Problem Description

Farmer John has received a noise complaint from his neighbor, Farmer Bob, stating that his cows are making too much noise.

FJ's N cows (1 <= N <= 10,000) all graze at various locations on a long one-dimensional pasture. The cows are very chatty animals. Every pair of cows simultaneously carries on a conversation (so every cow is simultaneously MOOing at all of the N-1 other cows). When cow i MOOs at cow j, the volume of this MOO must be equal to the distance between i and j, in order for j to be able to hear the MOO at all. Please help FJ compute the total volume of sound being generated by all N*(N-1) simultaneous MOOing sessions.

Input

* Line 1: N

* Lines 2..N+1: The location of each cow (in the range 0..1,000,000,000).

Output

There are five cows at locations 1, 5, 3, 2, and 4.

Sample Input

5
1
5
3
2
4

Sample Output

————————————————————————————————————————————————————

题意：给出 n 个栏位以及其编号，现将 c 头牛放到栏位中，要求任意两头牛的栏位编号的最小差值最大。

思路：

一开始题的每个字都认识就是不知道要求什么，样例也看不明白，看了题解才稍微有些思路。

样例排序后，栏位编号是 1 2 4 8 9，那么 1 放一头牛，4 放一头牛，它们的差值为 3，最后一头牛放在 8 或 9 位置均可，和 4 的差值分别为 4、5，与 1 的差值分别为 7、8，均不比 3 小，所以最大的最小值为 3

现将栏位排序，则最大距离不会超过两端的两头牛之间的差值，最小值为0，可以通过二分枚举最小值来求。

假设当前的最小值为 x，如果最小差值为 x 时就可以放下 c 头牛，那么就让 x 变大再进行判断，如果放不下，说明当前 x 值太大，那么就让 x 变小再进行判断，直到求出一个最大的 x 值。

Source Program

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<cstdlib>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<vector>
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define N 100001
#define MOD 123
#define E 1e-6
using namespace std;
int n,c;
int stall[N];
bool judge(int x)
{
    int temp=stall[0];
    int cnt=1;
    for(int i=1;i<n;i++)
    {
        if(stall[i]-temp>=x)
        {
            cnt++;
            temp=stall[i];
            if(cnt>=c)
                return true;
        }
    }
    return false;
}
int main()
{
	scanf("%d%d",&n,&c);
	for(int i=0;i<n;i++)
        scanf("%d",&stall[i]);

	sort(stall,stall+n);

	int left=0,right=stall[n-1]-stall[0];
	while(left<=right)
    {
        int mid=(left+right)/2;
        if(judge(mid))
            left=mid+1;
        else
            right=mid-1;
    }

	printf("%d\n",right);

	return 0;
}

Problem N：Aggressive cows（POJ-2456）