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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s157893759
|
p03992
|
u331464808
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 34 |
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[:3]+' '+s[4:])
|
s614170133
|
Accepted
| 18 | 2,940 | 34 |
s = input()
print(s[:4]+' '+s[4:])
|
s607417254
|
p03555
|
u152638361
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 100 |
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
a = input()
b = input()
if a[0]==b[2] and a[1]==b[2] and a[2]==b[0]:
print("YES")
else: print("NO")
|
s490582163
|
Accepted
| 17 | 2,940 | 100 |
a = input()
b = input()
if a[0]==b[2] and a[1]==b[1] and a[2]==b[0]:
print("YES")
else: print("NO")
|
s721374980
|
p02697
|
u511268447
| 2,000 | 1,048,576 |
Wrong Answer
| 106 | 9,292 | 784 |
You are going to hold a competition of one-to-one game called AtCoder Janken. _(Janken is the Japanese name for Rock-paper-scissors.)_ N players will participate in this competition, and they are given distinct integers from 1 through N. The arena has M playing fields for two players. You need to assign each playing field two distinct integers between 1 and N (inclusive). You cannot assign the same integer to multiple playing fields. The competition consists of N rounds, each of which proceeds as follows: * For each player, if there is a playing field that is assigned the player's integer, the player goes to that field and fight the other player who comes there. * Then, each player adds 1 to its integer. If it becomes N+1, change it to 1. You want to ensure that no player fights the same opponent more than once during the N rounds. Print an assignment of integers to the playing fields satisfying this condition. It can be proved that such an assignment always exists under the constraints given.
|
if __name__ == '__main__':
from sys import stdin
# from collections import defaultdict
import math
input = stdin.readline
# import random
n, m = map(int, input().split())
"""
a = random.randint(1, 10**6)
b = random.randint(1, 10**6)
n = random.randint(1, 10**12)
"""
# print(a, b, n)
for i in range(1,m+1):
if i%2 == 1:
if m%2 == 0:
print(((m-1)-i)//2+1, ((m-1)-i)//2+1+i)
if m%2 == 1:
print(m+((m-i)//2)-i+1, m+((m-i)//2+1))
if i%2 ==0:
if m%2 == 1:
print(((m-1)-i)//2+1, ((m-1)-i)//2+1+i)
if m%2 == 0:
print(m+((m-i)//2)-i+1, m+((m-i)//2)+1)
|
s890378706
|
Accepted
| 95 | 9,280 | 499 |
if __name__ == '__main__':
from sys import stdin
# from collections import defaultdict
import math
input = stdin.readline
# import random
n, m = map(int, input().split())
"""
a = random.randint(1, 10**6)
b = random.randint(1, 10**6)
n = random.randint(1, 10**12)
"""
# print(a, b, n)
for i in range(1,m+1):
if (m-i)%2 == 1:
print((m+1-i)//2, (m+1-i)//2+i)
else:
print(m+(m+2-i)//2, m+(m+2-i)//2+i)
|
s869923092
|
p03377
|
u207799478
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 186 |
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
def readints():
return list(map(int, input().split()))
A, B, X = map(int, input().split())
if A+B < X:
print("No")
exit()
if A > X:
print("No")
exit()
print("Yes")
|
s509628722
|
Accepted
| 17 | 2,940 | 186 |
def readints():
return list(map(int, input().split()))
A, B, X = map(int, input().split())
if A+B < X:
print("NO")
exit()
if A > X:
print("NO")
exit()
print("YES")
|
s969917800
|
p03827
|
u408958033
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 124 |
You have an integer variable x. Initially, x=0. Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`. Find the maximum value taken by x during the operations (including before the first operation, and after the last operation).
|
n = int(input())
s = input()
x = 0
i = 0
for i in range(len(s)):
i = i + (s[i]=='I') - (s[i]=='D')
x = max(x,i)
print(x)
|
s965367450
|
Accepted
| 17 | 2,940 | 125 |
n = int(input())
s = input()
x = 0
j = 0
for i in range(len(s)):
j = j + (s[i]=='I') - (s[i]=='D')
x = max(x,j)
print(x)
|
s949223792
|
p02262
|
u320121447
| 6,000 | 131,072 |
Wrong Answer
| 30 | 5,604 | 682 |
Shell Sort is a generalization of [Insertion Sort](http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_1_A) to arrange a list of $n$ elements $A$. 1 insertionSort(A, n, g) 2 for i = g to n-1 3 v = A[i] 4 j = i - g 5 while j >= 0 && A[j] > v 6 A[j+g] = A[j] 7 j = j - g 8 cnt++ 9 A[j+g] = v 10 11 shellSort(A, n) 12 cnt = 0 13 m = ? 14 G[] = {?, ?,..., ?} 15 for i = 0 to m-1 16 insertionSort(A, n, G[i]) A function shellSort(A, n) performs a function insertionSort(A, n, g), which considers every $g$-th elements. Beginning with large values of $g$, it repeats the insertion sort with smaller $g$. Your task is to complete the above program by filling ?. Write a program which reads an integer $n$ and a sequence $A$, and prints $m$, $G_i (i = 0, 1, ..., m − 1)$ in the pseudo code and the sequence $A$ in ascending order. The output of your program must meet the following requirements: * $1 \leq m \leq 100$ * $0 \leq G_i \leq n$ * cnt does not exceed $\lceil n^{1.5}\rceil$
|
def swap(A, i, j):
tmp = A[i]
A[i] = A[j]
A[j] = tmp
def insertionSort(A, n, g):
global cnt
for i in range(g, n):
v = A[i]
j = i - g
while j >= 0 and A[j] > v:
A[j+g] = A[j]
j = j - g
cnt += 1
A[j+g] = v
def shellSort(A, n):
global cnt
cnt = 0
m = 0
G = []
while 2 ** m < n:
G.append(2 ** m)
m += 1
G.reverse()
print(m)
print(' '.join(map(str, G)))
for i in range(m):
insertionSort(A, n, G[i])
n = int(input())
A = []
for i in range(n):
a = int(input())
A.append(a)
shellSort(A, n)
print(cnt)
for a in A:
print(a)
|
s633587482
|
Accepted
| 19,090 | 45,516 | 652 |
from sys import stdin
def insertionSort(A, n, g):
global cnt
for i in range(g, n):
v = A[i]
j = i - g
while j >= 0 and A[j] > v:
A[j+g] = A[j]
j = j - g
cnt += 1
A[j+g] = v
def shellSort(A, n):
global cnt
cnt = 0
g = 1
G = [g]
while 3 * g + 1 < n:
g = 3 * g + 1
G.append(g)
m = len(G)
G.reverse()
print(m)
print(' '.join(map(str, G)))
for i in range(m):
insertionSort(A, n, G[i])
n = int(stdin.readline())
A = [int(stdin.readline()) for i in range(n)]
shellSort(A, n)
print(cnt)
for a in A:
print(a)
|
s835158093
|
p02821
|
u106298129
| 2,000 | 1,048,576 |
Wrong Answer
| 2,136 | 535,116 | 288 |
Takahashi has come to a party as a special guest. There are N ordinary guests at the party. The i-th ordinary guest has a _power_ of A_i. Takahashi has decided to perform M _handshakes_ to increase the _happiness_ of the party (let the current happiness be 0). A handshake will be performed as follows: * Takahashi chooses one (ordinary) guest x for his left hand and another guest y for his right hand (x and y can be the same). * Then, he shakes the left hand of Guest x and the right hand of Guest y simultaneously to increase the happiness by A_x+A_y. However, Takahashi should not perform the same handshake more than once. Formally, the following condition must hold: * Assume that, in the k-th handshake, Takahashi shakes the left hand of Guest x_k and the right hand of Guest y_k. Then, there is no pair p, q (1 \leq p < q \leq M) such that (x_p,y_p)=(x_q,y_q). What is the maximum possible happiness after M handshakes?
|
N, M = map(int, input().split())
tmp = input().split()
A = [int(tmp[i]) for i in range(N)]
score = [0 for i in range(N*N)]
print(A)
for i in range(N):
for j in range(N):
score[i*N+j] = A[i] + A[j]
score.sort()
sum = 0
for i in range(M):
sum += score[N*N-1-i]
print(sum)
|
s020943345
|
Accepted
| 1,343 | 20,584 | 997 |
N, M = map(int, input().split())
tmp = input().split()
A = [int(tmp[i]) for i in range(N)]
A.sort()
cum_A = [None]*N
cum_A[N-1] = A[N-1]
for i in range(1,N):
cum_A[N-1-i] = cum_A[N-i] + A[N-1-i]
cum_A.append(0)
def geq(A, cum_A, X, N):
sum = 0
val = 0
index = 0
for i in range(N):
while True:
if index != N and A[i] + A[N - 1 - index] >= X:
index += 1
else:
sum += index
val += A[i] * index + cum_A[N-index]
break
return sum, val
low = 2
high = 2*A[N-1]
while low <= high:
X = (low + high)//2
sum, val = geq(A, cum_A, X, N)
if sum == M:
break
elif sum > M:
low = X+1
else:
high = X-1
if sum < M:
X -= 1
sum, val = geq(A, cum_A, X, N)
ans = val - X*(sum-M)
print(ans)
|
s480029031
|
p03339
|
u853900545
| 2,000 | 1,048,576 |
Wrong Answer
| 2,104 | 8,100 | 110 |
There are N people standing in a row from west to east. Each person is facing east or west. The directions of the people is given as a string S of length N. The i-th person from the west is facing east if S_i = `E`, and west if S_i = `W`. You will appoint one of the N people as the leader, then command the rest of them to face in the direction of the leader. Here, we do not care which direction the leader is facing. The people in the row hate to change their directions, so you would like to select the leader so that the number of people who have to change their directions is minimized. Find the minimum number of people who have to change their directions.
