Median - POJ 3579 二分_mediangiven n numbers, x1, x2, ... , xn, let us ca-优快云博客

本文介绍了一个高效算法来找出给定一组数的所有两两差的中位数。通过排序和二分查找，该算法能快速解决这一问题。

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Median

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3699		Accepted: 1056

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

Sample Output

1
8

题意：找到所有数两者差的集合中和第k小的数，k为（C(N,2)+1）/2，因为如果不能整除的话要+1再除以2。

思路：排序后二分第k小的数，每次枚举对于每个数，左边有多少个比它小的数可以使他们的差不大于二分出的第k小的数，这里还要再用一次二分。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,num[100010],k,mi;
int divi(int pos)
{ int l=1,r=pos-1,mid;
  while(l<=r)
  { mid=(l+r)/2;
    if(num[pos]-num[mid]<=mi)
     r=mid-1;
    else
     l=mid+1;
  }
  return l;
}
bool solve()
{ int ans=0,i;
  for(i=2;i<=n;i++)
   ans+=i-divi(i);
  if(ans>=k)
   return true;
  else
   return false;
}
int main()
{ int i,j,l,r,ans;
  while(~scanf("%d",&n))
  { for(i=1;i<=n;i++)
     scanf("%d",&num[i]);
    sort(num+1,num+1+n);
    k=n*(n-1)/2;
    if(k%2==0)
     k/=2;
    else
     k=(k+1)/2;
    l=1;r=1000000000;
    while(l<=r)
    { mi=(l+r)/2;
      if(solve())
       r=mi-1;
      else
       l=mi+1;
    }
    printf("%d\n",l);
  }
}