Problem:
Score : 400 points

Problem Statement

N people are arranged in a row from left to right.

You are given a string S of length N consisting of 0 and 1, and a positive integer K.

The i-th person from the left is standing on feet if the i-th character of S is 0, and standing on hands if that character is 1.

You will give the following direction at most K times (possibly zero):

Direction: Choose integers l and r satisfying 1 \leq l \leq r \leq N, and flip the l-th, (l+1)-th, ..., and r-th persons. That is, for each i = l, l+1, ..., r, the i-th person from the left now stands on hands if he/she was standing on feet, and stands on feet if he/she was standing on hands.

Find the maximum possible number of consecutive people standing on hands after at most K directions.

Constraints

N is an integer satisfying 1 \leq N \leq 10^5.

K is an integer satisfying 1 \leq K \leq 10^5.

The length of the string S is N.

Each character of the string S is 0 or 1.

Input

Input is given from Standard Input in the following format:

N K
S

Output

Print the maximum possible number of consecutive people standing on hands after at most K directions.

Sample Input 1

5 1
00010

Sample Output 1

4

We can have four consecutive people standing on hands, which is the maximum result, by giving the following direction:

Give the direction with l = 1, r = 3, which flips the first, second and third persons from the left.

Sample Input 2

14 2
11101010110011

Sample Output 2

8

Sample Input 3

1 1
1

Sample Output 3

1

No directions are necessary.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

int main(){
   
    int N,K;
    cin>>N>>K;
    vector<char> S(N+1,'5');//1から入れる
    vector<int> A(N,0),B(N,0);
    for(int i=0;i<N;i++){
        cin>>S[i+1];
    }
    int j=0,k=0;
    for(int i=1;i<=N;i++){
        if(S[i]=='1'){
            A[j+1]++;
            if(S[i-1]=='0') k++;
        }
        else if(S[i]=='0'){
            B[k+1]++;
            if(S[i-1]=='1') j++;
        }
    }
    int l=k;//0パートの個数
    if(S[N]=='0') l++;
    if(l<=K) cout<<N<<endl;
    else if(S[1]=='0'){
        int ans=0;
        for(int x=1;x<=K;x++){
            ans+=(A[x]+B[x]);
        }
        int now=ans;
        for(int x=1;x<=l-K;x++){
            now+=(A[x+K]+B[x+K]);
            now-=(A[x]+B[x-1]);
            ans=max(ans,now);
        }
        cout<<ans<<endl;
    }
    else if(S[1]=='1'){
        int ans=0;
        for(int x=1;x<=K;x++){
            ans+=(A[x]+B[x]);
        }
        ans+=A[K+1];
        int now=ans;
        for(int x=1;x<=l-K;x++){
            now+=(A[x+K+1]+B[x+K]);
            now-=(A[x]+B[x]);
            ans=max(ans,now);
        }
        cout<<ans<<endl;
    }
}