text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n athletes in front of you. Athletes are numbered from 1 to n from left to right. You know the strength of each athlete β the athlete number i has the strength s_i.
You want to split all athletes into two teams. Each team must have at least one athlete, and each athlete must be exactly in one team.
You want the strongest athlete from the first team to differ as little as possible from the weakest athlete from the second team. Formally, you want to split the athletes into two teams A and B so that the value |max(A) - min(B)| is as small as possible, where max(A) is the maximum strength of an athlete from team A, and min(B) is the minimum strength of an athlete from team B.
For example, if n=5 and the strength of the athletes is s=[3, 1, 2, 6, 4], then one of the possible split into teams is:
* first team: A = [1, 2, 4],
* second team: B = [3, 6].
In this case, the value |max(A) - min(B)| will be equal to |4-3|=1. This example illustrates one of the ways of optimal split into two teams.
Print the minimum value |max(A) - min(B)|.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Then t test cases follow.
Each test case consists of two lines.
The first line contains positive integer n (2 β€ n β€ 50) β number of athletes.
The second line contains n positive integers s_1, s_2, β¦, s_n (1 β€ s_i β€ 1000), where s_i β is the strength of the i-th athlete. Please note that s values may not be distinct.
Output
For each test case print one integer β the minimum value of |max(A) - min(B)| with the optimal split of all athletes into two teams. Each of the athletes must be a member of exactly one of the two teams.
Example
Input
5
5
3 1 2 6 4
6
2 1 3 2 4 3
4
7 9 3 1
2
1 1000
3
100 150 200
Output
1
0
2
999
50
Note
The first test case was explained in the statement. In the second test case, one of the optimal splits is A=[2, 1], B=[3, 2, 4, 3], so the answer is |2-2|=0.
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
L=list(map(int,input().split()))
L.sort()
p=L[1]-L[0]
for i in range(n-1):
if L[i+1]-L[i]<p:
p=L[i+1]-L[i]
if p==0:
print(p)
break
if p!=0:
print(p)
```
No
| 107,200 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and x (1 β€ n β€ 10^5; 1 β€ x β€ 10^9) β the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 β€ a_i β€ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer β the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
Tags: brute force, dp, greedy, implementation, sortings
Correct Solution:
```
n = int(input())
for _ in range(n):
a, b = list(map(int, input().split()))
arr = list(map(int, input().split()))
arr.sort(reverse=True)
index = 1
ans = 0
for i in range(a):
if(arr[i]*index < b):
index += 1
else:
ans += 1
index = 1
print(ans)
```
| 107,201 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and x (1 β€ n β€ 10^5; 1 β€ x β€ 10^9) β the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 β€ a_i β€ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer β the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
Tags: brute force, dp, greedy, implementation, sortings
Correct Solution:
```
from sys import stdin
def mp():return map(int,stdin.readline().split())
def ml():return list(map(int,stdin.readline().split()))
for _ in range(int(input())):
n,x=mp()
a=ml();a.sort(reverse=True)
ans=0;c=0
for num in a:
c+=1
if c*num >= x:
ans+=1;c=0
print(ans)
```
| 107,202 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and x (1 β€ n β€ 10^5; 1 β€ x β€ 10^9) β the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 β€ a_i β€ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer β the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
Tags: brute force, dp, greedy, implementation, sortings
Correct Solution:
```
from sys import stdin
input = stdin.readline
def solve():
_, x = map(int, input().split())
a = list(map(int, input().split()))
a.sort(reverse=True)
ans = 0
cnt = 0
for i in a:
k = (x + i - 1) // i
if cnt + 1 >= k:
cnt -= k - 1
ans += 1
else:
cnt += 1
print(ans)
def main():
t = int(input())
for _ in range(t):
solve()
if __name__ == "__main__":
main()
```
| 107,203 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and x (1 β€ n β€ 10^5; 1 β€ x β€ 10^9) β the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 β€ a_i β€ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer β the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
Tags: brute force, dp, greedy, implementation, sortings
Correct Solution:
```
import sys
input = sys.stdin.readline
for _ in range(int(input())):
n, x = map(int, input().split())
A = sorted(map(int, input().split()), reverse=True)
c = 0
r = 0
for v in A:
c += 1
if c * v >= x:
r += 1
c = 0
print(r)
```
| 107,204 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and x (1 β€ n β€ 10^5; 1 β€ x β€ 10^9) β the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 β€ a_i β€ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer β the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
Tags: brute force, dp, greedy, implementation, sortings
Correct Solution:
```
t = int(input())
for i in range(t):
n,x = map(int,input().split())
l = list(map(int,input().split()))
l.sort()
j = n-1
pointer = n
pointer2 = n-1
count = 0
while j >= 0:
if l[j]*(pointer-j) >= x:
pointer = j
count = count + 1
j = j - 1
print(count)
```
| 107,205 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and x (1 β€ n β€ 10^5; 1 β€ x β€ 10^9) β the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 β€ a_i β€ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer β the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
Tags: brute force, dp, greedy, implementation, sortings
Correct Solution:
```
t=int(input())
for _ in range(t):
n,x=map(int,input().split(" "))
a=list(map(int,input().split(" ")))
a.sort(reverse=True)
s=0
t=[]
count=0
for y in range(n):
s=s+1
if a[y]*s>=x:
count=count+1
s=0
print(count)
```
| 107,206 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and x (1 β€ n β€ 10^5; 1 β€ x β€ 10^9) β the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 β€ a_i β€ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer β the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
Tags: brute force, dp, greedy, implementation, sortings
Correct Solution:
```
t=int(input())
for _ in range(t):
n,x=list(map(int,input().split(" ")))
arr=list(map(int,input().split(" ")))
arr.sort(reverse=True)
team=0
count=0
for i in arr:
team+=1
if team*i>=x:
count+=1
team=0
print(count)
```
| 107,207 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and x (1 β€ n β€ 10^5; 1 β€ x β€ 10^9) β the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 β€ a_i β€ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer β the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
Tags: brute force, dp, greedy, implementation, sortings
Correct Solution:
```
import sys
def answer(n, x, a):
num = 0
lindx = -1
a.sort(reverse=True)
#print('a=', a)
for i in range(n):
if (i-lindx)*a[i] >= x:
num += 1
lindx = i
return num
def main():
t = int(sys.stdin.readline())
while t:
n, x = map(int, sys.stdin.readline().split())
a = list(map(int, sys.stdin.readline().split()))
print(answer(n, x, a))
t -= 1
return
main()
```
| 107,208 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and x (1 β€ n β€ 10^5; 1 β€ x β€ 10^9) β the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 β€ a_i β€ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer β the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
Submitted Solution:
```
from math import ceil
for _ in range(int(input())):
n , x =[int(i) for i in input().split()]
a = [int(i) for i in input().split()]
a.sort(reverse=True)
# print(a)
c = 0
k = 1
for i in range(n):
# print(a[i],k)
if a[i]*k >=x:
c = c+1
k = 1
else:
k = k+1
print(c)
```
Yes
| 107,209 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and x (1 β€ n β€ 10^5; 1 β€ x β€ 10^9) β the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 β€ a_i β€ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer β the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
Submitted Solution:
```
import math
for j in range(int(input())):
n,x=map(int,input().split())
l=list(map(int,input().split()))
p=0
z=0
k=0
t=[]
for i in range(n):
if l[i]>=x:
p+=1
elif l[i]>=math.ceil(x/2):
z+=1
if z==2:
z=0
k+=1
else:
t.append(l[i])
t.sort()
m=0
y=z
for i in range(len(t)-1,-1,-1):
u=t[i]*(y+1)
y+=1
if u>=x:
m+=1
y=0
print(m+k+p)
```
Yes
| 107,210 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and x (1 β€ n β€ 10^5; 1 β€ x β€ 10^9) β the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 β€ a_i β€ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer β the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
Submitted Solution:
```
if __name__ == "__main__":
T = int(input())
for t in range(T):
n, x = map(int, input().split())
a = list(map(int, input().split()))
a.sort(reverse = True)
cnt = 0
size = 1
for e in a:
if e*size >= x:
cnt += 1
size = 1
else: size += 1
print(cnt)
```
Yes
| 107,211 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and x (1 β€ n β€ 10^5; 1 β€ x β€ 10^9) β the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 β€ a_i β€ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer β the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
Submitted Solution:
```
from sys import stdin, stdout
# 2 5 7 9 11, 10
# 7 9, 11
# 2 2 3 4, 8
#
if __name__ == '__main__':
def dfs(idx, a, x, memo):
if idx >= len(a):
return 0
if memo[idx] != -1:
return memo[idx]
r = x // a[idx]
if x % a[idx] != 0:
r += 1
r1 = 0
if idx + r - 1 < len(a):
r1 = 1 + dfs(idx + r, a, x, memo)
r2 = dfs(idx + 1, a, x, memo)
memo[idx] = max(r1, r2)
return memo[idx]
# dfs
def create_the_teams2(n, x, a):
a.sort()
memo = [-1]*n
return dfs(0, a, x, memo)
# greedy
def create_the_teams(n, x, a):
a.sort(reverse=True)
res = 0
cnt = 0
for i in range(n):
cnt += 1
if a[i]*cnt >= x:
res += 1
cnt = 0
return res
t = int(stdin.readline())
for i in range(t):
n, x = map(int, stdin.readline().split())
a = list(map(int, stdin.readline().split()))
res = create_the_teams(n, x, a)
stdout.write(str(res) + '\n')
```
Yes
| 107,212 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and x (1 β€ n β€ 10^5; 1 β€ x β€ 10^9) β the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 β€ a_i β€ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer β the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
Submitted Solution:
```
from math import ceil
def team_maker(arr, maxlen, x):
i = 0
cnt = 0
while i + maxlen <= n:
if arr[i] * maxlen >= x:
i += min(ceil(x / arr[i]), maxlen)
cnt += 1
else:
i += 1
return cnt
for _ in range(int(input())):
n, x = map(int, input().split())
ls = sorted(list(map(int, input().split())))
ar = [0] * (n + 1)
for i in ls:
p = ceil(x / i)
if p <= n:
ar[p] += 1
mx = -1
minlen = 0
ans = 0
for i in range(1, n + 1):
ans = max(team_maker(ls,i, x), ans)
print(ans)
```
No
| 107,213 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and x (1 β€ n β€ 10^5; 1 β€ x β€ 10^9) β the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 β€ a_i β€ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer β the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
Submitted Solution:
```
from math import ceil
from collections import Counter
for _ in range(int(input())):
n,x = map(int,input().split())
l = list(map(int,input().split()))
l.sort()
ind,count = 0,0
for i in range(n-1,-1,-1):
if l[i] >= x:
count += 1
else:
ind = i
break
l = l[0:ind+1]
for i in range(len(l)):
l[i] = ceil(x/l[i])
l.sort()
num = list(Counter(l).items())
l.clear()
for i in range(len(num)):
count += num[i][1]//num[i][0]
for j in range(num[i][1]%num[i][0]):
l.append(num[i][0])
x = 1
for i in range(len(l)):
if l[i]==x:
x = 0
count += 1
x += 1
print(count)
```
No
| 107,214 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and x (1 β€ n β€ 10^5; 1 β€ x β€ 10^9) β the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 β€ a_i β€ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer β the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
Submitted Solution:
```
import math
for _ in range(int(input())):
n,x=map(int,input().split())
arr=list(map(int,input().split()))
arr.sort()
teams,members=0,1
for i in range(n):
if(arr[i]*members>=x):
teams+=1
members=0
members+=1
print(teams)
```
No
| 107,215 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.
Each programmer should belong to at most one team. Some programmers may be left without a team.
Calculate the maximum number of teams that you can assemble.
Input
The first line contains the integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and x (1 β€ n β€ 10^5; 1 β€ x β€ 10^9) β the number of programmers and the restriction of team skill respectively.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 β€ a_i β€ 10^9), where a_i is the skill of the i-th programmer.
The sum of n over all inputs does not exceed 10^5.
Output
For each test case print one integer β the maximum number of teams that you can assemble.
Example
Input
3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7
Output
2
1
0
Submitted Solution:
```
from collections import defaultdict as dd
from collections import deque
import bisect
import heapq
def ri():
return int(input())
def rl():
return list(map(int, input().split()))
q = ri()
for _ in range(q):
n, k = rl()
a = rl()
a.sort(reverse=True)
d = deque(a)
ans = 0
l=2
while (l<=len(d)):
if d[l-1]*l>=k:
ans+=1
for _ in range(l):
d.popleft()
else:
l+=1
print(ans)
```
No
| 107,216 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x β€ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 β€ |s| β€ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 β€ x β€ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
Tags: 2-sat, brute force, constructive algorithms, greedy
Correct Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz'
M=998244353
EPS=1e-6
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
for _ in range(Int()):
s=input()
x=Int()
n=len(s)
w=[1]*n
for i in range(n):
if(s[i]=='0'):
if(i+x<n):
w[i+x]=0
if(i-x>=0):
w[i-x]=0
ok=True
for i in range(n):
if(s[i]=='0'):
if(i+x<n and w[i+x]==1):
ok=False
if(i-x>=0 and w[i-x]==1):
ok=False
else:
if(i+x>=n or w[i+x]==0):
if(i-x<0 or w[i-x]==0):
ok=False
if(ok):
print(*w,sep="")
else:
print(-1)
```
| 107,217 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x β€ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 β€ |s| β€ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 β€ x β€ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
Tags: 2-sat, brute force, constructive algorithms, greedy
Correct Solution:
```
t = int(input())
for i in range(t):
s = input()
m = len(s)
x = int(input())
ANS = [1] * m
for i in range(m):
if s[i] == "0":
if i-x >= 0:
ANS[i-x] = 0
if i+x < m:
ANS[i+x] = 0
ng = 0
for i in range(m):
one = 0
if (i-x >= 0 and ANS[i-x] == 1) or (i+x < m and ANS[i+x] == 1):
one = 1
if (one == 1 and s[i] == "0") or (one == 0 and s[i] == "1"):
ng = 1
break
if ng == 1:
print(-1)
else:
print("".join([str(i) for i in ANS]))
```
| 107,218 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x β€ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 β€ |s| β€ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 β€ x β€ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
Tags: 2-sat, brute force, constructive algorithms, greedy
Correct Solution:
```
import sys;input=sys.stdin.readline
def process():
s = input().strip()
x, = map(int, input().split())
n = len(s)
t = [1] * n
for i in range(n):
if int(s[i]) == 0:
if i-x >= 0:
t[i-x] = 0
if i+x < n:
t[i+x] = 0
for i in range(n):
if int(s[i]) == 1:
r = 0
if i-x >= 0:
if t[i-x] == 1:
r = 1
if i+x < n:
if t[i+x] == 1:
r = 1
if not r:
print(-1)
return
print("".join(str(c) for c in t))
T, = map(int, input().split())
for _ in range(T):
process()
```
| 107,219 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x β€ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 β€ |s| β€ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 β€ x β€ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
Tags: 2-sat, brute force, constructive algorithms, greedy
Correct Solution:
```
from sys import stdin
inp = lambda : stdin.readline().strip()
t = int(inp())
for _ in range(t):
s = inp()
x = int(inp())
w = ['1']*len(s)
for i in range(len(s)):
if s[i] == '0':
if i - x > -1:
w[i-x] = '0'
if i + x < len(s):
w[i+x] = '0'
for i in range(len(s)):
if s[i] == '1':
if not ((i - x > -1 and w[i-x] == '1' ) or (i + x < len(s) and w[i+x] =='1')):
print(-1)
break
else:
print(''.join(map(str,w)))
```
| 107,220 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x β€ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 β€ |s| β€ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 β€ x β€ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
Tags: 2-sat, brute force, constructive algorithms, greedy
Correct Solution:
```
def get():
n=len(s)
w = [1] * n
for i in range(n):
if s[i] == 0:
if i - x >= 0: w[i - x] = 0
if i + x < n: w[i + x] = 0
for i in range(n):
if s[i]:
if i-x>=0 and w[i-x]==1:continue
elif i+x<n and w[i+x]==1:continue
else:return [-1]
return w
for _ in range(int(input())):
s=list(map(int,input()))
x=int(input())
for i in get():
print(i,end='')
print()
```
| 107,221 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x β€ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 β€ |s| β€ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 β€ x β€ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
Tags: 2-sat, brute force, constructive algorithms, greedy
Correct Solution:
```
def process(n, x):
l = [0 for i in range(len(n))]
for i in range(len(l)):
if i-x >= 0 and n[i-x] == 1: l[i] = 1
if i+x < len(l) and n[i+x] == 1: l[i] = 1
return l
for t in range(int(input())):
l = list(map(int, list(input())))
x = int(input())
n = [1 for i in range(len(l))]
#print(l, x)
for i in range(len(n)):
if l[i] == 0:
if i+x < len(n):
n[i+x] = 0
if i-x > -1:
n[i-x] = 0
#print(n)
if process(n, x) != l:
print(-1)
else:
print(''.join(map(str, n)))
```
| 107,222 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x β€ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 β€ |s| β€ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 β€ x β€ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
Tags: 2-sat, brute force, constructive algorithms, greedy
Correct Solution:
```
import sys
input = sys.stdin.readline
def const(S):
n=len(S)
ANS=["0"]*n
for i in range(n):
if i-x>=0 and S[i-x]=="1":
ANS[i]="1"
if i+x<n and S[i+x]=="1":
ANS[i]="1"
return "".join(ANS)
t=int(input())
for tests in range(t):
S=input().strip()
x=int(input())
n=len(S)
ANS=["1"]*n
for i in range(n):
if S[i]=="0":
if i-x>=0:
ANS[i-x]="0"
if i+x<n:
ANS[i+x]="0"
A="".join(ANS)
if const(A)==S:
print(A)
else:
print(-1)
```
| 107,223 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x β€ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 β€ |s| β€ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 β€ x β€ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
Tags: 2-sat, brute force, constructive algorithms, greedy
Correct Solution:
```
res = ''
for _ in range(int(input())):
s = input()
x = int(input())
ls = len(s)
a = ['1'] * ls
for i, k in enumerate(s):
if k == '0':
if i-x>=0: a[i-x] = '0'
if i + x < ls: a[i+x] = '0'
for i, k in enumerate(s):
if k == '1':
f = 0
if i-x>=0: f |= int(a[i-x])
if i + x < ls: f |= int(a[i+x])
if not f:
res += '-1\n'
break
else: res += ''.join(a) + '\n'
print(res)
```
| 107,224 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x β€ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 β€ |s| β€ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 β€ x β€ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
Submitted Solution:
```
from sys import stdin,stdout
# 010,1
for _ in range(int(stdin.readline())):
# n,mnl,mns=list(map(int, stdin.readline().split()))
s=input()
x=int(stdin.readline());n=len(s)
ans=[' ']*n;f=1
for i in range(n):
if s[i]=='0':
if i-x>=0:ans[i-x]='0'
if i+x<n:ans[i+x]='0'
# print(ans)
for i in range(n):
if s[i]=='1':
left=right=0
if i-x>=0:
if ans[i-x]=='1' or ans[i-x]==' ':
left=1
ans[i-x]='1'
if left==0 and i+x<n:
if ans[i+x]==' ':
right=1
ans[i+x]='1'
if right==left==0:
f=0
break
# print(ans)
ans=''.join(['0' if ch==' ' else ch for ch in ans])
print(ans if f else -1)
```
Yes
| 107,225 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x β€ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 β€ |s| β€ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 β€ x β€ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
Submitted Solution:
```
for t in range(int(input())):
s=input()
x=int(input())
n=len(s)
w=["1"]*len(s)
for i in range(n):
if s[i]=="0":
if i-x>=0:
w[i-x]="0"
if i+x<n:
w[i+x]="0"
ws=["0"]*n
for i in range(n):
if w[i]=="1":
if i-x>=0:
ws[i-x]="1"
if i+x<n:
ws[i+x]="1"
s=list(s)
if s==ws:
print(''.join(w))
else:
print(-1)
```
Yes
| 107,226 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x β€ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 β€ |s| β€ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 β€ x β€ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
Submitted Solution:
```
import math
from collections import deque
from sys import stdin, stdout
from string import ascii_letters
letters = ascii_letters[:26]
input = stdin.readline
#print = stdout.write
for _ in range(int(input())):
arr = input().strip()
x = int(input())
n = len(arr)
res = ['-'] * n
for i in range(n):
if arr[i] == '0':
if i + x < n:
res[i + x] = '0'
if i - x >= 0:
res[i - x] = '0'
can = True
for i in range(n):
if arr[i] != '1':
continue
bf = []
was = False
if i + x < n:
if res[i + x] == '-':
res[i + x] = '1'
if res[i + x] == '1':
was = True
if i - x >= 0 :
if res[i - x] == '-':
res[i - x] = '1'
if res[i - x] == '1':
was = True
if not was:
can = False
if not can:
print(-1)
else:
for i in range(n):
if res[i] == '-':
res[i] = '0'
print(''.join(res))
```
Yes
| 107,227 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x β€ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 β€ |s| β€ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 β€ x β€ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
Submitted Solution:
```
for _ in range(int(input())):
s=input()
x=int(input())
n=len(s)
flag=False
arr=['1']*n
for i in range(n):
if s[i]=='0':
if i+x<n:
arr[i+x]='0'
if i-x>=0:
arr[i-x]='0'
for i in range(n):
if s[i]=='1':
if i+x<n and i-x<0:
if arr[i+x]=='0':
flag=True
break
elif i-x>=0 and i+x>=n:
if arr[i-x]=='0':
flag=True
break
elif i+x<n and i-x>=0:
if arr[i-x]=='0' and arr[i+x]=='0':
flag=True
break
else:
flag=True
break
if flag:
print(-1)
continue
ans=''
for i in arr:
ans+=i
print(ans)
```
Yes
| 107,228 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x β€ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 β€ |s| β€ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 β€ x β€ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
Submitted Solution:
```
for ad in range(int(input())):
s=list(input())
n=len(s)
x=int(input())
w=[2]*n
w[x:]=s[:n-x]
t=1
for i in range(x,n):
if s[i]=="1":
if w[i-x]==2 or w[i-x]=="1":
w[i-x]="1"
else:
t=0
break
else:
if w[i-x]==2 or w[i-x]=="0":
w[i-x]="0"
else:
t=0
break
if t==0:
print(-1)
else:
for i in range(n):
if w[i]==2:
w[i]="0"
print("".join(w))
```
No
| 107,229 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x β€ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 β€ |s| β€ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 β€ x β€ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
Submitted Solution:
```
import os, sys
from io import IOBase, BytesIO
py2 = round(0.5)
if py2:
from future_builtins import ascii, filter, hex, map, oct, zip
range = xrange
BUFSIZE = 8192
class FastIO(BytesIO):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.writable = 'x' in file.mode or 'w' in file.mode
self.write = super(FastIO, self).write if self.writable else None
def _fill(self):
s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
return s
def read(self):
while self._fill(): pass
return super(FastIO,self).read()
def readline(self):
while self.newlines == 0:
s = self._fill(); self.newlines = s.count(b'\n') + (not s)
self.newlines -= 1
return super(FastIO, self).readline()
def flush(self):
if self.writable:
os.write(self._fd, self.getvalue())
self.truncate(0), self.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
if py2:
self.write = self.buffer.write
self.read = self.buffer.read
self.readline = self.buffer.readline
else:
self.write = lambda s:self.buffer.write(s.encode('ascii'))
self.read = lambda:self.buffer.read().decode('ascii')
self.readline = lambda:self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
# Cout implemented in Python
import sys
class ostream:
def __lshift__(self,a):
sys.stdout.write(str(a))
return self
cout = ostream()
endl = '\n'
def get_input(a=str):
return a(input())
def get_int_input():
return get_input(int)
def get_input_arr(a):
return list(map(a, input().split()))
def get_int_input_arr():
return get_input_arr(int)
def solve():
t = get_int_input()
for _ in range(t):
st = get_input()
n = len(st)
x = get_int_input()
w = ["X"] * n
for i in range(n):
if i + x < n and i < x:
w[i + x] = st[i]
elif i >= x and i + x < n:
val1 = w[i - x]
val_res = st[i]
if val_res == "1":
if val1 == "X":
w[i - x] = "1"
else:
w[i + x] = "1"
else:
if (val1 == "X" or val1 == "0") and w[i + x] == "X":
w[i + x] = "0"
else:
w = ["-1"]
break
else:
if w[i - x] == "1" and st[i] == "0":
w = ["-1"]
break
w[i - x] = st[i]
res = "".join(w).replace("X", "0")
cout<<res<<endl
def main():
solve()
if __name__ == "__main__":
main()
```
No
| 107,230 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x β€ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 β€ |s| β€ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 β€ x β€ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
Submitted Solution:
```
import sys
I = lambda: int(input())
RL = readline = lambda: sys.stdin.readline().strip('\n')
RM = readmap = lambda x = int: map(x,readline().split(' '))
#C
for _ in range(I()):
s,x = RL(),I();n = len(s)
w = ['1']*n
for ind,i in enumerate(s):
if i == '0':
i1,i2 = ind-x,ind+x
if i1 >=0: w[i1] = '0'
if i2 < n: w[i2] = '0'
for ind,i in enumerate(s):
if i == '1':
i1,i2 = ind-x,ind+x
flag = (i1 >=0 and w[i1] == '1') or (i2 < n and w[i2] == '1')
if not flag: break
print(''.join(w) if flag else -1)
```
No
| 107,231 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:
* if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} = 1, then s_i = 1);
* if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x β€ n and w_{i+x} = 1, then s_i = 1);
* if both of the aforementioned conditions are false, then s_i is 0.
