Description

Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and another rock at the end, L units away from the start (1 ≤ L

 ≤ 1,000,000,000). Along the river between the starting and ending rocks, N (0 ≤ N ≤ 50,000) more rocks appear, each at an integral distance Di from the start (0 < Di < L).

To play the game, each cow in turn starts at the starting rock and tries to reach the finish at the ending rock, jumping only from rock to rock. Of course, less agile cows never make it to the final rock, ending up instead in the river.

Farmer John is proud of his cows and watches this event each year. But as time goes by, he tires of watching the timid cows of the other farmers limp across the short distances between rocks placed too closely together. He plans to remove several rocks in order to increase the shortest distance a cow will have to jump to reach the end. He knows he cannot remove the starting and ending rocks, but he calculates that he has enough resources to remove up to M 

rocks (0 ≤ M ≤ N).

FJ wants to know exactly how much he can increase the shortest distance *before* he starts removing the rocks. Help Farmer John determine the greatest possible shortest distance a cow has to jump after removing the optimal set of M rocks.

Input

Line 1: Three space-separated integers: L, N, and M 
Lines 2..N+1: Each line contains a single integer indicating how far some rock is away from the starting rock. No two rocks share the same position.

Output

Line 1: A single integer that is the maximum of the shortest distance a cow has to jump after removing M rocks

Sample Input

25 5 2
2
14
11
21
17

Sample Output

4

Hint

Before removing any rocks, the shortest jump was a jump of 2 from 0 (the start) to 2. After removing the rocks at 2 and 14, the shortest required jump is a jump of 4 (from 17 to 21 or from 21 to 25).

題意：牛要到河對岸，在與河岸垂直的一條線上，河中有N塊石頭，給定河岸寬度L，以及每一塊石頭離牛所在河岸的距離，

          現在去掉M塊石頭，要求去掉M塊石頭後，剩下的石頭之間以及石頭與河岸的最小距離的最大值。

思路：簡單的二分查詢，注意不要漏掉終點

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int a[50005];
int end,n,m;
int judge(int x)   //這個函式是關鍵 
{
	int num=0;
	int index=0;
	for(int i=0;i<=n;i++)
	{
		if(a[i]-index<=x)   
		{
			num++;
		}
		else index=a[i];
		if(num>m) return 0;
	}
	return 1;
}
int solve()
{
	int l=1,r=end;
	while(l<=r)
	{
		int mid=(l+r)>>1;
		//cout<<mid<<endl;
		if(judge(mid)) l=mid+1;
		else r=mid-1;
	}
	cout<<l<<endl;
}
int main()
{
	scanf("%d%d%d",&end,&n,&m);
	for(int i=0;i<n;i++)
	scanf("%d",&a[i]);
	sort(a,a+n);
	a[n]=end;   //注意不要忘了加上終點 
	solve();
	return 0;
}