POJ 3579 Median

最新推荐文章于 2019-07-01 16:19:08 发布

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最新推荐文章于 2019-07-01 16:19:08 发布

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分类专栏：二分,三分

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本文介绍了一个算法问题：如何从给定的一组整数中快速计算所有数对差值的中位数。文章提供了一段C++代码实现，通过排序和二分查找的方法来高效解决这个问题。

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Median

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions:8914		Accepted: 3098

Description

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

Sample Output

1
8

#include<cstdio>
#include<algorithm>
using namespace std;
int a[100004];
int main()
{
    long long n;
    while(~scanf("%lld",&n))
    {
        int i,j;
        long long m=n*(n-1)/2;
        for(i=0; i<n; i++)
            scanf("%d",&a[i]);
        sort(a,a+n);
        int q=0,r=1000000000;
        while(r-q>1)
        {
            int mid=(r+q)/2;
            long long ans=0;
            for(i=0;i<n;i++)
            {
                ans+=n-(lower_bound(a+i,a+n,mid+a[i])-a);
            }
            if((m%2==1&&ans>=m/2+1)||(m%2==0&&ans>m/2))
                q=mid;
            else r=mid;
        }
        printf("%d\n",q);
    }
    return 0;
}