Shortest Distance of HDU 2083 simple Edition

Problem Description
During the winter vacation, ACBOY visited many friends. It happened that all his friends were on the X axis of the coordinate plane. ACBOY can select a friend's home to start access, but he must go back to the starting point after each visit before accessing the next friend.
For example, if you have four friends, the corresponding X axis coordinates are 1, 2, 3, and 4. When ACBOY selects a point whose coordinate is 2 as the starting point, the final time required by ACBOY is | 1-2 | + | 2-2 | + | 3-2 | + | 4-2 | = 4.
Now, how can ACBOY take the least time to get the coordinates of N friends?

 

Input
The input is a positive integer M, indicating M test instances. Each instance has two input rows. The first is a positive integer N (N <= 500), indicating that there are N friends and the next row is a positive integer, represents the specific coordinates (all data is <= 10000 ).

 

Output
For each test instance, please output the minimum time it takes to access all friends, and the output of each instance occupies one line.

 

Sample Input
2
2
2 4
3
2 4 6

Sample Output
2
4

import java.io.BufferedInputStream;import java.util.*;public class Main {public static void main(String[] args) {Scanner sc=new Scanner(new BufferedInputStream(System.in));int k,m,n;k=sc.nextInt();for(int i=0;i<k;i++){m=sc.nextInt();int a[]=new int[m];for(int j=0;j<a.length;j++){a[j]=sc.nextInt();}n=fun(m,a);System.out.println(n);}}public static int fun(int m,int a[]){int k,d=0;Arrays.sort(a);k=m/2;for(int i=0;i<a.length;i++){d+=Math.abs(a[i]-a[k]);}return d;}}