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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>
int main(void)
{
    int n,k,max=0,ans;
    scanf("%d%d", &n,&k);
    char s[n];
    scanf("%s", s);
    int i,j=0,block[n+2];
    for(i=0; i<n; i++) block[i]=1;
    for(i=0; i<n-1; i++)
    {
        if(strncmp(&s[i], &s[i+1],1)==0) block[j]++;
        else j++;
    } 
    if(strncmp(&s[0], "0",1)==0)//偶数ブロックを変える
    {
        if(k*2>j) ans=n;
        else
        {
            if(strncmp(&s[n-1], "0",1)==0)
            {
                for(i=j; i>=0; i--) block[i+1]=block[i];
                block[0]=0;
                block[j+2]=0;
                for(i=0; i<2*k+1; i++) max+=block[i];
                for(i=1; 2*i+2*k+1<=j+2; i++)
                {
                    if(max<max-block[2*i-1]-block[2*i-2]+block[2*i+2*k+1]+block[2*i+2*k+2])
                    {
                        max=max-block[2*i-1]-block[2*i-2]+block[2*i+2*k+1]+block[2*i+2*k+2];
                    }
                }
            }
            else
            {
                for(i=j; i>=0; i--) block[i+1]=block[i];
                block[0]=0;
                for(i=0; i<2*k+1; i++) max+=block[i];
                for(i=1; 2*i+2*k+1<=j+2; i++)
                {
                    if(max<max-block[2*i-1]-block[2*i-2]+block[2*i+2*k+1]+block[2*i+2*k+2])
                    {
                        max=max-block[2*i-1]-block[2*i-2]+block[2*i+2*k+1]+block[2*i+2*k+2];
                    }
                }
            }
            
        }
    }
    else//奇数ブロックを変える
    {
        if(k*2+1>=j) ans=n;
        else
        {
            if(strncmp(&s[n-1], "0",1)==0) block[j+1]=0;
            for(i=0; i<2*k+1; i++) max+=block[i];
                for(i=1; 2*i+2*k+1<=j+2; i++)
                {
                    if(max<max-block[2*i-1]-block[2*i-2]+block[2*i+2*k+1]+block[2*i+2*k+2])
                    {
                        max=max-block[2*i-1]-block[2*i-2]+block[2*i+2*k+1]+block[2*i+2*k+2];
                    }
                }
        }
        
    }
    printf("%d", max);
    return 0;
}