,000 | 262,144 |
Wrong Answer
| 28 | 9,168 | 111 |
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())
a = []
for i in range(n):
a.append(input())
a.sort()
for i in a:
print(i)
|
s236598680
|
Accepted
| 31 | 9,104 | 122 |
n,l = map(int,input().split())
a = []
for i in range(n):
a.append(input())
a.sort()
print("".join(str(i) for i in a))
|
s338152689
|
p03352
|
u318427318
| 2,000 | 1,048,576 |
Wrong Answer
| 117 | 27,276 | 727 |
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.
|
#-*-coding:utf-8-*-
import sys
input=sys.stdin.readline
import numpy as np
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
def main():
n = int(input())
if n==1:
print(1)
exit()
while True:
dp=np.array(factorization(n),dtype=np.int)
if np.all(dp[:,1:]!=1) and len(dp)==1:
print(n)
exit()
else:
n-=1
if __name__=="__main__":
main()
|
s241090677
|
Accepted
| 26 | 9,060 | 478 |
#-*-coding:utf-8-*-
import sys
input=sys.stdin.readline
def main():
n = int(input())
answers=[]
if n<=3:
print(n)
exit()
for b in range(2,n):
for p in range(2,11):
if b**p > n:
break
elif b**p <=n:
answers.append(b**p)
else:
continue
print(max(answers))
if __name__=="__main__":
main()
|
s573136524
|
p03672
|
u722535636
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 21 |
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.
|
print(len(input())-2)
|
s515035725
|
Accepted
| 17 | 2,940 | 96 |
s=input()
while 1:
s=s[:-2]
if s[:len(s)//2]==s[len(s)//2:]:
break
print(len(s))
|
s248425774
|
p03543
|
u136090046
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 212 |
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**?
|
num = list(input())
res = 0
tmp = num[0]
for i in range(1, 3):
if tmp == num[i]:
res += 1
else:
res = 0
tmp = num[i]
if res >= 2:
break
print("YES" if res >= 2 else "NO")
|
s846877057
|
Accepted
| 17 | 2,940 | 192 |
num = input()
array = ["111","222","333","444","555","666","777","888","999","000"]
res = 0
for i in array:
if i in num:
res =1
break
print("Yes" if res != 0 else "No")
|
s269367134
|
p02612
|
u812891913
| 2,000 | 1,048,576 |
Wrong Answer
| 31 | 9,148 | 51 |
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
print(N - (a*1000))
|
s846076831
|
Accepted
| 33 | 9,148 | 80 |
N = int(input())
b = N % 1000
if b == 0:
print(0)
else:
print(1000 - b)
|
s374633682
|
p03485
|
u100800700
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 97 |
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())
if (a + b) % 2 == 0:
print((a + b)/2)
else:
print((a + b +1)/2)
|
s047729012
|
Accepted
| 17 | 2,940 | 47 |
a,b=map(int,input().split());print(0--(a+b)//2)
|
s330888233
|
p02397
|
u169794024
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,572 | 72 |
Write a program which reads two integers x and y, and prints them in ascending order.
|
x,y=map(int,input().split())
if x==0 and y==0:
brake
print(sort(x,y))
|
s927406822
|
Accepted
| 60 | 7,632 | 97 |
while True:
x,y=sorted ([int(x) for x in input().split()])
if (x,y)==(0,0):
break
print(x,y)
|
s947811319
|
p02417
|
u387507798
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,560 | 247 |
Write a program which counts and reports the number of each alphabetical letter. Ignore the case of characters.
|
import sys
cs = {c: 0 for c in list("abcdefghijklmnopqrstuvwxys")}
for c in list(sys.stdin.read()):
c = c.lower()
if c in cs.keys():
cs[c] = cs[c] + 1
for (k, v) in sorted(list(cs.items())):
print('{0} : {1}'.format(k, v))
|
s898091463
|
Accepted
| 20 | 5,576 | 249 |
import sys
cs = { c : 0 for c in list("abcdefghijklmnopqrstuvwxyz")}
for c in list(sys.stdin.read()):
c = c.lower()
if c in cs.keys():
cs[c] = cs[c] + 1
for (c,cnt) in sorted(list(cs.items())):
print('{0} : {1}'.format(c,cnt))
|
s632652630
|
p03795
|
u601082779
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 37 |
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*800-200*n//15)
|
s468025116
|
Accepted
| 17 | 2,940 | 37 |
n=int(input());print(n*800-n//15*200)
|
s731543780
|
p03160
|
u436173409
| 2,000 | 1,048,576 |
Wrong Answer
| 137 | 13,980 | 192 |
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 = [int(e) for e in input().split()]
dp = [0, abs(h[1]-h[0])]
for i in range(2,n):
dp.append(min(abs(h[i]-h[i-2]),(abs(h[i]-h[i-1])+abs(h[i-1]-h[i-2]))))
print(dp[-1])
|
s811499197
|
Accepted
| 122 | 13,928 | 185 |
n = int(input())
h = [int(e) for e in input().split()]
dp = [0, abs(h[1]-h[0])]
for i in range(2,n):
dp.append(min(dp[-1]+abs(h[i]-h[i-1]),dp[-2]+abs(h[i]-h[i-2])))
print(dp[-1])
|
s497016087
|
p03167
|
u113971909
| 2,000 | 1,048,576 |
Wrong Answer
| 1,702 | 21,748 | 620 |
There is a grid with H horizontal rows and W vertical columns. Let (i, j) denote the square at the i-th row from the top and the j-th column from the left. For each i and j (1 \leq i \leq H, 1 \leq j \leq W), Square (i, j) is described by a character a_{i, j}. If a_{i, j} is `.`, Square (i, j) is an empty square; if a_{i, j} is `#`, Square (i, j) is a wall square. It is guaranteed that Squares (1, 1) and (H, W) are empty squares. Taro will start from Square (1, 1) and reach (H, W) by repeatedly moving right or down to an adjacent empty square. Find the number of Taro's paths from Square (1, 1) to (H, W). As the answer can be extremely large, find the count modulo 10^9 + 7.
