Problem:
Score : 300 points

Problem Statement

There are N towns in a coordinate plane. Town i is located at coordinates (x_i, y_i). The distance between Town i and Town j is \sqrt{\left(x_i-x_j\right)^2+\left(y_i-y_j\right)^2}.

There are N! possible paths to visit all of these towns once. Let the length of a path be the distance covered when we start at the first town in the path, visit the second, third, \dots, towns, and arrive at the last town (assume that we travel in a straight line from a town to another). Compute the average length of these N! paths.

Constraints

2 \leq N \leq 8

-1000 \leq x_i \leq 1000

-1000 \leq y_i \leq 1000

\left(x_i, y_i\right) \neq \left(x_j, y_j\right) (if i \neq j)

(Added 21:12 JST) All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
:
x_N y_N

Output

Print the average length of the paths.
Your output will be judges as correct when the absolute difference from the judge's output is at most 10^{-6}.

Sample Input 1

3
0 0
1 0
0 1

Sample Output 1

2.2761423749

There are six paths to visit the towns: 1 → 2 → 3, 1 → 3 → 2, 2 → 1 → 3, 2 → 3 → 1, 3 → 1 → 2, and 3 → 2 → 1.

The length of the path 1 → 2 → 3 is \sqrt{\left(0-1\right)^2+\left(0-0\right)^2} + \sqrt{\left(1-0\right)^2+\left(0-1\right)^2} = 1+\sqrt{2}.

By calculating the lengths of the other paths in this way, we see that the average length of all routes is:

\frac{\left(1+\sqrt{2}\right)+\left(1+\sqrt{2}\right)+\left(2\right)+\left(1+\sqrt{2}\right)+\left(2\right)+\left(1+\sqrt{2}\right)}{6} = 2.276142...

Sample Input 2

2
-879 981
-866 890

Sample Output 2

91.9238815543

There are two paths to visit the towns: 1 → 2 and 2 → 1. These paths have the same length.

Sample Input 3

8
-406 10
512 859
494 362
-955 -475
128 553
-986 -885
763 77
449 310

Sample Output 3

7641.9817824387

Solution in Rust (1.15.1) (judge verdict: Accepted)

//  ____            _                  _                       _       _
// |  _ \ _   _ ___| |_    ___ _ __   | |_ ___ _ __ ___  _ __ | | __ _| |_ ___
// | |_) | | | / __| __|  / __| '_ \  | __/ _ \ '_ ` _ \| '_ \| |/ _` | __/ _ \
// |  _ <| |_| \__ \ |_  | (__| |_) | | ||  __/ | | | | | |_) | | (_| | ||  __/
// |_| \_\\__,_|___/\__|  \___| .__/___\__\___|_| |_| |_| .__/|_|\__,_|\__\___|
//                            |_| |_____|               |_|

//https://github.com/manta1130/Competitive_Programming_Template_Rust

#[macro_use]
mod input {

    use std;
    use std::io;

    const SPLIT_DELIMITER: char = ' ';

    #[macro_export]
    #[allow(unused_macros)]
    macro_rules! input {
    ( $($x:expr ),*) => {
        {
            let temp_str = input_line_str();
            let mut split_result_iter = temp_str.split_whitespace();
                $(
                let buf_split_result = split_result_iter.next();
                let buf_split_result = buf_split_result.unwrap();
                    ($x) = buf_split_result.parse().unwrap();
                )*
        }
    };
}

    #[allow(dead_code)]
    pub fn input_line_str() -> String {
        let mut s = String::new();
        io::stdin().read_line(&mut s).unwrap();
        s.trim().to_string()
    }

    #[allow(dead_code)]
    pub fn p<T>(t: T)
    where
        T: std::fmt::Display,
    {
        println!("{}", t);
    }

    #[allow(dead_code)]
    pub fn input_vector2d<T>(line: usize) -> Vec<Vec<T>>
    where
        T: std::str::FromStr,
    {
        let mut v: Vec<Vec<T>> = Vec::new();

        for _ in 0..line {
            let vec_line = input_vector();
            v.push(vec_line);
        }
        v
    }

    #[allow(dead_code)]
    pub fn input_vector<T>() -> Vec<T>
    where
        T: std::str::FromStr,
    {
        let mut v: Vec<T> = Vec::new();

        let s = input_line_str();
        let split_result = s.split(SPLIT_DELIMITER);
        for z in split_result {
            let buf = match z.parse() {
                Ok(r) => r,
                Err(_) => panic!("Parse Error"),
            };
            v.push(buf);
        }
        v
    }

    #[allow(dead_code)]
    pub fn input_vector_row<T>(n: usize) -> Vec<T>
    where
        T: std::str::FromStr,
    {
        let mut v = Vec::with_capacity(n);
        for _ in 0..n {
            let buf = match input_line_str().parse() {
                Ok(r) => r,
                Err(_) => panic!("Parse Error"),
            };
            v.push(buf);
        }
        v
    }

    pub trait ToCharVec {
        fn to_charvec(&self) -> Vec<char>;
    }

    impl ToCharVec for String {
        fn to_charvec(&self) -> Vec<char> {
            self.to_string().chars().collect::<Vec<_>>()
        }
    }
}

mod vectools {

///C++のupper_boundと同等の実装
pub fn upper_bound<T>(v: &Vec<T>, k: T) -> usize
where
    T: PartialOrd,
{
    let mut from = 0;
    let mut to = v.len();

    while to - from > 1 {
        let mid = (from + to) / 2;
        if v[mid] <= k {
            from = mid;
        } else {
            to = mid;
        }
    }
    if v[from] > k {
        from
    } else {
        to
    }
}

///C++のlower_boundと同等の実装
pub fn lower_bound<T>(v: &Vec<T>, k: T) -> usize
where
    T: PartialOrd,
{
    let mut from = 0;
    let mut to = v.len();

    while to - from > 1 {
        let mid = (from + to) / 2;
        if v[mid] < k {
            from = mid;
        } else {
            to = mid;
        }
    }

    if v[from] < k {
        to
    } else {
        from
    }
}

///次の順列を求める。
pub fn next_permutation<T>(v: &mut Vec<T>) -> bool
where
    T: PartialOrd,
{
    let mut pivot = v.len() - 1;
    loop {
        if pivot == 0 {
            return false;
        }
        if v[pivot] > v[pivot - 1] {
            pivot -= 1;
            break;
        }
        pivot -= 1;
    }
    let mut pivot_swap = v.len() - 1;
    while v[pivot] >= v[pivot_swap] {
        pivot_swap -= 1;
    }
    v.swap(pivot, pivot_swap);
    v[pivot + 1..].reverse();
    true
}

}

use input::*;
use vectools::*;

fn main() {
    let n: usize;
    input!(n);
    let mut v = vec![];

    for _ in 0..n {
        let (a, b): (isize, isize);
        input!(a, b);
        v.push((a, b));
    }

    let mut table = vec![];
    for i in 0..n {
        table.push(i);
    }
    let mut ans = 0.0;
    let mut to = 1;
    for i in 1..n + 1 {
        to *= i;
    }
    for _ in 0..to {
        let mut buf = 0.0;
        for i in 0..n - 1 {
            let t1 = table[i];
            let t2 = table[i + 1];
            buf += ((v[t1].0 - v[t2].0).pow(2_u32) as f64 + (v[t1].1 - v[t2].1).pow(2_u32) as f64)
                .sqrt();
        }
        ans += buf;
        next_permutation(&mut table);
    }
    p(ans / to as f64);
}