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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

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


   String st = br.readLine();

   int n = Integer.parseInt( st );
   double[] abs = new double[n];
   String[] x = br.readLine().split(" ");
   String[] y = br.readLine().split(" ");

   double p1 = 0;
   for(int i = 0 ;i < n; i++){

	   abs[i] = Math.abs( Double.parseDouble(x[i]) - Double.parseDouble(y[i]) );
	   p1 += abs[i];

   }
   System.out.println(p1);

   double sum = 0;
   for(int i = 0 ;i < n; i++){

	   sum += Math.pow(Math.abs( Double.parseDouble(x[i]) - Double.parseDouble(y[i]) ) , 2);

   }
   double p2 = Math.sqrt(sum);
   System.out.println(p2);

   double sum2 = 0;
   for(int i = 0 ;i < n; i++){

	   sum2 += Math.pow(Math.abs( Double.parseDouble(x[i]) - Double.parseDouble(y[i]) ) , 3);

   }

  double triple = 1/3;
   double p3 = Math.pow(sum2 , 1.0 / 3.0);
   System.out.println(p3);


   double Ch = abs[0];
   for(int j = 0 ;j < n; j++){
     if(Ch < abs[j]){
       Ch = abs[j];
     }
   }
   System.out.println(Ch);

  }
}