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.*;

class Main {
	public static void main(String[] args) {
		BufferedReader kb = new BufferedReader( new InputStreamReader( System.in ) );
		try {
			int n;
			String str;
			double[] xi = new double[ 100+1 ];
			double[] yi = new double[ 100+1 ];
			double[] absxy = new double[ 100+1 ];
			int i;
			double Dxy = 0, max = 0;
			
			n = Integer.parseInt( kb.readLine() );
			i = 0;
			for( String nums: kb.readLine().split( " ", 0 ) ) {
				xi[i++] = Double.parseDouble( nums );
			}
			i = 0;
			for( String nums: kb.readLine().split( " ", 0 ) ) {
				yi[i++] = Double.parseDouble( nums );
			}
			Dxy = 0;
			for( i=0; i<n; i++ ) {
				absxy[i] = Math.abs( xi[i]-yi[i] );
				if( max < absxy[i] ) {
					max = absxy[i];
				}
				Dxy += absxy[i];
			}
			System.out.print( Dxy + "\n" );

			Dxy = 0;
			for( i=0; i<n; i++ ) {
				Dxy += Math.pow( absxy[i], 2 );
			}
			System.out.print( Math.sqrt( Dxy ) + "\n" );
			
			Dxy = 0;
			for( i=0; i<n; i++ ) {
				Dxy += Math.pow( absxy[i], 3 );
			}			
			System.out.print( Math.cbrt( Dxy ) + "\n" );
			
			System.out.print( max + "\n" );

		} catch( IOException e ) {
			System.err.println( e );
		}
	}
}