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: Runtime Error)

#include <stdio.h>

int sum(int *hairetsu,int n){
	int sum_add=0;
	for(int num=0;num<n;num++){
		sum_add += hairetsu[num];
	}
	return sum_add;
}

int main(){
	char s[100001];
	int n;
	int k;
	int a = scanf('%d', &n);
	int b = scanf('%d', &k);
	int c = scanf('%s', s);
	
	int cnt = 0;
	int init = s[0]=='0'?0:1;
	int h[100001] = {};
	int h_cnt = 0;
	
	for(int i=1;i<n;i++){
		if(s[i-1]!=s[i]){
			h[h_cnt++]+=1;
		}
		else{
			h[h_cnt]+=1;
		}
	}
	//h_cnt is finally length of h_cnt
	
	int zero_num = 0;
	int max = 0;
	if(init){ //start with 1
		zero_num = h_cnt/2;
		if(zero_num<=k){
			printf('%d', n);
		}
		else{
			int sum_0 = sum(h,k*2+1);
			int max = sum_0;
			for(int j=0;j<=zero_num-k+1;j++){ //j is 0's location
				sum_0 = sum_0 - h[j*2] - h[j*2+1] + h[(j+k)*2+1] + h[(j+k+1)*2];
				if(max<=sum_0){
					max = sum_0;
				}
			}
			printf('%d', max);
		}
	}
	
	else{ //start with 0
		zero_num = h_cnt/2 +1;
		if(zero_num<=k){
			printf('%d', n);
		}
		else{
			int sum_0 = sum(h,k*2);
			int max = sum_0;
			for(int j=0;j<=zero_num-k+1;j++){ //j is 0's location
				sum_0 = sum_0 - h[j*2] - h[j*2+1] + h[(j+k)*2] + h[(j+k)*2+1];
				if(max<=sum_0){
					max = sum_0;
				}
			}
			printf('%d', max);
		}
	}
	return 0;
}