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 <iostream>
#include <algorithm>
#include <string>
#define rep(i, x, n) for (int i = x; i < n; i++)
using namespace std;
int main()
{
    int N, K;
    cin >> N >> K;
    string S;
    cin >> S;
    int H[100000];
    rep(i, 0, N) { H[i] = 0; }
    int tmp = S[0] - '0';
    H[0] = 1;
    int tmpIdx = 0;
    int ind = 1;
    while (ind < N)
    {
        int si = S[ind] - '0';
        if (tmp == si)
        {
            H[tmpIdx]++;
        }
        else
        {
            H[tmpIdx + 1]++;
            tmpIdx++;
            tmp = si;
        }
        ind++;
    }
    int kukansu = tmpIdx + 1;
    int ruisekiwa[kukansu];
    rep(i, 0, kukansu)
    {
        if (i == 0)
        {
            ruisekiwa[i] = H[i];
        }
        else
        {
            ruisekiwa[i] = ruisekiwa[i - 1] + H[i];
        }
        //  cout << ruisekiwa[i] << endl;
    }
    int kukanlong = 2 * K + 1;
    int max = 0;
    rep(i, 0, kukansu)
    {
        int sum;
        if (i + kukanlong > N)
        {
            kukanlong = N - i;
        }
        if (i == 0)
        {
            if (S[0] == '0' && i + kukanlong < N)
            {
                sum = ruisekiwa[kukanlong -2];
            }
            else
            {
                sum = ruisekiwa[kukanlong-1];
            }
        }else{
            sum=ruisekiwa[i+kukanlong-1]-ruisekiwa[i-1];
        }
        if(max<sum){
            max=sum;
        }
    }
    cout << max << endl;
}