Codeforces--645C--Enduring Exodus（二分）

Enduring Exodus

Time Limit: 2000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

Submit Status

Description

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 n rooms 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.

Sample Input

Input

7 2
0100100

Output

2

Input

5 1
01010

Output

2

Input

3 2
000

Output

1

题意：有一个人去住旅店，随身带了k头牛，旅店有n个房间，0表示没人，1表示有人这个人想尽可能的让自己跟牛住得近，求最远的牛跟他之间的距离

思路：先记录前缀和，然后二分枚举最大长度，判断这个长度是否可以，因为求的是最短的距离所以判断稍微改一下

#include<cstdio>
#include<cstring>
#include<queue>
#include<cmath>
#include<algorithm>
using namespace std;
int n,k,dp[100000+100];
char s[100000+100];
bool judge(int x)
{
	for(int i=1;i<=n;i++)
	{
		if(s[i]=='0')
		{
			int l=i-x;
			int r=i+x;
			if(l<=1) l=1;
			if(r>=n) r=n;
			if(dp[r]-dp[l-1]-1>=k)
			{
				return true;
			}
		}
	}
	return false;
}
int erfen(int l,int r)
{
	int ans=0;
	while(l<=r)
	{
		int mid=(l+r)/2;
		if(judge(mid))
		{
			ans=mid;
			r=mid-1;
		}
		else
		l=mid+1;
	}
	return ans;
}
int main()
{
	while(scanf("%d%d",&n,&k)!=EOF)
	{
		memset(s,0,sizeof(s));
		scanf("%s",s+1);
		memset(dp,0,sizeof(dp));
		dp[0]=0;
		for(int i=1;i<=n;i++)
		{
			if(s[i]=='0')
				dp[i]=dp[i-1]+1;
			else 
				dp[i]=dp[i-1];
		}
		printf("%d\n",erfen(0,n));
	}
	return 0;
}