You are given the integer x and the resulting string s. Reconstruct the original string w.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
Each test case consists of two lines. The first line contains the resulting string s (2 β€ |s| β€ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 β€ x β€ |s| - 1).
The total length of all strings s in the input does not exceed 10^5.
Output
For each test case, print the answer on a separate line as follows:
* if no string w can produce the string s at the end of the process, print -1;
* otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them.
Example
Input
3
101110
2
01
1
110
1
Output
111011
10
-1
Submitted Solution:
```
from sys import stdin, stdout
input = stdin.readline
#print = stdout.write
for _ in range(int(input())):
s = input()
x = int(input())
n = len(s) - 1
ans = [''] * n
for i in range(n):
if i >= x and i + x < n:
if ans[i - x]:
if ans[i - x] == '1' and s[i] == '0':
print(-1)
break
if s[i] == '0':
ans[i - x] = ans[i + x] = '0'
elif ans[i - x] == '0':
ans[i + x] = '1'
else:
ans[i - x] = '1'
else:
ans[i - x] = s[i]
if s[i] == '0':
ans[i + x] = '0'
elif i >= x:
if (ans[i - x] == '1' and s[i] == '0') or (s[i] == '1' and ans[i - x] == '0'):
print(-1)
break
ans[i - x] = s[i]
elif i + x < n:
ans[i + x] = s[i]
else:
for i in range(n):
if not ans[i]:
ans[i] = '0'
print(''.join(ans))
```
No
| 107,232 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One drew a closed polyline on a plane, that consisted only of vertical and horizontal segments (parallel to the coordinate axes). The segments alternated between horizontal and vertical ones (a horizontal segment was always followed by a vertical one, and vice versa). The polyline did not contain strict self-intersections, which means that in case any two segments shared a common point, that point was an endpoint for both of them (please consult the examples in the notes section).
Unfortunately, the polyline was erased, and you only know the lengths of the horizonal and vertical segments. Please construct any polyline matching the description with such segments, or determine that it does not exist.
Input
The first line contains one integer t (1 β€ t β€ 200) βthe number of test cases.
The first line of each test case contains one integer h (1 β€ h β€ 1000) β the number of horizontal segments. The following line contains h integers l_1, l_2, ..., l_h (1 β€ l_i β€ 1000) β lengths of the horizontal segments of the polyline, in arbitrary order.
The following line contains an integer v (1 β€ v β€ 1000) β the number of vertical segments, which is followed by a line containing v integers p_1, p_2, ..., p_v (1 β€ p_i β€ 1000) β lengths of the vertical segments of the polyline, in arbitrary order.
Test cases are separated by a blank line, and the sum of values h + v over all test cases does not exceed 1000.
Output
For each test case output Yes, if there exists at least one polyline satisfying the requirements, or No otherwise. If it does exist, in the following n lines print the coordinates of the polyline vertices, in order of the polyline traversal: the i-th line should contain two integers x_i and y_i β coordinates of the i-th vertex.
Note that, each polyline segment must be either horizontal or vertical, and the segments should alternate between horizontal and vertical. The coordinates should not exceed 10^9 by their absolute value.
Examples
Input
2
2
1 1
2
1 1
2
1 2
2
3 3
Output
Yes
1 0
1 1
0 1
0 0
No
Input
2
4
1 1 1 1
4
1 1 1 1
3
2 1 1
3
2 1 1
Output
Yes
1 0
1 1
2 1
2 2
1 2
1 1
0 1
0 0
Yes
0 -2
2 -2
2 -1
1 -1
1 0
0 0
Input
2
4
1 4 1 2
4
3 4 5 12
4
1 2 3 6
2
1 3
Output
Yes
2 0
2 3
3 3
3 7
4 7
4 12
0 12
0 0
No
Note
In the first test case of the first example, the answer is Yes β for example, the following picture illustrates a square that satisfies the requirements:
<image>
In the first test case of the second example, the desired polyline also exists. Note that, the polyline contains self-intersections, but only in the endpoints:
<image>
In the second test case of the second example, the desired polyline could be like the one below:
<image>
Note that the following polyline is not a valid one, since it contains self-intersections that are not endpoints for some of the segments:
<image>
Tags: constructive algorithms, dp, geometry
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
def main():
t = int(input())
for _ in range(t):
if _ != 0:
input()
h = int(input())
l1 = list(map(int,input().split()))
v = int(input())
l2 = list(map(int,input().split()))
hKnap = [1]
vKnap = [1]
if h != v or sum(l1) % 2 != 0 or sum(l2) % 2 != 0:
print("No")
continue
for elem in l1:
hKnap.append((hKnap[-1] << elem) | hKnap[-1])
for elem in l2:
vKnap.append((vKnap[-1] << elem) | vKnap[-1])
hSet = []
hSet2 = []
vSet = []
vSet2 = []
if hKnap[-1] & 1 << (sum(l1) // 2):
curSum = sum(l1) // 2
for i in range(h - 1, -1, -1):
if curSum >= l1[i] and hKnap[i] & (1 << (curSum - l1[i])):
curSum -= l1[i]
hSet.append(l1[i])
else:
hSet2.append(l1[i])
if vKnap[-1] & 1 << (sum(l2) // 2):
curSum = sum(l2) // 2
for i in range(v - 1, -1, -1):
if curSum >= l2[i] and vKnap[i] & (1 << (curSum - l2[i])):
curSum -= l2[i]
vSet.append(l2[i])
else:
vSet2.append(l2[i])
if not hSet or not vSet:
print("No")
else:
print("Yes")
if len(hSet) < len(hSet2):
hTupleS = tuple(sorted(hSet))
hTupleL = tuple(sorted(hSet2))
else:
hTupleS = tuple(sorted(hSet2))
hTupleL = tuple(sorted(hSet))
if len(vSet) < len(vSet2):
vTupleS = tuple(sorted(vSet))
vTupleL = tuple(sorted(vSet2))
else:
vTupleS = tuple(sorted(vSet2))
vTupleL = tuple(sorted(vSet))
currentLoc = [0,0]
isHS = True
isHL = False
isVS = False
isVL = True
hIndex = len(hTupleS)-1
vIndex = 0
for i in range(h):
if isHS:
currentLoc[0] += hTupleS[hIndex]
hIndex -= 1
if hIndex < 0:
hIndex = len(hTupleL) - 1
isHS = False
isHL = True
elif isHL:
currentLoc[0] -= hTupleL[hIndex]
hIndex -= 1
print(str(currentLoc[0]) + " " + str(currentLoc[1]))
if isVL:
currentLoc[1] += vTupleL[vIndex]
vIndex += 1
if vIndex >= len(vTupleL):
vIndex = 0
isVL = False
isVH = True
elif isHL:
currentLoc[1] -= vTupleS[vIndex]
vIndex += 1
print(str(currentLoc[0]) + " " + str(currentLoc[1]))
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
```
| 107,233 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One drew a closed polyline on a plane, that consisted only of vertical and horizontal segments (parallel to the coordinate axes). The segments alternated between horizontal and vertical ones (a horizontal segment was always followed by a vertical one, and vice versa). The polyline did not contain strict self-intersections, which means that in case any two segments shared a common point, that point was an endpoint for both of them (please consult the examples in the notes section).
Unfortunately, the polyline was erased, and you only know the lengths of the horizonal and vertical segments. Please construct any polyline matching the description with such segments, or determine that it does not exist.
Input
The first line contains one integer t (1 β€ t β€ 200) βthe number of test cases.
The first line of each test case contains one integer h (1 β€ h β€ 1000) β the number of horizontal segments. The following line contains h integers l_1, l_2, ..., l_h (1 β€ l_i β€ 1000) β lengths of the horizontal segments of the polyline, in arbitrary order.
The following line contains an integer v (1 β€ v β€ 1000) β the number of vertical segments, which is followed by a line containing v integers p_1, p_2, ..., p_v (1 β€ p_i β€ 1000) β lengths of the vertical segments of the polyline, in arbitrary order.
Test cases are separated by a blank line, and the sum of values h + v over all test cases does not exceed 1000.
Output
For each test case output Yes, if there exists at least one polyline satisfying the requirements, or No otherwise. If it does exist, in the following n lines print the coordinates of the polyline vertices, in order of the polyline traversal: the i-th line should contain two integers x_i and y_i β coordinates of the i-th vertex.
Note that, each polyline segment must be either horizontal or vertical, and the segments should alternate between horizontal and vertical. The coordinates should not exceed 10^9 by their absolute value.
Examples
Input
2
2
1 1
2
1 1
2
1 2
2
3 3
Output
Yes
1 0
1 1
0 1
0 0
No
Input
2
4
1 1 1 1
4
1 1 1 1
3
2 1 1
3
2 1 1
Output
Yes
1 0
1 1
2 1
2 2
1 2
1 1
0 1
0 0
Yes
0 -2
2 -2
2 -1
1 -1
1 0
0 0
Input
2
4
1 4 1 2
4
3 4 5 12
4
1 2 3 6
2
1 3
Output
Yes
2 0
2 3
3 3
3 7
4 7
4 12
0 12
0 0
No
Note
In the first test case of the first example, the answer is Yes β for example, the following picture illustrates a square that satisfies the requirements:
<image>
In the first test case of the second example, the desired polyline also exists. Note that, the polyline contains self-intersections, but only in the endpoints:
<image>
In the second test case of the second example, the desired polyline could be like the one below:
<image>
Note that the following polyline is not a valid one, since it contains self-intersections that are not endpoints for some of the segments:
<image>
Submitted Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
def main():
t = int(input())
for _ in range(t):
if _ != 0:
input()
h = int(input())
l1 = list(map(int,input().split()))
v = int(input())
l2 = list(map(int,input().split()))
hKnap = [1]
vKnap = [1]
if h != v or sum(l1) % 2 != 0 or sum(l2) % 2 != 0:
print("No")
continue
for elem in l1:
hKnap.append((hKnap[-1] << elem) | hKnap[-1])
for elem in l2:
vKnap.append((vKnap[-1] << elem) | vKnap[-1])
hSet = []
hSet2 = []
vSet = []
vSet2 = []
if hKnap[-1] & 1 << (sum(l1) // 2):
curSum = sum(l1) // 2
for i in range(h - 1, -1, -1):
if curSum >= l1[i] and hKnap[i] & (1 << (curSum - l1[i])):
curSum -= l1[i]
hSet.append(l1[i])
else:
hSet2.append(l1[i])
if vKnap[-1] & 1 << (sum(l2) // 2):
curSum = sum(l2) // 2
for i in range(v - 1, -1, -1):
if curSum >= l2[i] and vKnap[i] & (1 << (curSum - l2[i])):
curSum -= l2[i]
vSet.append(l2[i])
else:
vSet2.append(l2[i])
if not hSet or not vSet:
print("No")
else:
print("Yes")
if len(hSet) < len(hSet2):
hTupleS = tuple(sorted(hSet))
hTupleL = tuple(sorted(hSet2))
else:
hTupleS = tuple(sorted(hSet2))
hTupleL = tuple(sorted(hSet))
if len(vSet) < len(vSet2):
vTupleS = tuple(sorted(vSet))
vTupleL = tuple(sorted(vSet2))
else:
vTupleS = tuple(sorted(vSet2))
vTupleL = tuple(sorted(vSet))
currentLoc = [0,0]
isHS = True
isHL = False
isVS = False
isVL = True
hIndex = len(hTupleS)-1
vIndex = 0
for i in range(h):
if isHS:
currentLoc[0] += hTupleS[hIndex]
hIndex -= 1
if hIndex < 0:
hIndex = 0
isHS = False
isHL = True
elif isHL:
currentLoc[0] -= hTupleL[hIndex]
hIndex += 1
print(str(currentLoc[0]) + " " + str(currentLoc[1]))
if isVL:
currentLoc[1] += vTupleL[vIndex]
vIndex += 1
if vIndex >= len(vTupleL):
vIndex = len(vTupleS) - 1
isVL = False
isVH = True
elif isHL:
currentLoc[1] -= vTupleS[vIndex]
vIndex -= 1
print(str(currentLoc[0]) + " " + str(currentLoc[1]))
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
```
No
| 107,234 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings a and b (both of length k) a bit similar if they have the same character in some position, i. e. there exists at least one i β [1, k] such that a_i = b_i.
You are given a binary string s of length n (a string of n characters 0 and/or 1) and an integer k. Let's denote the string s[i..j] as the substring of s starting from the i-th character and ending with the j-th character (that is, s[i..j] = s_i s_{i + 1} s_{i + 2} ... s_{j - 1} s_j).
Let's call a binary string t of length k beautiful if it is a bit similar to all substrings of s having length exactly k; that is, it is a bit similar to s[1..k], s[2..k+1], ..., s[n-k+1..n].
Your goal is to find the lexicographically smallest string t that is beautiful, or report that no such string exists. String x is lexicographically less than string y if either x is a prefix of y (and x β y), or there exists such i (1 β€ i β€ min(|x|, |y|)), that x_i < y_i, and for any j (1 β€ j < i) x_j = y_j.
Input
The first line contains one integer q (1 β€ q β€ 10000) β the number of test cases. Each test case consists of two lines.
The first line of each test case contains two integers n and k (1 β€ k β€ n β€ 10^6). The second line contains the string s, consisting of n characters (each character is either 0 or 1).
It is guaranteed that the sum of n over all test cases does not exceed 10^6.
Output
For each test case, print the answer as follows:
* if it is impossible to construct a beautiful string, print one line containing the string NO (note: exactly in upper case, you can't print No, for example);
* otherwise, print two lines. The first line should contain the string YES (exactly in upper case as well); the second line β the lexicographically smallest beautiful string, consisting of k characters 0 and/or 1.
Example
Input
7
4 2
0110
4 2
1001
9 3
010001110
9 3
101110001
10 3
0101110001
10 10
1111111111
11 10
11111111110
Output
YES
11
YES
00
YES
010
YES
101
NO
YES
0000000001
YES
0000000010
Tags: bitmasks, brute force, hashing, string suffix structures, strings, two pointers
Correct Solution:
```
from itertools import accumulate, chain
BITS = 20
for _ in range(int(input())):
n, k = map(int, input().split());s = input();antis = [1 - int(si) for si in s];antis_s = ''.join(map(str, antis));antis_sums = tuple(chain((0,), accumulate(antis)));fbd = set();ans = 0
for i in range(n + 1 - k):
if k < BITS or antis_sums[i] == antis_sums[i + k - BITS]:fbd.add(int(antis_s[max(i, i + k - BITS):i + k], base=2))
while ans in fbd:ans += 1
t = bin(ans)[2:]
if len(t) > k:print("NO")
else:print("YES");print(t.rjust(k, '0'))
```
| 107,235 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings a and b (both of length k) a bit similar if they have the same character in some position, i. e. there exists at least one i β [1, k] such that a_i = b_i.
You are given a binary string s of length n (a string of n characters 0 and/or 1) and an integer k. Let's denote the string s[i..j] as the substring of s starting from the i-th character and ending with the j-th character (that is, s[i..j] = s_i s_{i + 1} s_{i + 2} ... s_{j - 1} s_j).
Let's call a binary string t of length k beautiful if it is a bit similar to all substrings of s having length exactly k; that is, it is a bit similar to s[1..k], s[2..k+1], ..., s[n-k+1..n].
Your goal is to find the lexicographically smallest string t that is beautiful, or report that no such string exists. String x is lexicographically less than string y if either x is a prefix of y (and x β y), or there exists such i (1 β€ i β€ min(|x|, |y|)), that x_i < y_i, and for any j (1 β€ j < i) x_j = y_j.
Input
The first line contains one integer q (1 β€ q β€ 10000) β the number of test cases. Each test case consists of two lines.
The first line of each test case contains two integers n and k (1 β€ k β€ n β€ 10^6). The second line contains the string s, consisting of n characters (each character is either 0 or 1).
It is guaranteed that the sum of n over all test cases does not exceed 10^6.
Output
For each test case, print the answer as follows:
* if it is impossible to construct a beautiful string, print one line containing the string NO (note: exactly in upper case, you can't print No, for example);
* otherwise, print two lines. The first line should contain the string YES (exactly in upper case as well); the second line β the lexicographically smallest beautiful string, consisting of k characters 0 and/or 1.
Example
Input
7
4 2
0110
4 2
1001
9 3
010001110
9 3
101110001
10 3
0101110001
10 10
1111111111
11 10
11111111110
Output
YES
11
YES
00
YES
010
YES
101
NO
YES
0000000001
YES
0000000010
Tags: bitmasks, brute force, hashing, string suffix structures, strings, two pointers
Correct Solution:
```
from collections import deque
T = int(input())
r = 1
maxdigit = 20
def fastfrac(a,b,M):
numb = pow(b,M-2,M)
return ((a%M)*(numb%M))%M
def getnum(k,s):
n = len(s)
dic = {}
prefixnum = max(0,k-maxdigit)
real = k - prefixnum
maxreal = (1<<real) - 1
front = 0
pres = deque()
for i in range(prefixnum):
if s[i]=='1': pres.append(i)
num = int(s[prefixnum:prefixnum+real],2)
if len(pres)==prefixnum:
dic[maxreal-num] = 1
if maxreal-num==front: front += 1
# print(pres)
for i in range(n-k):
if prefixnum>0:
if pres and pres[0]==i: pres.popleft()
if s[i+prefixnum] =='1': pres.append(i+prefixnum)
# print(pres)
num = num % ((maxreal+1)//2)
num = 2*num + int(s[i+k])
if len(pres)<prefixnum: continue
dic[maxreal-num] = 1
while front in dic:
front += 1
# print(dic)
if front>maxreal: return -1
else: return front
while r<=T:
n,k = map(int,input().split())
s = input()
ans = getnum(k,s)
if ans>=0:
print("YES")
subs = bin(ans)[2:]
subs = '0'*(k-len(subs)) + subs
print(subs)
else:
print("NO")
r += 1
```
| 107,236 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings a and b (both of length k) a bit similar if they have the same character in some position, i. e. there exists at least one i β [1, k] such that a_i = b_i.
You are given a binary string s of length n (a string of n characters 0 and/or 1) and an integer k. Let's denote the string s[i..j] as the substring of s starting from the i-th character and ending with the j-th character (that is, s[i..j] = s_i s_{i + 1} s_{i + 2} ... s_{j - 1} s_j).
Let's call a binary string t of length k beautiful if it is a bit similar to all substrings of s having length exactly k; that is, it is a bit similar to s[1..k], s[2..k+1], ..., s[n-k+1..n].
Your goal is to find the lexicographically smallest string t that is beautiful, or report that no such string exists. String x is lexicographically less than string y if either x is a prefix of y (and x β y), or there exists such i (1 β€ i β€ min(|x|, |y|)), that x_i < y_i, and for any j (1 β€ j < i) x_j = y_j.
Input
The first line contains one integer q (1 β€ q β€ 10000) β the number of test cases. Each test case consists of two lines.
The first line of each test case contains two integers n and k (1 β€ k β€ n β€ 10^6). The second line contains the string s, consisting of n characters (each character is either 0 or 1).
It is guaranteed that the sum of n over all test cases does not exceed 10^6.
Output
For each test case, print the answer as follows:
* if it is impossible to construct a beautiful string, print one line containing the string NO (note: exactly in upper case, you can't print No, for example);
* otherwise, print two lines. The first line should contain the string YES (exactly in upper case as well); the second line β the lexicographically smallest beautiful string, consisting of k characters 0 and/or 1.
Example
Input
7
4 2
0110
4 2
1001
9 3
010001110
9 3
101110001
10 3
0101110001
10 10
1111111111
11 10
11111111110
Output
YES
11
YES
00
YES
010
YES
101
NO
YES
0000000001
YES
0000000010
Tags: bitmasks, brute force, hashing, string suffix structures, strings, two pointers
Correct Solution:
```
import sys
input = sys.stdin.readline
from collections import Counter
import math
t = int(input())
for _ in range(t):
n,m = list(map(int,input().split()))
k = min(32,m)
s = input().rstrip()
D = set()
v = 2**k - 1
for i in range(m-k, len(s)-k+1):
temp = s[i:i+k]
#print(temp)
D.add(v - int(temp,2))
ok = 0
cc = s[:m].count('0')
if cc>0:
on = 1
else:
on = 0
for i in range(len(s)-m):
if cc==0:
on = 0
if s[i]=='0':
cc -= 1
if s[i+m]=='0':
cc += 1
if cc==0:
on = 0
if on==1:
print('YES')
print(m*'0')
ok = 1
else:
for i in range(v+1):
if i not in D:
ans = (bin(i))[2:]
ans = (m - len(ans)) * '0' + ans
print('YES')
print(ans)
ok = 1
break
if ok==0:
print('NO')
```
| 107,237 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings a and b (both of length k) a bit similar if they have the same character in some position, i. e. there exists at least one i β [1, k] such that a_i = b_i.
You are given a binary string s of length n (a string of n characters 0 and/or 1) and an integer k. Let's denote the string s[i..j] as the substring of s starting from the i-th character and ending with the j-th character (that is, s[i..j] = s_i s_{i + 1} s_{i + 2} ... s_{j - 1} s_j).
Let's call a binary string t of length k beautiful if it is a bit similar to all substrings of s having length exactly k; that is, it is a bit similar to s[1..k], s[2..k+1], ..., s[n-k+1..n].
Your goal is to find the lexicographically smallest string t that is beautiful, or report that no such string exists. String x is lexicographically less than string y if either x is a prefix of y (and x β y), or there exists such i (1 β€ i β€ min(|x|, |y|)), that x_i < y_i, and for any j (1 β€ j < i) x_j = y_j.
Input
The first line contains one integer q (1 β€ q β€ 10000) β the number of test cases. Each test case consists of two lines.