|
n=int(input())
s=input()
c=[0]*n
for i in range(n):
c[i]=c[i+1:].count('E')+c[:i].count('W')
print(min(c))
|
s297630050
|
Accepted
| 175 | 3,672 | 157 |
n=int(input())
s=input()
c=s[1:].count('E')
a=c
for i in range(1,n):
if s[i-1]=='W':
c+=1
if s[i]=='E':
c-=1
a=min(a,c)
print(a)
|
s894017268
|
p03730
|
u540762794
| 2,000 | 262,144 |
Wrong Answer
| 28 | 9,172 | 266 |
We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
# -*- coding: utf-8 -*-
A,B,C = map(int, input().split())
mod = []
for i in range(max(B,C)):
res = A * i % B
if res == C:
ans = "Yes"
print(i)
break
if res in mod:
ans = "No"
break
mod.append(res)
print(ans)
|
s264067606
|
Accepted
| 28 | 9,136 | 174 |
# -*- coding: utf-8 -*-
A,B,C = map(int, input().split())
ans = "NO"
for i in range(B+1):
res = A * i % B
if res == C:
ans = "YES"
break
print(ans)
|
s373104059
|
p03407
|
u860546679
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 70 |
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=map(int,input().split())
print("Yes") if a+b<=c else print("No")
|
s094074745
|
Accepted
| 17 | 2,940 | 69 |
a,b,c=map(int,input().split())
print("No") if a+b<c else print("Yes")
|
s451158823
|
p03370
|
u107091170
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 133 |
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts. Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition: * For each of the N kinds of doughnuts, make at least one doughnut of that kind. At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
|
N,X=map(int, input().split())
mmin = 10000000
for i in range(N):
m = int(input())
X -= m
mmin = min(m, mmin)
ans = N + X//mmin
|
s959248909
|
Accepted
| 17 | 2,940 | 143 |
N,X=map(int, input().split())
mmin = 10000000
for i in range(N):
m = int(input())
X -= m
mmin = min(m, mmin)
ans = N + X//mmin
print(ans)
|
s484226576
|
p03471
|
u635958201
| 2,000 | 262,144 |
Wrong Answer
| 801 | 3,060 | 191 |
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=map(int,input().split())
Y=Y/1000
ans=[-1,-1,-1]
for i in range(N):
for j in range(N-i):
if 10*i+5*j+N-i-j==Y:
ans=[i,j,N-i-j]
break
print(ans)
|
s982010464
|
Accepted
| 660 | 3,060 | 213 |
N,Y=map(int,input().split())
Y=Y//1000
ans=[-1,-1,-1]
for i in range(N+1):
for j in range(N-i+1):
if 10*i+5*j+N-i-j==Y:
ans=[i,j,N-i-j]
break
print(ans[0],ans[1],ans[2])
|
s026614250
|
p04043
|
u779308281
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 374 |
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.
|
b = input().split()
c = [int(s) for s in b]
print(c)
num5 = 0
num7 = 0
for i in range(len(c)):
if c[i] == 5:
num5 += 1
continue
if c[i] == 7:
num7 +=1
continue
if num5 == 2 and num7 == 1:
print("YES")
else:
print("NO")
|
s673564733
|
Accepted
| 17 | 3,060 | 365 |
b = input().split()
c = [int(s) for s in b]
num5 = 0
num7 = 0
for i in range(len(c)):
if c[i] == 5:
num5 += 1
continue
if c[i] == 7:
num7 +=1
continue
if num5 == 2 and num7 == 1:
print("YES")
else:
print("NO")
|
s104870033
|
p02406
|
u494314211
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,576 | 110 |
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=input()
n=int(n)
s=""
for i in range(n):
if i%3==0:
s+=" "+str(i)
elif i%10==3:
s+=" "+str(i)
print(s)
|
s341260415
|
Accepted
| 20 | 7,532 | 126 |
n=input()
n=int(n)
s=""
for i in range(1,n+1):
if i%3==0:
s+=" "+str(i)
elif "3" in list(str(i)):
s+=" "+str(i)
print(s)
|
s789624802
|
p02612
|
u618953765
| 2,000 | 1,048,576 |
Wrong Answer
| 29 | 9,044 | 30 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n = int(input())
print(n%1000)
|
s590287350
|
Accepted
| 27 | 8,912 | 37 |
n = int(input())
print((1000-n)%1000)
|
s272722881
|
p03645
|
u026788530
| 2,000 | 262,144 |
Wrong Answer
| 667 | 4,596 | 407 |
In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands. For convenience, we will call them Island 1, Island 2, ..., Island N. There are M kinds of regular boat services between these islands. Each service connects two islands. The i-th service connects Island a_i and Island b_i. Cat Snuke is on Island 1 now, and wants to go to Island N. However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services. Help him.
|
N,M =[int(i) for i in input().split(' ')]
table = [0]*N
for i in range(M):
a,b = [int(i) for i in input().split(' ')]
if a==1:
table[b-1] +=1
if b==M:
table[b-1] +=1
if b==N:
table[a-1]+=1
# print(table)
flag = True
for i in table:
if i==2:
print("POSSIBLE")
flag = False
break
if flag:
print("IMPOSSIBLE")
#print(table)
|
s799382712
|
Accepted
| 665 | 4,596 | 407 |
N,M =[int(i) for i in input().split(' ')]
table = [0]*N
for i in range(M):
a,b = [int(i) for i in input().split(' ')]
if a==1:
table[b-1] +=1
if b==N:
table[b-1] +=1
if b==N:
table[a-1]+=1
# print(table)
flag = True
for i in table:
if i==2:
print("POSSIBLE")
flag = False
break
if flag:
print("IMPOSSIBLE")
#print(table)
|
s791497843
|
p02419
|
u213265973
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,368 | 152 |
Write a program which reads a word W and a text T, and prints the number of word W which appears in text T T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.
|
W = input()
count = 0
while True:
T = input()
if T == "END_OF_TEXT":
break
if W in T:
count += 1
print(count)
|
s657180258
|
Accepted
| 30 | 7,376 | 165 |
W = input().lower()
count = 0
while True:
T = input()
if T == "END_OF_TEXT":
break
count += T.lower().split(" ").count(W)
print(count)
|
s326552439
|
p03090
|
u977389981
| 2,000 | 1,048,576 |
Wrong Answer
| 23 | 3,612 | 365 |
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())
G = [[] for i in range(N + 1)]
for i in range(1, N + 1):
for j in range(i, N + 1):
if N % 2 == 1:
if i + j != N and i != j:
G[i].append(j)
else:
if i + j != N + 1 and i != j:
G[i].append(j)
for i in range(1, N + 1):
for g in G[i]:
print(i, g)
|
s580150254
|
Accepted
| 24 | 3,612 | 419 |
N = int(input())
cnt = 0
G = [[] for i in range(N + 1)]
for i in range(1, N + 1):
for j in range(i, N + 1):
if N % 2 == 1:
if i + j != N and i != j:
G[i].append(j)
cnt += 1
else:
if i + j != N + 1 and i != j:
G[i].append(j)
cnt += 1
print(cnt)
for i in range(1, N + 1):
for g in G[i]:
print(i, g)
|
s160362011
|
p03719
|
u493440568
| 2,000 | 262,144 |
Wrong Answer
| 26 | 8,856 | 79 |
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
A, B, C = map(int, input().split())
print("YES" if C >= A and C <= B else "No")
|
s832961656
|
Accepted
| 30 | 9,092 | 79 |
A, B, C = map(int, input().split())
print("Yes" if C >= A and C <= B else "No")
|
s327177585
|
p03416
|
u074338678
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 107 |
Find the number of _palindromic numbers_ among the integers between A and B (inclusive). Here, a palindromic number is a positive integer whose string representation in base 10 (without leading zeros) reads the same forward and backward.
|
n, m = map(int, input().split())
if n==1:
print(m-2)
elif m==1:
print(n-2)
else:
print((n-2)*(m-2))
|
s056276349
|
Accepted
| 72 | 2,940 | 136 |
a, b = map(int, input().split())
count = 0
for i in range(a, b+1):
if str(i)[0:2] == str(i)[:-3:-1]:
count += 1
print(count)
|
s076997390
|
p03369
|
u642418876
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 54 |
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
S=str(input())
X=S.count("○")
print(700+100*int(X))
|
s544023282
|
Accepted
| 17 | 2,940 | 47 |
s=str(input())
x=s.count('o')
print(700+100*x)
|
s291883176
|
p02418
|
u908651435
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,540 | 72 |
Write a program which finds a pattern $p$ in a ring shaped text $s$.
|
s=input()*2
p=input()
if s in p:
print('Yes')
else:
print('No')
|
s700940718
|
Accepted
| 20 | 5,548 | 73 |
s=input()*2
p=input()
if p in s:
print('Yes')
else:
print('No')
|
s913635994
|
p03795
|
u446711904
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 38 |
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(800*n-60//15*200)
|
s686504303
|
Accepted
| 17 | 2,940 | 37 |
n=int(input());print(800*n-n//15*200)
|
s175630687
|
p02418
|
u498511622
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,296 | 41 |
Write a program which finds a pattern $p$ in a ring shaped text $s$.
|
s=input()
p=input()
print(s.find(p) != p)
|
s106693984
|
Accepted
| 30 | 7,460 | 76 |
s=input()*2
p=input()
if s.find(p) != -1:
print('Yes')
else:
print('No')
|
s118842464
|
p03080
|
u619785253
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 146 |
There are N people numbered 1 to N. Each person wears a red hat or a blue hat. You are given a string s representing the colors of the people. Person i wears a red hat if s_i is `R`, and a blue hat if s_i is `B`. Determine if there are more people wearing a red hat than people wearing a blue hat.
|
i = int(input())
hat = list(input())
B_count = hat.count('B')
R_count = hat.count('R')
if B_count < R_count:
print("YES")
else:
print('NO')
|
s284906672
|
Accepted
| 18 | 2,940 | 145 |
i = int(input())
hat = list(input())
B_count = hat.count('B')
R_count = hat.count('R')
if B_count < R_count:
print("Yes")
else:
print('No')
|
s716906014
|
p03853
|
u694665829
| 2,000 | 262,144 |
Wrong Answer
| 33 | 9,252 | 224 |
There is an image with a height of H pixels and a width of W pixels. Each of the pixels is represented by either `.` or `*`. The character representing the pixel at the i-th row from the top and the j-th column from the left, is denoted by C_{i,j}. Extend this image vertically so that its height is doubled. That is, print a image with a height of 2H pixels and a width of W pixels where the pixel at the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division is rounded down).
