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)

// ITP1_10_D Distance2
import java.io.*;

public class Main {
   public static void main(String[] args) {
      BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

      try{
         int n = Integer.parseInt(br.readLine());
         int[] x = new int[n];
         int[] y = new int[n];
         String strx[] = br.readLine().split(" ");
         for (int i = 0; i < n ;i++)
            x[i] = Integer.parseInt(strx[i]);

         String stry[] = br.readLine().split(" ");
         for (int i = 0; i < n; i++)
            y[i] = Integer.parseInt(stry[i]);

         minkowski(n, x, y);
      }

      catch (Exception e) {
         e.printStackTrace();
      }
   }

   private static void minkowski(int n, int[] x, int[] y) {
      double D1 = 0;
      double D2 = 0;
      double D3 = 0;
      double Dmax = 0;

      for (int i = 0; i < n; i++) {
         D1 += Math.abs(x[i] - y[i]);
         D2 += Math.pow(Math.abs(x[i] - y[i]), 2);
         D3 += Math.pow(Math.abs(x[i] - y[i]), 3);
         if (Dmax < Math.abs(x[i] - y[i]))
            Dmax = Math.abs(x[i] - y[i]);
      }

      System.out.println(D1);
      System.out.println(Math.sqrt(D2));
      System.out.println(Math.cbrt(D3));
      System.out.println(Dmax);
   }
}