Problem:
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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.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n=sc.nextInt();
		double[] x=new double[n];
		double[] y=new double[n];
		for(int i=0; i<n; i++){
			x[i]=sc.nextDouble();
		}
		for(int i=0; i<n; i++){
			y[i]=sc.nextDouble();
		}
		d1(x,y);
		d2(x,y);
		d3(x,y);
		dmax(x,y);

    }
	public static void d1(double[] x,double[] y){
		double sum=0;
		for(int i=0; i<x.length; i++){
			sum+=Math.abs(x[i]-y[i]);
		}
		System.out.printf("%.6f\n",sum);
	}
	public static void d2(double[] x,double[] y){
		double sum=0;
		for(int i=0; i<x.length; i++){
			sum+=Math.pow(Math.abs(x[i]-y[i]), 2);

		}
		sum=Math.sqrt(sum);
		System.out.printf("%.6f\n",sum);
	}
	public static void d3(double[] x,double[] y){
		double sum=0;
		for(int i=0; i<x.length; i++){
			sum+=Math.pow(Math.abs(x[i]-y[i]), 3);

		}
		sum=Math.pow(sum, 1.0/3);
		System.out.printf("%.6f\n",sum);
	}
	public static void dmax(double[] x, double[] y){
		double max=0;
		for(int i=0; i<x.length; i++){
			if(Math.abs(x[i]-y[i])>max){
				max=Math.abs(x[i]-y[i]);
			}
		}
		System.out.printf("%.6f\n",max);
	}
}