The first line of each test case contains two integers n and k (1 β€ k β€ n β€ 10^6). The second line contains the string s, consisting of n characters (each character is either 0 or 1).
It is guaranteed that the sum of n over all test cases does not exceed 10^6.
Output
For each test case, print the answer as follows:
* if it is impossible to construct a beautiful string, print one line containing the string NO (note: exactly in upper case, you can't print No, for example);
* otherwise, print two lines. The first line should contain the string YES (exactly in upper case as well); the second line β the lexicographically smallest beautiful string, consisting of k characters 0 and/or 1.
Example
Input
7
4 2
0110
4 2
1001
9 3
010001110
9 3
101110001
10 3
0101110001
10 10
1111111111
11 10
11111111110
Output
YES
11
YES
00
YES
010
YES
101
NO
YES
0000000001
YES
0000000010
Tags: bitmasks, brute force, hashing, string suffix structures, strings, two pointers
Correct Solution:
```
"""
#If FastIO not needed, use this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os
import sys
from io import BytesIO, IOBase
import heapq as h
from bisect import bisect_left, bisect_right
import time
from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
"""
recursively make the set smaller
I need to remove from consideration anything with a 0 in its first K-20 digits
"""
out = []
for _ in range(int(input())):
N, K = [int(s) for s in input().split()]
S = input()
K2 = min(K,20)
start_str = -1
curr = 0
max_bit = pow(2,K2) - 1
no = set()
for i in range(N):
curr *= 2
if S[i] == '0': curr += 1
if curr > max_bit:
curr &= max_bit
start_str = i
if i >= K-1 and (i - start_str) >= (K-K2): no.add(curr)
flag = False
for i in range(pow(2,K2)):
if i not in no:
out.append("YES")
out.append('0'*(K-len(bin(i)[2:])) + bin(i)[2:])
flag = True
break
if not flag:
out.append("NO")
print('\n'.join(out))
#print(time.time()-start_time)
```
| 107,238 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings a and b (both of length k) a bit similar if they have the same character in some position, i. e. there exists at least one i β [1, k] such that a_i = b_i.
You are given a binary string s of length n (a string of n characters 0 and/or 1) and an integer k. Let's denote the string s[i..j] as the substring of s starting from the i-th character and ending with the j-th character (that is, s[i..j] = s_i s_{i + 1} s_{i + 2} ... s_{j - 1} s_j).
Let's call a binary string t of length k beautiful if it is a bit similar to all substrings of s having length exactly k; that is, it is a bit similar to s[1..k], s[2..k+1], ..., s[n-k+1..n].
Your goal is to find the lexicographically smallest string t that is beautiful, or report that no such string exists. String x is lexicographically less than string y if either x is a prefix of y (and x β y), or there exists such i (1 β€ i β€ min(|x|, |y|)), that x_i < y_i, and for any j (1 β€ j < i) x_j = y_j.
Input
The first line contains one integer q (1 β€ q β€ 10000) β the number of test cases. Each test case consists of two lines.
The first line of each test case contains two integers n and k (1 β€ k β€ n β€ 10^6). The second line contains the string s, consisting of n characters (each character is either 0 or 1).
It is guaranteed that the sum of n over all test cases does not exceed 10^6.
Output
For each test case, print the answer as follows:
* if it is impossible to construct a beautiful string, print one line containing the string NO (note: exactly in upper case, you can't print No, for example);
* otherwise, print two lines. The first line should contain the string YES (exactly in upper case as well); the second line β the lexicographically smallest beautiful string, consisting of k characters 0 and/or 1.
Example
Input
7
4 2
0110
4 2
1001
9 3
010001110
9 3
101110001
10 3
0101110001
10 10
1111111111
11 10
11111111110
Output
YES
11
YES
00
YES
010
YES
101
NO
YES
0000000001
YES
0000000010
Tags: bitmasks, brute force, hashing, string suffix structures, strings, two pointers
Correct Solution:
```
import sys
input = sys.stdin.readline
t = int(input())
outL = []
for _ in range(t):
n, k = map(int, input().split())
s = input().strip()
bad = set()
k2 = min(k, 20)
last = -1
curr = 0
mask = (1 << k2) - 1
for i in range(n):
curr *= 2
if s[i] == '0':
curr += 1
if curr > mask:
curr &= mask
last = i
if i >= k-1 and (i - last) >= (k-k2):
bad.add(curr)
for i in range(1 << k2):
if i not in bad:
outL.append('YES')
out = bin(i)[2:]
outL.append('0'*(k-len(out))+out)
break
else:
outL.append('NO')
print('\n'.join(outL))
```
| 107,239 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings a and b (both of length k) a bit similar if they have the same character in some position, i. e. there exists at least one i β [1, k] such that a_i = b_i.
You are given a binary string s of length n (a string of n characters 0 and/or 1) and an integer k. Let's denote the string s[i..j] as the substring of s starting from the i-th character and ending with the j-th character (that is, s[i..j] = s_i s_{i + 1} s_{i + 2} ... s_{j - 1} s_j).
Let's call a binary string t of length k beautiful if it is a bit similar to all substrings of s having length exactly k; that is, it is a bit similar to s[1..k], s[2..k+1], ..., s[n-k+1..n].
Your goal is to find the lexicographically smallest string t that is beautiful, or report that no such string exists. String x is lexicographically less than string y if either x is a prefix of y (and x β y), or there exists such i (1 β€ i β€ min(|x|, |y|)), that x_i < y_i, and for any j (1 β€ j < i) x_j = y_j.
Input
The first line contains one integer q (1 β€ q β€ 10000) β the number of test cases. Each test case consists of two lines.
The first line of each test case contains two integers n and k (1 β€ k β€ n β€ 10^6). The second line contains the string s, consisting of n characters (each character is either 0 or 1).
It is guaranteed that the sum of n over all test cases does not exceed 10^6.
Output
For each test case, print the answer as follows:
* if it is impossible to construct a beautiful string, print one line containing the string NO (note: exactly in upper case, you can't print No, for example);
* otherwise, print two lines. The first line should contain the string YES (exactly in upper case as well); the second line β the lexicographically smallest beautiful string, consisting of k characters 0 and/or 1.
Example
Input
7
4 2
0110
4 2
1001
9 3
010001110
9 3
101110001
10 3
0101110001
10 10
1111111111
11 10
11111111110
Output
YES
11
YES
00
YES
010
YES
101
NO
YES
0000000001
YES
0000000010
Tags: bitmasks, brute force, hashing, string suffix structures, strings, two pointers
Correct Solution:
```
import sys
input = sys.stdin.readline
a = int(input())
for case in range(a):
n, k0 = [int(x) for x in input().split(' ')]
start_inds = set([i for i in range(n-k0+1)])
s = input()
k = min(20, k0)
if k < k0:
next_0 = []
curr_0 = n
for i in range(n-1,-1,-1):
if s[i] == '0':
curr_0 = i
next_0.append(curr_0)
next_0.reverse()
for i in range(n-k0+1):
if next_0[i] < i + k0 - 20:
start_inds.remove(i)
Z = 1 << k
avoid = set()
for i in start_inds:
r = int(s[i + k0 - k:i + k0], 2)
avoid.add(Z - 1 - r)
j = 0
while j in avoid and j < Z:
j += 1
ans = bin(j)[2:]
if j == Z:
print("NO")
else:
print("YES")
ans = '0' * (k0 - len(ans)) + ans
print(ans)
```
| 107,240 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings a and b (both of length k) a bit similar if they have the same character in some position, i. e. there exists at least one i β [1, k] such that a_i = b_i.
You are given a binary string s of length n (a string of n characters 0 and/or 1) and an integer k. Let's denote the string s[i..j] as the substring of s starting from the i-th character and ending with the j-th character (that is, s[i..j] = s_i s_{i + 1} s_{i + 2} ... s_{j - 1} s_j).
Let's call a binary string t of length k beautiful if it is a bit similar to all substrings of s having length exactly k; that is, it is a bit similar to s[1..k], s[2..k+1], ..., s[n-k+1..n].
Your goal is to find the lexicographically smallest string t that is beautiful, or report that no such string exists. String x is lexicographically less than string y if either x is a prefix of y (and x β y), or there exists such i (1 β€ i β€ min(|x|, |y|)), that x_i < y_i, and for any j (1 β€ j < i) x_j = y_j.
Input
The first line contains one integer q (1 β€ q β€ 10000) β the number of test cases. Each test case consists of two lines.
The first line of each test case contains two integers n and k (1 β€ k β€ n β€ 10^6). The second line contains the string s, consisting of n characters (each character is either 0 or 1).
It is guaranteed that the sum of n over all test cases does not exceed 10^6.
Output
For each test case, print the answer as follows:
* if it is impossible to construct a beautiful string, print one line containing the string NO (note: exactly in upper case, you can't print No, for example);
* otherwise, print two lines. The first line should contain the string YES (exactly in upper case as well); the second line β the lexicographically smallest beautiful string, consisting of k characters 0 and/or 1.
Example
Input
7
4 2
0110
4 2
1001
9 3
010001110
9 3
101110001
10 3
0101110001
10 10
1111111111
11 10
11111111110
Output
YES
11
YES
00
YES
010
YES
101
NO
YES
0000000001
YES
0000000010
Tags: bitmasks, brute force, hashing, string suffix structures, strings, two pointers
Correct Solution:
```
import sys
input = sys.stdin.readline
t = int(input())
for case in range(t):
n, k0 = [int(x) for x in input().split(' ')]
starts = set([i for i in range(n - k0 + 1)])
s = input()
k = min(20, k0)
if k < k0:
nz = []
cz = n
for i in range(n - 1, -1, -1):
if s[i] == '0':
cz = i
nz.append(cz)
nz.reverse()
for i in range(n - k0 + 1):
if nz[i] < i + k0 - 20:
starts.remove(i)
Z = 1 << k
avoid = set()
for i in starts:
r = int(s[i + k0 - k:i + k0], 2)
avoid.add(Z - 1 - r)
j = 0
while j in avoid and j < Z:
j += 1
ans = bin(j)[2:]
if j == Z:
print("NO")
else:
print("YES")
ans = '0' * (k0 - len(ans)) + ans
print(ans)
```
| 107,241 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings a and b (both of length k) a bit similar if they have the same character in some position, i. e. there exists at least one i β [1, k] such that a_i = b_i.
You are given a binary string s of length n (a string of n characters 0 and/or 1) and an integer k. Let's denote the string s[i..j] as the substring of s starting from the i-th character and ending with the j-th character (that is, s[i..j] = s_i s_{i + 1} s_{i + 2} ... s_{j - 1} s_j).
Let's call a binary string t of length k beautiful if it is a bit similar to all substrings of s having length exactly k; that is, it is a bit similar to s[1..k], s[2..k+1], ..., s[n-k+1..n].
Your goal is to find the lexicographically smallest string t that is beautiful, or report that no such string exists. String x is lexicographically less than string y if either x is a prefix of y (and x β y), or there exists such i (1 β€ i β€ min(|x|, |y|)), that x_i < y_i, and for any j (1 β€ j < i) x_j = y_j.
Input
The first line contains one integer q (1 β€ q β€ 10000) β the number of test cases. Each test case consists of two lines.
The first line of each test case contains two integers n and k (1 β€ k β€ n β€ 10^6). The second line contains the string s, consisting of n characters (each character is either 0 or 1).
It is guaranteed that the sum of n over all test cases does not exceed 10^6.
Output
For each test case, print the answer as follows:
* if it is impossible to construct a beautiful string, print one line containing the string NO (note: exactly in upper case, you can't print No, for example);
* otherwise, print two lines. The first line should contain the string YES (exactly in upper case as well); the second line β the lexicographically smallest beautiful string, consisting of k characters 0 and/or 1.
Example
Input
7
4 2
0110
4 2
1001
9 3
010001110
9 3
101110001
10 3
0101110001
10 10
1111111111
11 10
11111111110
Output
YES
11
YES
00
YES
010
YES
101
NO
YES
0000000001
YES
0000000010
Tags: bitmasks, brute force, hashing, string suffix structures, strings, two pointers
Correct Solution:
```
# Author: yumtam
# Created at: 2020-12-30 16:53
def solve():
n, k = [int(t) for t in input().split()]
A = [int(c) for c in input()]
for i in range(n):
A[i] = 1 - A[i]
ctz = []
zeros = 0
for x in reversed(A):
if x == 0:
zeros += 1
else:
zeros = 0
ctz.append(zeros)
ctz = ctz[::-1]
S = n - k + 1
L = S.bit_length()
V = set()
for i in range(n-k+1):
if ctz[i] >= k - L:
v = 0
d = 1
pos = i+k-1
for _ in range(min(L, k)):
v += d * A[pos]
pos -= 1
d *= 2
V.add(v)
for ans in range(2**k):
if ans not in V:
print("YES")
res = bin(ans)[2:]
res = '0'*(k-len(res)) + res
print(res)
break
else:
print("NO")
import sys, os, io
input = lambda: sys.stdin.readline().rstrip('\r\n')
stdout = io.BytesIO()
sys.stdout.write = lambda s: stdout.write(s.encode("ascii"))
for _ in range(int(input())):
solve()
os.write(1, stdout.getvalue())
```
| 107,242 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings a and b (both of length k) a bit similar if they have the same character in some position, i. e. there exists at least one i β [1, k] such that a_i = b_i.
You are given a binary string s of length n (a string of n characters 0 and/or 1) and an integer k. Let's denote the string s[i..j] as the substring of s starting from the i-th character and ending with the j-th character (that is, s[i..j] = s_i s_{i + 1} s_{i + 2} ... s_{j - 1} s_j).
Let's call a binary string t of length k beautiful if it is a bit similar to all substrings of s having length exactly k; that is, it is a bit similar to s[1..k], s[2..k+1], ..., s[n-k+1..n].
Your goal is to find the lexicographically smallest string t that is beautiful, or report that no such string exists. String x is lexicographically less than string y if either x is a prefix of y (and x β y), or there exists such i (1 β€ i β€ min(|x|, |y|)), that x_i < y_i, and for any j (1 β€ j < i) x_j = y_j.
Input
The first line contains one integer q (1 β€ q β€ 10000) β the number of test cases. Each test case consists of two lines.
The first line of each test case contains two integers n and k (1 β€ k β€ n β€ 10^6). The second line contains the string s, consisting of n characters (each character is either 0 or 1).
It is guaranteed that the sum of n over all test cases does not exceed 10^6.
Output
For each test case, print the answer as follows:
* if it is impossible to construct a beautiful string, print one line containing the string NO (note: exactly in upper case, you can't print No, for example);
* otherwise, print two lines. The first line should contain the string YES (exactly in upper case as well); the second line β the lexicographically smallest beautiful string, consisting of k characters 0 and/or 1.
Example
Input
7
4 2
0110
4 2
1001
9 3
010001110
9 3
101110001
10 3
0101110001
10 10
1111111111
11 10
11111111110
Output
YES
11
YES
00
YES
010
YES
101
NO
YES
0000000001
YES
0000000010
Submitted Solution:
```
from sys import stdin
t = int(stdin.readline())
for _ in range(t):
n, k = map(int, stdin.readline().split())
maxx_digit = min(21, k)
maxx_so = 2 ** maxx_digit
arr = stdin.readline()
so = 0
to = 2 ** k
dct = {}
lientiep = 0
for i in range(k):
so = so * 2 + 1 - int(arr[i])
if so >= maxx_so:
so -= maxx_so
for j in range(k - maxx_digit):
if arr[j] == '1':
lientiep += 1
else:
lientiep = 0
dau = k - maxx_digit
if lientiep >= k - maxx_digit:
dct[so] = 1
for i in range(k, n):
so = so * 2 + 1 - int(arr[i])
if so >= maxx_so:
so -= maxx_so
if arr[dau] == '1':
lientiep += 1
else:
lientiep = 0
if lientiep >= k - maxx_digit:
dct[so] = 1
dau += 1
ans = 0
while ans in dct:
ans += 1
ans = bin(ans)[2:]
if k < len(ans):
print('NO')
else:
ans = '0' * (k - len(ans)) + ans
print('YES')
print(ans)
```
Yes
| 107,243 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings a and b (both of length k) a bit similar if they have the same character in some position, i. e. there exists at least one i β [1, k] such that a_i = b_i.
You are given a binary string s of length n (a string of n characters 0 and/or 1) and an integer k. Let's denote the string s[i..j] as the substring of s starting from the i-th character and ending with the j-th character (that is, s[i..j] = s_i s_{i + 1} s_{i + 2} ... s_{j - 1} s_j).
Let's call a binary string t of length k beautiful if it is a bit similar to all substrings of s having length exactly k; that is, it is a bit similar to s[1..k], s[2..k+1], ..., s[n-k+1..n].
Your goal is to find the lexicographically smallest string t that is beautiful, or report that no such string exists. String x is lexicographically less than string y if either x is a prefix of y (and x β y), or there exists such i (1 β€ i β€ min(|x|, |y|)), that x_i < y_i, and for any j (1 β€ j < i) x_j = y_j.
Input
The first line contains one integer q (1 β€ q β€ 10000) β the number of test cases. Each test case consists of two lines.
The first line of each test case contains two integers n and k (1 β€ k β€ n β€ 10^6). The second line contains the string s, consisting of n characters (each character is either 0 or 1).
It is guaranteed that the sum of n over all test cases does not exceed 10^6.
Output
For each test case, print the answer as follows:
* if it is impossible to construct a beautiful string, print one line containing the string NO (note: exactly in upper case, you can't print No, for example);
* otherwise, print two lines. The first line should contain the string YES (exactly in upper case as well); the second line β the lexicographically smallest beautiful string, consisting of k characters 0 and/or 1.
Example
Input
7
4 2
0110
4 2
1001
9 3
010001110
9 3
101110001
10 3
0101110001
10 10
1111111111
11 10
11111111110
Output
YES
11
YES
00
YES
010
YES
101
NO
YES
0000000001
YES
0000000010
Submitted Solution:
```
import sys
input = sys.stdin.readline
t = int(input())
for case in range(t):
n, k0 = [int(x) for x in input().split(' ')]
start_inds = set([i for i in range(n-k0+1)])
s = input()
k = min(20, k0)
if k < k0:
next_0 = []
curr_0 = n
for i in range(n-1,-1,-1):
if s[i] == '0':
curr_0 = i
next_0.append(curr_0)
next_0.reverse()
for i in range(n-k0+1):
if next_0[i] < i + k0 - 20:
start_inds.remove(i)
Z = 1 << k
avoid = set()
for i in start_inds:
r = int(s[i + k0 - k:i + k0], 2)
avoid.add(Z - 1 - r)
j = 0
while j in avoid and j < Z:
j += 1
ans = bin(j)[2:]
if j == Z:
print("NO")
else:
print("YES")
ans = '0' * (k0 - len(ans)) + ans
print(ans)
```
Yes
| 107,244 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings a and b (both of length k) a bit similar if they have the same character in some position, i. e. there exists at least one i β [1, k] such that a_i = b_i.
You are given a binary string s of length n (a string of n characters 0 and/or 1) and an integer k. Let's denote the string s[i..j] as the substring of s starting from the i-th character and ending with the j-th character (that is, s[i..j] = s_i s_{i + 1} s_{i + 2} ... s_{j - 1} s_j).
Let's call a binary string t of length k beautiful if it is a bit similar to all substrings of s having length exactly k; that is, it is a bit similar to s[1..k], s[2..k+1], ..., s[n-k+1..n].
Your goal is to find the lexicographically smallest string t that is beautiful, or report that no such string exists. String x is lexicographically less than string y if either x is a prefix of y (and x β y), or there exists such i (1 β€ i β€ min(|x|, |y|)), that x_i < y_i, and for any j (1 β€ j < i) x_j = y_j.
Input
The first line contains one integer q (1 β€ q β€ 10000) β the number of test cases. Each test case consists of two lines.
The first line of each test case contains two integers n and k (1 β€ k β€ n β€ 10^6). The second line contains the string s, consisting of n characters (each character is either 0 or 1).
It is guaranteed that the sum of n over all test cases does not exceed 10^6.
Output
For each test case, print the answer as follows:
* if it is impossible to construct a beautiful string, print one line containing the string NO (note: exactly in upper case, you can't print No, for example);
* otherwise, print two lines. The first line should contain the string YES (exactly in upper case as well); the second line β the lexicographically smallest beautiful string, consisting of k characters 0 and/or 1.
Example
Input
7
4 2
0110
4 2
1001
9 3
010001110
9 3
101110001
10 3
0101110001
10 10
1111111111
11 10
11111111110
Output
YES
11
YES
00
YES
010
YES
101
NO
YES
0000000001
YES
0000000010
Submitted Solution:
```
from itertools import accumulate, chain
BITS = 20
for _ in range(int(input())):
n, k = map(int, input().split())
s = input()
antis = [1 - int(si) for si in s]
antis_s = ''.join(map(str, antis))
antis_sums = tuple(chain((0,), accumulate(antis)))
fbd = set()
for i in range(n + 1 - k):
if k < BITS or antis_sums[i] == antis_sums[i + k - BITS]:
cur = int(antis_s[max(i, i + k - BITS):i + k], base=2)
fbd.add(cur)
ans = 0
while ans in fbd:
ans += 1
t = bin(ans)[2:]
if len(t) > k:
print("NO")
else:
print("YES")
print(t.rjust(k, '0'))
```
Yes
| 107,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings a and b (both of length k) a bit similar if they have the same character in some position, i. e. there exists at least one i β [1, k] such that a_i = b_i.
You are given a binary string s of length n (a string of n characters 0 and/or 1) and an integer k. Let's denote the string s[i..j] as the substring of s starting from the i-th character and ending with the j-th character (that is, s[i..j] = s_i s_{i + 1} s_{i + 2} ... s_{j - 1} s_j).
Let's call a binary string t of length k beautiful if it is a bit similar to all substrings of s having length exactly k; that is, it is a bit similar to s[1..k], s[2..k+1], ..., s[n-k+1..n].
Your goal is to find the lexicographically smallest string t that is beautiful, or report that no such string exists. String x is lexicographically less than string y if either x is a prefix of y (and x β y), or there exists such i (1 β€ i β€ min(|x|, |y|)), that x_i < y_i, and for any j (1 β€ j < i) x_j = y_j.
Input
The first line contains one integer q (1 β€ q β€ 10000) β the number of test cases. Each test case consists of two lines.
The first line of each test case contains two integers n and k (1 β€ k β€ n β€ 10^6). The second line contains the string s, consisting of n characters (each character is either 0 or 1).
It is guaranteed that the sum of n over all test cases does not exceed 10^6.
Output
For each test case, print the answer as follows:
* if it is impossible to construct a beautiful string, print one line containing the string NO (note: exactly in upper case, you can't print No, for example);
* otherwise, print two lines. The first line should contain the string YES (exactly in upper case as well); the second line β the lexicographically smallest beautiful string, consisting of k characters 0 and/or 1.
Example
Input
7
4 2
0110
4 2
1001
9 3
010001110
9 3
101110001
10 3
0101110001
10 10
1111111111
11 10
11111111110
Output
YES
11
YES
00
YES
010
YES
101
NO
YES
0000000001
YES
0000000010
Submitted Solution:
```
T = int(input())
ans = []
for t in range(T):
n,k = map(int,input().split())
s = list(input())
for i in range(n):
if s[i]=="0":
s[i]="1"
else:
s[i]="0"
count1 = 0
if k > 20:
count1 = s[0:k-20].count("1")
s = "".join(s)
dic = {}
for i in range(n-k+1):
if k <= 20:
dic[int(s[i:i+k],2)]=0
else:
if count1==0:
dic[int(s[i+k-20:i+k],2)]=0
count1 -= int(s[i])
count1 += int(s[i+k-20])
tmpans=10**10
for i in range(0,n+1):
if not(i in dic):
tmpans=i
break
if tmpans < pow(2,k):
ans.append("YES")
ans.append(format(tmpans,"0"+str(k)+"b"))
else:
ans.append("NO")
for i in range(len(ans)):
print(ans[i])
```
Yes
| 107,246 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings a and b (both of length k) a bit similar if they have the same character in some position, i. e. there exists at least one i β [1, k] such that a_i = b_i.