|
from collections import deque
mod=10**9+7
H,W = map(int,input().split())
Gd = [list(input()) for _ in range(H)]
INF = 0
direc = [[1,0],[0,1]]
def dfs(start):
Rt = [[INF]*W for _ in range(H)]
Rt[start[0]][start[1]]=1
q = deque([])
q.append(start)
while len(q)!=0:
h,w = q.pop()取りだし（後ろ）
for d in direc:
hs, ws = h + d[0], w + d[1]
if not (0<=hs<H and 0<=ws<W):
continue
if Gd[hs][ws]=='.':
if Rt[hs][ws]==INF:
q.append([hs,ws])
Rt[hs][ws] = (Rt[hs][ws] + Rt[h][w])%mod
return Rt[H-1][W-1]
print(dfs([0,0]))
|
s230031802
|
Accepted
| 1,920 | 51,848 | 926 |
#!/usr/bin python3
# -*- coding: utf-8 -*-
H, W = map(int,input().split())
sth, stw = 0, 0
glh, glw = H-1, W-1
INF = 0
Gmap = [list(input()) for _ in range(H)]
Dist = [[INF]*W for _ in range(H)]
direc = {(1,0), (0,1)}
mod = 10**9+7
from collections import deque
def bfs(init):
next_q = deque([])
for hi, hw in init:
next_q.append([hi,hw])
Dist[hi][hw] = 1
while len(next_q)!=0:
h,w = next_q.popleft()
for d in direc:
hs, ws = h + d[0], w + d[1]
if not (0<=hs<H and 0<=ws<W):
continue
if Gmap[hs][ws]=='.':
if Dist[hs][ws]==INF:
next_q.append([hs,ws])
Dist[hs][ws] += Dist[h][w]
Dist[hs][ws] %= mod
return Dist
def main():
ret = bfs([[sth, stw]])
print(ret[glh][glw])
if __name__ == '__main__':
main()
|
s026060769
|
p02972
|
u133936772
| 2,000 | 1,048,576 |
Wrong Answer
| 2,104 | 10,708 | 222 |
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
n=int(input())
a=list(map(int,input().split()))
l=[0]*n
for i in range(n):
if a[i]:
i+=1
for j in range(int(i**.5)):
j+=1
if i%j<1:
l[j-1]^=1
if j*j<i:
l[i//j-1]^=1
print(*l)
|
s678648610
|
Accepted
| 195 | 11,700 | 154 |
n=int(input())
l=list(map(int,input().split()))
for i in range(n//2,0,-1):
l[i-1]=sum(l[i-1::i])%2
print(sum(l))
print(*[i+1 for i in range(n) if l[i]])
|
s510594630
|
p04043
|
u308695322
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 195 |
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.
|
require = ['5','7','5']
data = input().split()
result = True
for d in data:
if d in require:
require.pop(require.index(d))
else:
result = False
break
print(result)
|
s087980733
|
Accepted
| 17 | 2,940 | 202 |
require = ['5','7','5']
data = input().split()
result = 'YES'
for d in data:
if d in require:
require.pop(require.index(d))
else:
result = 'NO'
break
print(result)
|
s471283153
|
p03407
|
u207707177
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 84 |
An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.
|
a,b,c = [int(i) for i in range(3)]
if a+b>=c:
print("Yes")
else:
print("No")
|
s593147257
|
Accepted
| 18 | 2,940 | 95 |
a,b,c = [int(i) for i in input().split()]
if a + b >= c:
print("Yes")
else:
print("No")
|
s539440265
|
p02865
|
u111559399
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 89 |
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
A = input()
if (int( A )%2) == 0:
print( int(A) / 2 )
else:
print( (int(A)-1)/2 )
|
s249175106
|
Accepted
| 17 | 2,940 | 102 |
N = input()
if (int( N )%2) == 0:
print( int(int(N) / 2 -1) )
else:
print( int((int(N)-1)/2) )
|
s855146905
|
p03494
|
u045953894
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,104 | 16,080 | 197 |
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())
nums = list(map(int,input().split()))
ans = 0
while True:
for i in range(n):
if nums[i] % 2 == 0:
nums[i] = nums[i] / 2
else:
print(ans)
break
ans += 1
|
s011838603
|
Accepted
| 19 | 2,940 | 163 |
n = int(input())
A = [int(x) for x in input().split()]
c = 0
while all(a % 2 == 0 for a in A):
A = [a / 2 for a in A]
c += 1
print(c)
|
s264183885
|
p02406
|
u482227082
| 1,000 | 131,072 |
Time Limit Exceeded
| 9,990 | 5,576 | 186 |
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())
i = 1
while True:
x = i
if x % 3 == 0 or x % 10 ==3 or x/10 == 0:
print(" %d" %i)
i+=1
if i > n:
print()
break
|
s058833421
|
Accepted
| 20 | 5,636 | 395 |
#
# 5d
#
def main():
n = int(input())
s = ""
for i in range(1, n+1):
x = i
if x % 3 == 0 or x % 10 == 3:
s += " " + str(i)
else:
x //= 10
while x:
if x % 10 == 3:
s += " " + str(i)
break
x //= 10
print(s)
if __name__ == '__main__':
main()
|
s087713253
|
p02844
|
u348868667
| 2,000 | 1,048,576 |
Wrong Answer
| 2,227 | 1,981,448 | 289 |
AtCoder Inc. has decided to lock the door of its office with a 3-digit PIN code. The company has an N-digit lucky number, S. Takahashi, the president, will erase N-3 digits from S and concatenate the remaining 3 digits without changing the order to set the PIN code. How many different PIN codes can he set this way? Both the lucky number and the PIN code may begin with a 0.
