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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>

//0:直立
//1:逆立ち
//今回は連続で逆立ちさせたい
//num[]の偶数番が0の個数

int main(void)
{
	
	char str[100002];
	int i, j;
	int n, k;
	int count_ichi;
	int count_zero;
	int count = 0;
	int num[300000] = {0};
	int max = 0;
	int sum = 0;
	
	scanf("%d%d%s", &n, &k, str);
	
	str[n] = '5';
	
	if(str[0] == '0') {
		count++;
		num[0] = 0;
	}
	
	for(i = 0; i < n; i++) {
		if(str[i] == '1') {
			count_ichi = 1;
			for(j = i + 1; j <= n; j++) {
				if(str[j] == '1') {
					count_ichi++;
				} else {
					num[count] = count_ichi;
					count++;
					break;
				}
			}
			i = j - 1;
		} else {
			count_zero = 1;
			for(j = i + 1; j <= n; j++) {
				if(str[j] == '0') {
					count_zero++;
				} else {
					num[count] = count_zero;
					count++;
					break;
				}
			}
			i = j - 1;
		}
	}
	
	for(i = 0; i < 2 * k + 1; i++) {
		sum += num[i];
	}
	
	for(i = 2 * k + 1; i < n; i = i + 2) {
		sum += (num[i] + num[i + 1]);
		sum -= (num[i - 2 * k - 1] + num[i - 2 * k]);
		
		if(max < sum) {
			max = sum;
		}
	}
	
	printf("%d\n", max);
	
	return 0;
}