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: Wrong Answer)

import java.io.*;
import java.util.*;
import java.lang.*;
import java.lang.Math;

class Main{
  public static void main(String[] args) throws IOException{
    Scanner scan=new Scanner(System.in);
    int n=Integer.parseInt(scan.next());
    int[] x=new int[n];
    int[] y=new int[n];
    for(int i=0;i<n;i++){
      x[i]=Integer.parseInt(scan.next());
    }
    for(int i=0;i<n;i++){
      y[i]=Integer.parseInt(scan.next());
    }
    double ans1=0;
    double ans2=0;
    double ans3=0;
    double ans4=0;
    for(int i=0;i<x.length;i++){
      ans1+=Math.abs(x[i]-y[i]);
      ans2+=Math.pow(x[i]-y[i],2);
      ans3+=Math.pow(Math.abs(x[i]-y[i]),3);
      try{
        ans4=Math.max(Math.abs(x[i]-y[i]),Math.abs(x[i+1]-y[i+1]));
      }catch(Exception e){
      }
      if(i==x.length-1){
        ans2=Math.sqrt(ans2);
        ans3=Math.cbrt(ans3);
      }
    }

    System.out.printf("%.8f\n",ans1);
    System.out.printf("%.8f\n",ans2);
    System.out.printf("%.8f\n",ans3);
    System.out.printf("%.8f\n",ans4);
  }
}