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)

#[allow(unused_imports)]
use proconio::{
    fastout, input,
    marker::{Bytes, Chars, Isize1, Usize1},
};

fn count_divx(x: usize, mut n: usize) -> usize {
    let mut r = 0;
    while n % x == 0 {
        r += 1;
        n /= x;
    }
    r
}

#[fastout]
fn main() {
    input! {
        n:usize,
        a:[f64;n]
    }
    let mut v = vec![];
    let mut table = vec![vec![0; 64]; 64];

    let mut ans = 0;
    let mut zero = 0;
    for (c, i) in a.iter().enumerate() {
        let buf = (i * 1000000000.0 + 0.2) as usize;
        if buf == 0 {
            ans += c;
            zero += 1;
        } else {
            let d2 = count_divx(2, buf);
            let d5 = count_divx(5, buf);
            v.push((d2, d5));
            ans +=
                table[if d2 > 18 { 0 } else { 18 - d2 }][if d5 > 18 { 0 } else { 18 - d5 }] + zero;
            for i in 0..=d2 {
                for j in 0..=d5 {
                    table[i][j] += 1;
                }
            }
        }
    }
    println!("{}", ans);
}