|
import itertools
N = int(input())
S = input()
comb = list(itertools.combinations(list(range(N)),3))
ans = []
for i in comb:
pw = []
for j in range(len(comb[0])):
pw.append(S[i[j]])
pw = "".join(pw)
if pw not in ans:
ans.append(pw)
print(len(ans))
print(ans)
|
s060032814
|
Accepted
| 741 | 4,012 | 436 |
N = int(input())
S = list(input())
pas = []
for i in range(10):
for j in range(10):
for k in range(10):
pas.append(str(i)+str(j)+str(k))
ans = 0
for pw in pas:
if pw[0] in S:
ind = S.index(pw[0])
tmp = S[ind+1:]
else:
continue
if pw[1] in tmp:
ind = tmp.index(pw[1])
tmp = tmp[ind+1:]
else:
continue
if pw[2] in tmp:
ans += 1
print(ans)
|
s809681459
|
p02742
|
u390901183
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 106 |
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
H, W = map(int, input().split())
if H * W % 2 == 0:
print(H * W / 2)
else:
print((H * W + 1) / 2)
|
s904758126
|
Accepted
| 17 | 2,940 | 153 |
H, W = map(int, input().split())
if H == 1 or W == 1:
print(1)
exit()
if H * W % 2 == 0:
print(H * W // 2)
else:
print((H * W + 1) // 2)
|
s266302746
|
p02936
|
u502731482
| 2,000 | 1,048,576 |
Wrong Answer
| 2,106 | 22,772 | 472 |
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations.
|
from collections import deque
n, q = map(int, input().split())
cnt = [0] * n
a, b = [0] * (n - 1), [0] * (n - 1)
for i in range(n - 1):
a[i], b[i] = map(int, input().split())
def dfs(a, b, s, x):
cnt[s - 1] += x
if s not in a:
return
for i in range(n - 1):
if a[i] == s:
dfs(a, b, b[i], x)
for i in range(q):
p, x = map(int, input().split())
dfs(a, b, p, x)
for i in range(n):
print(cnt[i], end = " ")
print()
|
s780151027
|
Accepted
| 1,852 | 230,932 | 506 |
import sys
sys.setrecursionlimit(10 ** 6)
input = sys.stdin.readline
n, q = map(int, input().split())
cnt = [0] * n
tree = [[] for i in range(n)]
for i in range(n - 1):
a, b = map(int, input().split())
tree[a - 1].append(b - 1)
tree[b - 1].append(a - 1)
for i in range(q):
p, x = map(int, input().split())
cnt[p - 1] += x
def dfs(cur, par):
for chi in tree[cur]:
if chi == par:
continue
cnt[chi] += cnt[cur]
dfs(chi, cur)
dfs(0, -1)
print(*cnt)
|
s809719042
|
p03737
|
u190178779
| 2,000 | 262,144 |
Wrong Answer
| 28 | 9,048 | 248 |
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.
|
import sys
S = list(map(str,input().split()))
for I in S:
if not I.islower():
sys.exit()
if len(I) < 0 or len(I) > 10:
sys.exit()
result = ""
for J in S:
print(J[0:1])
result = result + J[0:1]
print(result.upper())
|
s183916550
|
Accepted
| 27 | 9,052 | 230 |
import sys
S = list(map(str,input().split()))
for I in S:
if not I.islower():
sys.exit()
if len(I) < 0 or len(I) > 10:
sys.exit()
result = ""
for J in S:
result = result + J[0:1]
print(result.upper())
|
s485754935
|
p03860
|
u346474533
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 72 |
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
s = input().split()
print(s)
for t in s:
print(t[0], end='')
print()
|
s670228226
|
Accepted
| 17 | 2,940 | 65 |
s = input().split()
for t in s:
print(t[0], end='')
print('')
|
s194310734
|
p03485
|
u294376483
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 165 |
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 = input().split()
x = (int(a) + int(b)) / 2
print(x)
ans = 0
if (int(a) + int(b)) % 2 == 0:
ans = x
else:
ans = (int(a) + int(b)) // 2 + 1
print(ans)
|
s793430983
|
Accepted
| 19 | 3,060 | 150 |
a, b = input().split()
ans = 0
if (int(a) + int(b)) % 2 == 0:
ans = (int(a) + int(b)) // 2
else:
ans = (int(a) + int(b)) // 2 + 1
print(ans)
|
s471921047
|
p03557
|
u426572476
| 2,000 | 262,144 |
Wrong Answer
| 1,821 | 24,820 | 2,809 |
The season for Snuke Festival has come again this year. First of all, Ringo will perform a ritual to summon Snuke. For the ritual, he needs an altar, which consists of three parts, one in each of the three categories: upper, middle and lower. He has N parts for each of the three categories. The size of the i-th upper part is A_i, the size of the i-th middle part is B_i, and the size of the i-th lower part is C_i. To build an altar, the size of the middle part must be strictly greater than that of the upper part, and the size of the lower part must be strictly greater than that of the middle part. On the other hand, any three parts that satisfy these conditions can be combined to form an altar. How many different altars can Ringo build? Here, two altars are considered different when at least one of the three parts used is different.
