Problem:
Score : 300 points

Problem Statement

You are given N real values A_1, A_2, \ldots, A_N.
Compute the number of pairs of indices (i, j)
such that i < j and the product A_i \cdot A_j is integer.

Constraints

2 \leq N \leq 200\,000

0 < A_i < 10^4

A_i is given with at most 9 digits after the decimal.

Input

Input is given from Standard Input in the following format.

N
A_1
A_2
\vdots
A_N

Output

Print the number of pairs with integer product A_i \cdot A_j (and i < j).

Sample Input 1

5
7.5
2.4
17.000000001
17
16.000000000

Sample Output 1

3

There are 3 pairs with integer product:

7.5 \cdot 2.4 = 18

7.5 \cdot 16 = 120

17 \cdot 16 = 272

Sample Input 2

11
0.9
1
1
1.25
2.30000
5
70
0.000000001
9999.999999999
0.999999999
1.000000001

Sample Output 2

8

Solution in Rust (1.42.0) (judge verdict: Accepted)

use std::io;

fn div_count(mut x: u64, div: u64) -> usize {
    let mut count = 0;
    while x % div == 0 {
        x /= div;
        count += 1;
    }
    count
}

fn main() {
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).unwrap();
    let n: usize = buf.trim_end().parse().unwrap();
    let a: Vec<f64> = (0..n)
        .map(|_| {
            buf.clear();
            io::stdin().read_line(&mut buf).unwrap();
            buf.trim_end().parse().unwrap()
        })
        .collect();
    let a_proc: Vec<_> = a.into_iter().map(|x| (x * 10f64.powi(9)).round() as u64).collect();
    let div: Vec<_> = a_proc
        .iter()
        .map(|&x| (div_count(x, 2), div_count(x, 5)))
        .collect();
    let mut imos = [[0usize; 100]; 100];
    for (x, y) in div.iter().cloned() {
        imos[x][y] += 1;
    }
    for i in 0..100 {
        for j in (0..99).rev() {
            imos[i][j] += imos[i][j + 1];
        }
    }
    for i in (0..99).rev() {
        for j in 0..100 {
            imos[i][j] += imos[i + 1][j];
        }
    }
    let result: usize = div
        .into_iter()
        .map(|(x, y)| {
            if x >= 9 && y >= 9 {
                imos[18usize.saturating_sub(x)][18usize.saturating_sub(y)] - 1
            } else {
                imos[18usize.saturating_sub(x)][18usize.saturating_sub(y)]
            }
        })
        .sum();
    assert!(result % 2 == 0);
    println!("{}", result / 2);
}