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最新推荐文章于 2019-02-02 15:35:28 发布

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最新推荐文章于 2019-02-02 15:35:28 发布

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分类专栏：日场比赛 CF

本文链接：https://blog.youkuaiyun.com/qq_41524782/article/details/81448015

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本文介绍了一个关于查找连续可用座位的算法，旨在帮助用户判断是否能在电影院中为他们及其朋友找到足够的连续空位。该算法通过扫描座位布局来确定最大连续空座位数，并据此决定是否能满足需求。

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Rami and K of his friends are going to watch a movie tonight. There are only one row that contains some available seats, all other rows are taken. Given the description of this row, Rami wants to know if he can find consecutive seats available to sit with his friends.

Input

The first line of input contains a single integer T, the number of test cases.

The first line of each test case consists of two-separated integers C, K (1 ≤ C, K ≤ 105), the number of seats in the row and the number of friends going with Rami, respectively.

The second line contains a string of C digits, where 0 represents an empty seat, and 1represents taken one.

Output

For each test case, print a single line with yesif Rami can find a place for him and his friends, otherwise print no.

Example

Input

Output

no
yes

AC代码

Select Code

#include <iostream>
#include <bits/stdc++.h>
using namespace std;

char a[100000+10];

int main()
{

    int t, c, k, i, b;
    scanf("%d",&t);
    while(t--)
    {
       b = 0;
        memset(a, 0, sizeof(a));
        int  mini = 0;
        scanf("%d%d",&c, &k);
        cin>>a;
        for(i = 0;i<c;i++)
        {
            if(a[i]=='0')
                b++;
            else if(a[i]=='1')
            {
               b = 0;
            }
            if(mini<b)
            mini = b;
        }
        if(mini>=k+1)
            printf("yes\n");
        else
            printf("no\n");
    }
    return 0;
}