codeforces 676C C. Vasya and String(二分)

题目链接：

C. Vasya and String

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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：

问最多改变k个字母才能使相同字母组成的子串最长;

思路：

最长的那个字串可以是a组成的,也可能是b组成的,现在枚举最长的子串的左端点,二分右端点,找到最长的长度,对于a,b两种情况都这样处理;用尺取法也可以搞;

AC代码：

#include <bits/stdc++.h>
/*
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <cstring>
#include <algorithm>
#include <cstdio>
*/
using namespace std;
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
const LL mod=1e9+;
const double PI=acos(-1.0);
const int inf=0x3f3f3f3f;
const int N=1e5+;
int n,sum[N],k;
char s[N];
int check(int x,int y)
{
    if(sum[x]-sum[y]<=k)return ;
    return ;
}
int main()
{
       scanf("%d%d",&n,&k);
       scanf("%s",s);
       sum[]=;
       for(int i=;i<n;i++)
       {
           if(s[i]=='a')sum[i+]=sum[i];
           else sum[i+]=sum[i]+;
       }
       int ans=;
       for(int i=;i<=n;i++)
       {
           int l=i,r=n;
           while(l<=r)
           {
               int mid=(l+r)>>;
               if(check(mid,i-))l=mid+;
               else r=mid-;
           }
           ans=max(ans,l-i);
       }
       for(int i=;i<n;i++)
       {
           if(s[i]=='b')sum[i+]=sum[i];
           else sum[i+]=sum[i]+;
       }
       for(int i=;i<=n;i++)
       {
           int l=i,r=n;
           while(l<=r)
           {
               int mid=(l+r)>>;
               if(check(mid,i-))l=mid+;
               else r=mid-;
           }
           ans=max(ans,l-i);
       }
       cout<<ans<<"\n";
    return ;
}