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 <stdlib.h>
int main(void){
	int n,k,i;
	scanf("%d%d",&n,&k);
	char s[100002];
	int a[100002];
	int b[100002];
	int cnt;
	scanf("%s",s);
	//連続する1の数を正、0の数を負として配列aに保存する
	i=0;cnt=0;
	while(i<n){
		while(s[i]=='0'){
			a[cnt]--;
			i++;
			if(s[i]=='1'){
				cnt++;
			}
		}
		while(s[i]=='1'){
			a[cnt]++;
			i++;
			if(s[i]=='0'){
				cnt++;
			}
		}
	}
	cnt++;
	b[0]=0;
	for(i=1;i<=cnt;i++){
		b[i]+= b[i-1]+abs(a[i-1]);
	}
	for(i=0;i<=cnt;i++){
	//	printf("%d,",b[i]);
	}
	int ans = 0;
	int temp;
	//printf("cnt = %d",cnt);
	if(a[0]>0){
		for(i=1;i<=(cnt-2*k);i++){
			temp = b[i+2*k]-b[i-1];
			if(ans<temp)ans=temp;
		}
	}else{
		for(i=1;i<=(cnt-(2*k)+1);i++){
			temp = b[i+(2*k)-1]-b[i-1];
			if(ans<temp)ans=temp;
		}
	}
	if(cnt==1){
		ans = b[1];
	}
	printf("%d",ans);
	return 0;
}