题目链接：http://poj.org/problem?id=3258

River Hopscotch

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 22092		Accepted: 9138

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 D_i from the start (0 < D_i < 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 Mrocks.

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).

Source

USACO 2006 December Silver   题目大意：输入L N M  L代表所有的石头距离在起点的最大距离 N代表有N个石头  M代表你可以消去M个石头  问你消去M个石头之后   你走的每一步里面最小值的最大值是多少 思路: 赤裸裸的二分 但是自己没想到 真的菜   其实自己的思路是  先把所有的点里原点的距离从小到大排序一遍，然后取最小的。 但是这样不容易处理啊  因为你取了一个的话  后面一个的距离和前面一个的 距离就变了  如果要维护这一点的话  就算你记录了他们的下标  也改变了后面一个点和前面一个点的距离  然后呢？  你又得把新得到的数组重新从小到大排序  真的不好控制
但是二分来解是真的快！ 二分的是答案    思路真的简单  但是没有想到     仔细想想，答案肯定在0-L之间的   判断一个值能否成立也很容易  太菜了。。。 看代码：