Codeforces Round #354 (Div. 2) C. Vasya and String

Vasya and String time limit：1 second memory limit ：256 megabytes

题目连接

http://www.codeforces.com/contest/676/problem/C

Description

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.

Sample Input

4 2abba

Sample Output

4

Hint

题意

有一个长度为n的字符串，你可以改变最多k次，问你最长的全是一样字符的串是多长

这个字符串只含有a和b字符

题解：

第三次没看出用二分做这类题了，我要好好总结下二分做题的类型与技巧

枚举开头，然后暴力二分就好了。

用前缀和预处理一下，那么前缀和之差，就是你需要改变的个数……

代码

#include<bits/stdc++.h> using namespace std; const int maxn=100000+7; char s[maxn]; int a[maxn],b[maxn]; int n,k; bool check(int x) { for(int i=1;i+x-1<=n;i++){ int aa=a[i+x-1]-a[i-1]; int bb=b[i+x-1]-b[i-1]; if(aa<=k||bb<=k) return true; } return false; } int main() { //freopen("in.txt","r",stdin); scanf("%d %d",&n,&k); a[0]=b[0]=0; scanf("%s",s+1); for(int i=1;i<=n;i++){ a[i]=a[i-1]+(s[i]=='a'); b[i]=b[i-1]+(s[i]=='b'); } int l=1,r=n+1; while(l+1<r){ int mid=(l+r)/2; if(check(mid)) l=mid; else r=mid; } printf("%d",l); return 0; }

 

转载于:https://www.cnblogs.com/wangdongkai/p/5531851.html