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: Wrong Answer)

import java.util.Scanner;
public class Main {
    public static void main(String[] args){
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int i;
        double sum1 = 0;
        double sum2 = 0;
        double sum3 = 0;
        double sum4 = 0;
        double ans2,ans3,ans4;
        double max;
        double[] array_max;
        array_max = new double[1000];
        double[] array_x,array_y;
        array_x = new double[1000];
        array_y = new double[1000];
        for(i = 0; i < n; i++){
        	array_x[i] = scan.nextDouble();
        }
        for(i = 0; i < n; i++){
        	array_y[i] = scan.nextDouble();
        }
        for(i = 0; i < n; i++){
        	sum1 += Math.abs(array_x[i] - array_y[i]);
        	sum2 += (array_x[i] - array_y[i]) * (array_x[i] - array_y[i]);
        	sum3 += Math.abs(array_x[i] - array_y[i]) * Math.abs(array_x[i] - array_y[i]) * Math.abs(array_x[i] - array_y[i]);
        }
        for(i = 0; i < n; i++){
        	sum4 = Math.abs(array_x[i] - array_y[i]);
        	array_max[i] = sum4;
        }
        max = array_max[0];
        for(i = 0; i < n - 1; i++){
        	if(array_max[i] < array_max[i + 1])
        		max = array_max[i + 1];
        }
        ans2 = Math.sqrt(sum2);
        ans3 = Math.cbrt(sum3);
        System.out.println(sum1);
        System.out.println(ans2);
        System.out.println(ans3);
        System.out.println(max);
            }
}