You are given a binary string s of length n (a string of n characters 0 and/or 1) and an integer k. Let's denote the string s[i..j] as the substring of s starting from the i-th character and ending with the j-th character (that is, s[i..j] = s_i s_{i + 1} s_{i + 2} ... s_{j - 1} s_j).
Let's call a binary string t of length k beautiful if it is a bit similar to all substrings of s having length exactly k; that is, it is a bit similar to s[1..k], s[2..k+1], ..., s[n-k+1..n].
Your goal is to find the lexicographically smallest string t that is beautiful, or report that no such string exists. String x is lexicographically less than string y if either x is a prefix of y (and x β y), or there exists such i (1 β€ i β€ min(|x|, |y|)), that x_i < y_i, and for any j (1 β€ j < i) x_j = y_j.
Input
The first line contains one integer q (1 β€ q β€ 10000) β the number of test cases. Each test case consists of two lines.
The first line of each test case contains two integers n and k (1 β€ k β€ n β€ 10^6). The second line contains the string s, consisting of n characters (each character is either 0 or 1).
It is guaranteed that the sum of n over all test cases does not exceed 10^6.
Output
For each test case, print the answer as follows:
* if it is impossible to construct a beautiful string, print one line containing the string NO (note: exactly in upper case, you can't print No, for example);
* otherwise, print two lines. The first line should contain the string YES (exactly in upper case as well); the second line β the lexicographically smallest beautiful string, consisting of k characters 0 and/or 1.
Example
Input
7
4 2
0110
4 2
1001
9 3
010001110
9 3
101110001
10 3
0101110001
10 10
1111111111
11 10
11111111110
Output
YES
11
YES
00
YES
010
YES
101
NO
YES
0000000001
YES
0000000010
Submitted Solution:
```
import sys
input = sys.stdin.readline
from collections import Counter
import math
t = int(input())
for _ in range(t):
n,m = list(map(int,input().split()))
k = min(21,m)
s = input().rstrip()
D = set()
v = 2**k - 1
for i in range(len(s)-k+1):
temp = s[i:i+k]
#print(temp)
D.add(v - int(temp,2))
ok = 0
if s[:m].count('1')==0 and m>len(s)//2:
print('YES')
print('0' * m)
ok = 1
else:
for i in range(v+1):
if i not in D:
ans = (bin(i))[2:]
ans = (m - len(ans)) * '0' + ans
print('YES')
print(ans)
ok = 1
break
if ok==0:
print('NO')
```
No
| 107,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings a and b (both of length k) a bit similar if they have the same character in some position, i. e. there exists at least one i β [1, k] such that a_i = b_i.
You are given a binary string s of length n (a string of n characters 0 and/or 1) and an integer k. Let's denote the string s[i..j] as the substring of s starting from the i-th character and ending with the j-th character (that is, s[i..j] = s_i s_{i + 1} s_{i + 2} ... s_{j - 1} s_j).
Let's call a binary string t of length k beautiful if it is a bit similar to all substrings of s having length exactly k; that is, it is a bit similar to s[1..k], s[2..k+1], ..., s[n-k+1..n].
Your goal is to find the lexicographically smallest string t that is beautiful, or report that no such string exists. String x is lexicographically less than string y if either x is a prefix of y (and x β y), or there exists such i (1 β€ i β€ min(|x|, |y|)), that x_i < y_i, and for any j (1 β€ j < i) x_j = y_j.
Input
The first line contains one integer q (1 β€ q β€ 10000) β the number of test cases. Each test case consists of two lines.
The first line of each test case contains two integers n and k (1 β€ k β€ n β€ 10^6). The second line contains the string s, consisting of n characters (each character is either 0 or 1).
It is guaranteed that the sum of n over all test cases does not exceed 10^6.
Output
For each test case, print the answer as follows:
* if it is impossible to construct a beautiful string, print one line containing the string NO (note: exactly in upper case, you can't print No, for example);
* otherwise, print two lines. The first line should contain the string YES (exactly in upper case as well); the second line β the lexicographically smallest beautiful string, consisting of k characters 0 and/or 1.
Example
Input
7
4 2
0110
4 2
1001
9 3
010001110
9 3
101110001
10 3
0101110001
10 10
1111111111
11 10
11111111110
Output
YES
11
YES
00
YES
010
YES
101
NO
YES
0000000001
YES
0000000010
Submitted Solution:
```
from sys import stdin, gettrace, stdout
if gettrace():
def inputi():
return input()
else:
def input():
return next(stdin)[:-1]
def inputi():
return stdin.buffer.readline()
def solve(ob):
n, k = map(int, inputi().split())
ss = inputi()
kk = min((n - k + 1).bit_length(), k)
asize = 1<<kk
mask = asize - 1
found = bytearray(asize)
v = 0
for s in ss[0:kk-1]:
v <<=1
if s == 0x31:
v += 1
for s in ss[kk-1:n]:
v <<=1
v &= mask
if s == 0x31:
v += 1
found[v] = 1
ndx = found.rfind(0)
if ndx == -1:
ob.write(b'NO\n')
return
else:
ob.write(b'YES\n')
v = asize - ndx - 1
res = bytearray(k)
for i in range(k-1, -1, -1):
res[i] = (v&1) + 0x30
v >>= 1
ob.write(res)
ob.write(b'\n')
def main():
t = int(inputi())
ob = stdout.buffer
for _ in range(t):
solve(ob)
if __name__ == "__main__":
main()
```
No
| 107,248 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings a and b (both of length k) a bit similar if they have the same character in some position, i. e. there exists at least one i β [1, k] such that a_i = b_i.
You are given a binary string s of length n (a string of n characters 0 and/or 1) and an integer k. Let's denote the string s[i..j] as the substring of s starting from the i-th character and ending with the j-th character (that is, s[i..j] = s_i s_{i + 1} s_{i + 2} ... s_{j - 1} s_j).
Let's call a binary string t of length k beautiful if it is a bit similar to all substrings of s having length exactly k; that is, it is a bit similar to s[1..k], s[2..k+1], ..., s[n-k+1..n].
Your goal is to find the lexicographically smallest string t that is beautiful, or report that no such string exists. String x is lexicographically less than string y if either x is a prefix of y (and x β y), or there exists such i (1 β€ i β€ min(|x|, |y|)), that x_i < y_i, and for any j (1 β€ j < i) x_j = y_j.
Input
The first line contains one integer q (1 β€ q β€ 10000) β the number of test cases. Each test case consists of two lines.
The first line of each test case contains two integers n and k (1 β€ k β€ n β€ 10^6). The second line contains the string s, consisting of n characters (each character is either 0 or 1).
It is guaranteed that the sum of n over all test cases does not exceed 10^6.
Output
For each test case, print the answer as follows:
* if it is impossible to construct a beautiful string, print one line containing the string NO (note: exactly in upper case, you can't print No, for example);
* otherwise, print two lines. The first line should contain the string YES (exactly in upper case as well); the second line β the lexicographically smallest beautiful string, consisting of k characters 0 and/or 1.
Example
Input
7
4 2
0110
4 2
1001
9 3
010001110
9 3
101110001
10 3
0101110001
10 10
1111111111
11 10
11111111110
Output
YES
11
YES
00
YES
010
YES
101
NO
YES
0000000001
YES
0000000010
Submitted Solution:
```
import bisect
import sys
input=sys.stdin.readline
from math import ceil
t=int(input())
while t:
n,k=map(int,input().split())
s=input().split()[0]
z=[]
for i in range(n):
if(s[i]=='0'):
z.append(i)
co=0
cz=0
l=0
r=k-1
lol=0
for i in range(k):
if(s[i]=='0'):
cz+=1
else:
co+=1
lol1=0
comp=[0]*k
if(co==k):
lol=1
if(cz==k):
lol1=1
#print(cz,co)
if(cz==1):
ind=bisect.bisect_left(z,r)
ind-=1
#print(ind,"ind")
comp[ind]=1
while r<n:
if(s[l]=='0'):
cz-=1
else:
co-=1
l+=1
r+=1
if(r==n):
break
if(s[r]=='0'):
cz+=1
else:
co+=1
if(co==k):
lol=1
if(cz==k):
lol1=1
if(cz==1):
#print(r)
ind=bisect.bisect_right(z,r)
ind-=1
#print(ind,"ind")
ind = ind-l+1
comp[ind]=1
yes=0
for i in range(k):
if(comp[i]==1):
comp[i]='0'
else:
yes=1
comp[i]='1'
if(lol==1 and yes==0):
if(lol1==0):
print("YES")
print("1"*k)
else:
print("NO")
elif(lol==1 and yes==1):
print("YES")
print("".join(comp))
else:
print("YES")
print('0'*k)
t-=1
```
No
| 107,249 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings a and b (both of length k) a bit similar if they have the same character in some position, i. e. there exists at least one i β [1, k] such that a_i = b_i.
You are given a binary string s of length n (a string of n characters 0 and/or 1) and an integer k. Let's denote the string s[i..j] as the substring of s starting from the i-th character and ending with the j-th character (that is, s[i..j] = s_i s_{i + 1} s_{i + 2} ... s_{j - 1} s_j).
Let's call a binary string t of length k beautiful if it is a bit similar to all substrings of s having length exactly k; that is, it is a bit similar to s[1..k], s[2..k+1], ..., s[n-k+1..n].
Your goal is to find the lexicographically smallest string t that is beautiful, or report that no such string exists. String x is lexicographically less than string y if either x is a prefix of y (and x β y), or there exists such i (1 β€ i β€ min(|x|, |y|)), that x_i < y_i, and for any j (1 β€ j < i) x_j = y_j.
Input
The first line contains one integer q (1 β€ q β€ 10000) β the number of test cases. Each test case consists of two lines.
The first line of each test case contains two integers n and k (1 β€ k β€ n β€ 10^6). The second line contains the string s, consisting of n characters (each character is either 0 or 1).
It is guaranteed that the sum of n over all test cases does not exceed 10^6.
Output
For each test case, print the answer as follows:
* if it is impossible to construct a beautiful string, print one line containing the string NO (note: exactly in upper case, you can't print No, for example);
* otherwise, print two lines. The first line should contain the string YES (exactly in upper case as well); the second line β the lexicographically smallest beautiful string, consisting of k characters 0 and/or 1.
Example
Input
7
4 2
0110
4 2
1001
9 3
010001110
9 3
101110001
10 3
0101110001
10 10
1111111111
11 10
11111111110
Output
YES
11
YES
00
YES
010
YES
101
NO
YES
0000000001
YES
0000000010
Submitted Solution:
```
import sys
input = sys.stdin.readline
for _ in range(int(input())):
N, K = map(int, input().split())
S = list(map(int, list(input())[: -1]))
res = [0] * K
ln = min(K, 25)
s = set()
for i in range(N - K + 1):
t = S[i + K - ln: i + K]
x = 0
for j in range(ln): x |= t[j] << (ln - j - 1)
s.add(x)
for x in range((1 << ln) - 1, -1, -1):
if x in s: continue
print("YES")
bn = list(bin(x))[2: ]
bn = [0] * (ln - len(bn)) + bn
print("".join([str(int(i) ^ 1) for i in bn]).zfill(K))
break
else: print("NO")
```
No
| 107,250 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Diamond Miner is a game that is similar to Gold Miner, but there are n miners instead of 1 in this game.
The mining area can be described as a plane. The n miners can be regarded as n points on the y-axis. There are n diamond mines in the mining area. We can regard them as n points on the x-axis. For some reason, no miners or diamond mines can be at the origin (point (0, 0)).
Every miner should mine exactly one diamond mine. Every miner has a hook, which can be used to mine a diamond mine. If a miner at the point (a,b) uses his hook to mine a diamond mine at the point (c,d), he will spend β{(a-c)^2+(b-d)^2} energy to mine it (the distance between these points). The miners can't move or help each other.
The object of this game is to minimize the sum of the energy that miners spend. Can you find this minimum?
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 10) β the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of miners and mines.
Each of the next 2n lines contains two space-separated integers x (-10^8 β€ x β€ 10^8) and y (-10^8 β€ y β€ 10^8), which represent the point (x,y) to describe a miner's or a diamond mine's position. Either x = 0, meaning there is a miner at the point (0, y), or y = 0, meaning there is a diamond mine at the point (x, 0). There can be multiple miners or diamond mines at the same point.
It is guaranteed that no point is at the origin. It is guaranteed that the number of points on the x-axis is equal to n and the number of points on the y-axis is equal to n.
It's guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, print a single real number β the minimal sum of energy that should be spent.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-9}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} β€ 10^{-9}.
Example
Input
3
2
0 1
1 0
0 -1
-2 0
4
1 0
3 0
-5 0
6 0
0 3
0 1
0 2
0 4
5
3 0
0 4
0 -3
4 0
2 0
1 0
-3 0
0 -10
0 -2
0 -10
Output
3.650281539872885
18.061819283610362
32.052255376143336
Note
In the first test case, the miners are at (0,1) and (0,-1), while the diamond mines are at (1,0) and (-2,0). If you arrange the miners to get the diamond mines in the way, shown in the picture, you can get the sum of the energy β2 + β5.
<image>
Tags: geometry, greedy, math, sortings
Correct Solution:
```
import sys
import math
# working with masth.pow() instead of "**"
r = sys.stdin.readline
t = int(r())
for tests in range(t):
n = int(r())
xarr = []
yarr = []
summ = 0
for i in range(n * 2):
x, y = map(int, r().split())
if x:
xarr.append(math.pow(x, 2))
else:
yarr.append(math.pow(y, 2))
xarr.sort()
yarr.sort()
for i in range(-n, 0):
summ += math.sqrt(xarr[i] + yarr[i])
print(summ)
```
| 107,251 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Diamond Miner is a game that is similar to Gold Miner, but there are n miners instead of 1 in this game.
The mining area can be described as a plane. The n miners can be regarded as n points on the y-axis. There are n diamond mines in the mining area. We can regard them as n points on the x-axis. For some reason, no miners or diamond mines can be at the origin (point (0, 0)).
Every miner should mine exactly one diamond mine. Every miner has a hook, which can be used to mine a diamond mine. If a miner at the point (a,b) uses his hook to mine a diamond mine at the point (c,d), he will spend β{(a-c)^2+(b-d)^2} energy to mine it (the distance between these points). The miners can't move or help each other.
The object of this game is to minimize the sum of the energy that miners spend. Can you find this minimum?
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 10) β the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of miners and mines.
Each of the next 2n lines contains two space-separated integers x (-10^8 β€ x β€ 10^8) and y (-10^8 β€ y β€ 10^8), which represent the point (x,y) to describe a miner's or a diamond mine's position. Either x = 0, meaning there is a miner at the point (0, y), or y = 0, meaning there is a diamond mine at the point (x, 0). There can be multiple miners or diamond mines at the same point.
It is guaranteed that no point is at the origin. It is guaranteed that the number of points on the x-axis is equal to n and the number of points on the y-axis is equal to n.
It's guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, print a single real number β the minimal sum of energy that should be spent.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-9}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} β€ 10^{-9}.
Example
Input
3
2
0 1
1 0
0 -1
-2 0
4
1 0
3 0
-5 0
6 0
0 3
0 1
0 2
0 4
5
3 0
0 4
0 -3
4 0
2 0
1 0
-3 0
0 -10
0 -2
0 -10
Output
3.650281539872885
18.061819283610362
32.052255376143336
Note
In the first test case, the miners are at (0,1) and (0,-1), while the diamond mines are at (1,0) and (-2,0). If you arrange the miners to get the diamond mines in the way, shown in the picture, you can get the sum of the energy β2 + β5.
<image>
Tags: geometry, greedy, math, sortings
Correct Solution:
```
if __name__ == '__main__':
for _ in range (int(input())):
n = int(input())
x = []
y = []
for i in range (2*n):
X,Y = map(int,input().split())
if X==0:
y.append(abs(Y))
else:
x.append(abs(X))
x.sort()
y.sort()
ans = 0
for i in range (n):
ans = ans + ((x[i]*x[i])+(y[i]*y[i]))**0.5
print(ans)
```
| 107,252 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Diamond Miner is a game that is similar to Gold Miner, but there are n miners instead of 1 in this game.
The mining area can be described as a plane. The n miners can be regarded as n points on the y-axis. There are n diamond mines in the mining area. We can regard them as n points on the x-axis. For some reason, no miners or diamond mines can be at the origin (point (0, 0)).
Every miner should mine exactly one diamond mine. Every miner has a hook, which can be used to mine a diamond mine. If a miner at the point (a,b) uses his hook to mine a diamond mine at the point (c,d), he will spend β{(a-c)^2+(b-d)^2} energy to mine it (the distance between these points). The miners can't move or help each other.
The object of this game is to minimize the sum of the energy that miners spend. Can you find this minimum?
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 10) β the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of miners and mines.
Each of the next 2n lines contains two space-separated integers x (-10^8 β€ x β€ 10^8) and y (-10^8 β€ y β€ 10^8), which represent the point (x,y) to describe a miner's or a diamond mine's position. Either x = 0, meaning there is a miner at the point (0, y), or y = 0, meaning there is a diamond mine at the point (x, 0). There can be multiple miners or diamond mines at the same point.
It is guaranteed that no point is at the origin. It is guaranteed that the number of points on the x-axis is equal to n and the number of points on the y-axis is equal to n.
It's guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, print a single real number β the minimal sum of energy that should be spent.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-9}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} β€ 10^{-9}.
Example
Input
3
2
0 1
1 0
0 -1
-2 0
4
1 0
3 0
-5 0
6 0
0 3
0 1
0 2
0 4
5
3 0
0 4
0 -3
4 0
2 0
1 0
-3 0
0 -10
0 -2
0 -10
Output
3.650281539872885
18.061819283610362
32.052255376143336
Note
In the first test case, the miners are at (0,1) and (0,-1), while the diamond mines are at (1,0) and (-2,0). If you arrange the miners to get the diamond mines in the way, shown in the picture, you can get the sum of the energy β2 + β5.
<image>
Tags: geometry, greedy, math, sortings
Correct Solution:
```
import math
def main():
t = int(input())
for _ in range(t):
n = int(input())
miners = [-1] * n
mines = [-1] * n
i, j = 0, 0
for _ in range(2*n):
a, b = map(lambda x: abs(int(x)), input().split())
if a == 0:
miners[i] = b
i += 1
else:
mines[j] = a
j += 1
mines.sort()
miners.sort()
acc = 0
for i in range(n):
acc += math.sqrt(miners[i] ** 2 + mines[i] ** 2)
print(f"{acc:.15f}")
main()
```
| 107,253 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Diamond Miner is a game that is similar to Gold Miner, but there are n miners instead of 1 in this game.
The mining area can be described as a plane. The n miners can be regarded as n points on the y-axis. There are n diamond mines in the mining area. We can regard them as n points on the x-axis. For some reason, no miners or diamond mines can be at the origin (point (0, 0)).
Every miner should mine exactly one diamond mine. Every miner has a hook, which can be used to mine a diamond mine. If a miner at the point (a,b) uses his hook to mine a diamond mine at the point (c,d), he will spend β{(a-c)^2+(b-d)^2} energy to mine it (the distance between these points). The miners can't move or help each other.
The object of this game is to minimize the sum of the energy that miners spend. Can you find this minimum?
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 10) β the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of miners and mines.
Each of the next 2n lines contains two space-separated integers x (-10^8 β€ x β€ 10^8) and y (-10^8 β€ y β€ 10^8), which represent the point (x,y) to describe a miner's or a diamond mine's position. Either x = 0, meaning there is a miner at the point (0, y), or y = 0, meaning there is a diamond mine at the point (x, 0). There can be multiple miners or diamond mines at the same point.
It is guaranteed that no point is at the origin. It is guaranteed that the number of points on the x-axis is equal to n and the number of points on the y-axis is equal to n.
It's guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, print a single real number β the minimal sum of energy that should be spent.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-9}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} β€ 10^{-9}.
Example
Input
3
2
0 1
1 0
0 -1
-2 0
4
1 0
3 0
-5 0
6 0
0 3
0 1
0 2
0 4
5
3 0
0 4
0 -3
4 0
2 0
1 0
-3 0
0 -10
0 -2
0 -10
Output
3.650281539872885
18.061819283610362
32.052255376143336
Note
In the first test case, the miners are at (0,1) and (0,-1), while the diamond mines are at (1,0) and (-2,0). If you arrange the miners to get the diamond mines in the way, shown in the picture, you can get the sum of the energy β2 + β5.
<image>
Tags: geometry, greedy, math, sortings
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
xs, ys = [], []
for _ in range(2 * n):
x, y = map(int, input().split())
if x == 0:
ys.append(y * y)
else:
xs.append(x * x)
xs.sort()
ys.sort()
dist = 0
for a, b in zip(xs, ys):
dist += (a + b) ** 0.5
print(dist)
```
| 107,254 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Diamond Miner is a game that is similar to Gold Miner, but there are n miners instead of 1 in this game.
The mining area can be described as a plane. The n miners can be regarded as n points on the y-axis. There are n diamond mines in the mining area. We can regard them as n points on the x-axis. For some reason, no miners or diamond mines can be at the origin (point (0, 0)).
Every miner should mine exactly one diamond mine. Every miner has a hook, which can be used to mine a diamond mine. If a miner at the point (a,b) uses his hook to mine a diamond mine at the point (c,d), he will spend β{(a-c)^2+(b-d)^2} energy to mine it (the distance between these points). The miners can't move or help each other.
The object of this game is to minimize the sum of the energy that miners spend. Can you find this minimum?
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 10) β the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of miners and mines.
Each of the next 2n lines contains two space-separated integers x (-10^8 β€ x β€ 10^8) and y (-10^8 β€ y β€ 10^8), which represent the point (x,y) to describe a miner's or a diamond mine's position. Either x = 0, meaning there is a miner at the point (0, y), or y = 0, meaning there is a diamond mine at the point (x, 0). There can be multiple miners or diamond mines at the same point.
It is guaranteed that no point is at the origin. It is guaranteed that the number of points on the x-axis is equal to n and the number of points on the y-axis is equal to n.
It's guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, print a single real number β the minimal sum of energy that should be spent.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-9}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} β€ 10^{-9}.
Example
Input
3
2
0 1
1 0
0 -1
-2 0
4
1 0
3 0
-5 0
6 0
0 3
0 1
0 2
0 4
5
3 0
0 4
0 -3
4 0
2 0
1 0
-3 0
0 -10
0 -2
0 -10
Output
3.650281539872885
18.061819283610362
32.052255376143336
Note
In the first test case, the miners are at (0,1) and (0,-1), while the diamond mines are at (1,0) and (-2,0). If you arrange the miners to get the diamond mines in the way, shown in the picture, you can get the sum of the energy β2 + β5.
<image>
Tags: geometry, greedy, math, sortings
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
a = []
b = []
for i in range(2*n):
x,y = map(int, input().split())
if y == 0:
a.append(abs(x))
else:
b.append(abs(y))
a.sort()
b.sort()
res = 0
j = 0
for i in range(len(a)):
if j >= len(b):
break
res += (a[i]**2 + b[j]**2)**(1/2)
j += 1
print(res)
```
| 107,255 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Diamond Miner is a game that is similar to Gold Miner, but there are n miners instead of 1 in this game.
