Median(二分+二分)

Median

 http://poj.org/problem?id=3579

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11225		Accepted: 4016

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

#include<iostream>
#include<algorithm>
#include<string>
#include<map>
#include<vector>
#include<cmath>
#include<string.h>
#include<stdlib.h>
#include<stack>
#include<queue>
#include<cstdio>
#define ll long long
const long long MOD=1000000007;
#define maxn 100005
using namespace std;

int n;
int m;
int a[maxn];

bool Check(int mid){
    int sum=0;
    int p;
    for(int i=1;i<=n;i++){
        p=upper_bound(a+1,a+n+1,a[i]+mid)-a;
        sum+=p-i-1;//排除a[i]之前的那些元素,共有i+1;
    }
    if(sum>=m){
        return true;
    }
    return false;
}

int main(){
    while(~scanf("%d",&n)){
        m=n*(n-1)/2;
        m=(m+1)/2;
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
        }
        sort(a+1,a+n+1);
        ll L=0,R=1000000000,mid;
        while(L<=R){
            mid=L+R>>1;
            if(Check(mid)){
                R=mid-1;
            }
            else{
                L=mid+1;
            }
        }
        printf("%d
",L);
    }

}