POJ 3579 Median (二分）

                                                                                                     Median

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7423		Accepted: 2538

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

4
1 3 2 4
3
1 10 2

Sample Output

1
8

Source

POJ Founder Monthly Contest – 2008.04.13, Lei Tao

题意：一个序列  他们的差值的绝对值产生一个新的序列  问这个序列的中位数是那个

二分答案 

判断的地方我们也需要用二分 假如 中位数是x的话  我们判断最开始的那个数组a

假如我们知道a[i]+x在a数组的位置 那么我就知道 和a[i]的差值小于等于x的个数为n-t个 我们就可以知道有多少个数是小于这个x的 

 

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<cctype>
 #include<cmath>
 #include<cstring>
 #include<map>
 #include<stack>
 #include<set>
 #include<vector>
 #include<algorithm>
 #include<string.h>
 typedef long long ll;
 typedef unsigned long long LL;
 using namespace std;
 const int INF=0x3f3f3f3f;
 const double eps=0.0000000001;
 const int N=+;
 int a[N];int n,m;
 int judge(int x){
     int sum=;
     for(int i=;i<n;i++){
         int t=lower_bound(a,a+n,a[i]+x)-a;
         // 存在差值 x
         sum=sum+n-t;// 和a[i]的差值小于等于x的个数为n-t；
     }
     if(sum>m)return ;
     else{
         return ;
     }
 }
 int main(){

     while(scanf("%d",&n)!=EOF){
         int maxx=;
         for(int i=;i<n;i++){
             scanf("%d",&a[i]);
         }
         sort(a,a+n);
         m=n*(n-)/;
         int low=;
         int ans;
         int high=a[n-];
         while(low<=high){
             int mid=(low+high)>>;
             if(judge(mid)){
                 ans=mid;
                 low=mid+;
             }
             else
                 high=mid-;
         }
         cout<<ans<<endl;
     }
 }