The mining area can be described as a plane. The n miners can be regarded as n points on the y-axis. There are n diamond mines in the mining area. We can regard them as n points on the x-axis. For some reason, no miners or diamond mines can be at the origin (point (0, 0)).
Every miner should mine exactly one diamond mine. Every miner has a hook, which can be used to mine a diamond mine. If a miner at the point (a,b) uses his hook to mine a diamond mine at the point (c,d), he will spend β{(a-c)^2+(b-d)^2} energy to mine it (the distance between these points). The miners can't move or help each other.
The object of this game is to minimize the sum of the energy that miners spend. Can you find this minimum?
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 10) β the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of miners and mines.
Each of the next 2n lines contains two space-separated integers x (-10^8 β€ x β€ 10^8) and y (-10^8 β€ y β€ 10^8), which represent the point (x,y) to describe a miner's or a diamond mine's position. Either x = 0, meaning there is a miner at the point (0, y), or y = 0, meaning there is a diamond mine at the point (x, 0). There can be multiple miners or diamond mines at the same point.
It is guaranteed that no point is at the origin. It is guaranteed that the number of points on the x-axis is equal to n and the number of points on the y-axis is equal to n.
It's guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, print a single real number β the minimal sum of energy that should be spent.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-9}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} β€ 10^{-9}.
Example
Input
3
2
0 1
1 0
0 -1
-2 0
4
1 0
3 0
-5 0
6 0
0 3
0 1
0 2
0 4
5
3 0
0 4
0 -3
4 0
2 0
1 0
-3 0
0 -10
0 -2
0 -10
Output
3.650281539872885
18.061819283610362
32.052255376143336
Note
In the first test case, the miners are at (0,1) and (0,-1), while the diamond mines are at (1,0) and (-2,0). If you arrange the miners to get the diamond mines in the way, shown in the picture, you can get the sum of the energy β2 + β5.
<image>
Tags: geometry, greedy, math, sortings
Correct Solution:
```
import math
import sys;Z=sys.stdin.readline
for _ in range(int(Z())):
n=int(input())
x=[]
y=[]
for i in range(2*n):
a,b=map(int,Z().split())
if b==0:
x.append(abs(a))
else:
y.append(abs(b))
x.sort()
y.sort()
dist=0
## print(x,y)
for i in range(n):
dist+=math.sqrt(x[i]*x[i]+y[i]*y[i])
print(dist)
```
| 107,256 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Diamond Miner is a game that is similar to Gold Miner, but there are n miners instead of 1 in this game.
The mining area can be described as a plane. The n miners can be regarded as n points on the y-axis. There are n diamond mines in the mining area. We can regard them as n points on the x-axis. For some reason, no miners or diamond mines can be at the origin (point (0, 0)).
Every miner should mine exactly one diamond mine. Every miner has a hook, which can be used to mine a diamond mine. If a miner at the point (a,b) uses his hook to mine a diamond mine at the point (c,d), he will spend β{(a-c)^2+(b-d)^2} energy to mine it (the distance between these points). The miners can't move or help each other.
The object of this game is to minimize the sum of the energy that miners spend. Can you find this minimum?
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 10) β the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of miners and mines.
Each of the next 2n lines contains two space-separated integers x (-10^8 β€ x β€ 10^8) and y (-10^8 β€ y β€ 10^8), which represent the point (x,y) to describe a miner's or a diamond mine's position. Either x = 0, meaning there is a miner at the point (0, y), or y = 0, meaning there is a diamond mine at the point (x, 0). There can be multiple miners or diamond mines at the same point.
It is guaranteed that no point is at the origin. It is guaranteed that the number of points on the x-axis is equal to n and the number of points on the y-axis is equal to n.
It's guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, print a single real number β the minimal sum of energy that should be spent.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-9}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} β€ 10^{-9}.
Example
Input
3
2
0 1
1 0
0 -1
-2 0
4
1 0
3 0
-5 0
6 0
0 3
0 1
0 2
0 4
5
3 0
0 4
0 -3
4 0
2 0
1 0
-3 0
0 -10
0 -2
0 -10
Output
3.650281539872885
18.061819283610362
32.052255376143336
Note
In the first test case, the miners are at (0,1) and (0,-1), while the diamond mines are at (1,0) and (-2,0). If you arrange the miners to get the diamond mines in the way, shown in the picture, you can get the sum of the energy β2 + β5.
<image>
Tags: geometry, greedy, math, sortings
Correct Solution:
```
import sys
input = sys.stdin.readline
t = int(input())
for i in range(t):
n = int(input())
miners, mines = [], []
for i in range(2*n):
x, y = map(int, input().split())
if x:
mines.append(abs(x))
else:
miners.append(abs(y))
miners.sort(reverse=True)
mines.sort(reverse=True)
res = 0
for x, y in zip(miners, mines):
res += (x*x + y*y) ** (1/2)
print(res)
```
| 107,257 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Diamond Miner is a game that is similar to Gold Miner, but there are n miners instead of 1 in this game.
The mining area can be described as a plane. The n miners can be regarded as n points on the y-axis. There are n diamond mines in the mining area. We can regard them as n points on the x-axis. For some reason, no miners or diamond mines can be at the origin (point (0, 0)).
Every miner should mine exactly one diamond mine. Every miner has a hook, which can be used to mine a diamond mine. If a miner at the point (a,b) uses his hook to mine a diamond mine at the point (c,d), he will spend β{(a-c)^2+(b-d)^2} energy to mine it (the distance between these points). The miners can't move or help each other.
The object of this game is to minimize the sum of the energy that miners spend. Can you find this minimum?
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 10) β the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of miners and mines.
Each of the next 2n lines contains two space-separated integers x (-10^8 β€ x β€ 10^8) and y (-10^8 β€ y β€ 10^8), which represent the point (x,y) to describe a miner's or a diamond mine's position. Either x = 0, meaning there is a miner at the point (0, y), or y = 0, meaning there is a diamond mine at the point (x, 0). There can be multiple miners or diamond mines at the same point.
It is guaranteed that no point is at the origin. It is guaranteed that the number of points on the x-axis is equal to n and the number of points on the y-axis is equal to n.
It's guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, print a single real number β the minimal sum of energy that should be spent.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-9}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} β€ 10^{-9}.
Example
Input
3
2
0 1
1 0
0 -1
-2 0
4
1 0
3 0
-5 0
6 0
0 3
0 1
0 2
0 4
5
3 0
0 4
0 -3
4 0
2 0
1 0
-3 0
0 -10
0 -2
0 -10
Output
3.650281539872885
18.061819283610362
32.052255376143336
Note
In the first test case, the miners are at (0,1) and (0,-1), while the diamond mines are at (1,0) and (-2,0). If you arrange the miners to get the diamond mines in the way, shown in the picture, you can get the sum of the energy β2 + β5.
<image>
Tags: geometry, greedy, math, sortings
Correct Solution:
```
import sys
input = sys.stdin.readline
import math
import copy
from decimal import Decimal
t = int(input())
for _ in range(t):
n = int(input())
diamond = []
miner = []
for i in range(2*n):
x, y = map(int,input().split())
if x == 0:
miner.append(abs(y))
else:
diamond.append(abs(x))
temp = []
diamond.sort()
miner.sort()
for i in range(n):
temp.append(miner[i]**2+diamond[i]**2)
ans = 0
for i in temp:
ans += i**0.5
print(ans)
```
| 107,258 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Diamond Miner is a game that is similar to Gold Miner, but there are n miners instead of 1 in this game.
The mining area can be described as a plane. The n miners can be regarded as n points on the y-axis. There are n diamond mines in the mining area. We can regard them as n points on the x-axis. For some reason, no miners or diamond mines can be at the origin (point (0, 0)).
Every miner should mine exactly one diamond mine. Every miner has a hook, which can be used to mine a diamond mine. If a miner at the point (a,b) uses his hook to mine a diamond mine at the point (c,d), he will spend β{(a-c)^2+(b-d)^2} energy to mine it (the distance between these points). The miners can't move or help each other.
The object of this game is to minimize the sum of the energy that miners spend. Can you find this minimum?
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 10) β the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of miners and mines.
Each of the next 2n lines contains two space-separated integers x (-10^8 β€ x β€ 10^8) and y (-10^8 β€ y β€ 10^8), which represent the point (x,y) to describe a miner's or a diamond mine's position. Either x = 0, meaning there is a miner at the point (0, y), or y = 0, meaning there is a diamond mine at the point (x, 0). There can be multiple miners or diamond mines at the same point.
It is guaranteed that no point is at the origin. It is guaranteed that the number of points on the x-axis is equal to n and the number of points on the y-axis is equal to n.
It's guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, print a single real number β the minimal sum of energy that should be spent.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-9}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} β€ 10^{-9}.
Example
Input
3
2
0 1
1 0
0 -1
-2 0
4
1 0
3 0
-5 0
6 0
0 3
0 1
0 2
0 4
5
3 0
0 4
0 -3
4 0
2 0
1 0
-3 0
0 -10
0 -2
0 -10
Output
3.650281539872885
18.061819283610362
32.052255376143336
Note
In the first test case, the miners are at (0,1) and (0,-1), while the diamond mines are at (1,0) and (-2,0). If you arrange the miners to get the diamond mines in the way, shown in the picture, you can get the sum of the energy β2 + β5.
<image>
Submitted Solution:
```
import sys
input = sys.stdin.readline
from collections import *
for _ in range(int(input())):
n = int(input())
xs = []
ys = []
for _ in range(2*n):
x, y = map(int, input().split())
if y==0:
xs.append(x**2)
else:
ys.append(y**2)
xs.sort()
ys.sort()
ans = 0
for x, y in zip(xs, ys):
ans += (x+y)**0.5
print(ans)
```
Yes
| 107,259 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Diamond Miner is a game that is similar to Gold Miner, but there are n miners instead of 1 in this game.
The mining area can be described as a plane. The n miners can be regarded as n points on the y-axis. There are n diamond mines in the mining area. We can regard them as n points on the x-axis. For some reason, no miners or diamond mines can be at the origin (point (0, 0)).
Every miner should mine exactly one diamond mine. Every miner has a hook, which can be used to mine a diamond mine. If a miner at the point (a,b) uses his hook to mine a diamond mine at the point (c,d), he will spend β{(a-c)^2+(b-d)^2} energy to mine it (the distance between these points). The miners can't move or help each other.
The object of this game is to minimize the sum of the energy that miners spend. Can you find this minimum?
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 10) β the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of miners and mines.
Each of the next 2n lines contains two space-separated integers x (-10^8 β€ x β€ 10^8) and y (-10^8 β€ y β€ 10^8), which represent the point (x,y) to describe a miner's or a diamond mine's position. Either x = 0, meaning there is a miner at the point (0, y), or y = 0, meaning there is a diamond mine at the point (x, 0). There can be multiple miners or diamond mines at the same point.
It is guaranteed that no point is at the origin. It is guaranteed that the number of points on the x-axis is equal to n and the number of points on the y-axis is equal to n.
It's guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, print a single real number β the minimal sum of energy that should be spent.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-9}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} β€ 10^{-9}.
Example
Input
3
2
0 1
1 0
0 -1
-2 0
4
1 0
3 0
-5 0
6 0
0 3
0 1
0 2
0 4
5
3 0
0 4
0 -3
4 0
2 0
1 0
-3 0
0 -10
0 -2
0 -10
Output
3.650281539872885
18.061819283610362
32.052255376143336
Note
In the first test case, the miners are at (0,1) and (0,-1), while the diamond mines are at (1,0) and (-2,0). If you arrange the miners to get the diamond mines in the way, shown in the picture, you can get the sum of the energy β2 + β5.
<image>
Submitted Solution:
```
import sys
sys.setrecursionlimit(10**5)
int1 = lambda x: int(x)-1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.buffer.readline())
def MI(): return map(int, sys.stdin.buffer.readline().split())
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def BI(): return sys.stdin.buffer.readline().rstrip()
def SI(): return sys.stdin.buffer.readline().rstrip().decode()
inf = 10**16
md = 10**9+7
# md = 998244353
for _ in range(II()):
n=II()
xx,yy=[],[]
for _ in range(n*2):
x,y=MI()
if y==0:
xx.append(abs(x))
else:
yy.append(abs(y))
ans=0
xx.sort()
yy.sort()
for x,y in zip(xx,yy):
ans+=(x**2+y**2)**0.5
print(ans)
```
Yes
| 107,260 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Diamond Miner is a game that is similar to Gold Miner, but there are n miners instead of 1 in this game.
The mining area can be described as a plane. The n miners can be regarded as n points on the y-axis. There are n diamond mines in the mining area. We can regard them as n points on the x-axis. For some reason, no miners or diamond mines can be at the origin (point (0, 0)).
Every miner should mine exactly one diamond mine. Every miner has a hook, which can be used to mine a diamond mine. If a miner at the point (a,b) uses his hook to mine a diamond mine at the point (c,d), he will spend β{(a-c)^2+(b-d)^2} energy to mine it (the distance between these points). The miners can't move or help each other.
The object of this game is to minimize the sum of the energy that miners spend. Can you find this minimum?
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 10) β the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of miners and mines.
Each of the next 2n lines contains two space-separated integers x (-10^8 β€ x β€ 10^8) and y (-10^8 β€ y β€ 10^8), which represent the point (x,y) to describe a miner's or a diamond mine's position. Either x = 0, meaning there is a miner at the point (0, y), or y = 0, meaning there is a diamond mine at the point (x, 0). There can be multiple miners or diamond mines at the same point.
It is guaranteed that no point is at the origin. It is guaranteed that the number of points on the x-axis is equal to n and the number of points on the y-axis is equal to n.
It's guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, print a single real number β the minimal sum of energy that should be spent.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-9}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} β€ 10^{-9}.
Example
Input
3
2
0 1
1 0
0 -1
-2 0
4
1 0
3 0
-5 0
6 0
0 3
0 1
0 2
0 4
5
3 0
0 4
0 -3
4 0
2 0
1 0
-3 0
0 -10
0 -2
0 -10
Output
3.650281539872885
18.061819283610362
32.052255376143336
Note
In the first test case, the miners are at (0,1) and (0,-1), while the diamond mines are at (1,0) and (-2,0). If you arrange the miners to get the diamond mines in the way, shown in the picture, you can get the sum of the energy β2 + β5.
<image>
Submitted Solution:
```
from math import sqrt
import sys
input=sys.stdin.readline
t=int(input())
for _ in range(t):
n=int(input())
mi=[]
m=[]
for i in range(n*2):
a,b=list(map(int,input().split()))
if a==0:
mi.append(abs(b))
else:
m.append(abs(a))
mi.sort()
m.sort()
total=0
# print(m,mi)
for i in range(0,n):
total+=sqrt(m[i]*m[i]+mi[i]*mi[i])
print(total)
```
Yes
| 107,261 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Diamond Miner is a game that is similar to Gold Miner, but there are n miners instead of 1 in this game.
The mining area can be described as a plane. The n miners can be regarded as n points on the y-axis. There are n diamond mines in the mining area. We can regard them as n points on the x-axis. For some reason, no miners or diamond mines can be at the origin (point (0, 0)).
Every miner should mine exactly one diamond mine. Every miner has a hook, which can be used to mine a diamond mine. If a miner at the point (a,b) uses his hook to mine a diamond mine at the point (c,d), he will spend β{(a-c)^2+(b-d)^2} energy to mine it (the distance between these points). The miners can't move or help each other.
The object of this game is to minimize the sum of the energy that miners spend. Can you find this minimum?
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 10) β the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of miners and mines.
Each of the next 2n lines contains two space-separated integers x (-10^8 β€ x β€ 10^8) and y (-10^8 β€ y β€ 10^8), which represent the point (x,y) to describe a miner's or a diamond mine's position. Either x = 0, meaning there is a miner at the point (0, y), or y = 0, meaning there is a diamond mine at the point (x, 0). There can be multiple miners or diamond mines at the same point.
It is guaranteed that no point is at the origin. It is guaranteed that the number of points on the x-axis is equal to n and the number of points on the y-axis is equal to n.
It's guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, print a single real number β the minimal sum of energy that should be spent.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-9}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} β€ 10^{-9}.
Example
Input
3
2
0 1
1 0
0 -1
-2 0
4
1 0
3 0
-5 0
6 0
0 3
0 1
0 2
0 4
5
3 0
0 4
0 -3
4 0
2 0
1 0
-3 0
0 -10
0 -2
0 -10
Output
3.650281539872885
18.061819283610362
32.052255376143336
Note
In the first test case, the miners are at (0,1) and (0,-1), while the diamond mines are at (1,0) and (-2,0). If you arrange the miners to get the diamond mines in the way, shown in the picture, you can get the sum of the energy β2 + β5.
<image>
Submitted Solution:
```
import math
test_case = int(input())
for i in range(test_case):
n = int(input())
miner_pos = []
dimond_pos = []
for i in range(2*n):
r,c = map(int, input().split())
if r == 0:
miner_pos.append(c*c)
else:
dimond_pos.append(r*r)
min_sum = 0
miner_pos.sort()
dimond_pos.sort()
for i in range(n):
min_sum = min_sum + math.sqrt(miner_pos[i]+dimond_pos[i])
print(min_sum)
```
Yes
| 107,262 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Diamond Miner is a game that is similar to Gold Miner, but there are n miners instead of 1 in this game.
The mining area can be described as a plane. The n miners can be regarded as n points on the y-axis. There are n diamond mines in the mining area. We can regard them as n points on the x-axis. For some reason, no miners or diamond mines can be at the origin (point (0, 0)).
Every miner should mine exactly one diamond mine. Every miner has a hook, which can be used to mine a diamond mine. If a miner at the point (a,b) uses his hook to mine a diamond mine at the point (c,d), he will spend β{(a-c)^2+(b-d)^2} energy to mine it (the distance between these points). The miners can't move or help each other.
The object of this game is to minimize the sum of the energy that miners spend. Can you find this minimum?
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 10) β the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of miners and mines.
Each of the next 2n lines contains two space-separated integers x (-10^8 β€ x β€ 10^8) and y (-10^8 β€ y β€ 10^8), which represent the point (x,y) to describe a miner's or a diamond mine's position. Either x = 0, meaning there is a miner at the point (0, y), or y = 0, meaning there is a diamond mine at the point (x, 0). There can be multiple miners or diamond mines at the same point.
It is guaranteed that no point is at the origin. It is guaranteed that the number of points on the x-axis is equal to n and the number of points on the y-axis is equal to n.
It's guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, print a single real number β the minimal sum of energy that should be spent.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-9}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} β€ 10^{-9}.
Example
Input
3
2
0 1
1 0
0 -1
-2 0
4
1 0
3 0
-5 0
6 0
0 3
0 1
0 2
0 4
5
3 0
0 4
0 -3
4 0
2 0
1 0
-3 0
0 -10
0 -2
0 -10
Output
3.650281539872885
18.061819283610362
32.052255376143336
Note
In the first test case, the miners are at (0,1) and (0,-1), while the diamond mines are at (1,0) and (-2,0). If you arrange the miners to get the diamond mines in the way, shown in the picture, you can get the sum of the energy β2 + β5.
<image>
Submitted Solution:
```
import math
for _ in range(int(input())):
n=int(input())
lx=[]
ly=[]
for j in range(2*n):
x,y=list(map(int,input().split()))
if x==0:
lx.append(y)
else:
ly.append(x)
lx=list(map(abs,lx))
ly=list(map(abs,ly))
lx.sort()
ly.sort()
ans=0
for i in range(n):
energy=math.sqrt((lx[i])**2+(ly[i])**2)
ans+=energy
print(energy)
```
No
| 107,263 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Diamond Miner is a game that is similar to Gold Miner, but there are n miners instead of 1 in this game.
The mining area can be described as a plane. The n miners can be regarded as n points on the y-axis. There are n diamond mines in the mining area. We can regard them as n points on the x-axis. For some reason, no miners or diamond mines can be at the origin (point (0, 0)).
Every miner should mine exactly one diamond mine. Every miner has a hook, which can be used to mine a diamond mine. If a miner at the point (a,b) uses his hook to mine a diamond mine at the point (c,d), he will spend β{(a-c)^2+(b-d)^2} energy to mine it (the distance between these points). The miners can't move or help each other.
The object of this game is to minimize the sum of the energy that miners spend. Can you find this minimum?
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 10) β the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of miners and mines.
Each of the next 2n lines contains two space-separated integers x (-10^8 β€ x β€ 10^8) and y (-10^8 β€ y β€ 10^8), which represent the point (x,y) to describe a miner's or a diamond mine's position. Either x = 0, meaning there is a miner at the point (0, y), or y = 0, meaning there is a diamond mine at the point (x, 0). There can be multiple miners or diamond mines at the same point.
It is guaranteed that no point is at the origin. It is guaranteed that the number of points on the x-axis is equal to n and the number of points on the y-axis is equal to n.
It's guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, print a single real number β the minimal sum of energy that should be spent.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-9}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} β€ 10^{-9}.
Example
Input
3
2
0 1
1 0
0 -1
-2 0
4
1 0
3 0
-5 0
6 0
0 3
0 1
0 2
0 4
5
3 0
0 4
0 -3
4 0
2 0
1 0
-3 0
0 -10
0 -2
0 -10
Output
3.650281539872885
18.061819283610362
32.052255376143336
Note
In the first test case, the miners are at (0,1) and (0,-1), while the diamond mines are at (1,0) and (-2,0). If you arrange the miners to get the diamond mines in the way, shown in the picture, you can get the sum of the energy β2 + β5.
<image>
Submitted Solution:
```
def qsort(A, left, right):
while left < right:
q = partition(A, left, right)
if (q - 1) - left < right - (q + 1):
qsort(A, left, q-1)
left = q + 1
else:
qsort(A, q+1, right)
right = q - 1
def partition(A, left, right):
q = A[right]
i = left - 1
for j in range(left, right):
if abs(A[j]) < abs(q):
i += 1
A[i], A[j] = A[j], A[i]
A[i+1], A[right] = A[right], A[i+1]
return i+1
def solve(miners, mines, n):
# qsort(miners, 0, n - 1)
# qsort(mines, 0, n - 1)
miners.sort()
mines.sort()
total = 0
for i in range(n):
total += distance(miners[i], mines[i])
print(total)
def distance(miner, mine):
return (miner * miner + mine * mine) ** (1/2)
# import random
# import time
# start = time.time()
# A = [random.randint(0, 1000000) for _ in range(100000)]
# B = [random.randint(0, 1000000) for _ in range(100000)]
# solve(A, B, 100000)
# print(time.time() - start)
if __name__ == "__main__":
t = int(input())
for _ in range(t):
n = int(input())
mines = [0 for _ in range(n)]
miners = [0 for _ in range(n)]
mines_count = 0
miners_count = 0
for _ in range(2*n):
coords = [int(x) for x in input().split()]
if coords[0] == 0:
mines[mines_count] = coords[1]
mines_count += 1
else:
miners[miners_count] = coords[0]
miners_count += 1
solve(miners, mines, n)
```
No
| 107,264 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Diamond Miner is a game that is similar to Gold Miner, but there are n miners instead of 1 in this game.
The mining area can be described as a plane. The n miners can be regarded as n points on the y-axis. There are n diamond mines in the mining area. We can regard them as n points on the x-axis. For some reason, no miners or diamond mines can be at the origin (point (0, 0)).
