TT mystery gift

Topic: TT - the mysterious gift
that Italy:
TT is a severe cat lovers, cat channel on the daily indulge in B station.
One day, a friend ZJM decided to TT TT a problem, if TT can solve this problem, ZJM will buy a cute cat gave TT.
SUMMARY is given array cat [i] a number N, and generate a new array ans [i] with the array. It is defined as the new array for any i, j and i = j, are ans [] = abs (cat [ i] - cat [j])!, 1 <= i <j <= N. This new array. Determine the median, then the median is the sort (len + 1) / 2 corresponding to the position number, '/' is a rounding.
TT desperately wanting bird cute cat, can you help him? ~

Input:
multiple sets of inputs, each input one N, the number N expressed, after the input of a sequence of length N cat, cat [i] <= 1e9, 3 <= n <= 1e5 ~

Output:
Output new array ans median ~

Example:

Problem-solving ideas: First of violence, O (n2) to white (violence also possible, csp could get dotted, of course, to vj white on the right). Then, the optimization; first, sorting the original array, so that it must xj-xi is greater than 0; the use after xj-xi <= p, find the two points p, find that there are several xj satisfies inequality (enumeration XI), when there xj (n * (n-1) / 2 + 1) / 2 th when p has been described distance close to the median (likely less than the median, but are smaller than p (len + 1) / 2, when p is not the median, must continue to increase until the last 1 p plus the condition is not satisfied, then p is the median). Of course, the search process is to be xj dichotomy, which is approximately the time complexity of O (nlogn + nlogn * logn);

Code:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
using namespace std;
int a[100005];
void gengxin()
{
    memset(a,0,sizeof(a));
}
int panduan(int k,int n)
{
    int total=0,m=n*(n-1)/2;
    m=(m+1)/2;
    for(int i=0;i<n-1;i++)//通过枚举xi找xj总个数
    {
        int h=a[i]+k;
        int l=0,r=n;
        while(l+1<r)//二分
        {
            int mid=(l+r)/2;
            if(a[mid]>h)
            {
                r=mid;
            }else
            {
                l=mid;
            }
        }
        total=total+(l-i);
    }
    if(total>=m)//根据xj的总数返回不同的值，0说明p得小点，1说明p得大点
    {
        return 0;
    }else
    {
        return 1;
    }
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        gengxin();
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        sort(a,a+n);//排序
        int l=0,r=a[n-1];
        while(l+1<r)
        {
            int mid=(r+l)/2;
            int flag=panduan(mid,n);
            if(flag==0)//判断
            {
                r=mid;
            }else
            {
                l=mid;
            }
        }
        printf("%d\n",r);
    }
}