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.io.*;
import java.util.Arrays;
import java.util.Scanner;

public class Main{
	public static void main(String[] args) throws IOException{ 
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String fst = br.readLine();
		int n = Integer.parseInt(fst);
		double[] x = new double[n]; 
		double[] y = new double[n]; 
		String[] linex = (br.readLine()).split(" ");
		String[] liney = (br.readLine()).split(" ");
		for (int i = 0; i<n ; i++ ) {
			x[i] = Double.parseDouble(linex[i]);
			y[i] = Double.parseDouble(liney[i]);
		}
		Double sum = 0.0;
		double p = 1;
		for (int i =0; i<n ;i++ ) {
			sum += Math.pow(Math.abs(x[i]-y[i]),p);
		}

		Double d = Math.pow(sum,(1/p));
		System.out.printf("%.6f\n",d);

		sum = 0.0;
		p = 2;
		for (int i =0; i<n ;i++ ) {
			sum += Math.pow(Math.abs(x[i]-y[i]),p);
		}

		d = Math.pow(sum,(1/p));
		System.out.printf("%.6f\n",d);

		sum = 0.0;
		p = 3;
		for (int i =0; i<n ;i++ ) {
			sum += Math.pow(Math.abs(x[i]-y[i]),p);
		}

		d = Math.pow(sum,(1/p));
		System.out.printf("%.6f\n",d);

		double max = 0.0;
		for (int i =0; i<n ;i++ ) {
			max = Math.max(max,Math.abs(x[i]-y[i]));
		}
		System.out.printf("%.6f\n",max);
	}
}