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: Time Limit Exceeded)

use proconio::input;

fn main() {
    input! {
        n: usize,
        a: [String; n],
    }
    let a: Vec<_> = a
        .iter()
        .map(|ai| {
            if ai.contains('.') {
                ai.trim_end_matches('0').trim_end_matches('.')
            } else {
                ai
            }
        })
        .collect();

    let mut count = 0;
    for i in 0..n {
        for j in i + 1..n {
            let (ai, aj) = (a[i].replace(".", ""), a[j].replace(".", ""));
            let (ai_dec, aj_dec) = (ai.parse::<u128>().unwrap(), aj.parse::<u128>().unwrap());
            let mul = (ai_dec * aj_dec).to_string();
            let zeros_ai = a[i].len() - 1 - a[i].find('.').unwrap_or(a[i].len() - 1);
            let zeros_aj = a[j].len() - 1 - a[j].find('.').unwrap_or(a[j].len() - 1);
            count += mul.ends_with(&"0".repeat(zeros_ai + zeros_aj)) as usize;
        }
    }
    println!("{}", count);
}