Codeforces 645C：Enduring Exodus

C. Enduring Exodus

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other side of Bovinia. During the journey, he and his k cows
have decided to stay at the luxurious Grand Moo-dapest Hotel. The hotel consists of nrooms located in a row, some of which are occupied.

Farmer John wants to book a set of k + 1 currently unoccupied rooms for him and his cows. He wants his cows to stay as safe as possible,
so he wishes to minimize the maximum distance from his room to the room of his cow. The distance between rooms i and j is
defined as |j - i|. Help Farmer John protect his cows by calculating this minimum possible distance.

Input

The first line of the input contains two integers n and k (1 ≤ k < n ≤ 100 000) —
the number of rooms in the hotel and the number of cows travelling with Farmer John.

The second line contains a string of length n describing the rooms. The i-th
character of the string will be '0' if the i-th
room is free, and '1' if the i-th
room is occupied. It is guaranteed that at least k + 1 characters of this string are '0',
so there exists at least one possible choice of k + 1 rooms for Farmer John and his cows to stay in.

Output

Print the minimum possible distance between Farmer John's room and his farthest cow.

Examples

input

7 2
0100100

output

2

input

5 1
01010

output

2

input

3 2000

output

1

Note

In the first sample, Farmer John can book room 3 for himself, and rooms 1 and 4 for
his cows. The distance to the farthest cow is 2. Note that it is impossible to make this distance 1,
as there is no block of three consecutive unoccupied rooms.

In the second sample, Farmer John can book room 1 for himself and room 3 for
his single cow. The distance between him and his cow is 2.

In the third sample, Farmer John books all three available rooms, taking the middle room for himself so that both cows are next to him. His distance from the farthest cow is 1.

题意是john带着k个牛进旅馆，旅馆房间的情况是0代表是空房间，1代表是有人住了。问把john和牛都安排好，john和牛的最远距离。

枚举john所在位置，二分其最大距离。用前缀和判断两头有没有k+1个房间。

代码：

#pragma warning(disable:4996)
#include <iostream>
#include <functional>
#include <algorithm>
#include <cstring>
#include <vector>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <deque>
#include <set>
#include <map>
using namespace std;
typedef long long ll;

#define INF 0x333f3f3f

const ll mod = 1000000007;
const int maxn = 100005;
const double PI = acos(-1.0);

int n, k;
int pre[maxn], pos[maxn];
char x[maxn];

bool check(int pos, int mid)
{
int ri = min(n, pos + mid);
int le = max(1, pos - mid);

return pre[ri] - pre[le - 1] >= k + 1;
}

void solve()
{
int i, j, nx;
cin >> n >> k >> x + 1;

for (i = 1; i <= n; i++)
{
if (x[i] == '0')
{
pre[i] = pre[i - 1] + 1;
}
else
{
pre[i] = pre[i - 1];
}
}
int le, ri, mid;
int ans = n;
for (i = 1; i <= n; i++)
{
if (x[i] == '1')continue;
le = 1, ri = n;
while (le < ri)
{
mid = (le + ri) >> 1;
if (check(i, mid))
{
ri = mid;
}
else
{
le = mid + 1;
}
}
ans = min(ans, ri);
}
cout << ans;
}

int main()
{
#ifndef ONLINE_JUDGE
freopen("i.txt", "r", stdin);
freopen("o.txt", "w", stdout);
#endif
solve();
//system("pause");
return 0;
}