Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

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.util.Scanner;
public class Main {

	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		
		int[] x = new int[N];
		int[] y = new int[N];
		
		double sum = 0;
		
		for(int i=0;i<2;i++) {
			for(int j=0;j<N;j++) {
				if(i==0) {
					x[j] = sc.nextInt();
				}else {
					y[j] = sc.nextInt();
				}
			}
		}
		
		for(int i=0;i<N;i++) { //p=1
			sum += Math.abs((x[i] - y[i]));
		}
		System.out.printf("%6f\n",sum);
		sum = 0;
		
		for(int i=0;i<N;i++) { //p=2
			sum += Math.pow(Math.abs(x[i] - y[i]),2);
		}
		sum = Math.sqrt(sum);
		
		System.out.printf("%6f\n",sum);
		sum = 0;
		
		for(int i=0;i<N;i++) { //p=3
			sum += Math.pow(Math.abs(x[i] - y[i]),3);
		}
		sum = Math.cbrt(sum);
		
		System.out.printf("%6f\n",sum);
		sum = 0;
		
		double max = -7000;
		
		for(int i=0;i<N;i++) { //p = infinity
			sum = Math.abs((x[i] - y[i]));
				if(sum > max) max = sum;
		}
		System.out.printf("%6f\n",max);
	}

}