Time
Limit: 3 sec
Background
The
world-known gangster Vito Deadstone is moving to New
York. He has a very big family there, all of them living in Lamafia Avenue. Since he will visit all his relatives very
often, he is trying to find a house close to them.
Description
Vito
wants to minimize the total distance to all of them and has blackmailed you to
write a program that solves his problem.
Input
The
input consists of several test cases. The first line contains the number of test
cases.
For
each test case you will be given the integer number of relatives r ( 0 < r < 500) and the street numbers (also
integers) where they live ( 0
< si < 30000 ). Note that
several relatives could live in the same street number.
Output
For
each test case your program must write the minimal sum of distances from the
optimal Vito's house to each one of his relatives. The distance between two
street numbers si and sj is dij= |si-sj|.
Sample
Input
|
Sample
Output
|
2
2
2 4
3
2 4 6
|
2
4
|
import java.util.*;
public class main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int test_time = sc.nextInt(); //輸入測試次數
for (int c=0; c<test_time; c++)
{
int[] r = new int[sc.nextInt()]; //輸入親戚數量
for (int i=0; i<r.length; i++)
{
r[i] = sc.nextInt(); //輸入親戚住在哪條街上
}
Arrays.sort(r); //排序( Array.sort() ),將號碼從小排到大
int m = r[r.length/2]; //找出陣列中點
int sum=0;
for (int i=0; i<r.length; i++)
{
sum = sum + Math.abs(r[i]-m); // Math.abs絕對值
}
System.out.println(sum);
}
}
}
public class main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int test_time = sc.nextInt(); //輸入測試次數
for (int c=0; c<test_time; c++)
{
int[] r = new int[sc.nextInt()]; //輸入親戚數量
for (int i=0; i<r.length; i++)
{
r[i] = sc.nextInt(); //輸入親戚住在哪條街上
}
Arrays.sort(r); //排序( Array.sort() ),將號碼從小排到大
int m = r[r.length/2]; //找出陣列中點
int sum=0;
for (int i=0; i<r.length; i++)
{
sum = sum + Math.abs(r[i]-m); // Math.abs絕對值
}
System.out.println(sum);
}
}
}
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