|
from itertools import permutations
import sys
import heapq
from collections import Counter
from collections import deque
from fractions import gcd
from math import factorial
from math import sqrt
INF = 1 << 60
sys.setrecursionlimit(10 ** 6)
class UnionFind():
def __init__(self, n):
self.n = n
self.parents = [-1] * n
def find(self, x):
if self.parents[x] < 0:
return x
else:
self.parents[x] = self.find(self.parents[x])
return self.parents[x]
def union(self, x, y):
x = self.find(x)
y = self.find(y)
if x == y:
return
if self.parents[x] > self.parents[y]:
x, y = y, x
self.parents[x] += self.parents[y]
self.parents[y] = x
def size(self, x):
return -self.parents[self.find(x)]
def same(self, x, y):
return self.find(x) == self.find(y)
def members(self, x):
root = self.find(x)
return [i for i in range(self.n) if self.find(i) == root]
def roots(self):
return [i for i, x in enumerate(self.parents) if x < 0]
def group_count(self):
return len(self.roots())
def all_group_members(self):
return {r: self.members(r) for r in self.roots()}
def __str__(self):
return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())
# def is_ok(arg):
# '''
# '''
# while abs(ok - ng) > 1:
# mid = (ok + ng) // 2
# if is_ok(mid):
# ok = mid
# else:
# ng = mid
# return ok
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = list(map(int, input().split()))
a.sort()
c.sort()
def is_ok(index, key, x):
if x[index] >= key:
return True
return False
def meguru_bisect(ng, ok, key, x):
while abs(ok - ng) > 1:
mid = (ok + ng) // 2
if is_ok(mid, key, x):
ok = mid
else:
ng = mid
return ok
ans = 0
# print("a =", a)
# print("b =", b)
# print("c =", c)
for i in b:
total1 = 0
total2 = 0
if i > a[0]:
total1 = meguru_bisect(-1, n, i, a)
if i < c[-1]:
total2 = n - meguru_bisect(-1, n, i, c)
if i == c[meguru_bisect(-1, n, i, c)]:
total2 -= 1
ans += total1 * total2
print(ans)
|
s149170785
|
Accepted
| 1,317 | 31,248 | 5,119 |
import sys
import heapq
import re
from itertools import permutations
from bisect import bisect_left, bisect_right
from collections import Counter, deque
from math import factorial, sqrt, ceil, gcd
from functools import lru_cache, reduce
from decimal import Decimal
from operator import mul
INF = 1 << 60
MOD = 1000000007
sys.setrecursionlimit(10 ** 7)
# UnionFind
class UnionFind():
def __init__(self, n):
self.n = n
self.parents = [-1] * n
def find(self, x):
if self.parents[x] < 0:
return x
else:
self.parents[x] = self.find(self.parents[x])
return self.parents[x]
def union(self, x, y):
x = self.find(x)
y = self.find(y)
if x == y:
return
if self.parents[x] > self.parents[y]:
x, y = y, x
self.parents[x] += self.parents[y]
self.parents[y] = x
def size(self, x):
return -self.parents[self.find(x)]
def same(self, x, y):
return self.find(x) == self.find(y)
def members(self, x):
root = self.find(x)
return [i for i in range(self.n) if self.find(i) == root]
def roots(self):
return [i for i, x in enumerate(self.parents) if x < 0]
def group_count(self):
return len(self.roots())
def all_group_members(self):
return {r: self.members(r) for r in self.roots()}
def __str__(self):
return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())
def dijkstra_heap(s, edge, n):
d = [10**20] * n
used = [True] * n
d[s] = 0
used[s] = False
edgelist = []
for a,b in edge[s]:
heapq.heappush(edgelist,a*(10**6)+b)
while len(edgelist):
minedge = heapq.heappop(edgelist)
if not used[minedge%(10**6)]:
continue
v = minedge%(10**6)
d[v] = minedge//(10**6)
used[v] = False
for e in edge[v]:
if used[e[1]]:
heapq.heappush(edgelist,(e[0]+d[v])*(10**6)+e[1])
return d
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
def combinations_count(n, r):
if n < r:
return 0
r = min(r, n - r)
numer = reduce(mul, range(n, n - r, -1), 1)
denom = reduce(mul, range(1, r + 1), 1)
return numer // denom
def lcm(x, y):
return (x * y) // gcd(x, y)
def lcm_list(numbers):
return reduce(lcm, numbers, 1)
def gcd_list(numbers):
return reduce(gcd, numbers)
def is_prime(n):
if n <= 1:
return False
p = 2
while True:
if p ** 2 > n:
break
if n % p == 0:
return False
p += 1
return True
def eratosthenes(limit):
A = [i for i in range(2, limit+1)]
P = []
while True:
prime = min(A)
if prime > sqrt(limit):
break
P.append(prime)
i = 0
while i < len(A):
if A[i] % prime == 0:
A.pop(i)
continue
i += 1
for a in A:
P.append(a)
return P
def permutation_with_duplicates(L):
if L == []:
return [[]]
else:
ret = []
S = sorted(set(L))
for i in S:
data = L[:]
data.remove(i)
for j in permutation_with_duplicates(data):
ret.append([i] + j)
return ret
def make_divisors(n):
lower_divisors , upper_divisors = [], []
i = 1
while i*i <= n:
if n % i == 0:
lower_divisors.append(i)
if i != n // i:
upper_divisors.append(n//i)
i += 1
return lower_divisors + upper_divisors[::-1]
n = int(input())
a = sorted(list(map(int, input().split())), reverse=True)
b = list(map(int, input().split()))
c = sorted(list(map(int, input().split())))
ans = 0
for i in range(n):
ng = -1
ok = n
while abs(ok - ng) > 1:
mid = (ok + ng) // 2
if a[mid] < b[i]:
ok = mid
else:
ng = mid
ng2 = -1
ok2 = n
while abs(ok2 - ng2) > 1:
mid2 = (ok2 + ng2) // 2
if c[mid2] > b[i]:
ok2 = mid2
else:
ng2 = mid2
# print("ok =", ok)
# print("ok2 =", ok2)
ans += (n - ok) * (n - ok2)
print(ans)
|
s381474274
|
p03471
|
u259861571
| 2,000 | 262,144 |
Wrong Answer
| 21 | 3,060 | 363 |
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.