Every miner should mine exactly one diamond mine. Every miner has a hook, which can be used to mine a diamond mine. If a miner at the point (a,b) uses his hook to mine a diamond mine at the point (c,d), he will spend β{(a-c)^2+(b-d)^2} energy to mine it (the distance between these points). The miners can't move or help each other.
The object of this game is to minimize the sum of the energy that miners spend. Can you find this minimum?
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 10) β the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of miners and mines.
Each of the next 2n lines contains two space-separated integers x (-10^8 β€ x β€ 10^8) and y (-10^8 β€ y β€ 10^8), which represent the point (x,y) to describe a miner's or a diamond mine's position. Either x = 0, meaning there is a miner at the point (0, y), or y = 0, meaning there is a diamond mine at the point (x, 0). There can be multiple miners or diamond mines at the same point.
It is guaranteed that no point is at the origin. It is guaranteed that the number of points on the x-axis is equal to n and the number of points on the y-axis is equal to n.
It's guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, print a single real number β the minimal sum of energy that should be spent.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-9}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} β€ 10^{-9}.
Example
Input
3
2
0 1
1 0
0 -1
-2 0
4
1 0
3 0
-5 0
6 0
0 3
0 1
0 2
0 4
5
3 0
0 4
0 -3
4 0
2 0
1 0
-3 0
0 -10
0 -2
0 -10
Output
3.650281539872885
18.061819283610362
32.052255376143336
Note
In the first test case, the miners are at (0,1) and (0,-1), while the diamond mines are at (1,0) and (-2,0). If you arrange the miners to get the diamond mines in the way, shown in the picture, you can get the sum of the energy β2 + β5.
<image>
Submitted Solution:
```
import math
test = int(input())
for _ in range(test):
n = int(input())
miners = []
mines = []
for i in range(2*n):
x, y = map(int, input().split())
if x == 0:
miners.append([x,y])
elif y == 0:
mines.append([x,y])
result = []
for i in miners:
energy = []
for j in mines:
energy.append([math.sqrt(((i[0]-j[0])**2)+((i[1]-j[1])**2)), mines.index(j)])
small = min(energy)
result.append(small[0])
mines.pop(small[1])
print(sum(result))
```
No
| 107,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Diamond Miner is a game that is similar to Gold Miner, but there are n miners instead of 1 in this game.
The mining area can be described as a plane. The n miners can be regarded as n points on the y-axis. There are n diamond mines in the mining area. We can regard them as n points on the x-axis. For some reason, no miners or diamond mines can be at the origin (point (0, 0)).
Every miner should mine exactly one diamond mine. Every miner has a hook, which can be used to mine a diamond mine. If a miner at the point (a,b) uses his hook to mine a diamond mine at the point (c,d), he will spend β{(a-c)^2+(b-d)^2} energy to mine it (the distance between these points). The miners can't move or help each other.
The object of this game is to minimize the sum of the energy that miners spend. Can you find this minimum?
Input
The input consists of multiple test cases. The first line contains a single integer t (1β€ tβ€ 10) β the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of miners and mines.
Each of the next 2n lines contains two space-separated integers x (-10^8 β€ x β€ 10^8) and y (-10^8 β€ y β€ 10^8), which represent the point (x,y) to describe a miner's or a diamond mine's position. Either x = 0, meaning there is a miner at the point (0, y), or y = 0, meaning there is a diamond mine at the point (x, 0). There can be multiple miners or diamond mines at the same point.
It is guaranteed that no point is at the origin. It is guaranteed that the number of points on the x-axis is equal to n and the number of points on the y-axis is equal to n.
It's guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, print a single real number β the minimal sum of energy that should be spent.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-9}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} β€ 10^{-9}.
Example
Input
3
2
0 1
1 0
0 -1
-2 0
4
1 0
3 0
-5 0
6 0
0 3
0 1
0 2
0 4
5
3 0
0 4
0 -3
4 0
2 0
1 0
-3 0
0 -10
0 -2
0 -10
Output
3.650281539872885
18.061819283610362
32.052255376143336
Note
In the first test case, the miners are at (0,1) and (0,-1), while the diamond mines are at (1,0) and (-2,0). If you arrange the miners to get the diamond mines in the way, shown in the picture, you can get the sum of the energy β2 + β5.
<image>
Submitted Solution:
```
from sys import stdout,stdin
from math import sqrt
def Solve() :
result = 0.0
n = int(input())
while n > 0 :
x1,x2 = stdin.readline().split()
y1,y2 = stdin.readline().split()
result += sqrt(pow(abs( int(x1)-int(y1)),2) + pow(abs( int(x2)-int(y2)),2))
n-=1
stdout.write(str(result)+"\n")
if __name__ == "__main__":
T = stdin.readline()
for i in range(int(T)) :
Solve()
```
No
| 107,266 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Reforms have started in Berland again! At this time, the Parliament is discussing the reform of the calendar. To make the lives of citizens of Berland more varied, it was decided to change the calendar. As more and more people are complaining that "the years fly by...", it was decided that starting from the next year the number of days per year will begin to grow. So the coming year will have exactly a days, the next after coming year will have a + 1 days, the next one will have a + 2 days and so on. This schedule is planned for the coming n years (in the n-th year the length of the year will be equal a + n - 1 day).
No one has yet decided what will become of months. An MP Palevny made the following proposal.
* The calendar for each month is comfortable to be printed on a square sheet of paper. We are proposed to make the number of days in each month be the square of some integer. The number of days per month should be the same for each month of any year, but may be different for different years.
* The number of days in each year must be divisible by the number of days per month in this year. This rule ensures that the number of months in each year is an integer.
* The number of days per month for each year must be chosen so as to save the maximum amount of paper to print the calendars. In other words, the number of days per month should be as much as possible.
These rules provide an unambiguous method for choosing the number of days in each month for any given year length. For example, according to Palevny's proposition, a year that consists of 108 days will have three months, 36 days each. The year that consists of 99 days will have 11 months, 9 days each, and a year of 365 days will have 365 months, one day each.
The proposal provoked heated discussion in the community, the famous mathematician Perelmanov quickly calculated that if the proposal is supported, then in a period of n years, beginning with the year that has a days, the country will spend p sheets of paper to print a set of calendars for these years. Perelmanov's calculations take into account the fact that the set will contain one calendar for each year and each month will be printed on a separate sheet.
Repeat Perelmanov's achievement and print the required number p. You are given positive integers a and n. Perelmanov warns you that your program should not work longer than four seconds at the maximum test.
Input
The only input line contains a pair of integers a, n (1 β€ a, n β€ 107; a + n - 1 β€ 107).
Output
Print the required number p.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use cin, cout streams or the %I64d specifier.
Examples
Input
25 3
Output
30
Input
50 5
Output
125
Note
A note to the first sample test. A year of 25 days will consist of one month containing 25 days. A year of 26 days will consist of 26 months, one day each. A year of 27 days will have three months, 9 days each.
Tags: *special, number theory
Correct Solution:
```
F = {}
def f(k):
if not k in F:
s, i, j = 0, 4, 4
while i <= k:
s += i * f(k // i)
i += j + 1
j += 2
F[k] = (k * (k + 1)) // 2 - s
return F[k]
def g(k):
s, i, j = 0, 4, 4
while i <= k:
s += (i - 1) * f(k // i)
i += j + 1
j += 2
return (k * (k + 1)) // 2 - s
a, n = map(int, input().split())
print(g(a + n - 1) - g(a - 1))
```
| 107,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reforms have started in Berland again! At this time, the Parliament is discussing the reform of the calendar. To make the lives of citizens of Berland more varied, it was decided to change the calendar. As more and more people are complaining that "the years fly by...", it was decided that starting from the next year the number of days per year will begin to grow. So the coming year will have exactly a days, the next after coming year will have a + 1 days, the next one will have a + 2 days and so on. This schedule is planned for the coming n years (in the n-th year the length of the year will be equal a + n - 1 day).
No one has yet decided what will become of months. An MP Palevny made the following proposal.
* The calendar for each month is comfortable to be printed on a square sheet of paper. We are proposed to make the number of days in each month be the square of some integer. The number of days per month should be the same for each month of any year, but may be different for different years.
* The number of days in each year must be divisible by the number of days per month in this year. This rule ensures that the number of months in each year is an integer.
* The number of days per month for each year must be chosen so as to save the maximum amount of paper to print the calendars. In other words, the number of days per month should be as much as possible.
These rules provide an unambiguous method for choosing the number of days in each month for any given year length. For example, according to Palevny's proposition, a year that consists of 108 days will have three months, 36 days each. The year that consists of 99 days will have 11 months, 9 days each, and a year of 365 days will have 365 months, one day each.
The proposal provoked heated discussion in the community, the famous mathematician Perelmanov quickly calculated that if the proposal is supported, then in a period of n years, beginning with the year that has a days, the country will spend p sheets of paper to print a set of calendars for these years. Perelmanov's calculations take into account the fact that the set will contain one calendar for each year and each month will be printed on a separate sheet.
Repeat Perelmanov's achievement and print the required number p. You are given positive integers a and n. Perelmanov warns you that your program should not work longer than four seconds at the maximum test.
Input
The only input line contains a pair of integers a, n (1 β€ a, n β€ 107; a + n - 1 β€ 107).
Output
Print the required number p.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use cin, cout streams or the %I64d specifier.
Examples
Input
25 3
Output
30
Input
50 5
Output
125
Note
A note to the first sample test. A year of 25 days will consist of one month containing 25 days. A year of 26 days will consist of 26 months, one day each. A year of 27 days will have three months, 9 days each.
Submitted Solution:
```
a,n=[int(i) for i in input().split()]
i=1
soma=0
respostas=dict()
while i*i<(a+n):
ii=i*i
j=ii
while j<(a+n):
respostas[j]=ii
j+=ii
i+=1
i=a
while i<a+n:
soma+=i/respostas[i]
i+=1
print(soma)
```
No
| 107,268 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You know that the Martians use a number system with base k. Digit b (0 β€ b < k) is considered lucky, as the first contact between the Martians and the Earthlings occurred in year b (by Martian chronology).
A digital root d(x) of number x is a number that consists of a single digit, resulting after cascading summing of all digits of number x. Word "cascading" means that if the first summing gives us a number that consists of several digits, then we sum up all digits again, and again, until we get a one digit number.
For example, d(35047) = d((3 + 5 + 0 + 4)7) = d(157) = d((1 + 5)7) = d(67) = 67. In this sample the calculations are performed in the 7-base notation.
If a number's digital root equals b, the Martians also call this number lucky.
You have string s, which consists of n digits in the k-base notation system. Your task is to find, how many distinct substrings of the given string are lucky numbers. Leading zeroes are permitted in the numbers.
Note that substring s[i... j] of the string s = a1a2... an (1 β€ i β€ j β€ n) is the string aiai + 1... aj. Two substrings s[i1... j1] and s[i2... j2] of the string s are different if either i1 β i2 or j1 β j2.
Input
The first line contains three integers k, b and n (2 β€ k β€ 109, 0 β€ b < k, 1 β€ n β€ 105).
The second line contains string s as a sequence of n integers, representing digits in the k-base notation: the i-th integer equals ai (0 β€ ai < k) β the i-th digit of string s. The numbers in the lines are space-separated.
Output
Print a single integer β the number of substrings that are lucky numbers.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
10 5 6
3 2 0 5 6 1
Output
5
Input
7 6 4
3 5 0 4
Output
1
Input
257 0 3
0 0 256
Output
3
Note
In the first sample the following substrings have the sought digital root: s[1... 2] = "3 2", s[1... 3] = "3 2 0", s[3... 4] = "0 5", s[4... 4] = "5" and s[2... 6] = "2 0 5 6 1".
Tags: math, number theory
Correct Solution:
```
Line1 = list(map(int,input().split()))
List = list(map(int,input().split()))
def MartianLuck(k,b,n,digit_list):
if b == 0:
return Zero(n,digit_list)
Subarray = dict()
Subarray[0]=1
LuckyNumbersCounter = 0
ActualNumber = 0
for Digit in digit_list :
ActualNumber = (ActualNumber + Digit) % (k-1)
Diference = (ActualNumber - b) % (k-1)
LuckyNumbersCounter += Subarray.get(Diference,0)
Subarray[ActualNumber] = Subarray.get(ActualNumber,0) + 1
if b == k-1:
return LuckyNumbersCounter - Zero(n,digit_list)
return LuckyNumbersCounter
def Zero(n,digit_list):
Digit_index = 0
LuckyZeroNumbersCounter =0
while Digit_index < n:
count = 0
while Digit_index + count <n and digit_list[Digit_index + count] == 0:
count += 1
LuckyZeroNumbersCounter += count*(count +1) //2
Digit_index += count + 1
return LuckyZeroNumbersCounter
print(MartianLuck(Line1[0],Line1[1],Line1[2],List))
```
| 107,269 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You know that the Martians use a number system with base k. Digit b (0 β€ b < k) is considered lucky, as the first contact between the Martians and the Earthlings occurred in year b (by Martian chronology).
A digital root d(x) of number x is a number that consists of a single digit, resulting after cascading summing of all digits of number x. Word "cascading" means that if the first summing gives us a number that consists of several digits, then we sum up all digits again, and again, until we get a one digit number.
For example, d(35047) = d((3 + 5 + 0 + 4)7) = d(157) = d((1 + 5)7) = d(67) = 67. In this sample the calculations are performed in the 7-base notation.
If a number's digital root equals b, the Martians also call this number lucky.
You have string s, which consists of n digits in the k-base notation system. Your task is to find, how many distinct substrings of the given string are lucky numbers. Leading zeroes are permitted in the numbers.
Note that substring s[i... j] of the string s = a1a2... an (1 β€ i β€ j β€ n) is the string aiai + 1... aj. Two substrings s[i1... j1] and s[i2... j2] of the string s are different if either i1 β i2 or j1 β j2.
Input
The first line contains three integers k, b and n (2 β€ k β€ 109, 0 β€ b < k, 1 β€ n β€ 105).
The second line contains string s as a sequence of n integers, representing digits in the k-base notation: the i-th integer equals ai (0 β€ ai < k) β the i-th digit of string s. The numbers in the lines are space-separated.
Output
Print a single integer β the number of substrings that are lucky numbers.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
10 5 6
3 2 0 5 6 1
Output
5
Input
7 6 4
3 5 0 4
Output
1
Input
257 0 3
0 0 256
Output
3
Note
In the first sample the following substrings have the sought digital root: s[1... 2] = "3 2", s[1... 3] = "3 2 0", s[3... 4] = "0 5", s[4... 4] = "5" and s[2... 6] = "2 0 5 6 1".
Tags: math, number theory
Correct Solution:
```
k, b, n = map(int, input().split())
digits = list(map(int, input().split()))
def ans0():
j = -1
answer = 0
for i in range(n):
if digits[i] != 0 or i < j:
continue
j = i
while j < n and digits[j] == 0:
j += 1
r = j - i
answer += r * (r + 1) // 2
return answer
if b == 0:
print(ans0())
else:
count = dict()
count[0] = 1
pref_sum = 0
answer = 0
if b == k - 1:
b = 0
answer -= ans0()
for d in digits:
pref_sum = (pref_sum + d) % (k - 1)
need = (pref_sum - b) % (k - 1)
answer += count.get(need, 0)
count[pref_sum] = count.get(pref_sum, 0) + 1
print(answer)
```
| 107,270 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You know that the Martians use a number system with base k. Digit b (0 β€ b < k) is considered lucky, as the first contact between the Martians and the Earthlings occurred in year b (by Martian chronology).
A digital root d(x) of number x is a number that consists of a single digit, resulting after cascading summing of all digits of number x. Word "cascading" means that if the first summing gives us a number that consists of several digits, then we sum up all digits again, and again, until we get a one digit number.
For example, d(35047) = d((3 + 5 + 0 + 4)7) = d(157) = d((1 + 5)7) = d(67) = 67. In this sample the calculations are performed in the 7-base notation.
If a number's digital root equals b, the Martians also call this number lucky.
You have string s, which consists of n digits in the k-base notation system. Your task is to find, how many distinct substrings of the given string are lucky numbers. Leading zeroes are permitted in the numbers.
Note that substring s[i... j] of the string s = a1a2... an (1 β€ i β€ j β€ n) is the string aiai + 1... aj. Two substrings s[i1... j1] and s[i2... j2] of the string s are different if either i1 β i2 or j1 β j2.
Input
The first line contains three integers k, b and n (2 β€ k β€ 109, 0 β€ b < k, 1 β€ n β€ 105).
The second line contains string s as a sequence of n integers, representing digits in the k-base notation: the i-th integer equals ai (0 β€ ai < k) β the i-th digit of string s. The numbers in the lines are space-separated.
Output
Print a single integer β the number of substrings that are lucky numbers.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
10 5 6
3 2 0 5 6 1
Output
5
Input
7 6 4
3 5 0 4
Output
1
Input
257 0 3
0 0 256
Output
3
Note
In the first sample the following substrings have the sought digital root: s[1... 2] = "3 2", s[1... 3] = "3 2 0", s[3... 4] = "0 5", s[4... 4] = "5" and s[2... 6] = "2 0 5 6 1".
Submitted Solution:
```
Line1 = list(map(int,input().split()))
List = list(map(int,input().split()))
def MartianLuck2(k,b,n,digit_list):
reminder = b % (k-1)
LuckyNumbersCounter = 0
ActualNumber = 0
Count = 0
while Count < n :
for digit_index in range(Count,n):
ActualNumber = (ActualNumber + digit_list[digit_index]) % (k-1)
if b == 0:
if ActualNumber == 0:
LuckyNumbersCounter += 1
elif (ActualNumber == reminder and not ((digit_index - Count) == 0 and ActualNumber == 0)):
LuckyNumbersCounter +=1
ActualNumber = 0
Count +=1
return LuckyNumbersCounter
print(MartianLuck2(Line1[0],Line1[1],Line1[2],List))
```
No
| 107,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You know that the Martians use a number system with base k. Digit b (0 β€ b < k) is considered lucky, as the first contact between the Martians and the Earthlings occurred in year b (by Martian chronology).
A digital root d(x) of number x is a number that consists of a single digit, resulting after cascading summing of all digits of number x. Word "cascading" means that if the first summing gives us a number that consists of several digits, then we sum up all digits again, and again, until we get a one digit number.
For example, d(35047) = d((3 + 5 + 0 + 4)7) = d(157) = d((1 + 5)7) = d(67) = 67. In this sample the calculations are performed in the 7-base notation.
If a number's digital root equals b, the Martians also call this number lucky.
You have string s, which consists of n digits in the k-base notation system. Your task is to find, how many distinct substrings of the given string are lucky numbers. Leading zeroes are permitted in the numbers.
Note that substring s[i... j] of the string s = a1a2... an (1 β€ i β€ j β€ n) is the string aiai + 1... aj. Two substrings s[i1... j1] and s[i2... j2] of the string s are different if either i1 β i2 or j1 β j2.
Input
The first line contains three integers k, b and n (2 β€ k β€ 109, 0 β€ b < k, 1 β€ n β€ 105).
The second line contains string s as a sequence of n integers, representing digits in the k-base notation: the i-th integer equals ai (0 β€ ai < k) β the i-th digit of string s. The numbers in the lines are space-separated.
Output
Print a single integer β the number of substrings that are lucky numbers.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
10 5 6
3 2 0 5 6 1
Output
5
Input
7 6 4
3 5 0 4
Output
1
Input
257 0 3
0 0 256
Output
3
Note
In the first sample the following substrings have the sought digital root: s[1... 2] = "3 2", s[1... 3] = "3 2 0", s[3... 4] = "0 5", s[4... 4] = "5" and s[2... 6] = "2 0 5 6 1".
Submitted Solution:
```
k, b, n = map(int, input().split())
digits = list(map(int, input().split()))
def conv(t):
if t == 0:
return k - 1
return t
if b == 0:
j = -1
answer = 0
for i in range(n):
if digits[i] != 0 or i < j:
continue
j = i
while j < n and digits[j] == 0:
j += 1
r = j - i
answer += r * (r + 1) // 2
print(answer)
else:
count = dict()
count[0] = 1
pref_sum = 0
answer = 0
while len(digits) > 0 and digits[0] == 0:
digits.pop(0)
zeroes = 0
for d in digits:
pref_sum = (pref_sum + d) % (k - 1)
real_val = conv(pref_sum)
need = (real_val - b) % (k - 1)
# print(pref_sum, need)
answer += count.get(need, 0)
if need == 0:
answer -= zeroes
if d == 0:
zeroes += 1
count[pref_sum] = count.get(pref_sum, 0) + 1
print(answer)
```
No
| 107,272 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You know that the Martians use a number system with base k. Digit b (0 β€ b < k) is considered lucky, as the first contact between the Martians and the Earthlings occurred in year b (by Martian chronology).
A digital root d(x) of number x is a number that consists of a single digit, resulting after cascading summing of all digits of number x. Word "cascading" means that if the first summing gives us a number that consists of several digits, then we sum up all digits again, and again, until we get a one digit number.
For example, d(35047) = d((3 + 5 + 0 + 4)7) = d(157) = d((1 + 5)7) = d(67) = 67. In this sample the calculations are performed in the 7-base notation.
If a number's digital root equals b, the Martians also call this number lucky.
You have string s, which consists of n digits in the k-base notation system. Your task is to find, how many distinct substrings of the given string are lucky numbers. Leading zeroes are permitted in the numbers.
Note that substring s[i... j] of the string s = a1a2... an (1 β€ i β€ j β€ n) is the string aiai + 1... aj. Two substrings s[i1... j1] and s[i2... j2] of the string s are different if either i1 β i2 or j1 β j2.
Input
The first line contains three integers k, b and n (2 β€ k β€ 109, 0 β€ b < k, 1 β€ n β€ 105).
The second line contains string s as a sequence of n integers, representing digits in the k-base notation: the i-th integer equals ai (0 β€ ai < k) β the i-th digit of string s. The numbers in the lines are space-separated.
Output
Print a single integer β the number of substrings that are lucky numbers.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
10 5 6
3 2 0 5 6 1
Output
5
Input
7 6 4
3 5 0 4
Output
1
Input
257 0 3
0 0 256
Output
3
Note
In the first sample the following substrings have the sought digital root: s[1... 2] = "3 2", s[1... 3] = "3 2 0", s[3... 4] = "0 5", s[4... 4] = "5" and s[2... 6] = "2 0 5 6 1".
Submitted Solution:
```
Line1 = list(map(int,input().split()))
List = list(input().split())
def MartianLuck(k,b,n,digit_list):
LuckyNumbersCounter = 0
ActualNumber = ""
Count = 0
while Count < n :
for digit_index in range(Count,n):
ActualNumber += digit_list[digit_index]
if int(ActualNumber) % (k-1) == b :
LuckyNumbersCounter +=1
ActualNumber = ""
Count +=1
return LuckyNumbersCounter
print(MartianLuck(Line1[0],Line1[1],Line1[2],List))
```
No
| 107,273 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You know that the Martians use a number system with base k. Digit b (0 β€ b < k) is considered lucky, as the first contact between the Martians and the Earthlings occurred in year b (by Martian chronology).
A digital root d(x) of number x is a number that consists of a single digit, resulting after cascading summing of all digits of number x. Word "cascading" means that if the first summing gives us a number that consists of several digits, then we sum up all digits again, and again, until we get a one digit number.
For example, d(35047) = d((3 + 5 + 0 + 4)7) = d(157) = d((1 + 5)7) = d(67) = 67. In this sample the calculations are performed in the 7-base notation.
If a number's digital root equals b, the Martians also call this number lucky.
