poj 3579 Median 中间值（二分搜索）
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 ifm,the amount of the differences, is even. For example, you have to find the third smallest one in the case of?m?= 6.
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 )
For each test case, output the median in a separate line.
Sample Output
1
8
#include <iostream> #include <cstdio> #include <algorithm> #define INF 1000000 using namespace std; int main() {int n;int a[100000];while(cin>>n){for(int i=0;i<n;++i)scanf("%d",&a[i]);int m,temp=n*(n-1)/2;if(temp&1)m=temp/2+1;elsem=temp/2; //有n個數，按照順序排列，涼涼相減，則有n-1個數，問這n-1個數的中間值所在的位置（1......n-1排列中的位 置)m，其1、2....m的和為多少？sort(a,a+n);int low=0,high=a[n-1]-a[0];int mid,value;while(high>=low)//{int ans=0;mid=low+(high-low)/2;int pos;for(int i=0;i<n-1;++i){pos=upper_bound(a+i,a+n,mid+a[i])-a;//mid(兩數之差，設定為中間值）則a[i]+mid為數組a中的中間值,pos為中間值在數組中的位置。//ans+=pos-i-1;}if(ans>=m){value=mid;high=mid-1;}elselow=mid+1;}cout<<high+1<<endl;} }
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