|
n, y = [int(i) for i in input().split()]
ans_l = -1
ans_c = -1
ans_r = -1
for i in range(0, n):
for j in range(0, n):
k = n - i - j
total = i + j + k
if total == n:
ans_l = str(i)
ans_c = str(j)
ans_r = str(k)
break
print(ans_l + " " + ans_c + " " + ans_r)
|
s754305502
|
Accepted
| 896 | 3,064 | 293 |
n, y = [int(i) for i in input().split()]
ans = [-1, -1, -1]
for i in range(n+1):
for j in range(n-i+1):
k = n - i - j
total = 10000*i+5000*j+1000*k
if total == y:
ans = [i, j, k]
break
print(ans[0],ans[1],ans[2])
|
s286209158
|
p02833
|
u982591663
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,064 | 803 |
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())
ans = 0
def check_ten(N):
count = 0
i = 10
exp = 1
while True:
divided = N // i
if divided == 0:
break
count += divided
exp += 1
i = i ** exp
print("check_ten", count)
return count
def check_five(N):
count = 0
i = 5
exp = 1
while True:
divided = N // i
if divided == 0:
break
count += divided
exp += 1
i = i ** exp
print("check_five", count)
return count
if N % 2 == 0:
ans = check_ten(N)
else:
ans = check_five(N) - check_ten(N)
if N < 9:
ans = 0
print(ans)
# if N % 2 == 1:
# else:
|
s127106101
|
Accepted
| 17 | 3,064 | 318 |
N = int(input())
ans = 0
def check_five(N):
count = 0
i = 5
exp = 1
while True:
divided = N // i
if divided == 0:
break
count += divided
exp += 1
i = 5 ** exp
return count
if N % 2 == 0:
ans = check_five(N//2)
else:
ans = 0
print(ans)
|
s957268340
|
p03672
|
u777028980
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 200 |
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.
|
hoge=input()
for i in range(len(hoge)):
n=int(len(hoge)/2)
print(hoge[n:])
print(hoge[:n])
if(hoge[n:]==hoge[:n]):
print(2*n)
break
else:
hoge=hoge[:len(hoge)-1]
print(hoge)
|
s544671523
|
Accepted
| 18 | 2,940 | 138 |
hoge=input()
for i in range(len(hoge)):
hoge=hoge[:len(hoge)-1]
n=int(len(hoge)/2)
if(hoge[n:]==hoge[:n]):
print(2*n)
break
|
s701021266
|
p02663
|
u558242240
| 2,000 | 1,048,576 |
Wrong Answer
| 24 | 9,048 | 98 |
In this problem, we use the 24-hour clock. Takahashi gets up exactly at the time H_1 : M_1 and goes to bed exactly at the time H_2 : M_2. (See Sample Inputs below for clarity.) He has decided to study for K consecutive minutes while he is up. What is the length of the period in which he can start studying?
|
h1, m1, h2, m2, k = map(int, input().split())
print(min(0, (h2 * 60 + m2 - k) - (h1 * 60 + m1)))
|
s104445998
|
Accepted
| 21 | 9,164 | 90 |
h1, m1, h2, m2, k = map(int, input().split())
print((h2 * 60 + m2) - (h1 * 60 + m1) - k)
|
s027980412
|
p02613
|
u112007848
| 2,000 | 1,048,576 |
Wrong Answer
| 152 | 9,220 | 340 |
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.
