Aggressive cows

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 14977	Accepted: 7219

Description

Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,...,xN (0 <= xi <= 1,000,000,000). 

His C (2 <= C <= N) cows don't like this barn layout and become aggressive towards each other once put into a stall. To prevent the cows from hurting each other, FJ want to assign the cows to the stalls, such that the minimum distance between any two of them is as large as possible. What is the largest minimum distance?

Input

* Line 1: Two space-separated integers: N and C 

* Lines 2..N+1: Line i+1 contains an integer stall location, xi

Output

* Line 1: One integer: the largest minimum distance

Sample Input

5 3
1
2
8
4
9

Sample Output

3

Hint

OUTPUT DETAILS: 

FJ can put his 3 cows in the stalls at positions 1, 4 and 8, resulting in a minimum distance of 3. 

Huge input data,scanf is recommended.

題目分析：先對隔間編號從小到大排序，則最大距離不會超過兩端的兩頭牛之間的差值，最小值為0。所以我們可以通過二分列舉最小值來求。假設當前的最小值為x，如果判斷出最小差值為x時可以放下C頭牛，就先讓x變大再判斷；如果放不下，說明當前的x太大了，就先讓x變小然後再進行判斷。直到求出一個最大的x就是最終的答案。

程式碼如下：

//
// Created by kzl on 17-7-25.
//

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxx = 1e5+500;
int n,c;
int cun[maxx];

bool flag(int num){
    int aa = cun[0];
    int temp = 1;
    for(int i=1;i<n;i++)
    {
        if(cun[i]-aa>=num){
            temp++; aa = cun[i];
        }
if(temp==c)return true;
    }
    return false;
}


int main(){
while (scanf("%d%d",&n,&c)!=EOF)
{
    for(int i=0;i<n;i++)scanf("%d",&cun[i]);
    sort(cun,cun+n);
    int l = 1,r = cun[n-1],mid = 0,ans = r+1;
    while(l<=r){
        mid = ((r-l)>>1)+ l;
        if(flag(mid))l = mid + 1;
        else  r = mid - 1;
    }
    printf("%d\n",l-1);
}
    return 0;
}