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.*;
class Main {
 
    public static void main(String[] args) {
        Scanner stdIn = new Scanner(System.in);
        int n = stdIn.nextInt();
        int[] x = new int[n];
        int[] y = new int[n];
        for(int i = 0; i < n; i++) {
            x[i] = stdIn.nextInt();
        }
        for(int i = 0; i < n; i++) {
            y[i] = stdIn.nextInt();
        }
        for(int p = 1; p <= 3; p++) {
            double dis = 0.0;
            for(int i = 0; i < n; i++) {
                dis += Math.pow(Math.abs(x[i] - y[i]),p);
            }
            System.out.println(Math.pow(dis, 1.0/p));
        }
        int max = 0;
        for(int i = 0; i < n; i++) {
            max = Math.max(max, Math.abs(x[i] - y[i]));
        }
        System.out.println(max);
         
    }
}