|
h,w=map(int,input().split())
C=[list(input()) for i in range(h)]
A=[[None for i in range(w)] for i in range(2*h)]
for i in range(2*h):
for j in range(w):
A[i][j]=C[(i+1)//2-1][j]
for a in A:
print(''.join(a))
|
s083063736
|
Accepted
| 25 | 9,144 | 120 |
h,w=map(int,input().split())
C=[list(input()) for i in range(h)]
for c in C:
print(''.join(c))
print(''.join(c))
|
s602616480
|
p03386
|
u918601425
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 26,448 | 182 |
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
A,B,K=[int(s) for s in input().split()]
if (B-A+1)>=2*K:
for i in range(A,B+1):
print(i)
else:
for i in range(K):
print(A+i)
for i in range(K):
print(B+i-K+1)
|
s421172330
|
Accepted
| 17 | 3,060 | 183 |
A,B,K=[int(s) for s in input().split()]
if (B-A+1)<=2*K:
for i in range(A,B+1):
print(i)
else:
for i in range(K):
print(A+i)
for i in range(K):
print(B+i-K+1)
|
s249183332
|
p03457
|
u600608564
| 2,000 | 262,144 |
Wrong Answer
| 388 | 11,636 | 382 |
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
N = int(input())
t = [0] * (N + 1)
x = [0] * (N + 1)
y = [0] * (N + 1)
for i in range(N):
t[i + 1], x[i + 1], y[i + 1] = map(int, input().split())
for i in range(N):
dt = t[i + 1] - t[i]
dist = abs(x[i + 1] - x[i]) + abs(y[i + 1] - y[i])
if dt < dist:
print("NO")
exit(0)
if dist % 2 != dt % 2:
print("NO")
exit(0)
print("YES")
|
s869802340
|
Accepted
| 388 | 11,636 | 382 |
N = int(input())
t = [0] * (N + 1)
x = [0] * (N + 1)
y = [0] * (N + 1)
for i in range(N):
t[i + 1], x[i + 1], y[i + 1] = map(int, input().split())
for i in range(N):
dt = t[i + 1] - t[i]
dist = abs(x[i + 1] - x[i]) + abs(y[i + 1] - y[i])
if dt < dist:
print("No")
exit(0)
if dist % 2 != dt % 2:
print("No")
exit(0)
print("Yes")
|
s464109887
|
p02697
|
u886747123
| 2,000 | 1,048,576 |
Wrong Answer
| 72 | 9,284 | 101 |
You are going to hold a competition of one-to-one game called AtCoder Janken. _(Janken is the Japanese name for Rock-paper-scissors.)_ N players will participate in this competition, and they are given distinct integers from 1 through N. The arena has M playing fields for two players. You need to assign each playing field two distinct integers between 1 and N (inclusive). You cannot assign the same integer to multiple playing fields. The competition consists of N rounds, each of which proceeds as follows: * For each player, if there is a playing field that is assigned the player's integer, the player goes to that field and fight the other player who comes there. * Then, each player adds 1 to its integer. If it becomes N+1, change it to 1. You want to ensure that no player fights the same opponent more than once during the N rounds. Print an assignment of integers to the playing fields satisfying this condition. It can be proved that such an assignment always exists under the constraints given.
|
N, M = map(int, input().split())
a = 1
b = N
for i in range(M):
print(a,b)
a += 1
b -= 1
|
s864062915
|
Accepted
| 76 | 9,284 | 458 |
N, M = map(int, input().split())
if M%2 == 1:
a = 1
b = M
for _ in range(M//2):
print(a,b)
a += 1
b -= 1
a = M+1
b = 2*M + 1
for _ in range(M - M//2):
print(a,b)
a += 1
b -= 1
else:
a = 1
b = M+1
for _ in range(M//2):
print(a,b)
a += 1
b -= 1
a = M+2
b = 2*M + 1
for _ in range(M - M//2):
print(a,b)
a += 1
b -= 1
|
s942615966
|
p02409
|
u692415695
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,656 | 367 |
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.
|
# (c) midandfeed
q = []
for i in range(4):
b = []
for j in range(3):
a = [0 for x in range(10)]
b.append(a)
q.append(b)
for _ in range(int(input())):
b, f, r, v = [int(x) for x in input().split()]
q[b-1][f-1][r-1] += v
for x in q:
for y in x:
for z in range(len(y)):
print(" {}".format(y[z]), end='')
if (z==(len(y))-1):
print()
print("#"*20)
|
s008593771
|
Accepted
| 50 | 7,744 | 401 |
# (c) midandfeed
q = []
for i in range(4):
b = []
for j in range(3):
a = [0 for x in range(10)]
b.append(a)
q.append(b)
t = int(input())
for _ in range(t):
b, f, r, v = [int(x) for x in input().split()]
q[b-1][f-1][r-1] += v
a = 0
for x in q:
a += 1
for y in x:
for z in range(len(y)):
print(" {}".format(y[z]), end='')
if (z==(len(y))-1):
print()
if (a!=4):
print("#"*20)
|
s761320292
|
p03485
|
u220904910
| 2,000 | 262,144 |
Wrong Answer
| 152 | 12,396 | 109 |
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
import numpy as np
if __name__ == '__main__':
a,b=map(int, input().split())
print(np.ceil((a+b)/2))
|
s247097699
|
Accepted
| 151 | 12,392 | 114 |
import numpy as np
if __name__ == '__main__':
a,b=map(int, input().split())
print(int(np.ceil((a+b)/2)))
|
s403764400
|
p03543
|
u395237353
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 172 |
We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**?
|
N = input()
num = "0"
count = 1
for i in N:
if (num == i):
count += 1
elif count > 2:
break
else:
count = 1
num = i
print(num, count)
if count > 2:
print("yes")
|
s608429728
|
Accepted
| 17 | 3,060 | 173 |
N = input()
num = "0"
count = 1
for i in N:
if (num == i):
count += 1
elif count > 2:
break
else:
count = 1
num = i
if count > 2:
print("Yes")
else:
print("No")
|
s299164986
|
p03505
|
u126823513
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 216 |
_ButCoder Inc._ runs a programming competition site called _ButCoder_. In this site, a user is given an integer value called rating that represents his/her skill, which changes each time he/she participates in a contest. The initial value of a new user's rating is 0, and a user whose rating reaches K or higher is called _Kaiden_ ("total transmission"). Note that a user's rating may become negative. Hikuhashi is a new user in ButCoder. It is estimated that, his rating increases by A in each of his odd-numbered contests (first, third, fifth, ...), and decreases by B in each of his even-numbered contests (second, fourth, sixth, ...). According to this estimate, after how many contests will he become Kaiden for the first time, or will he never become Kaiden?
|
k, a, b = map(int, input().split())
if a <= b:
if a < k:
print(-1)
else:
print(1)
else:
if a > k:
print(1)
else:
diff = a - b
print(2 * ((k - a) // diff) + 1)
|
s554631530
|
Accepted
| 17 | 2,940 | 226 |
k, a, b = map(int, input().split())
if a <= b:
if a < k:
print(-1)
else:
print(1)
else:
if a > k:
print(1)
else:
answer = ((k - b - 1) // (a - b) * 2) + 1
print(answer)
|
s169073709
|
p03370
|
u673281957
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 192 |
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts. Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition: * For each of the N kinds of doughnuts, make at least one doughnut of that kind. At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
|
m,n = map(int,input().split())
dlis = []
zan = n
cnt = int(0)
for i in range(m):
dlis.append(int(input()))
zan = zan - dlis[i]
cnt += 1
print(zan)
x = zan//int(min(dlis))
print(cnt+x)
|
s595850679
|
Accepted
| 17 | 3,060 | 179 |
m,n = map(int,input().split())
dlis = []
zan = n
cnt = int(0)
for i in range(m):
dlis.append(int(input()))
zan = zan - dlis[i]
cnt += 1
x = zan//int(min(dlis))
print(cnt+x)
|
s520495504
|
p03110
|
u489315616
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 185 |
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?
|
sum = 0.0
n = int(input())
for i in range(n):
x, u = map(str, input().split())
if u == 'JPY':
sum += int(x)
else:
sum += 380000.0 * float(x)
print(int(sum))
|
s546296609
|
Accepted
| 17 | 2,940 | 191 |
sum = 0.0
n = int(input())
for i in range(n):
x, u = map(str, input().split())
if u == 'JPY':
sum += int(x)
elif u == 'BTC':
sum += 380000.0 * float(x)
print(sum)
|
s923255644
|
p02694
|
u282676559
| 2,000 | 1,048,576 |
Wrong Answer
| 20 | 9,140 | 114 |
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
X = int(input())
money = 100
count = 0
while money <= X:
count += 1
money = int(money * 1.01)
print(count)
|
s839921962
|
Accepted
| 24 | 9,068 | 113 |
X = int(input())
money = 100
count = 0
while money < X:
count += 1
money = int(money * 1.01)
print(count)
|
s099665355
|
p03448
|
u205580583
| 2,000 | 262,144 |
Wrong Answer
| 58 | 3,064 | 320 |
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()) #500
b = int(input()) #100
c = int(input()) #50
x = int(input())
cnt = 0
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
total = 500 * i + 100 * j + 50 * k
if total == x:
print("cnt!")
cnt += 1
print(cnt)
|
s629232040
|
Accepted
| 53 | 3,060 | 290 |
a = int(input()) #500
b = int(input()) #100
c = int(input()) #50
x = int(input())
cnt = 0
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
total = 500 * i + 100 * j + 50 * k
if total == x:
cnt += 1
print(cnt)
|
s272156364
|
p03149
|
u419686324
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 57 |
You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974".
|
print("YES" if sorted(input()) == list("1479") else "NO")
|
s650091934
|
Accepted
| 18 | 2,940 | 65 |
print("YES" if sorted(input().split()) == list("1479") else "NO")
|
s975899642
|
p03472
|
u017415492
| 2,000 | 262,144 |
Wrong Answer
| 412 | 28,564 | 409 |
You are going out for a walk, when you suddenly encounter a monster. Fortunately, you have N katana (swords), Katana 1, Katana 2, …, Katana N, and can perform the following two kinds of attacks in any order: * Wield one of the katana you have. When you wield Katana i (1 ≤ i ≤ N), the monster receives a_i points of damage. The same katana can be wielded any number of times. * Throw one of the katana you have. When you throw Katana i (1 ≤ i ≤ N) at the monster, it receives b_i points of damage, and you lose the katana. That is, you can no longer wield or throw that katana. The monster will vanish when the total damage it has received is H points or more. At least how many attacks do you need in order to vanish it in total?
