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 MAX(int a,int b){
    return a>b?a:b;
}
int main()
{   int n,k;
    scanf("%d%d",&n,&k);
    char s[n+1];
    scanf("%s",s);
    int area[100001]={0},p[100001]={0},cnt=0,t=0;
    for(int i=0;i<n;i++){
        if(s[i]=='0'&&s[i-1]=='1'){
            cnt=1;
        }else if(s[i]=='0'){
            cnt++;
        }
        if(s[i]=='0'&&s[i+1]=='1'){
            area[t]=cnt;
            p[t]=i;
            t++;
        }
        if(i==n-1&&s[i]=='0'){
            area[t]=cnt;
            p[t]=i;
            t++;
        }
        
    }
    if(t<k){
        printf("%d",n);
        return 0;
    }
    int max=0;
    for(int i=0;i<t-k;i++){
        max=MAX(p[i+k+1]-p[i],max);
    }
    printf("%d",max);

    return 0;
}