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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
/**
 *
 * @author Y
 */
public class Main {
    public static void main(String[] args) throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        int[] x = new int[n];
        int[] y = new int[n];
        String[] s = new String[n];
        
        s = br.readLine().split(" ");
        for(int i=0;i<n;i++){
            x[i] = Integer.parseInt(s[i]);
        }
        
        s = br.readLine().split(" ");
        for(int i=0;i<n;i++){
            y[i] = Integer.parseInt(s[i]);
        }
        
        for(int i = 1;i<=3;i++){
            minkowski(i,x,y);
        }
        
        int[] z = new int[n];
        double D = 0;
        for(int i = 0;i<x.length;i++){
            z[i] = Math.abs(x[i]-y[i]);
            
            D = Math.max(z[i],D);
            
        }
        System.out.println(D);
        
        
        
    }
    
    static void minkowski(int p, int[] x, int[] y){
        int n = x.length;
        double D=0;
        for(int i = 0;i<n;i++){
            D += Math.pow( Math.abs(x[i]-y[i]), p );
        }
        D = Math.pow(D,1.0/p);
        System.out.println(D);
    }
             
}