|
n,h=map(int,input().split())
ab=[list(map(int,input().split())) for i in range(n)]
a_max=0
b=[]
ans=0
kaisuu=0
for i in range(n):
if a_max<ab[i][0]:
a_max=ab[i][0]
for i in range(n):
b.append(ab[i][1])
b.sort(reverse=True)
for i in range(n):
if a_max<b[i]:
h-=b[i]
kaisuu+=1
else:
if h%a_max==0:
kaisuu+=(h//a_max)
else:
kaisuu+=((h//a_max)+1)
break
print(kaisuu)
|
s791770340
|
Accepted
| 235 | 16,836 | 357 |
n,h=map(int,input().split())
A=[]
B=[]
for i in range(n):
a,b=map(int,input().split())
A.append(a)
B.append(b)
A=max(A)
B.sort(reverse=True)
count=0
for i in B:
if A>i:
break
else:
if h>0:
h-=i
count+=1
else:
break
if h%A==0 and h>0:
print(count+h//A)
elif h%A!=0 and h>0:
print(count+h//A+1)
else:
print(count)
|
s528937576
|
p02399
|
u344890307
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,596 | 84 |
Write a program which reads two integers a and b, and calculates the following values: * a ÷ b: d (in integer) * remainder of a ÷ b: r (in integer) * a ÷ b: f (in real number)
|
a,b=map(int,input().split())
print('{0} {1} {2}'.format(int(a/b),a % b,float(a/b)))
|
s884869959
|
Accepted
| 20 | 5,604 | 89 |
a,b=map(int,input().split())
print('{:d} {:d} {:.5f}'.format(int(a/b),a % b,a/float(b)))
|
s131759958
|
p03448
|
u582614471
| 2,000 | 262,144 |
Wrong Answer
| 50 | 2,940 | 197 |
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,b,c,x =[int(input(i)) for i in range(4)]
ans = 0
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
if 500*i+100*j+50*k==x:
ans+=1
print(ans)
|
s357711508
|
Accepted
| 50 | 2,940 | 197 |
a,b,c,x =[int(input()) for i in range(4)]
ans = 0
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
if 500*i+100*j+50*k==x:
ans+=1
print(ans)
|
s009714235
|
p03193
|
u652656291
| 2,000 | 1,048,576 |
Wrong Answer
| 20 | 3,060 | 140 |
There are N rectangular plate materials made of special metal called AtCoder Alloy. The dimensions of the i-th material are A_i \times B_i (A_i vertically and B_i horizontally). Takahashi wants a rectangular plate made of AtCoder Alloy whose dimensions are exactly H \times W. He is trying to obtain such a plate by choosing one of the N materials and cutting it if necessary. When cutting a material, the cuts must be parallel to one of the sides of the material. Also, the materials have fixed directions and cannot be rotated. For example, a 5 \times 3 material cannot be used as a 3 \times 5 plate. Out of the N materials, how many can produce an H \times W plate if properly cut?
|
n,h,w = map(int,input().split())
ans = 0
for i in range(n):
a,b = map(int,input().split())
if a <= h and b <= w:
ans += 1
print(ans)
|
s268258090
|
Accepted
| 20 | 3,060 | 140 |
n,h,w = map(int,input().split())
ans = 0
for i in range(n):
a,b = map(int,input().split())
if a >= h and b >= w:
ans += 1
print(ans)
|
s553059780
|
p03997
|
u922952729
| 2,000 | 262,144 |
Wrong Answer
| 23 | 3,064 | 64 |
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())
h=int(input())
print((a+b)*h/2)
|
s782720555
|
Accepted
| 23 | 3,064 | 68 |
a=int(input())
b=int(input())
h=int(input())
print(int((a+b)*h/2))
|
s875398245
|
p03673
|
u300579805
| 2,000 | 262,144 |
Wrong Answer
| 161 | 25,156 | 388 |
You are given an integer sequence of length n, a_1, ..., a_n. Let us consider performing the following n operations on an empty sequence b. The i-th operation is as follows: 1. Append a_i to the end of b. 2. Reverse the order of the elements in b. Find the sequence b obtained after these n operations.
|
import math
n = int(input())
A=list(map(int, input().split()))
ans = [0] * n
if (n % 2 == 0):
for i in range(n):
if (i%2 == 0):
ans[n//2+i//2] = A[i]
else:
ans[n//2-(i+2)//2] = A[i]
if (n % 2 != 0):
for i in range(n):
if (i%2 == 0):
ans[n//2-i//2] = A[i]
else:
ans[n//2+(i+2)//2] = A[i]
print(ans)
|
s001780902
|
Accepted
| 233 | 25,156 | 389 |
import math
n = int(input())
A=list(map(int, input().split()))
ans = [0] * n
if (n % 2 == 0):
for i in range(n):
if (i%2 == 0):
ans[n//2+i//2] = A[i]
else:
ans[n//2-(i+2)//2] = A[i]
if (n % 2 != 0):
for i in range(n):
if (i%2 == 0):
ans[n//2-i//2] = A[i]
else:
ans[n//2+(i+2)//2] = A[i]
print(*ans)
|
s164140214
|
p03438
|
u729707098
| 2,000 | 262,144 |
Wrong Answer
| 28 | 4,660 | 205 |
You are given two integer sequences of length N: a_1,a_2,..,a_N and b_1,b_2,..,b_N. Determine if we can repeat the following operation zero or more times so that the sequences a and b become equal. Operation: Choose two integers i and j (possibly the same) between 1 and N (inclusive), then perform the following two actions **simultaneously** : * Add 2 to a_i. * Add 1 to b_j.
|
n = int(input())
a = [int(i) for i in input().split()]
b = [int(i) for i in input().split()]
num,A,B = 0,sum(a),sum(b)
for i in range(n): num = max(num,b[i]-a[i])
if num>B-A: print("No")
else: print("Yes")
|
s404295249
|
Accepted
| 30 | 4,596 | 223 |
n = int(input())
a = [int(i) for i in input().split()]
b = [int(i) for i in input().split()]
num,A,B = 0,sum(a),sum(b)
for i in range(n): num+=max(0,((b[i]-a[i])+(b[i]-a[i])%2)//2)
if num>B-A: print("No")
else: print("Yes")
|
s172334555
|
p03964
|
u911575040
| 2,000 | 262,144 |
Wrong Answer
| 20 | 2,940 | 128 |
AtCoDeer the deer is seeing a quick report of election results on TV. Two candidates are standing for the election: Takahashi and Aoki. The report shows the ratio of the current numbers of votes the two candidates have obtained, but not the actual numbers of votes. AtCoDeer has checked the report N times, and when he checked it for the i-th (1≦i≦N) time, the ratio was T_i:A_i. It is known that each candidate had at least one vote when he checked the report for the first time. Find the minimum possible total number of votes obtained by the two candidates when he checked the report for the N-th time. It can be assumed that the number of votes obtained by each candidate never decreases.
|
N=int(input())
t=a=1
for i in range(N):
T,A=map(int,input().split())
num=max(-t//T,a//A)
t,a=num*T,num*A
print(t+a)
|
s958405679
|
Accepted
| 21 | 2,940 | 135 |
N=int(input())
t=a=1
for i in range(N):
T,A=map(int,input().split())
num=max(-(-t//T),-(-a//A))
t,a=num*T,num*A
print(t+a)
|
s683219225
|
p03610
|
u068538925
| 2,000 | 262,144 |
Wrong Answer
| 41 | 10,944 | 138 |
You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.
|
s = input()
#print(s[1])
#print(s[2])
i = list(range(0,len(s),2))
temp =""
for t in range(len(s)//2):
temp = temp+s[i[t]]
print(temp)
|
s422647742
|
Accepted
| 42 | 10,864 | 145 |
s = input()
#print(s[1])
#print(s[2])
i = list(range(0,len(s)+1,2))
temp =""
for t in range((len(s)+1)//2):
temp = temp+s[i[t]]
print(temp)
|
s052559549
|
p02936
|
u905329882
| 2,000 | 1,048,576 |
Wrong Answer
| 2,117 | 227,016 | 466 |
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.
|
import sys
sys.setrecursionlimit(1000000)
n,q=map(int,input().split())
g=[[]for i in range(n)]
for i in range(n-1):
a,b=map(int,input().split())
g[a-1].append(b-1)
g[b-1].append(a-1)
lis = [0 for i in range(n)]
for i in range(q):
p,x=map(int,input().split())
lis[p-1]+=x
def dfs(i,j,k):
nex=g[i]
lis[i]=k
for d in nex:
if d==j: continue
print(d)
nk=k+lis[d]
dfs(d,i,nk)
dfs(0,-1,lis[0])
print(*lis)
|
s263105672
|
Accepted
| 1,730 | 230,920 | 495 |
import sys
def input():
return sys.stdin.readline()[:-1]
sys.setrecursionlimit(1000000)
n,q=map(int,input().split())
g=[[]for i in range(n)]
for i in range(n-1):
a,b=map(int,input().split())
g[a-1].append(b-1)
g[b-1].append(a-1)
lis = [0 for i in range(n)]
for i in range(q):
p,x=map(int,input().split())
lis[p-1]+=x
def dfs(i,j):
nex=g[i]
for d in nex:
if d==j: continue
#print(d)
lis[d]+=lis[i]
dfs(d,i)
dfs(0,-1)
print(*lis)
|
s614164857
|
p02678
|
u309141201
| 2,000 | 1,048,576 |
Wrong Answer
| 23 | 9,144 | 11 |
There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.
|
print('No')
|
s345113285
|
Accepted
| 753 | 34,976 | 576 |
from collections import deque
n, m = map(int, input().split())
g = [[] for _ in range(n)]
for _ in range(m):
a, b = map(int, input().split())
a -= 1
b -= 1
g[a].append(b)
g[b].append(a)
ans = [-1]*n
used = [0]*n
used[0] = 1
que = deque()
que.appendleft(0)
while que:
v = que.pop()
# print(v)
for nv in g[v]:
if used[nv]:
continue
used[nv] = 1
que.appendleft(nv)
if ans[nv] == -1:
ans[nv] = v
# exit()
# print(ans)
# print(g)
print('Yes')
for i in range(1, n):
print(ans[i]+1)
|
s660236026
|
p03502
|
u584529823
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,104 | 2,940 | 130 |
An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10. Given an integer N, determine whether it is a Harshad number.