|
num = (int)(input())
ans = [0, 0, 0, 0]
for i in range(num):
temp = input()
if temp == "AC":
ans[0] += 1
elif temp == "WA":
ans[1] += 1
elif temp == "TLE":
ans[2] += 1
else:
ans[3] += 1
print("AC × " + (str)(ans[0]))
print("WA × " + (str)(ans[1]))
print("TLE × " + (str)(ans[2]))
print("RE × " + (str)(ans[3]))
|
s843465522
|
Accepted
| 152 | 9,216 | 336 |
num = (int)(input())
ans = [0, 0, 0, 0]
for i in range(num):
temp = input()
if temp == "AC":
ans[0] += 1
elif temp == "WA":
ans[1] += 1
elif temp == "TLE":
ans[2] += 1
else:
ans[3] += 1
print("AC x " + (str)(ans[0]))
print("WA x " + (str)(ans[1]))
print("TLE x " + (str)(ans[2]))
print("RE x " + (str)(ans[3]))
|
s688011155
|
p03399
|
u663710122
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 77 |
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(min(int(input()), int(input())), min(int(input()), int(input()))))
|
s954605054
|
Accepted
| 17 | 2,940 | 72 |
print(min(int(input()), int(input())) + min(int(input()), int(input())))
|
s984097629
|
p03598
|
u260764792
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 201 |
There are N balls in the xy-plane. The coordinates of the i-th of them is (x_i, i). Thus, we have one ball on each of the N lines y = 1, y = 2, ..., y = N. In order to collect these balls, Snuke prepared 2N robots, N of type A and N of type B. Then, he placed the i-th type-A robot at coordinates (0, i), and the i-th type-B robot at coordinates (K, i). Thus, now we have one type-A robot and one type-B robot on each of the N lines y = 1, y = 2, ..., y = N. When activated, each type of robot will operate as follows. * When a type-A robot is activated at coordinates (0, a), it will move to the position of the ball on the line y = a, collect the ball, move back to its original position (0, a) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. * When a type-B robot is activated at coordinates (K, b), it will move to the position of the ball on the line y = b, collect the ball, move back to its original position (K, b) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. Snuke will activate some of the 2N robots to collect all of the balls. Find the minimum possible total distance covered by robots.
|
N=int(input())
K=int(input())
c=list(map(int, input().split()))
l=[K]*N
sub=[]
res=[]
for i in range(0, N):
sub.append(l[i]-c[i])
if (sub[i]<=c[i]):
res.append(2*sub[i])
else:
res.append(2*c[i])
|
s005692759
|
Accepted
| 17 | 3,064 | 217 |
N=int(input())
K=int(input())
c=list(map(int, input().split()))
l=[K]*N
sub=[]
res=[]
for i in range(0, N):
sub.append(l[i]-c[i])
if (sub[i]<=c[i]):
res.append(2*sub[i])
else:
res.append(2*c[i])
print(sum(res))
|
s634883860
|
p02665
|
u541610817
| 2,000 | 1,048,576 |
Wrong Answer
| 983 | 682,396 | 876 |
Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.
|
def Z(): return int(input())
def ZZ(): return [int(_) for _ in input().split()]
def main():
N = Z()
A = ZZ()
if A[0] != 0:
print(-1)
return
v = [[1, 1] for _ in range(N+1)]
for i in range(N):
x = v[i][1] - A[i]
if x < 0:
print(-1)
return
v[i+1][1] = 2 * x
if v[N][0] > A[N] or A[N] > v[N][1]:
print(-1)
return
v[N] = [A[N], A[N]]
for i in range(N):
x = (v[N-i][0]+1)//2 + A[N-i-1]
y = v[N-i][1] + A[N-i-1]
v[N-i-1][0] = max(v[N-i-1][0], x)
v[N-i-1][1] = min(v[N-i-1][1], y)
v[N-i-1][0] += A[N-i-1]
if v[N-i-1][0] > v[N-i-1][1]:
print(-1)
break
v[0] = [1, 1]
output = 0
for i in range(N+1): output += v[i][1]
print(output)
return
if __name__ == '__main__':
main()
|
s099746232
|
Accepted
| 870 | 682,172 | 904 |
def Z(): return int(input())
def ZZ(): return [int(_) for _ in input().split()]
def main():
N, A = Z(), ZZ()
if N == 0 and A[0] == 1:
print(1)
return
if A[0] != 0:
print(-1)
return
v = [[1, 1] for _ in range(N+1)]
for i in range(N):
x = v[i][1] - A[i]
if x < 0:
print(-1)
return
v[i+1][1] = 2 * x
if v[N][0] > A[N] or A[N] > v[N][1]:
print(-1)
return
v[N] = [A[N], A[N]]
for i in range(N):
x = (v[N-i][0]+1)//2 + A[N-i-1]
y = v[N-i][1] + A[N-i-1]
v[N-i-1][0] = max(v[N-i-1][0], x)
v[N-i-1][1] = min(v[N-i-1][1], y)
if v[N-i-1][0] > v[N-i-1][1]:
print(-1)
return
v[0] = [1, 1]
output = 0
for i in range(N+1): output += v[i][1]
print(output)
return
if __name__ == '__main__':
main()
|
s442678744
|
p03720
|
u543954314
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 140 |
There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
|
n, m = map(int, input().split())
a = []
for i in range(m):
a += list(map(int, input().split()))
for i in range(n):
print(a.count(i))
|
s877016856
|
Accepted
| 18 | 2,940 | 139 |
n, m = map(int, input().split())
a = []
for i in range(m):
a += map(int, input().split())
for i in range(1, n+1):
print(a.count(i))
|
s875570038
|
p03598
|
u171366497
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 92 |
There are N balls in the xy-plane. The coordinates of the i-th of them is (x_i, i). Thus, we have one ball on each of the N lines y = 1, y = 2, ..., y = N. In order to collect these balls, Snuke prepared 2N robots, N of type A and N of type B. Then, he placed the i-th type-A robot at coordinates (0, i), and the i-th type-B robot at coordinates (K, i). Thus, now we have one type-A robot and one type-B robot on each of the N lines y = 1, y = 2, ..., y = N. When activated, each type of robot will operate as follows. * When a type-A robot is activated at coordinates (0, a), it will move to the position of the ball on the line y = a, collect the ball, move back to its original position (0, a) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. * When a type-B robot is activated at coordinates (K, b), it will move to the position of the ball on the line y = b, collect the ball, move back to its original position (K, b) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. Snuke will activate some of the 2N robots to collect all of the balls. Find the minimum possible total distance covered by robots.
