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 {
	private static Scanner sc=new Scanner(System.in);
	private static int n=sc.nextInt();
	private static int[] x=new int[n];
	private static int[] y=new int[n];
	public static void main (String[] args){
		for(int i=0;i<n;i++) x[i]=sc.nextInt();
		for(int i=0;i<n;i++) y[i]=sc.nextInt();
		System.out.println(cal(1));
		System.out.println(cal(2));
		System.out.println(cal(3));
		System.out.println(calf());
	}
	private static double cal(int k){
		double sum=0;
		for(int i=0;i<n;i++){
			sum=sum+Math.abs(Math.pow(x[i]-y[i],k));
		}
		return Math.pow(sum,1.0/k);
	}
	private static double calf(){
		double currentMax=Math.abs(x[0]-y[0]);
		for(int i=1;i<n;i++){
			currentMax=Math.max(Math.abs(x[i]-y[i]),currentMax);
		}
		return currentMax;
	}
}