|
N = int(input())
sum = 0
tmp = N
while tmp > 0:
sum += tmp % 10
tmp // 10
if N % sum == 0:
print("Yes")
else:
print("No")
|
s198225856
|
Accepted
| 17 | 2,940 | 132 |
N = int(input())
sum = 0
tmp = N
while tmp > 0:
sum += tmp % 10
tmp //= 10
if N % sum == 0:
print("Yes")
else:
print("No")
|
s463660836
|
p03997
|
u646336933
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 69 |
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())
h = int(input())
print((a+b)*h/2)
|
s240184489
|
Accepted
| 17 | 2,940 | 70 |
a = int(input())
b = int(input())
h = int(input())
print((a+b)*h//2)
|
s416201144
|
p02902
|
u875291233
| 2,000 | 1,048,576 |
Wrong Answer
| 21 | 3,316 | 2,259 |
Given is a directed graph G with N vertices and M edges. The vertices are numbered 1 to N, and the i-th edge is directed from Vertex A_i to Vertex B_i. It is guaranteed that the graph contains no self-loops or multiple edges. Determine whether there exists an induced subgraph (see Notes) of G such that the in-degree and out-degree of every vertex are both 1. If the answer is yes, show one such subgraph. Here the null graph is not considered as a subgraph.
|
# find a cycle
def find_cycle(g):
n = len(g)
used = [0]*n #0:not yet 1: visiting 2: visited
for v in range(n):
if used[v] == 2: continue
stack = [v]
hist =[]
while stack:
v = stack[-1]
if used[v] == 1:
used[v] = 2
stack.pop()
hist.pop()
continue
hist.append(v)
used[v] = 1
for c in g[v]:
if used[c] == 2: continue
elif used[c] == 1:
return hist[hist.index(c):]
else:
stack.append(c)
return None
def find_minimal_cycle(g,cycle):
n = len(g)
is_in_cycle = [0]*n
nxt = [-1]*n
l = len(cycle)
for i,c in enumerate(cycle):
is_in_cycle[c] = 1
nxt[c] = cycle[i+1-l]
for v in cycle:
if is_in_cycle[v]:
for c in g[v]:
if is_in_cycle[c] == 1:
v0 = nxt[v]
while v0 != c:
is_in_cycle[v0] = 0
v0 = nxt[v0]
nxt[v] = c
i = is_in_cycle.index(1)
v = nxt[i]
hist = [i]
while v != i:
hist.append(v)
v = nxt[v]
return hist
#########################################################
##########################################################
# coding: utf-8
# Your code here!
import sys
sys.setrecursionlimit(10**6)
readline = sys.stdin.readline
n,m = [int(i) for i in readline().split()]
g = [[] for _ in range(n)]
for _ in range(m):
a,b = [int(i)-1 for i in readline().split()]
g[a].append(b)
cycle = find_cycle(g)
if cycle == None: print(-1)
else:
print(cycle)
res = find_minimal_cycle(g,cycle)
print(len(res))
for i in res:
print(i+1)
|
s396475898
|
Accepted
| 21 | 3,316 | 2,242 |
# find a cycle
def find_cycle(g):
n = len(g)
used = [0]*n #0:not yet 1: visiting 2: visited
for v in range(n):
if used[v] == 2: continue
stack = [v]
hist =[]
while stack:
v = stack[-1]
if used[v] == 1:
used[v] = 2
stack.pop()
hist.pop()
continue
hist.append(v)
used[v] = 1
for c in g[v]:
if used[c] == 2: continue
elif used[c] == 1:
return hist[hist.index(c):]
else:
stack.append(c)
return None
def find_minimal_cycle(g,cycle):
n = len(g)
is_in_cycle = [0]*n
nxt = [-1]*n
l = len(cycle)
for i,c in enumerate(cycle):
is_in_cycle[c] = 1
nxt[c] = cycle[i+1-l]
for v in cycle:
if is_in_cycle[v]:
for c in g[v]:
if is_in_cycle[c] == 1:
v0 = nxt[v]
while v0 != c:
is_in_cycle[v0] = 0
v0 = nxt[v0]
nxt[v] = c
i = is_in_cycle.index(1)
v = nxt[i]
hist = [i]
while v != i:
hist.append(v)
v = nxt[v]
return hist
#########################################################
##########################################################
# coding: utf-8
# Your code here!
import sys
sys.setrecursionlimit(10**6)
readline = sys.stdin.readline
n,m = [int(i) for i in readline().split()]
g = [[] for _ in range(n)]
for _ in range(m):
a,b = [int(i)-1 for i in readline().split()]
g[a].append(b)
cycle = find_cycle(g)
if cycle == None: print(-1)
else:
res = find_minimal_cycle(g,cycle)
print(len(res))
for i in res:
print(i+1)
|
s756402826
|
p03759
|
u749770850
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 89 |
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())
if b - a == c - b :
print("Yes")
else:
print("No")
|
s330332852
|
Accepted
| 17 | 2,940 | 89 |
a, b, c = map(int, input().split())
if b - a == c - b :
print("YES")
else:
print("NO")
|
s371207244
|
p03574
|
u001024152
| 2,000 | 262,144 |
Wrong Answer
| 26 | 3,064 | 869 |
You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process.
|
H, W = map(int, input().split())
S = []
for h in range(H):
S.append(list(input()))
for h in range(H):
for w in range(W):
cnt = 0
if S[h][w]==".":
if w > 0:
cnt += int(S[h][w-1] == "#")
if w < W-1:
cnt += int(S[h][w+1] == "#")
if h > 0:
cnt += int(S[h-1][w] == "#")
if w > 0:
cnt += int(S[h-1][w-1] == "#")
if w < W-1:
cnt += int(S[h-1][w+1] == "#")
if h < H-1:
cnt += int(S[h+1][w] == "#")
if w > 0:
cnt += int(S[h+1][w-1] == "#")
if w < W-1:
cnt += int(S[h+1][w+1] == "#")
S[h][w] = cnt
print(S)
|
s458502790
|
Accepted
| 25 | 3,188 | 899 |
H, W = map(int, input().split())
S = []
for h in range(H):
S.append(list(input()))
for h in range(H):
for w in range(W):
cnt = 0
if S[h][w]==".":
if w > 0:
cnt += int(S[h][w-1] == "#")
if w < W-1:
cnt += int(S[h][w+1] == "#")
if h > 0:
cnt += int(S[h-1][w] == "#")
if w > 0:
cnt += int(S[h-1][w-1] == "#")
if w < W-1:
cnt += int(S[h-1][w+1] == "#")
if h < H-1:
cnt += int(S[h+1][w] == "#")
if w > 0:
cnt += int(S[h+1][w-1] == "#")
if w < W-1:
cnt += int(S[h+1][w+1] == "#")
S[h][w] = str(cnt)
for s in S:
print("".join(s))
|
s157676183
|
p03760
|
u140251125
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 238 |
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.
|
# input
O = input()
E = input()
password = [0 for _ in range(len(O) + len(E))]
for i in range(len(O)):
password[2 * i] = O[i]
for i in range(len(E)):
password[2 * i + 1] = E[i]
password_str = str(password)
print(password_str)
|
s204494818
|
Accepted
| 17 | 3,060 | 241 |
# input
O = input()
E = input()
password = [0 for _ in range(len(O) + len(E))]
for i in range(len(O)):
password[2 * i] = O[i]
for i in range(len(E)):
password[2 * i + 1] = E[i]
password_str = ''.join(password)
print(password_str)
|
s070310422
|
p02742
|
u998082063
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 170 |
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())
wid = 0
if w % 2 == 0:
wid = w // 2
else:
wid = w // 2 + 1
if h % 2 == 0:
print(h*wid)
else:
print(h*wid - h//2*(wid-1))
|
s525686676
|
Accepted
| 17 | 2,940 | 111 |
h, w = map(int,input().split())
if h == 1 or w == 1:
print(1)
exit()
else:
print((h * w + 1) // 2)
|
s996684638
|
p04029
|
u895536055
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 41 |
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)
|
s953945992
|
Accepted
| 20 | 3,060 | 43 |
N = int(input())
print((N * (N + 1)) // 2)
|
s504687470
|
p03644
|
u511457539
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 402 |
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
N = int(input())
count = 0
if N == 1:
print(1)
elif N == 2 or N == 3:
print(2)
else:
even = [i for i in range(2, N+1, 2)]
while len(even) >1:
i = 0
count +=1
while i < len(even):
if even[i]%2 == 0:
even[i] = even[i]/2
i +=1
else:
del even[i-1]
i +=1
print(2**count)
|
s456164380
|
Accepted
| 17 | 2,940 | 209 |
N = int(input())
if N == 1:
print(1)
elif N == 2 or N == 3:
print(2)
elif N < 8:
print(4)
elif N < 16:
print(8)
elif N < 32:
print(16)
elif N < 64:
print(32)
else:
print(64)
|
s594595514
|
p03696
|
u027929618
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 166 |
You are given a string S of length N consisting of `(` and `)`. Your task is to insert some number of `(` and `)` into S to obtain a _correct bracket sequence_. Here, a correct bracket sequence is defined as follows: * `()` is a correct bracket sequence. * If X is a correct bracket sequence, the concatenation of `(`, X and `)` in this order is also a correct bracket sequence. * If X and Y are correct bracket sequences, the concatenation of X and Y in this order is also a correct bracket sequence. * Every correct bracket sequence can be derived from the rules above. Find the shortest correct bracket sequence that can be obtained. If there is more than one such sequence, find the lexicographically smallest one.
|
N = int(input())
S = input()
cnt = [0, 0]
ans = ""
for i in range(N):
if S[i] == ")":
cnt[0] += 1
else:
cnt[1] += 1
print("(" * cnt[0] + S + ")" * cnt[1])
|
s351920500
|
Accepted
| 17 | 3,060 | 143 |
N = int(input())
S = input()
s = S[:]
while "()" in s:
s = s.replace("()", "")
r = s.count("(")
l = s.count(")")
print("(" * l + S + ")" * r)
|
s769825463
|
p03637
|
u652057333
| 2,000 | 262,144 |
Wrong Answer
| 71 | 14,252 | 333 |
We have a sequence of length N, a = (a_1, a_2, ..., a_N). Each a_i is a positive integer. Snuke's objective is to permute the element in a so that the following condition is satisfied: * For each 1 ≤ i ≤ N - 1, the product of a_i and a_{i + 1} is a multiple of 4. Determine whether Snuke can achieve his objective.