|
N=int(input())
K=int(input())
ans=0
for x in map(int,input().split()):
ans+=2*min(x,K-x)
|
s434026357
|
Accepted
| 17 | 2,940 | 103 |
N=int(input())
K=int(input())
ans=0
for x in map(int,input().split()):
ans+=2*min(x,K-x)
print(ans)
|
s493921689
|
p03992
|
u969190727
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 32 |
This contest is `CODE FESTIVAL`. However, Mr. Takahashi always writes it `CODEFESTIVAL`, omitting the single space between `CODE` and `FESTIVAL`. So he has decided to make a program that puts the single space he omitted. You are given a string s with 12 letters. Output the string putting a single space between the first 4 letters and last 8 letters in the string s.
|
s=input()
print(s[:5]+" "+s[4:])
|
s997895538
|
Accepted
| 17 | 2,940 | 32 |
s=input()
print(s[:4]+" "+s[4:])
|
s517355285
|
p02646
|
u219494936
| 2,000 | 1,048,576 |
Wrong Answer
| 23 | 9,176 | 188 |
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
A, V = map(int, input().split(" "))
B, W = map(int, input().split(" "))
T = int(input())
D = B - A
S = W - V
if S <= 0:
print("NO")
elif D / S <= T:
print("YES")
else:
print("NO")
|
s812265777
|
Accepted
| 19 | 9,144 | 717 |
A, V = map(int, input().split(" "))
B, W = map(int, input().split(" "))
T = int(input())
if W > V:
print("NO")
else:
if abs(A - B) <= T * (V - W):
print("YES")
else:
print("NO")
# if A - B > 0:
# if (B + 10**9) / W <= T:
# # print("left reach")
# Bdist = (B + 10**9)
# else:
# Bdist = W * T
# else:
# if (10**9 - B) / W <= T:
# # print("right reach")
# else:
# # print("run right")
# Bdist = W * T
# Adist = V * T
# if Bdist + abs(A - B) <= Adist:
# print("YES")
# else:
# print("NO")
|
s043353163
|
p03583
|
u934442292
| 2,000 | 262,144 |
Wrong Answer
| 1,012 | 9,128 | 357 |
You are given an integer N. Find a triple of positive integers h, n and w such that 4/N = 1/h + 1/n + 1/w. If there are multiple solutions, any of them will be accepted.
|
import sys
input = sys.stdin.readline
def main():
N = int(input())
for h in range(1, 3501):
for n in range(1, 3501):
a = h * n
b = 4 * h * n - (n + h) * N
if b > 0 and a % b == 0:
w = a // b
print(h, n, w)
exit()
if __name__ == "__main__":
main()
|
s207382573
|
Accepted
| 1,158 | 9,180 | 361 |
import sys
input = sys.stdin.readline
def main():
N = int(input())
for h in range(1, 3501):
for n in range(1, 3501):
a = h * n * N
b = 4 * h * n - (n + h) * N
if b > 0 and a % b == 0:
w = a // b
print(h, n, w)
exit()
if __name__ == "__main__":
main()
|
s249610575
|
p03997
|
u344959959
| 2,000 | 262,144 |
Wrong Answer
| 26 | 8,996 | 75 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a = int(input())
b = int(input())
c = int(input())
S = ((a+b)*c)/2
print(S)
|
s711087832
|
Accepted
| 24 | 9,048 | 80 |
a = int(input())
b = int(input())
h = int(input())
S = (a+b)*h/2
print(round(S))
|
s835958799
|
p03605
|
u808593466
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 77 |
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
N = input()
if N[0] == 9 or N[1] == 9:
print("Yes")
else:
print("No")
|
s321273677
|
Accepted
| 17 | 2,940 | 81 |
N = input()
if N[0] == "9" or N[1] == "9":
print("Yes")
else:
print("No")
|
s168133422
|
p03493
|
u257226830
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 51 |
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.
|
s1,s2,s3 = map(int,input())
s=[s1,s2,s3]
s.count(1)
|
s218998576
|
Accepted
| 17 | 2,940 | 58 |
s1,s2,s3 = map(int,input())
s=[s1,s2,s3]
print(s.count(1))
|
s443156452
|
p03024
|
u908349502
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 93 |
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()
win = s.count('o')
if win + (15-len(s)) >= 8:
print('Yes')
else:
print('No')
|
s286781248
|
Accepted
| 18 | 2,940 | 94 |
s = input()
win = s.count('o')
if win + (15-len(s)) >= 8:
print('YES')
else:
print('NO')
|
s461884270
|
p02936
|
u971091945
| 2,000 | 1,048,576 |
Wrong Answer
| 1,988 | 63,560 | 284 |
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations.
