M - Median - POJ-3575 ——mystical_curve

最新推荐文章于 2018-07-19 15:50:49 发布

原创最新推荐文章于 2018-07-19 15:50:49 发布 · 287 阅读

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本文介绍了一个算法问题：给定一系列整数，快速找出所有可能的两两数值差值的中位数。文章提供了一段C++代码实现，通过排序和二分查找的方法高效地解决了这个问题。

Given N numbers, X1, X2, … , XN, let us calculate the difference of every pair of numbers: ∣Xi - Xj∣ (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, … , XN, ( Xi ≤ 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

水题

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
using namespace std;
typedef long long ll;
const long long INF = 1000500;
ll n,s[100050],m;
bool fun(int mid)
{
    ll t=0;
    for(int i=0;i<n;i++)
    {
        ll c= upper_bound(s+i,s+n,s[i]+mid)-s-i-1;
        t+=c;
    }
    return t<m;
}
int main()
{
    while(~scanf("%d",&n))
    {
         m=n*(n-1)/2;
         if(m%2)
            m=m/2+1;
         else
            m=m/2;
        for(int i=0;i<n;i++)
            scanf("%d",s+i);
        sort(s,s+n);
        ll l=0,r=s[n-1]-s[0];
        while(r-l>1)
        {
            ll mid=(l+r)/2;
            if(fun(mid))
                l=mid;
            else
                r=mid;

        }
        cout<<r<<endl;
    }
    return 0;
}