B. Partial Replacement

题面：

You are given a number k and a string s of length n, consisting of the characters ‘.’ and ‘’. You want to replace some of the '’ characters with ‘x’ characters so that the following conditions are met:

The first character ‘’ in the original string should be replaced with ‘x’;
The last character '’ in the original string should be replaced with ‘x’;
The distance between two neighboring replaced characters ‘x’ must not exceed k (more formally, if you replaced characters at positions i and j (i<j) and at positions [i+1,j−1] there is no “x” symbol, then j−i must be no more than k).
For example, if n=7, s=..* and k=3, then the following strings will satisfy the conditions above:

.xx.xx;
.x.xx;
.xx.xxx.
But, for example, the following strings will not meet the conditions:
.**.xx (the first character '’ should be replaced with ‘x’);
.x.xx* (the last character ‘’ should be replaced with ‘x’);
.x.xx (the distance between characters at positions 2 and 6 is greater than k=3).
Given n, k, and s, find the minimum number of '’ characters that must be replaced with ‘x’ in order to meet the above conditions.

Input

The first line contains one integer t (1≤t≤500). Then t test cases follow.

The first line of each test case contains two integers n and k (1≤k≤n≤50).

The second line of each test case contains a string s of length n, consisting of the characters ‘.’ and ‘*’.

It is guaranteed that there is at least one ‘*’ in the string s.

It is guaranteed that the distance between any two neighboring ‘*’ characters does not exceed k.

Output

For each test case output the minimum number of ‘*’ characters that must be replaced with ‘x’ characters in order to satisfy the conditions above.

题解：

本体首先可以先找出"*“的个数如果其个数小于等于0或1就输出0或1，当其个数大于1的时候可以记录下第一个” * ",并且记录其位置，在让其变成x，再从最后找出第一个” * “，也让其变成x，再从记录第一个“ * ”点后面开始寻找x，如果是“ * ”则更新记录的位置如果为x，则输出ans+2，下面上代码。

代码：

#include<iostream>
using namespace std;int main()
{int t;cin>>t;while(t--){string s;int n,ans=0,cut=0,m,l;int p=-1;cin>>n>>m>>s;for(int i=0;i<n;i++){if(s[i]=='*')cut++;}if(cut==0||cut==1){cout<<(cut==0?"0":"1")<<endl;continue;}for(int i=0;i<n;i++)if(s[i]=='*'){s[i]='x';p=i;break;}for(int i=n-1;i>=0;i--)if(s[i]=='*'){s[i]='x';break;}int b=p;l=p;int f=0;for(int i=p;i<n;i++){for(int j=i+1;j<=i+m;j++){if(s[j]=='*')b=j;else if(s[j]=='x'){cout<<ans+2<<endl;f=1;break;}}if(f==1) break;i=b-1;ans++;}}
}

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B. Partial Replacement

B. Partial Replacement

题面：

Input

Output

题解：

代码：

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