|
n = int(input())
a = list(map(int, input().split()))
count = [0] * 3
for i in range(n):
if a[i] % 4 == 0:
count[2] += 1
elif a[i] % 2 == 0:
count[1] += 1
else:
count[0] += 1
ans = False
if count[0] + (count[1] % 2) <= count[2] + 1:
ans = True
if ans:
print("YES")
else:
print("NO")
|
s825803751
|
Accepted
| 71 | 14,252 | 333 |
n = int(input())
a = list(map(int, input().split()))
count = [0] * 3
for i in range(n):
if a[i] % 4 == 0:
count[2] += 1
elif a[i] % 2 == 0:
count[1] += 1
else:
count[0] += 1
ans = False
if count[0] + (count[1] % 2) <= count[2] + 1:
ans = True
if ans:
print("Yes")
else:
print("No")
|
s453280768
|
p03455
|
u421499233
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 93 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a,b=map(int,input().split())
p = a*b % 2
if p == 1 :
print("Even")
else:
print("Odd")
|
s891804870
|
Accepted
| 17 | 2,940 | 98 |
a,b = map(int,input().split())
ans = a*b
if ans % 2 == 0:
print("Even")
else:
print("Odd")
|
s422339992
|
p02612
|
u633203155
| 2,000 | 1,048,576 |
Wrong Answer
| 31 | 9,076 | 31 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
print(N%1000)
|
s063468113
|
Accepted
| 32 | 9,132 | 89 |
N = int(input())
amari = N%1000
if amari == 0:
print(0)
else:
print(1000-amari)
|
s333033951
|
p03693
|
u246820565
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,060 | 89 |
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
rgb = list(map(str,input().split()))
print('Yes' if int(''.join(rgb)) % 4 == 0 else 'NO')
|
s659128524
|
Accepted
| 17 | 2,940 | 89 |
rgb = list(map(str,input().split()))
print('YES' if int(''.join(rgb)) % 4 == 0 else 'NO')
|
s151044502
|
p02399
|
u477464845
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,600 | 84 |
Write a program which reads two integers a and b, and calculates the following values: * a ÷ b: d (in integer) * remainder of a ÷ b: r (in integer) * a ÷ b: f (in real number)
|
a,b = map(int,input().split())
d = a // b
r = a % b
f = a / b
print(d,r,f,sep=' ')
|
s316801812
|
Accepted
| 20 | 5,608 | 95 |
a,b = map(int,input().split(" "))
d = a // b
r = a % b
f = a / b
print(d,r,"{:.5f}".format(f))
|
s205960808
|
p02261
|
u777299405
| 1,000 | 131,072 |
Wrong Answer
| 40 | 6,724 | 423 |
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).
|
n = int(input())
c = input().split()
d = c[:]
for i in range(n):
for j in range(n - 1, i, -1):
if d[j] < d[j - 1]:
d[j], d[j - 1] = d[j - 1], d[j]
print(*d)
print("Stable")
# selectionsort
for i in range(n):
minj = i
for j in range(i, n):
if c[j] < c[minj]:
minj = j
c[i], c[minj] = c[minj], c[i]
print(*c)
print("Stable" if c == d else "Not stable")
|
s536912191
|
Accepted
| 30 | 6,724 | 434 |
n = int(input())
c = input().split()
d = c[:]
for i in range(n):
for j in range(n - 1, i, -1):
if d[j][1] < d[j - 1][1]:
d[j], d[j - 1] = d[j - 1], d[j]
print(*d)
print("Stable")
# selectionsort
for i in range(n):
minj = i
for j in range(i, n):
if c[j][1] < c[minj][1]:
minj = j
c[i], c[minj] = c[minj], c[i]
print(*c)
print("Stable" if c == d else "Not stable")
|
s522409342
|
p03855
|
u112317104
| 2,000 | 262,144 |
Wrong Answer
| 860 | 32,296 | 921 |
There are N cities. There are also K roads and L railways, extending between the cities. The i-th road bidirectionally connects the p_i-th and q_i-th cities, and the i-th railway bidirectionally connects the r_i-th and s_i-th cities. No two roads connect the same pair of cities. Similarly, no two railways connect the same pair of cities. We will say city A and B are _connected by roads_ if city B is reachable from city A by traversing some number of roads. Here, any city is considered to be connected to itself by roads. We will also define _connectivity by railways_ similarly. For each city, find the number of the cities connected to that city by both roads and railways.
|
def find(x, parent):
if parent[x] == x:
return x
else:
parent[x] = find(parent[x], parent)
return parent[x]
def unite(x, y, parent):
x = find(x, parent)
y = find(y, parent)
if x == y: return
parent[y] = x
def solve():
N, K, L = map(int, input().split())
street = [i for i in range(N+1)]
train = [i for i in range(N+1)]
for _ in range(K):
a, b = map(int, input().split())
unite(0, a, street)
unite(0, b, street)
for _ in range(L):
a, b = map(int, input().split())
unite(0, a, train)
unite(0, b, train)
ans = []
for i in range(1, N+1):
c = 0
if street[i] == 0:
c += 1
if train[i] == 0:
c += 1
ans.append(str(c))
print(' '.join(ans))
print(solve())
|
s740798732
|
Accepted
| 1,255 | 55,924 | 1,010 |
from collections import defaultdict
def find(x, parent):
if parent[x] == x:
return x
else:
parent[x] = find(parent[x], parent)
return parent[x]
def unite(x, y, parent):
x = find(x, parent)
y = find(y, parent)
if x == y: return
parent[y] = x
def solve():
N, K, L = map(int, input().split())
street = [i for i in range(N)]
train = [i for i in range(N)]
for _ in range(K):
a, b = map(int, input().split())
unite(a-1, b-1, street)
for _ in range(L):
a, b = map(int, input().split())
unite(a-1, b-1, train)
d = defaultdict(int)
for i in range(N):
a = find(street[i], street)
b = find(train[i], train)
d[(a, b)] += 1
ans = []
for i in range(N):
a = find(street[i], street)
b = find(train[i], train)
ans.append(str(d[(a, b)]))
print(' '.join(ans))
solve()
|
s281911261
|
p03456
|
u077898957
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 103 |
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.
|
import math
a,b = input().split()
print('YES' if (int(a+b))**.5 == int(math.sqrt(int(a+b))) else 'No')
|
s184206213
|
Accepted
| 17 | 3,064 | 103 |
import math
a,b=map(str,input().split())
c=int(a+b)
print('Yes' if (math.sqrt(c)//1)**2==c else 'No')
|
s310817254
|
p03555
|
u923794601
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 172 |
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
#A_Rotation
def rotation(c1, c2):
if str(c1) == str(c2)[:-1]:
print("YES")
else:
print("NO")
if __name__ == "__main__":
c1 = input()
c2 = input()
rotation(c1, c2)
|
s965212598
|
Accepted
| 17 | 2,940 | 173 |
#A_Rotation
def rotation(c1, c2):
if str(c1) == str(c2)[::-1]:
print("YES")
else:
print("NO")
if __name__ == "__main__":
c1 = input()
c2 = input()
rotation(c1, c2)
|
s237317154
|
p00311
|
u546285759
| 1,000 | 262,144 |
Wrong Answer
| 20 | 7,584 | 344 |
浩と健次郎の兄弟は猪苗代湖に魚釣りをしに来ました。二人は以下のように点数を決め、釣り上げた魚の合計得点で勝負することにしました。 * イワナは１匹 a 点 * ヤマメは１匹 b 点 * イワナ１０匹ごとに c 点追加 * ヤマメ２０匹ごとに d点追加 浩と健次郎が釣り上げた魚の数をもとに、どちらが勝ちか、あるいは引き分けか判定するプログラムを作成せよ。
|
h1, h2 = map(int, input().split())
k1, k2 = map(int, input().split())
a, b, c, d = map(int, input().split())
hiroshi = (a * h1) + (b * h2) + (divmod(h1, 10)[0] * c) + (divmod(h2, 20)[0] * d)
kenjiro = (a * k1) + (b * k2) + (divmod(k1, 10)[0] * c) + (divmod(k2, 20)[0] * d)
print("even" if hiroshi else ["hiroshi", "kenjiro"][hiroshi < kenjiro])
|
s673827449
|
Accepted
| 50 | 7,736 | 355 |
h1, h2 = map(int, input().split())
k1, k2 = map(int, input().split())
a, b, c, d = map(int, input().split())
hiroshi = (a * h1) + (b * h2) + (divmod(h1, 10)[0] * c) + (divmod(h2, 20)[0] * d)
kenjiro = (a * k1) + (b * k2) + (divmod(k1, 10)[0] * c) + (divmod(k2, 20)[0] * d)
print("even" if hiroshi == kenjiro else ["hiroshi", "kenjiro"][hiroshi < kenjiro])
|
s157505184
|
p02694
|
u749742659
| 2,000 | 1,048,576 |
Wrong Answer
| 24 | 9,240 | 158 |
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
x = int(input())
year = 0
money = 100
while(True):
money *= 1.01
money = money // 1
year += 1
if(money > x):
break
print(str(year))
|
s133055438
|
Accepted
| 24 | 9,088 | 159 |
x = int(input())
year = 0
money = 100
while(True):
money *= 1.01
money = money // 1
year += 1
if(money >= x):
break
print(str(year))
|
s489479849
|
p02392
|
u092736322
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,588 | 119 |
Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No".
|
ins=input().split()
a=int(ins[0])
b=int(ins[1])
c=int(ins[2])
if a<b and b<c:
print("Yse")
else:
print("No")
|
s284654008
|
Accepted
| 20 | 5,592 | 119 |
ins=input().split()
a=int(ins[0])
b=int(ins[1])
c=int(ins[2])
if a<b and b<c:
print("Yes")
else:
print("No")
|
s524613325
|
p03962
|
u516242950
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 136 |
AtCoDeer the deer recently bought three paint cans. The color of the one he bought two days ago is a, the color of the one he bought yesterday is b, and the color of the one he bought today is c. Here, the color of each paint can is represented by an integer between 1 and 100, inclusive. Since he is forgetful, he might have bought more than one paint can in the same color. Count the number of different kinds of colors of these paint cans and tell him.
|
iro = list(map(int, input().split()))
shu = 1
iro.sort()
for i in range(1, len(iro)):
if iro[i - 1] != iro[i]:
shu += 1
print(iro)
|
s652538825
|
Accepted
| 17 | 2,940 | 136 |
iro = list(map(int, input().split()))
shu = 1
iro.sort()
for i in range(1, len(iro)):
if iro[i - 1] != iro[i]:
shu += 1
print(shu)
|
s918841966
|
p03730
|
u703823201
| 2,000 | 262,144 |
Wrong Answer
| 26 | 8,980 | 157 |
We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
A, B, C = map(int, input().split())
ans = "No"
for i in range(1, 1000):
a = A*i
if a % B == C:
ans = "Yes"
break
print(ans)
|
s222402207
|
Accepted
| 27 | 9,100 | 157 |
A, B, C = map(int, input().split())
ans = "NO"
for i in range(1, 1000):
a = A*i
if a % B == C:
ans = "YES"
break
print(ans)
|
s568010850
|
p03565
|
u210827208
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,192 | 616 |
E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One more thing he found is a sheet of paper with the following facts written on it: * Condition 1: The string S contains a string T as a contiguous substring. * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1. Print the string S. If such a string does not exist, print `UNRESTORABLE`.
