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 final int BIG_NUM  = 2000000000;
    public static final int MOD  = 1000000007;
 
    public static void main(String[] args) {
 
        Scanner scanner = new Scanner(System.in);
 
        double x[] = new double[100];
        double y[] = new double[100];
        double p_1 = 0,p_2 = 0,p_3 = 0,p_infinity = -1;
 
        int N = scanner.nextInt();
 
        for(int i = 0; i < N; i++)x[i] = scanner.nextDouble();
        for(int i = 0; i < N; i++)y[i] = scanner.nextDouble();
 
        for(int i = 0; i < N; i++){
            p_1 += Math.abs(x[i]-y[i]);
            p_2 += (x[i]-y[i])*(x[i]-y[i]);
            p_3 += Math.abs((x[i]-y[i])*(x[i]-y[i])*(x[i]-y[i]));
            p_infinity = Math.max(p_infinity, Math.abs(x[i]-y[i]));
        }
 
        System.out.printf("%.10f\n", p_1);
        System.out.printf("%.10f\n", Math.sqrt(p_2));
        System.out.printf("%.10f\n", Math.cbrt(p_3));
        System.out.printf("%.10f\n", p_infinity);
    }
}