Codeforces 676C Vasya and String 尺取法-优快云博客

本文探讨了如何通过改变不多于k个字符来最大化字符串中相同字符的最大连续子串长度。利用滑动窗口技巧实现高效求解。

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High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotesbeauty 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.

Examples

input
4 2
abba

output
4

input
8 1
aabaabaa

output
5

题意:给你一个长度为n字符串，然后通过k次操作(a->b或者b->a).问你能得到最长连续相同的串的长度；

思路:直接尺取就好；

代码:

#include<cstdio>  
#include<cstring>  
#include<cstdlib>  
#include<cmath>  
#include<iostream>  
#include<algorithm>  
#include<vector>  
#include<map>  
#include<set>  
#include<queue>  
#include<string>  
#include<bitset>  
#include<utility>  
#include<functional>  
#include<iomanip>  
#include<sstream>  
#include<ctime>  
using namespace std;  

#define maxn int(1e5+50)  
#define inf int(0x3f3f3f3f) 
#define mod int(1e9+7)
#define PI acos(-1.0)
typedef long long ll;
char s[maxn];
int main()  
{   
#ifdef CDZSC_June  
	freopen("t.txt", "r", stdin);   
#endif 
	int n,m,suma,sumb;
	while(~scanf("%d%d",&n,&m))
	{
		suma = sumb = 0;//suma表示a的数量,sumb表示b的数量
		scanf("%s",s);
		int L =0 ,R = 0;
		int max1 = 0;
		while(1)
		{
			while(R < n && min(suma,sumb) <= m){
				if(s[R] == 'a') suma++;
				if(s[R] == 'b') sumb++;
				R++;
			}
			if(min(suma,sumb) <= m) break;
			max1 = max(max1,R- L-1);
			while(min(suma,sumb) > m){
				if(s[L] == 'a') suma--;
				if(s[L] == 'b') sumb--;
				L++;
			}
		}
		max1 = max(max1,R- L);
		printf("%d\n",max1);
	}
	return 0;                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            
}