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 C (judge verdict: Accepted)

#include<stdio.h>
#include<math.h>

double cubicRoot( double ab )
{
	double cr,a;
	double i;
	
	if( 0==ab ) { return 0; }
	
	for( i=1; ; i+=0.00001 ) {
		if( ab<=(i*i*i)  ) {
			a=i-0.00001;
			break;
		}
	}
	
	cr = a+(ab-(a*a*a))/(3*a*a);
	
	return cr;
}
int main( void ) 
{
	int n;
	int i;
	double x[1001],y[1001];
	double tmp;
	double Dxy,Dxy2,Dxy3,Dmax;
	
	scanf( "%d", &n );
	for( i=0; i<n; i++ ) {
    	scanf( "%lf", &x[i] );
    }
    Dxy=Dxy2=Dxy3=Dmax=0;
	for( i=0; i<n; i++ ) {
    	scanf( "%lf", &y[i] );
    	tmp=fabs( x[i]-y[i] );
    	Dxy+=tmp;
    	Dxy2+=(tmp*tmp);
    	Dxy3+=(tmp*tmp*tmp);
    	if( Dmax < tmp ) {
    		Dmax=tmp;
    	}
    }
	printf( "%.5lf\n", Dxy );
	printf( "%.5lf\n", sqrt( Dxy2 ) );
	printf( "%.5lf\n", cubicRoot( Dxy3 ) );
	printf( "%.5lf\n", Dmax );

		
	return 0;
}