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?
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.
Print the only integer — the maximum beauty of the string Vasya can achieve by changing no more than k characters.
4 2
abba
4
8 1
aabaabaa
5
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".
解题思路 :
维护前缀和。枚举每一个位置,二分查找能够延续的最远距离。
(1)
a->0,b->1
如样例 aabaabaa n=8 k=2
变成01序列 00100100 相当于把b替换成a
(2)同理a->1 b->0。就相当与a替换成b
/* *********************************************** Author :guanjun Created Time :2016/5/26 15:30:33 File Name :cf354c.cpp ************************************************ */ #include <iostream> #include <cstring> #include <cstdlib> #include <stdio.h> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <iomanip> #include <list> #include <deque> #include <stack> #define ull unsigned long long #define ll long long #define mod 90001 #define INF 0x3f3f3f3f #define maxn 100010 #define cle(a) memset(a,0,sizeof(a)) const ull inf = 1LL << 61; const double eps=1e-5; using namespace std; priority_queue<int,vector<int>,greater<int> >pq; struct Node{ int x,y; }; struct cmp{ bool operator()(Node a,Node b){ if(a.x==b.x) return a.y> b.y; return a.x>b.x; } }; bool cmp(int a,int b){ return a>b; } int a[maxn]; int b[maxn]; int sum[maxn]; int n,k; int solve(int *sum){ int len=-1; for(int i=1;i<=n;i++){ int x=upper_bound(sum+1,sum+1+n,sum[i-1]+k)-sum; x--; if(x-i+1>len)len=x-i+1; } return len; } int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); #endif //freopen("out.txt","w",stdout); char s[maxn]; while(cin>>n>>k){ scanf("%s",s+1); cle(sum); cle(a),cle(b); for(int i=1;i<=n;i++){ if(s[i]=='a')a[i]=1; else b[i]=1; } for(int i=1;i<=n;i++){ sum[i]=sum[i-1]+b[i]; } int x=solve(sum); for(int i=1;i<=n;i++){ sum[i]=sum[i-1]+a[i]; } int y=solve(sum); cout<<max(x,y)<<endl; } return 0; }