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发表于 2017-07-17 22:32:54 | 查看: 576 | 回复: 1
Given N numbers, X1, X2, … , XN, let us calculate the difference of every pair of numbers: ∣Xi - Xj∣ (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, … , XN, ( Xi ≤ 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

题意:给N个数,求他们之间差值绝对值的中间值。

做法:二分两次,第一次二分差值,看这个差值能不能达到差值总数的一半,这个判断方法是对于每个xi,查找有多少个值大于等于xi加上这个差值,也可以用二分。

注意:当二分对象是离散的情况的处理方式,上界和下界的确定,
upperbound函数的应用。

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#define maxn 100010using namespace std;int x[maxn];int n,m;bool solve(int d){    int ans=0;    int t;    for(int i=0;i<n;i++)    {        t=upper_bound(x,x+n,x[i]+d)-x;        ans+=t-1-i;    }    return ans>=(m+1)/2;}int main(){    while(scanf("%d",&n)==1)    {        m=(n-1)*n/2;        for(int i=0;i<n;i++)        {            scanf("%d",&x[i]);        }        sort(x,x+n);        int st=0;        int end=x[n-1]-x[0];        int mid=0;        int ans;//这种处理方式很重要,保证了ans一定满足solve(ans)        while(end>=st)        {            mid=(end+st)/2;            if(solve(mid)){end=mid-1;ans=mid;}            else st=mid+1;        }        cout<<ans<<endl;    }    return 0;}

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发表于 2017-07-17 22:33:37

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