POJ3579：Median（二分）

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本文介绍了一种通过二分查找及二分判断的方法来高效计算一组数两两差值的中位数。针对大规模数据集，该算法能够在规定的时间限制内得出结果。

Median

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6929		Accepted: 2321

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

Source

POJ Founder Monthly Contest – 2008.04.13, Lei Tao

题意：给出n个数，将他们的两两差值排序，找到中位数。

思路：二分答案+二分判断。

# include <iostream>
# include <cstdio>
# include <algorithm>

using namespace std;
int a[100001];
long long n, tar;

bool fun(int x)
{
    long long sum = 0;
    for(int i=0; i<n-1; ++i)
    {
        int pos = upper_bound(a+i, a+n, a[i]+x)-a;
        sum += pos - i - 1;
        if(sum >= tar)
            return true;
    }
    return false;
}
int main()
{
    while(~scanf("%lld",&n))
    {
        for(int i=0; i<n; ++i)
            scanf("%d",&a[i]);
        long long tmp = n*(n-1)>>1;
        if(tmp&1)
            tar = (tmp>>1)+1;
        else
            tar = tmp>>1;
        sort(a, a+n);
        int l=0, r=a[n-1]-a[0], mid;
        while(l<r)
        {
            int mid = (l + r)>>1;
            if(fun(mid))
                r = mid;
            else
                l = mid+1;
        }
        printf("%d\n",r);
    }
    return 0;
}