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: Wrong Answer)

#include<iostream>
#include<vector>
#include<cstring>

using namespace std;

struct Range {
	int left, right, cnt;

	Range() {
		left = 0, right = 0, cnt = 0;
	}

	Range(int left, int right) {
		this->left = left;
		this->right = right;
		cnt = right - left + 1;
	}
};

int main() {
	int n, k;
	cin >> n >> k;
	char s[n + 1];
	cin >> s;

	int one[n + 1];
	vector<Range> range;
	memset(one, 0, sizeof(one));
	range.push_back(Range());

	int st = -1;
	for (int i = 0; i < n; i++) {
		if (s[i] == '0' && st == -1)
			st = i + 1;
		if (s[i] == '1' && st != -1) {
			range.push_back(Range(st, i));
			st = -1;
		}
		one[i + 1] = (s[i] == '1');
	}
	if (st != -1)
		range.push_back(Range(st, n - 1));

	if (k >= (int) range.size() - 1) {
		cout << n;
		return 0;
	}

	for (int i = 1; i <= n; i++) {
		one[i] += one[i - 1];
	}

	for (int i = 1; i < (int) range.size(); i++) {
		range[i].cnt += range[i - 1].cnt;
	}

	int ma = 0;
	for (int i = 1, j = k, cur = 0; j < (int) range.size(); ++i, ++j) {
		cur = range[j].cnt - range[i - 1].cnt;
		cur += one[range[j].left] - one[range[i].right - 1];
		if (range[j].right < n) {
			if (j < (int) range.size() - 1) {
				cur += one[range[j + 1].left - 1] - one[range[j].right];
			} else {
				cur += one[n] - one[range[j].right];
			}
		}
		if (range[i].left > 1) {
			if (i > 0) {
				cur += one[range[i].left - 1] - one[range[i - 1].right];
			} else {
				cur += one[range[i].left - 1];
			}
		}
		ma = max(ma, cur);
	}
	cout << ma;
	return 0;
}