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: Wrong Answer)

import java.util.Scanner;

class Main {
	public static void main(String args[]) {
		Scanner scanner = new Scanner(System.in);
		int  num = scanner.nextInt();	//要素数
		double array_x[] = new double[num];
		double array_y[] = new double[num];
		double max = 0;
		double dif;
		double abs;
		double pow;
		double sum = 0;
		double sum_min[] = new double[5];
		
		//xの要素
		for(int i = 0 ; i < num ; i ++){
			array_x[i] = scanner.nextDouble();
		}
		//yの要素
		for(int i = 0 ; i < num ; i ++){
			array_y[i] = scanner.nextDouble();
		}
		
		////距離を求める
		//３まで
		for(int i = 1; i < 4 ; i++ ){
			for( int j = 0; j < num ; j ++ ){
				dif = ( array_x[j] - array_y[j] );
				abs = Math.abs(dif);
				pow = Math.pow(abs, i);
				sum += pow;
				
				if( max < dif ){
					max = dif;
				}
			}
			double man = (double)i;
			sum_min[i] = Math.pow(sum, 1/man);
			sum = 0;
		}
		sum_min[4] = max;
		
		for(int i = 1 ; i <= 4 ; i ++){
			System.out.println(sum_min[i]);
		}
		
		scanner.close();
	}
}