You have string s, which consists of n digits in the k-base notation system. Your task is to find, how many distinct substrings of the given string are lucky numbers. Leading zeroes are permitted in the numbers.
Note that substring s[i... j] of the string s = a1a2... an (1 β€ i β€ j β€ n) is the string aiai + 1... aj. Two substrings s[i1... j1] and s[i2... j2] of the string s are different if either i1 β i2 or j1 β j2.
Input
The first line contains three integers k, b and n (2 β€ k β€ 109, 0 β€ b < k, 1 β€ n β€ 105).
The second line contains string s as a sequence of n integers, representing digits in the k-base notation: the i-th integer equals ai (0 β€ ai < k) β the i-th digit of string s. The numbers in the lines are space-separated.
Output
Print a single integer β the number of substrings that are lucky numbers.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
10 5 6
3 2 0 5 6 1
Output
5
Input
7 6 4
3 5 0 4
Output
1
Input
257 0 3
0 0 256
Output
3
Note
In the first sample the following substrings have the sought digital root: s[1... 2] = "3 2", s[1... 3] = "3 2 0", s[3... 4] = "0 5", s[4... 4] = "5" and s[2... 6] = "2 0 5 6 1".
Submitted Solution:
```
Line1 = list(map(int,input().split()))
List = list(map(int,input().split()))
def MartianLuck2(k,b,n,digit_list):
Subarray = dict()
Subarray[0]=1
reminder = b % (k-1)
LuckyNumbersCounter = 0
ActualNumber = 0
Digit_index = 0
while Digit_index < n :
if b == 0:
count = 0
while digit_list[Digit_index + count] == 0:
count += 1
LuckyNumbersCounter += count*(count +1) //2
Digit_index += count
else:
ActualNumber = (ActualNumber + digit_list[Digit_index]) % (k-1)
Diference = (ActualNumber - b) % (k-1)
if b == k-1 and digit_list[Digit_index]==0 and Diference == 0 and LuckyNumbersCounter > 0:
LuckyNumbersCounter+=1
elif not (b==k-1 and digit_list[Digit_index]==0):
LuckyNumbersCounter += Subarray.get(Diference,0)
Subarray[ActualNumber] = Subarray.get(ActualNumber,0) + 1
Digit_index +=1
return LuckyNumbersCounter
print(MartianLuck2(Line1[0],Line1[1],Line1[2],List))
```
No
| 107,274 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Bob invented a new game with a tree (we should remind you, that a tree is a connected graph without cycles): he deletes any (possibly, zero) amount of edges of the tree, and counts the product of sizes of the connected components left after the deletion. Your task is to find out the maximum number that Bob can get in his new game for a given tree.
Input
The first input line contains integer number n (1 β€ n β€ 700) β amount of vertices in the tree. The following n - 1 lines contain the description of the edges. Each line contains the pair of vertices' indexes, joined by an edge, ai, bi (1 β€ ai, bi β€ n). It's guaranteed that the graph described in the input is a tree.
Output
Output the only number β the maximum product of sizes of the connected components, that Bob can get after deleting some of the tree's edges.
Examples
Input
5
1 2
2 3
3 4
4 5
Output
6
Input
8
1 2
1 3
2 4
2 5
3 6
3 7
6 8
Output
18
Input
3
1 2
1 3
Output
3
Tags: dp
Correct Solution:
```
def dfs(v, pr = -1):
dp[v][1] = 1
sz[v] = 1
for to in g[v]:
if to == pr:
continue
dfs(to, v)
for x in range(sz[v], 0, -1):
for y in range(sz[to], -1, -1):
#print(v, x, y, dp[v][x], dp[to][y])
dp[v][x + y] = max(dp[v][x + y], dp[v][x] * dp[to][y])
sz[v] += sz[to]
for i in range(1, sz[v] + 1):
dp[v][0] = max(dp[v][0], dp[v][i] * i)
n = int(input())
sz = [0 for i in range(n)]
g = [[] for i in range(n)]
dp = [[0 for j in range(n + 1)] for i in range(n + 1)]
for i in range(n - 1):
x, y = map(int, input().split())
x -= 1
y -= 1
g[x].append(y)
g[y].append(x)
dfs(0)
print(dp[0][0])
```
| 107,275 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Bob invented a new game with a tree (we should remind you, that a tree is a connected graph without cycles): he deletes any (possibly, zero) amount of edges of the tree, and counts the product of sizes of the connected components left after the deletion. Your task is to find out the maximum number that Bob can get in his new game for a given tree.
Input
The first input line contains integer number n (1 β€ n β€ 700) β amount of vertices in the tree. The following n - 1 lines contain the description of the edges. Each line contains the pair of vertices' indexes, joined by an edge, ai, bi (1 β€ ai, bi β€ n). It's guaranteed that the graph described in the input is a tree.
Output
Output the only number β the maximum product of sizes of the connected components, that Bob can get after deleting some of the tree's edges.
Examples
Input
5
1 2
2 3
3 4
4 5
Output
6
Input
8
1 2
1 3
2 4
2 5
3 6
3 7
6 8
Output
18
Input
3
1 2
1 3
Output
3
Tags: dp
Correct Solution:
```
from fractions import Fraction
n = int(input())
adj = [list() for x in range(n)]
H = [0] * n
F = [0] * n
FoH = [list() for x in range(n)]
sz = 0
order = [0] * n
pi = [-1] * n
def dfs(u, p = -1):
global pi, order, sz
pi[u] = p
for v in adj[u]:
if v != p:
dfs(v, u)
order[sz] = u
sz += 1
T1 = [0] * n
T2 = [0] * n
T3 = [0] * n
def solve(u, p = -1):
global H, F, FoH
F[u] = 1
for v in adj[u]:
if v != p:
F[u] *= H[v]
FoH[u].append(Fraction(F[v], H[v]))
ans = F[u]
FoH[u].sort()
FoH[u].reverse()
pd = 1
s = 0
for x in FoH[u]:
pd *= x
s += 1
ans = max(ans, int(pd * F[u]) * (s+1))
for v in adj[u]:
if v != p:
pd = 1
s = 0
for x in FoH[v]:
pd *= x
s += 1
ans = max(ans, int(pd * F[u] * F[v]) // H[v] * (s+2))
#print(u+1, ':', FoH[u], ans)
H[u] = ans
for i in range(1, n):
u, v = [int(x) for x in input().split()]
u -= 1
v -= 1
adj[u].append(v)
adj[v].append(u)
dfs(0)
for x in order:
solve(x, pi[x])
print(H[0])
```
| 107,276 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Bob invented a new game with a tree (we should remind you, that a tree is a connected graph without cycles): he deletes any (possibly, zero) amount of edges of the tree, and counts the product of sizes of the connected components left after the deletion. Your task is to find out the maximum number that Bob can get in his new game for a given tree.
Input
The first input line contains integer number n (1 β€ n β€ 700) β amount of vertices in the tree. The following n - 1 lines contain the description of the edges. Each line contains the pair of vertices' indexes, joined by an edge, ai, bi (1 β€ ai, bi β€ n). It's guaranteed that the graph described in the input is a tree.
Output
Output the only number β the maximum product of sizes of the connected components, that Bob can get after deleting some of the tree's edges.
Examples
Input
5
1 2
2 3
3 4
4 5
Output
6
Input
8
1 2
1 3
2 4
2 5
3 6
3 7
6 8
Output
18
Input
3
1 2
1 3
Output
3
Tags: dp
Correct Solution:
```
n=int(input())
w=[[] for i in range(n+1)]
sz=[0]*(n+1)
f=[[0]*(n+1) for i in range(n+1)]
def dfs(u,p):
f[u][1]=sz[u]=1
for v in w[u]:
if v!=p:
dfs(v,u)
for j in range(sz[u],-1,-1):
for k in range(sz[v],-1,-1):f[u][j+k]=max(f[u][j+k],f[u][j]*f[v][k])
sz[u]+=sz[v]
for i in range(1,sz[u]+1):f[u][0]=max(f[u][0],f[u][i]*i)
for i in range(n-1):
x,y=map(int,input().split())
w[x]+=[y]
w[y]+=[x]
dfs(1,0)
print(f[1][0])
```
| 107,277 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Bob invented a new game with a tree (we should remind you, that a tree is a connected graph without cycles): he deletes any (possibly, zero) amount of edges of the tree, and counts the product of sizes of the connected components left after the deletion. Your task is to find out the maximum number that Bob can get in his new game for a given tree.
Input
The first input line contains integer number n (1 β€ n β€ 700) β amount of vertices in the tree. The following n - 1 lines contain the description of the edges. Each line contains the pair of vertices' indexes, joined by an edge, ai, bi (1 β€ ai, bi β€ n). It's guaranteed that the graph described in the input is a tree.
Output
Output the only number β the maximum product of sizes of the connected components, that Bob can get after deleting some of the tree's edges.
Examples
Input
5
1 2
2 3
3 4
4 5
Output
6
Input
8
1 2
1 3
2 4
2 5
3 6
3 7
6 8
Output
18
Input
3
1 2
1 3
Output
3
Tags: dp
Correct Solution:
```
maxn = 710
to = [[] for i in range(maxn)]
dp = [[0 for i in range(maxn)] for i in range(maxn)]
size = [0 for i in range(maxn)]
n = int(input())
def dfs(now, f):
size[now] = 1
dp[now][1] = 1
for i in to[now]:
if i != f:
dfs(i, now)
size[now] += size[i]
for j in reversed(range(0, size[now] + 1)):
for k in range(max(0, j - size[now] + size[i]), min(j, size[i]) + 1):
dp[now][j] = max(dp[now][j], dp[now][j - k] * dp[i][k])
for i in range(1, n + 1):
dp[now][0] = max(dp[now][0], dp[now][i] * i)
for i in range(1, n):
u, v = map(int, input().split(' '))
to[u].append(v)
to[v].append(u)
dfs(1, 0)
print(dp[1][0])
```
| 107,278 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Bob invented a new game with a tree (we should remind you, that a tree is a connected graph without cycles): he deletes any (possibly, zero) amount of edges of the tree, and counts the product of sizes of the connected components left after the deletion. Your task is to find out the maximum number that Bob can get in his new game for a given tree.
Input
The first input line contains integer number n (1 β€ n β€ 700) β amount of vertices in the tree. The following n - 1 lines contain the description of the edges. Each line contains the pair of vertices' indexes, joined by an edge, ai, bi (1 β€ ai, bi β€ n). It's guaranteed that the graph described in the input is a tree.
Output
Output the only number β the maximum product of sizes of the connected components, that Bob can get after deleting some of the tree's edges.
Examples
Input
5
1 2
2 3
3 4
4 5
Output
6
Input
8
1 2
1 3
2 4
2 5
3 6
3 7
6 8
Output
18
Input
3
1 2
1 3
Output
3
Tags: dp
Correct Solution:
```
n = int(input())
edges = [[] for i in range(n)]
ch = [[] for i in range(n)]
for i in range(n - 1):
a, b = map(int, input().split())
a -= 1
b -= 1
edges[a].append(b)
edges[b].append(a)
dp = [1 for i in range(n)]
prod = [1 for i in range(n)]
def dfs(v, pr):
prod[v] = 1
for to in edges[v]:
if to != pr:
dfs(to, v)
ch[v].append(to)
prod[v] *= dp[to]
ch[v].sort(key = lambda i : dp[i] / prod[i])
dp[v] = prod[v]
now = prod[v]
cnt = 1
for to in ch[v]:
cnt += 1
now //= dp[to]
now *= prod[to]
dp[v] = max(dp[v], now * cnt)
for to in ch[v]:
now = prod[v] // dp[to] * prod[to]
cnt = 2
for gr in ch[to]:
cnt += 1
now //= dp[gr]
now *= prod[gr]
dp[v] = max(dp[v], now * cnt)
dfs(0, -1)
print(dp[0])
```
| 107,279 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Bob invented a new game with a tree (we should remind you, that a tree is a connected graph without cycles): he deletes any (possibly, zero) amount of edges of the tree, and counts the product of sizes of the connected components left after the deletion. Your task is to find out the maximum number that Bob can get in his new game for a given tree.
Input
The first input line contains integer number n (1 β€ n β€ 700) β amount of vertices in the tree. The following n - 1 lines contain the description of the edges. Each line contains the pair of vertices' indexes, joined by an edge, ai, bi (1 β€ ai, bi β€ n). It's guaranteed that the graph described in the input is a tree.
Output
Output the only number β the maximum product of sizes of the connected components, that Bob can get after deleting some of the tree's edges.
Examples
Input
5
1 2
2 3
3 4
4 5
Output
6
Input
8
1 2
1 3
2 4
2 5
3 6
3 7
6 8
Output
18
Input
3
1 2
1 3
Output
3
Tags: dp
Correct Solution:
```
n=int(input())
sz=[0]*(n+1)
g=[[] for _ in range(n+1)]
dp=[[0]*(n+1) for _ in range(n+1)]
def dfs(u,pr):
sz[u]=dp[u][1]=1
for v in g[u]:
if(v==pr):continue
dfs(v,u)
for i in range(sz[u],-1,-1):
for j in range(sz[v],-1,-1):
dp[u][i+j]=max(dp[u][i+j],dp[u][i]*dp[v][j])
sz[u]+=sz[v]
for i in range(sz[u]+1):
dp[u][0]=max(dp[u][0],dp[u][i]*i);
for _ in range(n-1):
a,b=map(int,input().split())
g[a].append(b)
g[b].append(a)
dfs(1,-1)
print(dp[1][0])
```
| 107,280 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Bob invented a new game with a tree (we should remind you, that a tree is a connected graph without cycles): he deletes any (possibly, zero) amount of edges of the tree, and counts the product of sizes of the connected components left after the deletion. Your task is to find out the maximum number that Bob can get in his new game for a given tree.
Input
The first input line contains integer number n (1 β€ n β€ 700) β amount of vertices in the tree. The following n - 1 lines contain the description of the edges. Each line contains the pair of vertices' indexes, joined by an edge, ai, bi (1 β€ ai, bi β€ n). It's guaranteed that the graph described in the input is a tree.
Output
Output the only number β the maximum product of sizes of the connected components, that Bob can get after deleting some of the tree's edges.
Examples
Input
5
1 2
2 3
3 4
4 5
Output
6
Input
8
1 2
1 3
2 4
2 5
3 6
3 7
6 8
Output
18
Input
3
1 2
1 3
Output
3
Tags: dp
Correct Solution:
```
from fractions import Fraction
n = int(input())
G = [[] for i in range(n)]
for i in range(n-1):
a, b = (int(x)-1 for x in input().split())
G[a].append(b)
G[b].append(a)
f, g, h = [1]*n, [1]*n, [1]*n
def dfs(v, p):
if p != -1:
G[v].remove(p)
for e in G[v]:
dfs(e, v)
h[v] *= f[e]
tmp = []
f[v], g[v] = h[v], h[v]*2
for e in G[v]:
f[v] = max(f[v], h[v] // f[e] * g[e])
tmp.append((f[e], h[e]))
fPart, hPart = h[v], 1
tmp.sort(key=lambda x: Fraction(*x))
for i in range(len(tmp)):
fPart //= tmp[i][0]
hPart *= tmp[i][1]
f[v] = max(f[v], fPart*hPart*(i+2))
g[v] = max(g[v], fPart*hPart*(i+3))
dfs(0, -1)
print(f[0])
```
| 107,281 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Bob invented a new game with a tree (we should remind you, that a tree is a connected graph without cycles): he deletes any (possibly, zero) amount of edges of the tree, and counts the product of sizes of the connected components left after the deletion. Your task is to find out the maximum number that Bob can get in his new game for a given tree.
Input
The first input line contains integer number n (1 β€ n β€ 700) β amount of vertices in the tree. The following n - 1 lines contain the description of the edges. Each line contains the pair of vertices' indexes, joined by an edge, ai, bi (1 β€ ai, bi β€ n). It's guaranteed that the graph described in the input is a tree.
Output
Output the only number β the maximum product of sizes of the connected components, that Bob can get after deleting some of the tree's edges.
Examples
Input
5
1 2
2 3
3 4
4 5
Output
6
Input
8
1 2
1 3
2 4
2 5
3 6
3 7
6 8
Output
18
Input
3
1 2
1 3
Output
3
Tags: dp
Correct Solution:
```
def dfs(u,pa,gr):
szz[u]=1
for i in gr[u]:
if i ==pa:
continue
dfs(i,u,gr)
szz[u]+=szz[i]
first=True
n=szz[u]
temp=[0 for i in range(0,n)]
cs=0
for it in gr[u]:
if it==pa:
continue
if first:
first=False
for i in range(0,szz[it]+1):
temp[i]=dp[it][i]
if i!=0:
temp[i]//=i
cs+=szz[it]
else:
cs+=szz[it]
for i in reversed(range(0,cs+1)):
mx=0
for j in range(max(0,i-cs+szz[it]),min(i,szz[it])+1):
mx1=temp[i-j]*dp[it][j]
if j!=0:
mx1//=j
mx=max(mx,mx1)
temp[i]=mx
if first:
dp[u][0]=1
dp[u][1]=1
else:
for i in range(1,n+1):
dp[u][i]=temp[i-1]*i
dp[u][0]=max(dp[u][0],dp[u][i])
n1 = int(input())
gr=[[]for i in range(0,n1)]
for i in range(0,n1-1):
a,b=input().split()
a=int(a)
b=int(b)
a-=1
b-=1
gr[a].append(b)
gr[b].append(a)
szz=[0 for i in range(0,n1)]
dp= [[0 for i in range(0,n1+1)] for j in range(0,n1+1)]
dfs(0,-1,gr)
print(int(dp[0][0]))
```
| 107,282 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Bob invented a new game with a tree (we should remind you, that a tree is a connected graph without cycles): he deletes any (possibly, zero) amount of edges of the tree, and counts the product of sizes of the connected components left after the deletion. Your task is to find out the maximum number that Bob can get in his new game for a given tree.
Input
The first input line contains integer number n (1 β€ n β€ 700) β amount of vertices in the tree. The following n - 1 lines contain the description of the edges. Each line contains the pair of vertices' indexes, joined by an edge, ai, bi (1 β€ ai, bi β€ n). It's guaranteed that the graph described in the input is a tree.
Output
Output the only number β the maximum product of sizes of the connected components, that Bob can get after deleting some of the tree's edges.
Examples
Input
5
1 2
2 3
3 4
4 5
Output
6
Input
8
1 2
1 3
2 4
2 5
3 6
3 7
6 8
Output
18
Input
3
1 2
1 3
Output
3
Submitted Solution:
```
n = int(input())
node = [[] for i in range(n)]
for i in range(1,n):
a,b = map(int,input().split())
node[a-1].append(b-1)
node[b-1].append(a-1)
dp = [[] for i in range(n)]
def combine(a, b):
c = [0]*(len(a)+len(b)-1)
for i,va in enumerate(a):
for j,vb in enumerate(b):
c[i+j] = max(c[i+j],va*vb)
return c
def dfs(p,par=-1):
dp[p] = [0,1]
for i in node[p]:
if i == par: continue
dfs(i,p)
dp[p] = combine(dp[i],dp[p])
ma = 0
for i,v in enumerate(dp[p]):
ma = max(ma, i*v)
dp[p][0] = ma
dfs(0)
print(dp[0][0])
```
Yes
| 107,283 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Bob invented a new game with a tree (we should remind you, that a tree is a connected graph without cycles): he deletes any (possibly, zero) amount of edges of the tree, and counts the product of sizes of the connected components left after the deletion. Your task is to find out the maximum number that Bob can get in his new game for a given tree.
Input
The first input line contains integer number n (1 β€ n β€ 700) β amount of vertices in the tree. The following n - 1 lines contain the description of the edges. Each line contains the pair of vertices' indexes, joined by an edge, ai, bi (1 β€ ai, bi β€ n). It's guaranteed that the graph described in the input is a tree.
Output
Output the only number β the maximum product of sizes of the connected components, that Bob can get after deleting some of the tree's edges.
Examples
Input
5
1 2
2 3
3 4
4 5
Output
6
Input
8
1 2
1 3
2 4
2 5
3 6
3 7
6 8
Output
18
Input
3
1 2
1 3
Output
3
Submitted Solution:
```
from fractions import Fraction
import sys
#print(sys.getrecursionlimit())
sys.setrecursionlimit(10000)
c = int(input())
edges=dict((i,[]) for i in range(1,c+1))
for i in range(0,c-1):
a,b=tuple(map(int,input().split()))
edges[a].append(b)
edges[b].append(a)
#print(edges.get(1,[]))
#exit()
dp=[None for i in range(c+1)]
def dfs(r,p):
#print(r)
if dp[r] is not None:
return dp[r]
children=filter(lambda x: x != p, edges[r])
cs=[dfs(i,r) for i in children]
#print(r,cs)
cs.sort(key=lambda x:Fraction(x[0],x[1]),reverse=True)
f=1
for c in cs:
f*=c[1]
h=f
k=1
m=f
for c in cs:
m=m//c[1]*c[0]
k+=1
#if m*k==24:
#print("aaa")
h=max(h,m*k)
m=f
for c in cs:
k=2
a=f//c[1]*c[0]
h=max(h,a*k)
for d in c[2]:
a=a//d[1]*d[0]
k+=1
#if a*k==24:
#print("bbb",a,k,c,d)
h=max(h,a*k)
dp[r]=(f,h,cs)
#print(r,dp[r])
return dp[r]
print(dfs(1,0)[1])
```
Yes
| 107,284 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Bob invented a new game with a tree (we should remind you, that a tree is a connected graph without cycles): he deletes any (possibly, zero) amount of edges of the tree, and counts the product of sizes of the connected components left after the deletion. Your task is to find out the maximum number that Bob can get in his new game for a given tree.
Input
The first input line contains integer number n (1 β€ n β€ 700) β amount of vertices in the tree. The following n - 1 lines contain the description of the edges. Each line contains the pair of vertices' indexes, joined by an edge, ai, bi (1 β€ ai, bi β€ n). It's guaranteed that the graph described in the input is a tree.
Output
Output the only number β the maximum product of sizes of the connected components, that Bob can get after deleting some of the tree's edges.
Examples
Input
5
1 2
2 3
3 4
4 5
Output
6
Input
8
1 2
1 3
2 4
2 5
3 6
3 7
6 8
Output
18
Input
3
1 2
1 3
Output
3
Submitted Solution:
```
from fractions import Fraction
import sys
#print(sys.getrecursionlimit())
sys.setrecursionlimit(1000*100)
c = int(input())
edges=dict((i,[]) for i in range(1,c+1))
for i in range(0,c-1):
a,b=tuple(map(int,input().split()))
edges[a].append(b)
edges[b].append(a)
#print(edges.get(1,[]))
#exit()
dp=[None for i in range(c+1)]
def dfs(r,p):
#print(r)
if dp[r] is not None:
return dp[r]
children=filter(lambda x: x != p, edges[r])
cs=[dfs(i,r) for i in children]
#print(r,cs)
cs.sort(key=lambda x:Fraction(x[0],x[1]),reverse=True)
f=1
for c in cs:
f*=c[1]
h=f
k=1
m=f
for c in cs:
m=m//c[1]*c[0]
k+=1
#if m*k==24:
#print("aaa")
h=max(h,m*k)
m=f
for c in cs:
k=2
a=f//c[1]*c[0]
h=max(h,a*k)
for d in c[2]:
a=a//d[1]*d[0]
k+=1
#if a*k==24:
#print("bbb",a,k,c,d)
h=max(h,a*k)
dp[r]=(f,h,cs)
#print(r,dp[r])
return dp[r]
print(dfs(1,0)[1])
```
Yes
| 107,285 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Bob invented a new game with a tree (we should remind you, that a tree is a connected graph without cycles): he deletes any (possibly, zero) amount of edges of the tree, and counts the product of sizes of the connected components left after the deletion. Your task is to find out the maximum number that Bob can get in his new game for a given tree.
