Median POJ

Given N numbers, X₁, X₂, ... , X_N, let us calculate the difference of every pair of numbers: ∣X_i - X_j∣ (1 ≤ i ＜ j ≤ N). We can get C(N,2) differences through this work, and now your task is to find the median of the differences as quickly as you can!

Note in this problem, the median is defined as the (m/2)-th  smallest number if m,the amount of the differences, is even. For example, you have to find the third smallest one in the case of m = 6.

Input

The input consists of several test cases.
In each test case, N will be given in the first line. Then N numbers are given, representing X₁, X₂, ... , X_N, ( X_i ≤ 1,000,000,000  3 ≤ N ≤ 1,00,000 )

Output

For each test case, output the median in a separate line.

Sample Input

4
1 3 2 4
3
1 10 2

Sample Output

1
8

1.二分答案和二分验证。
2.验证的时候要找大于中间值的差值。
3.这是大神（不是本人）的代码~~~

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#define MAX 100005
using namespace std;
typedef long long ll;

int n,m;
ll a[MAX];

bool check(ll ans){
    int cnt=0;
    for(int i=1;i<=n;i++)
        cnt+=n-(lower_bound(a+1,a+n+1,a[i]+ans)-a-1);
    if(cnt>m) return true;
    else return false;
}
int main()
{  while(~scanf("%d",&n)){
      
      for(int i=1;i<=n;i++) scanf("%d",&a[i]);
      sort(a+1,a+n+1);
      m=n*(n-1)/4;
      ll l=0,r=a[n]-a[1];
      ll mid,ans;
      while(l<=r){
           mid=(l+r)>>1;
           if(check(mid)){
               ans=mid;
               l=mid+1;
           }
           else r=mid-1;
      }
      printf("%d
",ans);
   } 
   return 0;
}