poj3579 二分

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如题：http://poj.org/problem?id=3579

Median

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4518		Accepted: 1389

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

Source

POJ Founder Monthly Contest – 2008.04.13, Lei Tao

思路：对于所有的组合一共C（n,2）种，不可能一一枚举，因此二分C（x）：当前x作为中位数偏大？如果偏大，在左边找，否则，在右边找。

对a数组排序，对于a[i],如果>=a[i]+x的个数<=C(n,2)/2，则偏大注意相等时也是偏大，因为中位数落在了右侧区间。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define MAXN 100005
#define max(a,b)(a>b?a:b)
int a[MAXN];

int C(int x,int n)
{
int cnt=0;
int i;
for(i=0;i<n;i++)
{
cnt+=a+n-lower_bound(a+i,a+n,x+a[i]);
}
return cnt<=n*(n-1)/4;
}

int main()
{
//freopen("C:\\1.txt","r",stdin);
int n;
while(cin>>n)
{
  int i;
  int l=0,r=0;
  for(i=0;i<n;i++)
  {
   scanf("%d",&a[i]);
   r=max(r,a[i]);
  }
  sort(a,a+n);
  while(r-l>1)
  {
   int mid=(l+r)/2;
   if(C(mid,n))
    r=mid;
   else
    l=mid;
  }
  cout<<l<<endl;
}
return 0;
}