题意: 长为n,由a,b构成的串,可以做k次改变. 问最后最长的相等字串长度是多少.
思路: k个肯定连续改的. 想了好久..........
#include<bits/stdc++.h>
#define PI acos(-1.0)
#define pb push_back
#define F first
#define S second
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N=1e5+5;
const int MOD=1e9+7;
template <class T>
bool sf(T &ret){ //Faster Input
char c; int sgn; T bit=0.1;
if(c=getchar(),c==EOF) return 0;
while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar();
sgn=(c=='-')?-1:1;
ret=(c=='-')?0:(c-'0');
while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
if(c==' '||c=='\n'){ ret*=sgn; return 1; }
while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10;
ret*=sgn;
return 1;
}
char s[N];
int main(void){
int n,k;
sf(n),sf(k);
scanf("%s",s+1);
int ans=0;
queue<int> q;
int cnt=0;
int st=1;
for(int i=1;i<=n;i++){
if(s[i]=='b') {q.push(i);cnt++;}
if(cnt>k){
st=q.front()+1;
q.pop();
cnt--;
}
ans=max(ans,i-st+1);
// cout <<i<<" -> "<< ans << endl;
}
while(!q.empty()) q.pop();
// cout <<"here" << endl;
cnt=0;
st=1;
for(int i=1;i<=n;i++){
if(s[i]=='a') {q.push(i);cnt++;}
if(cnt>k){
st=q.front()+1;
q.pop();
cnt--;
}
ans=max(ans,i-st+1);
}
printf("%d\n",ans);
return 0;
}