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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;

struct s
{
	int a; //连续0的起点
	int b; //连续个数
};

vector
 vec;
string str;
int n, k; 
int M = 0;

bool cmp(const s& x, const s& y)
{
	return x.b > y.b;
}

void change(s p)
{
	string str2 = str;
	int start_p = p.a; int end_p = start_p + p.b;
	for(int i = p.a; i < end_p; i++)  str2[i] = '1';
	int flag = 0, cnt= 0;
	for(int i = 0; i < str2.length(); i++) 
	{
		if(str2[i] == '1') { cnt++; flag = 1; }
		else 
		{
			if(flag) 
			{
				if(cnt > M) M = cnt;
				flag = 0;
			}
		}
	}
	if(cnt > M) cnt = M;
}

int main()
{
	cin >> n >> k; getchar();
	getline(cin, str);
	int flag = 0;
	s p; p.b = 0;

	for(int i = 0; i < n; i++) 
	{
		if(str[i] == '0')
		{
			p.b++;
			if(flag == 0)
			{
				p.a = i; //记录起点 	
				flag = 1;
			}
			if(i == n-1) vec.push_back(p);
		}
		else 
		{
			if(flag) 
			{
				flag = 0;
				vec.push_back(p);
				p.b = 0;
			}
		}
	}

	sort(vec.begin(), vec.end(), cmp);

	for(int i = 0; i < k; i++)
	{
		change(vec[i]);
	}
	cout << M << endl;
	//for(auto i = vec.begin(); i != vec.end(); i++) cout << (*i).b << endl;

}