H - Median

H - Median

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

        之前一直都是在怕超時，後來竟然沒有超，還是特別開心的啦！

        思路：我在這裏是用0~（ai[n-1]-ai[0]）之間的差值進行的二分的方法，之後用了一個雙重的循環，來判斷小於m的個數，                     方法還是比較簡單易懂的，相信可以理解的！

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
long long n,m;
int ai[100005];
int fun(int mid)
{
    int sum=0,r=1;
    for(int i=0;i<n;i++)
    {
        while(ai[r]-ai[i]<=mid&&r<n) r++;
        sum+=r-i-1;//減1 是因爲上面多加了一個 1；
    }
    if(sum>=m) return 1;
    return 0;
}
int main()
{

    while(scanf("%lld",&n)==1)
    {
        for(int i=0;i<n;i++)
            scanf("%d",&ai[i]);
        m=n*(n-1)/2;
        if(m%2==0) m=m/2;
        else    m=(m+1)/2;
        sort(ai,ai+n);
        int q=0,w=ai[n-1]-ai[0],mid;
        while(q<=w)
        {
            mid=(q+w)/2;
            if(fun(mid))
                w=mid-1;/*爲什麼是把 mid-1 賦值給 w 呢？是因爲，如果滿足了那個 m 的話，說明小於 這個差是比較多的，所以                                        二分是要把這個範圍縮小，變成 [q,mid-1]*/
            else
                q=mid+1;
        }
        printf("%d\n",w+1);
    }
}