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.InputStreamReader;

class Main {
    
	public static void main(String[] args) throws Exception {

		double d1=0,d2=0,d3=0,d4=0,xy;

		BufferedReader br = new BufferedReader(new InputStreamReader(System.in) );

		int n = Integer.parseInt(br.readLine() );

		int[] x = new int[n];
		int[] y = new int[n];

		String[] strX = br.readLine().split(" ", n);
		String[] strY = br.readLine().split(" ", n);
		
		for(int i=0; i<n; i++)
		{
			x[i] = Integer.parseInt(strX[i]);
			y[i] = Integer.parseInt(strY[i]);
		}
		
		for(int i=0; i<n; i++)
		{
			xy = Math.abs(x[i] - y[i]);
			
			d1 = d1 + Math.pow(xy, 1);
			d2 = d2 + Math.pow(xy, 2);
			d3 = d3 + Math.pow(xy, 3);
			d4 = Math.max(xy, d4);
			
		}
		
		System.out.printf("%.5f\n",d1);
		System.out.printf("%.5f\n",Math.sqrt(d2) );
		System.out.printf("%.5f\n",Math.cbrt(d3) );
		System.out.printf("%.5f\n",d4);

	}
}