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();
        double s2 = 0;
        double D1 = 0;
        double D2 = 0;
        double D3 = 0;
        double Di = 0;
        int[] x = new int[n];
        int[] y = new int[n];
        for(int i = 0; i < n; i++){
            x[i] = scan.nextInt();
        }
        for(int j = 0; j < n; j++){
            y[j] = scan.nextInt();
        }

        //マンハッタン距離
        for(int z = 0; z < n; z++){
            if((x[z] - y[z]) > 0){
                D1 += x[z] - y[z];
            }else{
                D1 += y[z] - x[z];
            }
        }
        System.out.println(D1);

        //ユークリッド距離
        for(int z = 0; z < n; z++){
            D2 += Math.pow(x[z]-y[z],2);
        }
        D2 = Math.sqrt(D2);
        System.out.println(D2);

        //3乗
        for(int z = 0; z < n; z++){
            if((x[z] - y[z]) > 0){
                D3 += Math.pow(x[z]-y[z],3);
            }else{
                D3 += Math.pow(y[z]-x[z],3);
            }
        }
        D3 = Math.cbrt(D3);
        System.out.println(D3);


        //チェビシェフ　1/p -> ０　に収束し、最大値の要素 xk0 - yk0 が残る
        Di = x[0] - y[0];
        for(int i = 1; i < n; i++){
            if(x[i]-y[i] > 0){
                Di = Math.max(Di,x[i]-y[i]);
            }else{
                Di = Math.max(Di,y[i]-x[i]);
            }
        }
        System.out.println(Di);


    }
}