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(){
  char s[100001];
  int ary[100000]={0},j=0,max=0,sto=0,n,k;
  scanf("%d%d",&n,&k);
  scanf("%s",s);
  if (s[0]=='0'){
    j++;
  }
  ary[j]++;  
  for (int i=1;i<n;i++){
    if (s[i]==s[i-1]) ary[j]++;
    else {
      j++;
      ary[j]++;
    }
  }
  if ((j+1)/2<k){
    for (int i=0;i<j;i++) max+=ary[j];
  }
  else {
    for (int i=0;i<2*k+1;i++) max+=ary[i];
    sto=max;
    for (int i=1;i+2*k<j;i++){
      sto=sto-ary[i-1]-ary[i-2]+ary[i+2*k-1]+ary[i+2*k];
      if (sto>max) max=sto;
    }
  }
  printf("%d\n",max);
  return 0;
}