|
n, q = map(int, input().split())
ab = [list(map(int,input().split())) for _ in range(n-1)]
ab.sort()
li = [0]*n
for i in range(q):
p, x = map(int, input().split())
li[p-1] += x
for i in range(n-1):
a, b = ab[i]
li[b-1] += li[a-1]
for j in li:
print(j, end=" ")
|
s797306249
|
Accepted
| 1,589 | 268,400 | 456 |
import sys
input = sys.stdin.buffer.readline
sys.setrecursionlimit(10 ** 6)
n, q = map(int, input().split())
e = [[] for _ in range(n)]
for _ in range(n - 1):
a, b = map(int, input().split())
e[a-1].append(b-1)
e[b-1].append(a-1)
li = [0]*n
for i in range(q):
p, x = map(int, input().split())
li[p-1] += x
def dfs(i=0, r=-1):
for j in e[i]:
if j != r:
li[j] += li[i]
dfs(j, i)
dfs()
print(*li)
|
s178112697
|
p03377
|
u773246942
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 119 |
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 = list(map(int, input().split()))
if A > X:
print("No")
elif A + B < X:
print("No")
else:
print("Yes")
|
s367552707
|
Accepted
| 17 | 2,940 | 119 |
A,B,X = list(map(int, input().split()))
if A > X:
print("NO")
elif A + B < X:
print("NO")
else:
print("YES")
|
s159649877
|
p03759
|
u910358825
| 2,000 | 262,144 |
Wrong Answer
| 24 | 9,056 | 71 |
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")
|
s805317907
|
Accepted
| 26 | 8,952 | 71 |
a,b,c=map(int, input().split())
print("YES" if b-a==c-b else "NO")
|
s248331446
|
p03455
|
u391475811
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 146 |
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(str,input().split())
C=A+B
res=False
for i in range(2,1000):
if i*i==int(C):
res=True
if res:
print("Yes")
else:
print("No")
|
s158606066
|
Accepted
| 17 | 2,940 | 89 |
A,B=map(int,input().split())
if A%2==1 and B%2==1:
print("Odd")
else:
print("Even")
|
s314950972
|
p02850
|
u683134447
| 2,000 | 1,048,576 |
Wrong Answer
| 993 | 87,672 | 679 |
Given is a tree G with N vertices. The vertices are numbered 1 through N, and the i-th edge connects Vertex a_i and Vertex b_i. Consider painting the edges in G with some number of colors. We want to paint them so that, for each vertex, the colors of the edges incident to that vertex are all different. Among the colorings satisfying the condition above, construct one that uses the minimum number of colors.
|
import sys
sys.setrecursionlimit(10**9)
n = int(input())
abl = []
nodes = [[] for i in range(n)]
for _ in range(n-1):
a,b = map(int,input().split())
abl.append((a,b))
nodes[a-1] += [b-1]
nodes[b-1] += [a-1]
m_node = 0
for e in nodes:
m_node = max(len(e), m_node)
use_node = [0 for i in range(n)]
dic = {}
def dfs(node, start_color):
use_node[node] = 1
for e in nodes[node]:
if use_node[e] == 0:
dic[(node,e)] = (start_color%m_node)+1
dic[(e,node)] = (start_color%m_node)+1
dfs(e, (start_color%m_node)+1)
start_color += 1
dfs(0,0)
for a,b in abl:
print(dic[(a-1,b-1)])
|
s465290086
|
Accepted
| 795 | 87,676 | 692 |
import sys
sys.setrecursionlimit(10**9)
n = int(input())
abl = []
nodes = [[] for i in range(n)]
for _ in range(n-1):
a,b = map(int,input().split())
abl.append((a,b))
nodes[a-1] += [b-1]
nodes[b-1] += [a-1]
m_node = 0
for e in nodes:
m_node = max(len(e), m_node)
use_node = [0 for i in range(n)]
dic = {}
def dfs(node, start_color):
use_node[node] = 1
for e in nodes[node]:
if use_node[e] == 0:
dic[(node,e)] = (start_color%m_node)+1
dic[(e,node)] = (start_color%m_node)+1
dfs(e, (start_color%m_node)+1)
start_color += 1
dfs(0,0)
print(m_node)
for a,b in abl:
print(dic[(a-1,b-1)])
|
s523481093
|
p03494
|
u552510302
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 124 |
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.
|
a = int(input())
b = list(map(int,input().split()))
b.sort()
ans = 0
n = b[0]
while n % 2 ==0:
ans +=1
n = n//2
print(ans)
|
s806327801
|
Accepted
| 20 | 3,060 | 203 |
a = int(input())
b = list(map(int,input().split()))
ans = 0
x = True
while x:
for i in range(a):
if b[i] %2 ==1:
x=False
break
b[i] = b[i] /2
ans +=1
ans = ans//a
print(ans)
|
s240374843
|
p03487
|
u300579805
| 2,000 | 262,144 |
Wrong Answer
| 99 | 21,744 | 224 |
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.
|
import collections
N = int(input())
A=list(map(int, input().split()))
C = dict(collections.Counter(A))
print(C)
ans = 0
for k, v in C.items():
if (k>v):
ans += v
elif(k<v):
ans += v-k
print(ans)
|
s779716838
|
Accepted
| 82 | 21,744 | 215 |
import collections
N = int(input())
A=list(map(int, input().split()))
C = dict(collections.Counter(A))
ans = 0
for k, v in C.items():
if (k>v):
ans += v
elif(k<v):
ans += v-k
print(ans)
|
s123094318
|
p02578
|
u120789505
| 2,000 | 1,048,576 |
Wrong Answer
| 33 | 9,128 | 511 |
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.
|
n = 2 * 10**5 # = int(input())
data = list(map(int, input().split(' ')))
before_max = 0
dai = 0
for i in range(len(data)):
if(i == 0):
before_max = data[i]
else:
if(before_max < data[i]):
before_max = data[i]
continue
elif(before_max > data[i]):
dai += before_max - data[i]
print(dai)
|
s474104286
|
Accepted
| 135 | 32,228 | 497 |
n = int(input())
data = list(map(int, input().split(' ')))
before_max = 0
dai = 0
for i in range(len(data)):
if(i == 0):
before_max = data[i]
else:
if(before_max < data[i]):
before_max = data[i]
continue
elif(before_max > data[i]):
dai += before_max - data[i]
print(dai)
|
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