Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

Distance II

Your task is to calculate the distance between two $n$ dimensional vectors $x = \{x_1, x_2, ..., x_n\}$ and $y = \{y_1, y_2, ..., y_n\}$.

The Minkowski's distance defined below is a metric which is a generalization of both the Manhattan distance and the Euclidean distance.

\[
D_{xy} = (\sum_{i=1}^n |x_i - y_i|^p)^{\frac{1}{p}}
\]

It can be the Manhattan distance

\[
D_{xy} = |x_1 - y_1| + |x_2 - y_2| + ... + |x_n - y_n|
\]

where $p = 1 $.

It can be the Euclidean distance

\[
D_{xy} = \sqrt{(|x_1 - y_1|)^{2} + (|x_2 - y_2|)^{2} + ... + (|x_n - y_n|)^{2}}
\]

where $p = 2 $.

Also, it can be the Chebyshev distance

\[
D_{xy} = max_{i=1}^n (|x_i - y_i|)
\]

where $p = \infty$

Write a program which reads two $n$ dimensional vectors $x$ and $y$, and calculates Minkowski's distance where $p = 1, 2, 3, \infty$ respectively.

Input

In the first line, an integer $n$ is given. In the second and third line, $x = \{x_1, x_2, ... x_n\}$ and $y = \{y_1, y_2, ... y_n\}$ are given respectively. The elements in $x$ and $y$ are given in integers.

Output

Print the distance where $p = 1, 2, 3$ and $\infty$ in a line respectively.
The output should not contain an absolute error greater than 10-5.

Constraints

$1 \leq n \leq 100$

$0 \leq x_i, y_i \leq 1000$

Sample Input

3
1 2 3
2 0 4

Sample Output

4.000000
2.449490
2.154435
2.000000

Solution in JAVA (judge verdict: Accepted)

import java.util.*;
public class Main {
	public static void main(String...args) {
    	Scanner scan = new Scanner(System.in);
    	int times = scan.nextInt();
    	int[] x = new int[times];
    	int[] y = new int[times];
    	double p1   = 0;
    	double p2   = 0;
    	double p3   = 0;
    	List<Double> pMax = new ArrayList<>();
    	
    	for(int i = 0; i < 2; i++) {
    		for(int n = 0; n < times ; n++) {
    			if(i == 0) {
    				x[n] = scan.nextInt();
    			} else if(i == 1) {
    				y[n] = scan.nextInt();
    			}
    		}
    	}
		for(int m = 0; m < times ; m++) {
			double test = (x[m] - y[m]);
			test = Math.abs(test);
			pMax.add(Math.abs(test));
			p1 += Math.abs(test) ;
			p2 +=  Math.pow(Math.abs(x[m] - y[m]),2);
			p3 +=  Math.pow(Math.abs(x[m] - y[m]),3);
		}
		p2 = Math.sqrt(p2);
		p3 = Math.pow(p3 , 1.0/3.0);
		Collections.sort(pMax);
		System.out.println(p1);
		System.out.println(p2);
		System.out.println(p3);
		System.out.println(pMax.get(pMax.size()-1));
	}
}