B二分法

<span style="color:#330099;">/*
B - 二分法 基金会
Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u
SubmitStatus
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
By Grant Yuan
2014.7.15
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
int a[100002];
int n,c;
int d;
int l,r;
int top;
bool can(int k)
{  //cout<<"k"<<k<<endl;
    int last,sum,sta;
    sta=0;
    sum=1;
    for(int i=1;i<n;i++)
      {
          if(a[i]-a[sta]>=k)
           {//cout<<" i"<<i<<" sta"<<sta<<endl;
                sum++;
               sta=i;
           }
      }
      //cout<<"sum"<<sum<<endl;
      if(sum>=c)
         return true;
     else
         return false;
}int main()
{   int i;
    int num;
    int last;
    while(~scanf("%d%d",&n,&c)){
   // cin>>n>>c;
    for(i=0;i<n;i++)
       cin>>a[i];
    //top=i;
    sort(a,a+n);
    l=a[0];r=a[n-1]+1;
    int mid;
    while(r-l>1){
        mid=(l+r)/2;
       if(can(mid)){
           l=mid;
           }
        else
        {
             r=mid;
        }
  }
cout<<l<<endl;}  }
</span>

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