Median POJ - 3579 （二分）

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

题意：任意两个数的差值的绝对值，让你找出这些值中的中位数，如果数有6个那么你找出第3大的

显然有一个值满足大于等于这个值的个数大于等于k个（比如总共6个差值，那么k为3 3个差值，k为 2）,如果这个值-1 就小于k个那么说明这个值是中位数

#include<stdio.h>
#include<string.h>
#include<map>
#include<vector>
#include<iostream>
#include<algorithm>
#include<queue>
#include<deque>
#include<math.h>
#define ll long long
const ll inf=1e10;
using namespace std;
ll a[1000005];
ll k,n;
int aa(ll x)
{
   ll uu=0;
   for(ll i=0;i<n-1;i++)
   {
       ll o=upper_bound(a+i+1,a+n,x+a[i])-a; //因为排过序 ，所以 x+a[i]就是 差值+a[i] ，可以找到有几个差值在x以下
       uu+=o-i-1;
       if(uu>=k) return 1;//如果大于等于k个  返回 1
   }
   return 0;
}
int main()
{
  while(~scanf("%lld",&n))
  {
      k=n*(n-1)/2;
      k=(k+1)/2;
      for(int i=0;i<n;i++)
        scanf("%lld",&a[i]);
      sort(a,a+n);
      ll r=inf,l=0,mid,ans;//二分中位数的值
      while(l<=r)
      {
          mid=(l+r)/2;
//          printf("%lld\n",mid);
          if(aa(mid))
          {
              ans=mid;
              r=mid-1;
          }
          else
            l=mid+1;
      }
      printf("%lld\n",ans);
  }
}

Median POJ - 3579 （二分）

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