Input
The first input line contains integer number n (1 β€ n β€ 700) β amount of vertices in the tree. The following n - 1 lines contain the description of the edges. Each line contains the pair of vertices' indexes, joined by an edge, ai, bi (1 β€ ai, bi β€ n). It's guaranteed that the graph described in the input is a tree.
Output
Output the only number β the maximum product of sizes of the connected components, that Bob can get after deleting some of the tree's edges.
Examples
Input
5
1 2
2 3
3 4
4 5
Output
6
Input
8
1 2
1 3
2 4
2 5
3 6
3 7
6 8
Output
18
Input
3
1 2
1 3
Output
3
Submitted Solution:
```
def run(E, F, G, curr, prev):
for nex in E[curr]:
if nex == prev:
continue
run(E, F, G, nex, curr)
G[curr].append(F[nex])
G[curr].sort(key=lambda x: x[0] / x[1], reverse=True)
F[curr][0] = 1
for i in range(len(G[curr])):
F[curr][0] *= G[curr][i][1]
x = F[curr][0]
for i in range(len(G[curr])):
x = x * G[curr][i][0] // G[curr][i][1]
F[curr][1] = max(F[curr][1], x * (i + 2))
for i in range(len(E[curr])):
p = E[curr][i]
if p == prev:
continue
x = F[curr][0] * F[p][0] // F[p][1]
for j in range(len(G[p])):
x = x * G[p][j][0] // G[p][j][1]
F[curr][1] = max(F[curr][1], x * (j + 3))
n = int(input())
E = [[] for i in range(n)]
for i in range(n - 1):
a, b = map(lambda x: int(x) - 1, input().split())
E[a].append(b)
E[b].append(a)
F = [[1] * 2 for i in range(n)]
G = [[] for i in range(n)]
run(E, F, G, 0, -1)
print(max(F[0]))
```
Yes
| 107,286 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Bob invented a new game with a tree (we should remind you, that a tree is a connected graph without cycles): he deletes any (possibly, zero) amount of edges of the tree, and counts the product of sizes of the connected components left after the deletion. Your task is to find out the maximum number that Bob can get in his new game for a given tree.
Input
The first input line contains integer number n (1 β€ n β€ 700) β amount of vertices in the tree. The following n - 1 lines contain the description of the edges. Each line contains the pair of vertices' indexes, joined by an edge, ai, bi (1 β€ ai, bi β€ n). It's guaranteed that the graph described in the input is a tree.
Output
Output the only number β the maximum product of sizes of the connected components, that Bob can get after deleting some of the tree's edges.
Examples
Input
5
1 2
2 3
3 4
4 5
Output
6
Input
8
1 2
1 3
2 4
2 5
3 6
3 7
6 8
Output
18
Input
3
1 2
1 3
Output
3
Submitted Solution:
```
#fuck python
# 1:root estefade 2:!1
import sys
def dfs(x,par):
a=1
dp2[x]=1
dp1[x][1]=1
ans[x]=1
for i in (g[x]):
if(i!=par):
dfs(i,x)
dp2[x]*=ans[i]
for i in (g[x]):
if(i!=par):
for j in range(1,len(g[i])+2):
ans[x]=max(int((dp2[x]/ans[i]*dp1[i][j]*(j+1))/j),ans[x])
for i in (g[x]):
if(i!=par):
for j in range(len(g[x])+2,0,-1):
dp1[x][j]=dp1[x][j]*ans[i]
dp1[x][j]=max(dp1[x][j],dp1[x][j-1]*dp2[i])
for i in range(len(g[x])+3):
dp1[x][i]=dp1[x][i]*i
ans[x]=max(ans[x],dp1[x][i])
N=709
g=[]
ans=[]
dp1=[]
dp2=[]
for i in range(0,N):
g.append([])
dp1.append([])
dp2.append(0)
ans.append(0)
for i in range(0,N):
for j in range(0,N):
dp1[i].append(0)
n=int(input())
for i in range(1,n):
x,y=map(int,input().split())
g[x].append(y)
g[y].append(x)
dfs(1,0)
print(ans[1])
```
No
| 107,287 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Bob invented a new game with a tree (we should remind you, that a tree is a connected graph without cycles): he deletes any (possibly, zero) amount of edges of the tree, and counts the product of sizes of the connected components left after the deletion. Your task is to find out the maximum number that Bob can get in his new game for a given tree.
Input
The first input line contains integer number n (1 β€ n β€ 700) β amount of vertices in the tree. The following n - 1 lines contain the description of the edges. Each line contains the pair of vertices' indexes, joined by an edge, ai, bi (1 β€ ai, bi β€ n). It's guaranteed that the graph described in the input is a tree.
Output
Output the only number β the maximum product of sizes of the connected components, that Bob can get after deleting some of the tree's edges.
Examples
Input
5
1 2
2 3
3 4
4 5
Output
6
Input
8
1 2
1 3
2 4
2 5
3 6
3 7
6 8
Output
18
Input
3
1 2
1 3
Output
3
Submitted Solution:
```
#fuck python
# 1:root estefade 2:!1
import sys
def dfs(x,par):
a=1
dp2[x]=1
dp1[x][1]=1
ans[x]=1
for i in (g[x]):
if(i!=par):
dfs(i,x)
dp2[x]*=ans[i]
for i in (g[x]):
if(i!=par):
for j in range(1,len(g[i])+2):
if(dp2[x]%ans[i]>0):
print("ey vay")
ans[x]=int(max(dp2[x]/ans[i]*c[i][j]*(j+1),ans[x]))
for i in (g[x]):
if(i!=par):
for j in range(len(g[x])+2,0,-1):
dp1[x][j]=dp1[x][j]*ans[i]
dp1[x][j]=max(dp1[x][j],dp1[x][j-1]*dp2[i])
for i in range(len(g[x])+3):
c[x][i]=dp1[x][i]
dp1[x][i]=dp1[x][i]*i
ans[x]=max(ans[x],dp1[x][i])
N=709
g=[]
ans=[]
dp1=[]
dp2=[]
c=[]
for i in range(0,N):
g.append([])
dp1.append([])
c.append([])
dp2.append(0)
ans.append(0)
for i in range(0,N):
for j in range(0,N):
dp1[i].append(0)
c[i].append(0)
n=int(input())
for i in range(1,n):
x,y=map(int,input().split())
g[x].append(y)
g[y].append(x)
dfs(1,0)
print(int(ans[1]))
```
No
| 107,288 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Bob invented a new game with a tree (we should remind you, that a tree is a connected graph without cycles): he deletes any (possibly, zero) amount of edges of the tree, and counts the product of sizes of the connected components left after the deletion. Your task is to find out the maximum number that Bob can get in his new game for a given tree.
Input
The first input line contains integer number n (1 β€ n β€ 700) β amount of vertices in the tree. The following n - 1 lines contain the description of the edges. Each line contains the pair of vertices' indexes, joined by an edge, ai, bi (1 β€ ai, bi β€ n). It's guaranteed that the graph described in the input is a tree.
Output
Output the only number β the maximum product of sizes of the connected components, that Bob can get after deleting some of the tree's edges.
Examples
Input
5
1 2
2 3
3 4
4 5
Output
6
Input
8
1 2
1 3
2 4
2 5
3 6
3 7
6 8
Output
18
Input
3
1 2
1 3
Output
3
Submitted Solution:
```
def dfs(u,pa,gr):
szz[u]=1
for i in gr[u]:
if i ==pa:
continue
dfs(i,u,gr)
szz[u]+=szz[i]
first=True
# bool first=true;
n=szz[u]
# int n=szz[u];
# temp=[0 for i in range(0,n)]
temp=[0 for i in range(0,n)]
# v temp(n,0);
for it in gr[u]:
if it==pa:
continue
if first:
first=False
for i in range(0,szz[it]+1):
temp[i]=dp[it][i]
if i!=0:
temp[i]//=i
else:
for i in reversed(range(0,n)):
mx=0
for j in range(min(i,szz[it])+1):
mx1=temp[i-j]*dp[it][j]
# print(type(j))
if j!=0:
mx1//=j
mx=max(mx,mx1)
temp[i]=mx
if first:
dp[u][0]=1
dp[u][1]=1
else:
for i in range(1,n+1):
# print(type(i))
dp[u][i]=temp[i-1]*i
dp[u][0]=max(dp[u][0],dp[u][i])
n1 = int(input())
gr=[[]for i in range(0,n1)]
for i in range(0,n1-1):
a,b=input().split()
a=int(a)
b=int(b)
a-=1
b-=1
gr[a].append(b)
gr[b].append(a)
szz=[0 for i in range(0,n1)]
dp= [[0 for i in range(0,n1+1)] for j in range(0,n1+1)]
dfs(0,-1,gr)
print(int(dp[0][0]))
print(dp)
```
No
| 107,289 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Bob invented a new game with a tree (we should remind you, that a tree is a connected graph without cycles): he deletes any (possibly, zero) amount of edges of the tree, and counts the product of sizes of the connected components left after the deletion. Your task is to find out the maximum number that Bob can get in his new game for a given tree.
Input
The first input line contains integer number n (1 β€ n β€ 700) β amount of vertices in the tree. The following n - 1 lines contain the description of the edges. Each line contains the pair of vertices' indexes, joined by an edge, ai, bi (1 β€ ai, bi β€ n). It's guaranteed that the graph described in the input is a tree.
Output
Output the only number β the maximum product of sizes of the connected components, that Bob can get after deleting some of the tree's edges.
Examples
Input
5
1 2
2 3
3 4
4 5
Output
6
Input
8
1 2
1 3
2 4
2 5
3 6
3 7
6 8
Output
18
Input
3
1 2
1 3
Output
3
Submitted Solution:
```
maxn = 710
to = [[] for i in range(maxn)]
dp = [[0 for i in range(maxn)] for i in range(maxn)]
size = [0 for i in range(maxn)]
n = int(input())
def dfs(now, f):
size[now] = 1
dp[now][1] = 1
for i in to[now]:
if i != f:
dfs(i, now)
size[now] += size[i]
for j in reversed(range(0, 12)):
for k in range(0, min(j, size[i]) + 1):
dp[now][j] = max(dp[now][j], dp[now][j - k] * dp[i][k])
for i in range(1, n + 1):
dp[now][0] = max(dp[now][0], dp[now][i] * i)
for i in range(1, n):
u, v = map(int, input().split(' '))
to[u].append(v)
to[v].append(u)
dfs(1, 0)
print(dp[1][0])
# print(dp)
```
No
| 107,290 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0, 1]. Next, n stones will fall and Liss will escape from the stones. The stones are numbered from 1 to n in order.
The stones always fall to the center of Liss's interval. When Liss occupies the interval [k - d, k + d] and a stone falls to k, she will escape to the left or to the right. If she escapes to the left, her new interval will be [k - d, k]. If she escapes to the right, her new interval will be [k, k + d].
You are given a string s of length n. If the i-th character of s is "l" or "r", when the i-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the n stones falls.
Input
The input consists of only one line. The only line contains the string s (1 β€ |s| β€ 106). Each character in s will be either "l" or "r".
Output
Output n lines β on the i-th line you should print the i-th stone's number from the left.
Examples
Input
llrlr
Output
3
5
4
2
1
Input
rrlll
Output
1
2
5
4
3
Input
lrlrr
Output
2
4
5
3
1
Note
In the first example, the positions of stones 1, 2, 3, 4, 5 will be <image>, respectively. So you should print the sequence: 3, 5, 4, 2, 1.
Tags: constructive algorithms, data structures, implementation, two pointers
Correct Solution:
```
st = input()
a = []
b = []
for i in range(len(st)):
if st[i] == 'r':
a.append(i+1)
else:
b.append(i+1)
c = a+b[::-1]
for i in c:
print(i)
```
| 107,291 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0, 1]. Next, n stones will fall and Liss will escape from the stones. The stones are numbered from 1 to n in order.
The stones always fall to the center of Liss's interval. When Liss occupies the interval [k - d, k + d] and a stone falls to k, she will escape to the left or to the right. If she escapes to the left, her new interval will be [k - d, k]. If she escapes to the right, her new interval will be [k, k + d].
You are given a string s of length n. If the i-th character of s is "l" or "r", when the i-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the n stones falls.
Input
The input consists of only one line. The only line contains the string s (1 β€ |s| β€ 106). Each character in s will be either "l" or "r".
Output
Output n lines β on the i-th line you should print the i-th stone's number from the left.
Examples
Input
llrlr
Output
3
5
4
2
1
Input
rrlll
Output
1
2
5
4
3
Input
lrlrr
Output
2
4
5
3
1
Note
In the first example, the positions of stones 1, 2, 3, 4, 5 will be <image>, respectively. So you should print the sequence: 3, 5, 4, 2, 1.
Tags: constructive algorithms, data structures, implementation, two pointers
Correct Solution:
```
s=input()
r=[]
l=[]
num=1
for i in s:
if i=='r':
r.append(num)
elif i=='l':
l.append(num)
num+=1
res=r+l[::-1]
for i in res:
print(i)
```
| 107,292 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0, 1]. Next, n stones will fall and Liss will escape from the stones. The stones are numbered from 1 to n in order.
The stones always fall to the center of Liss's interval. When Liss occupies the interval [k - d, k + d] and a stone falls to k, she will escape to the left or to the right. If she escapes to the left, her new interval will be [k - d, k]. If she escapes to the right, her new interval will be [k, k + d].
You are given a string s of length n. If the i-th character of s is "l" or "r", when the i-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the n stones falls.
Input
The input consists of only one line. The only line contains the string s (1 β€ |s| β€ 106). Each character in s will be either "l" or "r".
Output
Output n lines β on the i-th line you should print the i-th stone's number from the left.
Examples
Input
llrlr
Output
3
5
4
2
1
Input
rrlll
Output
1
2
5
4
3
Input
lrlrr
Output
2
4
5
3
1
Note
In the first example, the positions of stones 1, 2, 3, 4, 5 will be <image>, respectively. So you should print the sequence: 3, 5, 4, 2, 1.
Tags: constructive algorithms, data structures, implementation, two pointers
Correct Solution:
```
from collections import *
s, right, left = input(), deque(), deque()
for i in range(len(s)):
if s[i] == 'l':
left.appendleft(str(i + 1))
else:
right.append(str(i + 1))
print('\n'.join(right))
print('\n'.join(left))
```
| 107,293 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0, 1]. Next, n stones will fall and Liss will escape from the stones. The stones are numbered from 1 to n in order.
The stones always fall to the center of Liss's interval. When Liss occupies the interval [k - d, k + d] and a stone falls to k, she will escape to the left or to the right. If she escapes to the left, her new interval will be [k - d, k]. If she escapes to the right, her new interval will be [k, k + d].
You are given a string s of length n. If the i-th character of s is "l" or "r", when the i-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the n stones falls.
Input
The input consists of only one line. The only line contains the string s (1 β€ |s| β€ 106). Each character in s will be either "l" or "r".
Output
Output n lines β on the i-th line you should print the i-th stone's number from the left.
Examples
Input
llrlr
Output
3
5
4
2
1
Input
rrlll
Output
1
2
5
4
3
Input
lrlrr
Output
2
4
5
3
1
Note
In the first example, the positions of stones 1, 2, 3, 4, 5 will be <image>, respectively. So you should print the sequence: 3, 5, 4, 2, 1.
Tags: constructive algorithms, data structures, implementation, two pointers
Correct Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
# from fractions import *
# from heapq import*
from bisect import *
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz'
M=10**9+7
EPS=1e-6
def Ceil(a,b): return a//b+int(a%b>0)
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
s=input()
n=len(s)
low=0
high=n-1
ans=[0]*n
for i in range(n):
if(s[i]=='r'):
ans[low]=i+1
low+=1
else:
ans[high]=i+1
high-=1
for i in ans:print(i)
```
| 107,294 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0, 1]. Next, n stones will fall and Liss will escape from the stones. The stones are numbered from 1 to n in order.
The stones always fall to the center of Liss's interval. When Liss occupies the interval [k - d, k + d] and a stone falls to k, she will escape to the left or to the right. If she escapes to the left, her new interval will be [k - d, k]. If she escapes to the right, her new interval will be [k, k + d].
You are given a string s of length n. If the i-th character of s is "l" or "r", when the i-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the n stones falls.
Input
The input consists of only one line. The only line contains the string s (1 β€ |s| β€ 106). Each character in s will be either "l" or "r".
Output
Output n lines β on the i-th line you should print the i-th stone's number from the left.
Examples
Input
llrlr
Output
3
5
4
2
1
Input
rrlll
Output
1
2
5
4
3
Input
lrlrr
Output
2
4
5
3
1
Note
In the first example, the positions of stones 1, 2, 3, 4, 5 will be <image>, respectively. So you should print the sequence: 3, 5, 4, 2, 1.
Tags: constructive algorithms, data structures, implementation, two pointers
Correct Solution:
```
# Key: the nth 'l' marks the nth stone from the right
directions = input()
r_stones = []
l_stones = []
stone = 1
for i in directions:
if i == 'r':
r_stones.append(stone)
if i == 'l':
l_stones.append(stone)
stone += 1
r_stones.sort()
l_stones.sort(reverse=True)
for i in r_stones + l_stones:
print(i)
```
| 107,295 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0, 1]. Next, n stones will fall and Liss will escape from the stones. The stones are numbered from 1 to n in order.
The stones always fall to the center of Liss's interval. When Liss occupies the interval [k - d, k + d] and a stone falls to k, she will escape to the left or to the right. If she escapes to the left, her new interval will be [k - d, k]. If she escapes to the right, her new interval will be [k, k + d].
You are given a string s of length n. If the i-th character of s is "l" or "r", when the i-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the n stones falls.
Input
The input consists of only one line. The only line contains the string s (1 β€ |s| β€ 106). Each character in s will be either "l" or "r".
Output
Output n lines β on the i-th line you should print the i-th stone's number from the left.
Examples
Input
llrlr
Output
3
5
4
2
1
Input
rrlll
Output
1
2
5
4
3
Input
lrlrr
Output
2
4
5
3
1
Note
In the first example, the positions of stones 1, 2, 3, 4, 5 will be <image>, respectively. So you should print the sequence: 3, 5, 4, 2, 1.
Tags: constructive algorithms, data structures, implementation, two pointers
Correct Solution:
```
__autor__ = 'Alex239'
s = input().rstrip()
print("\n".join([str(i + 1) for i in range(len(s)) if s[i] == 'r']), "\n".join([str(i + 1) for i in range(len(s) - 1, -1, -1) if s[i] == 'l']), sep="\n")
```
| 107,296 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0, 1]. Next, n stones will fall and Liss will escape from the stones. The stones are numbered from 1 to n in order.
The stones always fall to the center of Liss's interval. When Liss occupies the interval [k - d, k + d] and a stone falls to k, she will escape to the left or to the right. If she escapes to the left, her new interval will be [k - d, k]. If she escapes to the right, her new interval will be [k, k + d].
You are given a string s of length n. If the i-th character of s is "l" or "r", when the i-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the n stones falls.
Input
The input consists of only one line. The only line contains the string s (1 β€ |s| β€ 106). Each character in s will be either "l" or "r".
Output
Output n lines β on the i-th line you should print the i-th stone's number from the left.
Examples
Input
llrlr
Output
3
5
4
2
1
Input
rrlll
Output
1
2
5
4
3
Input
lrlrr
Output
2
4
5
3
1
Note
In the first example, the positions of stones 1, 2, 3, 4, 5 will be <image>, respectively. So you should print the sequence: 3, 5, 4, 2, 1.
Tags: constructive algorithms, data structures, implementation, two pointers
Correct Solution:
```
s=input()
l=[]
r=[]
for i in range(len(s)):
if s[i]=='l':
l.append(i+1)
else:
r.append(i+1)
for i in r:
print(i)
for i in reversed(l):
print(i)
```
| 107,297 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0, 1]. Next, n stones will fall and Liss will escape from the stones. The stones are numbered from 1 to n in order.
The stones always fall to the center of Liss's interval. When Liss occupies the interval [k - d, k + d] and a stone falls to k, she will escape to the left or to the right. If she escapes to the left, her new interval will be [k - d, k]. If she escapes to the right, her new interval will be [k, k + d].
You are given a string s of length n. If the i-th character of s is "l" or "r", when the i-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the n stones falls.
Input
The input consists of only one line. The only line contains the string s (1 β€ |s| β€ 106). Each character in s will be either "l" or "r".
Output
Output n lines β on the i-th line you should print the i-th stone's number from the left.
Examples
Input
llrlr
Output
3
5
4
2
1
Input
rrlll
Output
1
2
5
4
3
Input
lrlrr
Output
2
4
5
3
1
Note
In the first example, the positions of stones 1, 2, 3, 4, 5 will be <image>, respectively. So you should print the sequence: 3, 5, 4, 2, 1.
Tags: constructive algorithms, data structures, implementation, two pointers
Correct Solution:
```
# JAI SHREE RAM
import math; from collections import *
import sys; from functools import reduce
# sys.setrecursionlimit(10**6)
def get_ints(): return map(int, input().strip().split())
def get_list(): return list(get_ints())
def get_string(): return list(input().strip().split())
def printxsp(*args): return print(*args, end="")
def printsp(*args): return print(*args, end=" ")
UGLYMOD = int(1e9)+7; SEXYMOD = 998244353; MAXN = int(1e5)
# sys.stdin=open("input.txt","r");sys.stdout=open("output.txt","w")
# for _testcases_ in range(int(input())):
s = input()
li1 = []
li2 = []
for i, v in enumerate(s):
if v == 'l':
li1.append(i+1)
elif v == 'r':
li2.append(i+1)
print(*li2, sep='\n')
print(*li1[::-1], sep='\n')
'''
>>> COMMENT THE STDIN!! CHANGE ONLINE JUDGE !!
THE LOGIC AND APPROACH IS MINE @luctivud ( UDIT GUPTA )
Link may be copy-pasted here if it's taken from other source.
DO NOT PLAGIARISE.
>>> COMMENT THE STDIN!! CHANGE ONLINE JUDGE !!
'''
```
| 107,298 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0, 1]. Next, n stones will fall and Liss will escape from the stones. The stones are numbered from 1 to n in order.
The stones always fall to the center of Liss's interval. When Liss occupies the interval [k - d, k + d] and a stone falls to k, she will escape to the left or to the right. If she escapes to the left, her new interval will be [k - d, k]. If she escapes to the right, her new interval will be [k, k + d].
You are given a string s of length n. If the i-th character of s is "l" or "r", when the i-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the n stones falls.
Input
The input consists of only one line. The only line contains the string s (1 β€ |s| β€ 106). Each character in s will be either "l" or "r".
Output
Output n lines β on the i-th line you should print the i-th stone's number from the left.
Examples
Input
llrlr
Output
3
5
4
2
1
Input
rrlll
Output
1
2
5
4
3
Input
lrlrr
Output
2
4
5
3
1
Note
In the first example, the positions of stones 1, 2, 3, 4, 5 will be <image>, respectively. So you should print the sequence: 3, 5, 4, 2, 1.
Submitted Solution:
```
s=input()
l=[];r=[]
for i,j in enumerate(s):
if j=='l':l.append(i)
else:r.append(i)
for i in r:
print(i+1)
for i in reversed(l):
print(i+1)
```
Yes
| 107,299 |
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