|
s=input()
t=input()
ans=1
table=str.maketrans({'?':'a'})
for i in range(len(s),len(t),-1):
x=list(s[i-len(t):i])
if set(x)=={'?'}:
s=s.translate(table)
ans=2
break
else:
cnt=0
for j in range(len(t)):
if x[j]=='?' or x[j]==t[j]:
cnt+=1
if cnt==len(t):
ans=2
s.replace(s[i-len(t):i],t)
print(s[i-len(t):i])
print(s)
s=s.translate(table)
break
else:
ans=1
break
if ans==1:
print('UNRESTORABLE')
else:
print(s)
|
s182165369
|
Accepted
| 21 | 3,188 | 223 |
import re
s=input().replace('?','.')
t=input()
ans='UNRESTORABLE'
for i in range(len(s)-len(t),-1,-1):
if re.match(s[i:i+len(t)],t):
s=s.replace('.','a')
ans=s[:i]+t+s[i+len(t):]
break
print(ans)
|
s589279864
|
p03544
|
u594956556
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 126 |
It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers. You are given an integer N. Find the N-th Lucas number. Here, the i-th Lucas number L_i is defined as follows: * L_0=2 * L_1=1 * L_i=L_{i-1}+L_{i-2} (i≥2)
|
N = int(input())
if N == 1:
print(1)
exit()
L = [0]*(N+1)
for i in range(2, N+1):
L[i] = L[i-1] + L[i-2]
print(L[N])
|
s653417208
|
Accepted
| 18 | 3,060 | 144 |
N = int(input())
if N == 1:
print(1)
exit()
L = [0]*(N+1)
L[0] = 2
L[1] = 1
for i in range(2, N+1):
L[i] = L[i-1] + L[i-2]
print(L[N])
|
s845266200
|
p02389
|
u477717106
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,584 | 81 |
Write a program which calculates the area and perimeter of a given rectangle.
|
line=input().split()
a=int(line[0])
b=int(line[1])
print(a*b)
print((2*a)+(2*b))
|
s133791819
|
Accepted
| 20 | 5,588 | 71 |
line=input().split()
a=int(line[0])
b=int(line[1])
print(a*b,2*(a+b))
|
s769408416
|
p03360
|
u411858517
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 130 |
There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations?
|
l = list(map(int, input().split()))
K = int(input())
l.sort()
for i in range(K):
l[0] = 2*l[0]
print(l[0] + l[1] + l[2])
|
s788050537
|
Accepted
| 17 | 2,940 | 130 |
l = list(map(int, input().split()))
K = int(input())
l.sort()
for i in range(K):
l[2] = 2*l[2]
print(l[0] + l[1] + l[2])
|
s065528876
|
p03910
|
u276204978
| 2,000 | 262,144 |
Wrong Answer
| 1,801 | 399,640 | 199 |
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.
|
def solve():
N = int(input())
s = sum([i for i in range(1, N+1)])
for i in reversed(range(1, N+1)):
if s - i <= N:
return i
s -= i
print(solve())
|
s261457645
|
Accepted
| 19 | 3,444 | 261 |
def solve():
N = int(input())
score = 0
for i in range(1, N+1):
score += i
if score >= N:
break
x = score - N
res = list(range(1, x)) + list(range(x+1, i+1))
return '\n'.join(map(str, res))
print(solve())
|
s550041217
|
p03711
|
u360515075
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 192 |
Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group.
|
x, y = map(int, input().split())
g1 = [1, 3, 5, 7, 8, 10, 12]
g2 = [4, 6, 9, 11]
g3 = [2]
print ("YES" if x in g1 and y in g1 or
x in g2 and y in g2 or x in g3 and y in g3 else "NO")
|
s229423726
|
Accepted
| 17 | 3,060 | 192 |
x, y = map(int, input().split())
g1 = [1, 3, 5, 7, 8, 10, 12]
g2 = [4, 6, 9, 11]
g3 = [2]
print ("Yes" if x in g1 and y in g1 or
x in g2 and y in g2 or x in g3 and y in g3 else "No")
|
s664276425
|
p03377
|
u977855674
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 85 |
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, str(input()).split(" "))
print("Yes" if A <= X <= A + B else "No")
|
s037909762
|
Accepted
| 18 | 2,940 | 85 |
A, B, X = map(int, str(input()).split(" "))
print("YES" if A <= X <= A + B else "NO")
|
s807966405
|
p03407
|
u274045692
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 89 |
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 = map(int, input().split())
if A + B == C:
print("Yes")
else:
print("No")
|
s697617397
|
Accepted
| 19 | 3,060 | 90 |
A, B, C = map(int, input().split())
if A + B >= C:
print("Yes")
else:
print("No")
|
s803738236
|
p03557
|
u190079347
| 2,000 | 262,144 |
Wrong Answer
| 2,105 | 22,720 | 297 |
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.
|
import bisect
n = int(input())
a = sorted(list(map(int,input().split())))
b = sorted(list(map(int,input().split())))
c = sorted(list(map(int,input().split())))
count = 0
for i in range(n):
for s in range(bisect.bisect_right(b, a[i]),n):
count += n-bisect.bisect_right(c, a[s])-1
print(count)
|
s819424354
|
Accepted
| 339 | 23,328 | 271 |
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
c=list(map(int,input().split()))
a.sort()
b.sort()
c.sort()
import bisect
ans=0
for i in b:
q=bisect.bisect_left(a,i)
s=bisect.bisect_right(c,i)
ans=ans+q*(n-s)
print(ans)
|
s237055825
|
p03573
|
u967835038
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 101 |
You are given three integers, A, B and C. Among them, two are the same, but the remaining one is different from the rest. For example, when A=5,B=7,C=5, A and C are the same, but B is different. Find the one that is different from the rest among the given three integers.
|
a,b,c=map(int,input().split())
if a==b:
print('c')
elif b==c:
print('a')
else:
print('b')
|
s453156063
|
Accepted
| 17 | 2,940 | 95 |
a,b,c=map(int,input().split())
if a==b:
print(c)
elif b==c:
print(a)
else:
print(b)
|
s581291939
|
p02578
|
u736479342
| 2,000 | 1,048,576 |
Wrong Answer
| 167 | 32,328 | 194 |
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 = int(input())
a = [int(i) for i in input().split()]
print(a)
ans=0
for i in range(n-1):
d = a[i-1] - a[i]
if d<0:
ans+= d+1
a[i]=a[i]+d+1
else:
continue
print(abs(ans))
|
s106957668
|
Accepted
| 155 | 32,092 | 181 |
n = int(input())
a = [int(i) for i in input().split()]
ans=0
for i in range(1,n):
d = a[i-1] - a[i]
if d>0:
ans+= d
a[i]=a[i]+d
else:
continue
print(abs(ans))
|
s561302154
|
p03149
|
u405256066
| 2,000 | 1,048,576 |
Wrong Answer
| 20 | 3,188 | 253 |
You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974".
|
import re
from sys import stdin
N=(stdin.readline().rstrip())
cnt=0
for i in range(7):
mae="keyence"[0:i+1]
ato="keyence"[i+1:8]
if re.match("%s.*%s"%(mae, ato),N):
print("YES")
cnt+=1
break
if cnt==0:
print("NO")
|
s627209062
|
Accepted
| 17 | 2,940 | 167 |
from sys import stdin
N=(stdin.readline().rstrip())
if N.count("1")>0 and N.count("9")>0 and N.count("7")>0 and N.count("4") >0:
print("YES")
else:
print("NO")
|
s346916651
|
p03943
|
u514894322
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 103 |
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
candy = set(map(int,input().split()))
if max(candy)*2 == sum(candy):
print('YES')
else:
print('NO')
|
s645423333
|
Accepted
| 17 | 2,940 | 104 |
candy = list(map(int,input().split()))
if max(candy)*2 == sum(candy):
print('Yes')
else:
print('No')
|
s776856357
|
p02694
|
u553600587
| 2,000 | 1,048,576 |
Wrong Answer
| 23 | 9,180 | 115 |
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
import math
X = int(input().strip())
v = 100
y = 0
while v <= X:
y += 1
v = math.floor(v * 1.01)
print(y)
|
s945888361
|
Accepted
| 22 | 9,156 | 142 |
import math
X = int(input().strip())
v = 100
y = 0
while True:
y += 1
v = math.floor(v * 1.01)
if v >= X:
break
print(y)
|
s375842640
|
p03699
|
u371686382
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 318 |
You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well. However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade?
|
N = int(input())
S = sorted([int(input()) for _ in range(N)])
ans = sum(S)
if ans % 10 == 0:
# find numbers that are not multiples of 10 in order of decreasing.
for s in S:
if s % 10 != 0:
ans -= s
break
# if all the numbers in the list are multiples of 10, the ans is 0.
ans = 0
print(ans)
|
s108705168
|
Accepted
| 17 | 2,940 | 340 |
N = int(input())
S = sorted([int(input()) for _ in range(N)])
ans = sum(S)
if ans % 10 == 0:
# find numbers that are not multiples of 10 in order of decreasing.
for s in S:
if s % 10 != 0:
