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.IOException;
import java.io.InputStreamReader;
import java.util.function.Function;
import java.util.stream.DoubleStream;
import java.util.stream.IntStream;
import java.util.stream.Stream;

public class Main {

  public static void main(String[] args) throws IOException {
    
    try(BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {

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

      Function<String,double[]> lineSpliteThenArray =
        s -> Stream.of(s.split(" "))
          .mapToDouble(Double::parseDouble)
          .toArray();

      double[][] vecter =
        br.lines()
          .limit(2)
          .map( lineSpliteThenArray )
          .toArray(double[][]::new);

      double p1 =
        IntStream.range(0, n)
          .mapToDouble(i -> Math.abs(vecter[0][i] - vecter[1][i]) )
          .sum();

      double p2 = Math.sqrt(
        IntStream.range(0, n)
          .mapToDouble(i -> Math.pow( vecter[0][i] - vecter[1][i], 2))
          .sum() );

      double p3 = Math.pow(
        IntStream.range(0, n)
        .mapToDouble(i -> Math.pow( Math.abs(vecter[0][i] - vecter[1][i]), 3d))
        .sum() , 1d/3d );

      double p00 =
        IntStream.range(0, n)
          .mapToDouble(i -> Math.abs(vecter[0][i] - vecter[1][i]))
          .max()
          .getAsDouble();

      System.out.printf("%.8f\n%.8f\n%.8f\n%.8f\n", p1,p2,p3,p00);
    }
  }

}