Vasya and String(尺取法)

作者: zqqian 分类: ACM 发布时间: 2017-07-14 16:28

 

High school student Vasya got a string of length n as a birthday present. This string consists of letters ‘a‘ and ‘b‘ only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters.

Vasya can change no more than k characters of the original string. What is the maximum beauty of the string he can achieve?

Input

The first line of the input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n) — the length of the string and the maximum number of characters to change.

The second line contains the string, consisting of letters ‘a‘ and ‘b‘ only.

Output

Print the only integer — the maximum beauty of the string Vasya can achieve by changing no more than k characters.

Example

Input
4 2
abba
Output
4
Input
8 1
aabaabaa
Output
5

Note

In the first sample, Vasya can obtain both strings “aaaa” and “bbbb“.

In the second sample, the optimal answer is obtained with the string “aaaaabaa” or with the string “aabaaaaa“.

 

这个题使用尺取法即可。先用数组保存每一个b出现的位置,x[j]表示第j个b出现的位置是i+1,然后x[0]是0,x[num]是n+1

先计算b翻转到a的次数,再计算a翻转到b的次数,取其中最大的次数。

 

#include<iostream>
#include<cstdio>
#include<string.h>
using namespace std;
int main(){
	int n,k;
	char a[100009];
	int x[100009];
	
	while(~scanf("%d%d",&n,&k)){
		
		scanf("%s",&a);
		int num=1;
		x[0]=0;
		for(int i=0;i<n;i++){
			if(a[i]=='b'){
				x[num++]=i+1;
			}
		}
		
		x[num]=n+1;
		
		int maxa=0;
		if(k+1>=num){
			maxa=n;
		}else{
			
			int s;
			for(int i=k+1;i<=num;i++){
				
			s=x[i]-1-x[i-k-1];
			maxa=maxa>s?maxa:s;
			}
			
		
			
		}
		//开始计算a翻转到b
		 num=1;
		x[0]=0;
		for(int i=0;i<n;i++){
			if(a[i]=='a'){
				x[num++]=i+1;
			}
		}
		
		x[num]=n+1;
		
		int maxb=0;
		if(k+1>=num){
			maxb=n;
		}else{
			
			int s;
			for(int i=k+1;i<=num;i++){
				
			s=x[i]-1-x[i-k-1];
			maxb=maxb>s?maxb:s;
			}
			
		
		
		}
	printf("%d\n",maxa>maxb?maxa:maxb);	
}
	return 0;
} 

 

 

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