ans -= s
break
# if all the numbers in the list are multiples of 10, the ans is 0.
if ans == sum(S):
ans = 0
print(ans)
|
s657469908
|
p02608
|
u929996201
| 2,000 | 1,048,576 |
Wrong Answer
| 2,206 | 9,208 | 302 |
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
import itertools
import math
a = int(input())
def calc(x,y,z,n):
if(n==x*x+y*y+z*z+x*y+y*z+z*x):
return True
for i in range(1,a+1):
l = list(range(int(math.sqrt(a+1))))
cnt = 0;
for x,y,z in itertools.combinations_with_replacement(l, 3):
if(calc(x,y,z,i)):
cnt+=1;
print(cnt)
|
s496383097
|
Accepted
| 480 | 9,836 | 229 |
import itertools
import math
a = int(input())
cnt = 0
ans = [0]*100000
l = list(range(1,int(math.sqrt(a))))
for x,y,z in itertools.product(l, repeat=3):
ans[x*x+y*y+z*z+x*y+y*z+z*x]+=1
for i in range(1,a+1):
print(ans[i])
|
s835586464
|
p03023
|
u474514603
| 2,000 | 1,048,576 |
Wrong Answer
| 22 | 3,688 | 1,143 |
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
# -*-
"""
oj dl https://atcoder.jp/contests/abc116/tasks/abc116_d -d test-d
oj test -d test-d -c "python abc116d.py"
oj test -d test-d -c "python abc116d.py" test-d/sample-3.in
"""
from collections import defaultdict
import sys
import math
from datetime import datetime
def sol(n):
# 3 -> 3 * 60 = 180
# 4 -> 4 * 90 = 360
# 5 -> 5 *
# 360 / 5 = 180 - 72 = 108 = 54 * 2 * 5
# (180 - (360 / 5)) * 5
return (180 - (360 / n)) * n
do_submit = True
def input_parse(input_str):
lines = [x.strip() for x in input_str.split("\n") if x.strip()]
parsed_lines = [list(map(str, line.split())) for line in lines]
print(parsed_lines)
n = int(parsed_lines[0][0])
# S = parsed_lines[1][0]
# return n, k, S
return n
if not do_submit:
n = input_parse("""
3
""")
print(sol(n))
n = input_parse("""
100
""")
print(sol(n))
else:
# S = input().split()
# print(sol(n, k, S))
n = int(input().strip())
# S = input().strip()
print(sol(n))
|
s656280816
|
Accepted
| 23 | 3,688 | 1,148 |
#-*- coding: utf-8
# -*-
"""
oj dl https://atcoder.jp/contests/abc116/tasks/abc116_d -d test-d
oj test -d test-d -c "python abc116d.py"
oj test -d test-d -c "python abc116d.py" test-d/sample-3.in
"""
from collections import defaultdict
import sys
import math
from datetime import datetime
def sol(n):
# 3 -> 3 * 60 = 180
# 4 -> 4 * 90 = 360
# 5 -> 5 *
# 360 / 5 = 180 - 72 = 108 = 54 * 2 * 5
# (180 - (360 / 5)) * 5
return int((180 - (360 / n)) * n)
do_submit = True
def input_parse(input_str):
lines = [x.strip() for x in input_str.split("\n") if x.strip()]
parsed_lines = [list(map(str, line.split())) for line in lines]
print(parsed_lines)
n = int(parsed_lines[0][0])
# S = parsed_lines[1][0]
# return n, k, S
return n
if not do_submit:
n = input_parse("""
3
""")
print(sol(n))
n = input_parse("""
100
""")
print(sol(n))
else:
# S = input().split()
# print(sol(n, k, S))
n = int(input().strip())
# S = input().strip()
print(sol(n))
|
s823359512
|
p03737
|
u319612498
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 54 |
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.
|
a,b,c=map(str,input().split())
print("a[0]+b[0]+c[0]")
|
s160723307
|
Accepted
| 17 | 2,940 | 76 |
a,b,c=map(str,input().split())
print(a[0].upper()+b[0].upper()+c[0].upper())
|
s327491459
|
p02361
|
u279546122
| 3,000 | 131,072 |
Wrong Answer
| 30 | 8,024 | 793 |
For a given weighted graph G(V, E) and a source r, find the source shortest path to each vertex from the source (SSSP: Single Source Shortest Path).
|
#!usr/bin/env python3
from collections import defaultdict
import time
def main():
#Read stdin
start = time.clock()
fl = input().split(" ")
V = int(fl[0])
E = int(fl[1])
R = int(fl[2])
#Adjacency list
G = defaultdict(list)
for i in range(int(E)):
s, t, w = [int(x) for x in input().split(" ")]
G[s].append((t,w))
print(G)
#initialized
visited = []
#unseen = [R]
d = {}
d[0] = 0
q = [R]
while q != []:
u = q.pop(0)
if u not in visited:
for v in G[u]:
if v[0] not in visited:
d[v[0]] = d[u] + v[1]
q.append(v[0])
visited.append(u)
for k in d.keys():
print(d[k])
if __name__ == '__main__':
main()
|
s706372755
|
Accepted
| 2,430 | 104,916 | 722 |
from collections import defaultdict
from collections import deque
from sys import stdin
#time 2.49
def sp(G,R,V):
d = {}
INF = float('inf')
for i in range(V):
d[i] = INF
d[R] = 0
q = deque([(0, R)])
while q:
(cost, v) = q.popleft()
for (u, c) in G[v].items():
if d[u] > d[v] + c:
d[u] = d[v] + c
q.append((d[u], u))
return d
V, E, R = [int(x) for x in stdin.readline().split()]
G = defaultdict(dict)
for case in range(E):
s, t, w = [int(x) for x in stdin.readline().split()]
G[s][t] = w
d = sp(G, R, V)
for k in range(V):
if d[k] == float('inf'):
print("INF")
else:
print(d[k])
|
s951636818
|
p03251
|
u252828980
| 2,000 | 1,048,576 |
Wrong Answer
| 19 | 3,064 | 393 |
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())
li1 = list(map(int,input().split()))
li2 = list(map(int,input().split()))
for z in range(x,y+1):
if all([i < z for i in li1]):
flag = True
else:
flag = False
if all([i >= z for i in li2]):
flag = True
else:
flag = False
if flag == True:
print("No War")
elif flag == False:
print("War")
|
s930980130
|
Accepted
| 18 | 3,060 | 255 |
n,m,x,y = map(int,input().split())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
for i in range(x,y+1):
if all([p < i for p in a]) and all([q >=i for q in b]) and x < i <= y:
print("No War")
exit()
print("War")
|
s105154148
|
p03251
|
u730318873
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,064 | 289 |
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()))
ans = 0
for z in range(X+1,Y+1):
if max(x) < z and min(y) >= z:
ans = 1
if ans == 1:
j = z
break
if ans == 0:
print('War')
else:
print('No war')
|
s604390586
|
Accepted
| 18 | 3,060 | 289 |
N,M,X,Y = map(int,input().split())
x = list(map(int,input().split()))
y = list(map(int,input().split()))
ans = 0
for z in range(X+1,Y+1):
if max(x) < z and min(y) >= z:
ans = 1
if ans == 1:
j = z
break
if ans == 0:
print('War')
else:
print('No War')
|
s463065120
|
p02743
|
u117541450
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,064 | 185 |
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
|
l=[int(e) for e in input().split(' ')]
a=l.pop(0)
b=l.pop(0)
c=l.pop(0)
lt=4*a*b
rt=c**2 - 2*(a+b)*c + (a+b)**2
print(lt)
print(rt)
if lt < rt:
print('Yes')
else:
print('No')
|
s698920211
|
Accepted
| 18 | 3,064 | 207 |
l=[int(e) for e in input().split(' ')]
a=l.pop(0)
b=l.pop(0)
c=l.pop(0)
if c < (a+b):
print('No')
exit()
lt=4*a*b + 2*(a+b)*c
rt=c**2 + (a+b)**2
if lt < rt:
print('Yes')
else:
print('No')
|
s503498602
|
p03693
|
u086503932
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 79 |
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
a,b,c=map(int,input().split())
print('YES') if b*10+c % 4 == 0 else print('NO')
|
s578520072
|
Accepted
| 17 | 2,940 | 82 |
a,b,c=map(int,input().split())
print('YES') if (b*10+c) % 4 == 0 else print('NO')
|
s030697821
|
p03623
|
u881472317
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 116 |
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x, a, b = map(int, input().split())
xa = abs(x - a)
xb = abs(x - b)
if xa > xb:
print("A")
else:
print("B")
|
s305561451
|
Accepted
| 17 | 2,940 | 116 |
x, a, b = map(int, input().split())
xa = abs(x - a)
xb = abs(x - b)
if xa > xb:
print("B")
else:
print("A")
|
s493603194
|
p04029
|
u132974957
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 130 |
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("あめ"))
def main(x):
y = 0
for i in range(x+1):
y = y + i
return y
print(main(x))
|
s850772320
|
Accepted
| 17 | 2,940 | 68 |
x = int(input())
y = 0
for i in range(x+1):
y += i
print(y)
|
s168041240
|
p03401
|
u036340997
| 2,000 | 262,144 |
Wrong Answer
| 221 | 14,048 | 310 |
There are N sightseeing spots on the x-axis, numbered 1, 2, ..., N. Spot i is at the point with coordinate A_i. It costs |a - b| yen (the currency of Japan) to travel from a point with coordinate a to another point with coordinate b along the axis. You planned a trip along the axis. In this plan, you first depart from the point with coordinate 0, then visit the N spots in the order they are numbered, and finally return to the point with coordinate 0. However, something came up just before the trip, and you no longer have enough time to visit all the N spots, so you decided to choose some i and cancel the visit to Spot i. You will visit the remaining spots as planned in the order they are numbered. You will also depart from and return to the point with coordinate 0 at the beginning and the end, as planned. For each i = 1, 2, ..., N, find the total cost of travel during the trip when the visit to Spot i is canceled.
|
n = int(input())
a = [0]
a += list(map(int, input().split()))
a.append(0)
a_sum = 0
for i in range(n + 1):
a_sum += abs(a[i] - a[i+1])
for i in range(1, n+1):
if (a[i-1] <= a[i] and a[i] <= a[i+1]) or (a[i-1] >= a[i] and a[i] >= a[i+1]):
print(a_sum)
else:
print(a_sum - 2 * abs(a[i] - a[i+1]))
|
s167659619
|
Accepted
| 250 | 14,048 | 336 |
n = int(input())
a = [0]
a += list(map(int, input().split()))
a.append(0)
a_sum = 0
for i in range(n + 1):
a_sum += abs(a[i] - a[i+1])
for i in range(1, n+1):
if (a[i-1] <= a[i] and a[i] <= a[i+1]) or (a[i-1] >= a[i] and a[i] >= a[i+1]):
print(a_sum)
else:
print(a_sum - 2 * min(abs(a[i] - a[i+1]), abs(a[i-1] - a[i])))
|
s325103765
|
p03386
|
u859897687
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 98 |
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
a,b,c=map(int,input().split())
for i in range(a,a+c):
print(i)
for i in range(b-c,b):
print(i)
|
s240835608
|
Accepted
| 18 | 3,060 | 168 |
a,b,c=map(int,input().split())
if b-a+1>c*2:
for i in range(a,a+c):
print(i)
for i in range(b-c+1,b+1):
print(i)
else:
for i in range(a,b+1):
print(i)
|
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