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, non_snake_case, dead_code, non_upper_case_globals)]
use {
    proconio::{marker::*, *},
    std::{cmp::*, collections::*, mem::*},
};

#[fastout]
fn main() {
    input! {//
        n:usize,
        a:[f64;n],
    }
    //    mut  a:[f64;n],

    let mut b = vec![0i128; n];
    for i in 0..n {
        b[i] = (a[i] * 1e9 + 0.5f64) as i128;
    }
    let mut c = vec![];
    let mut tfs = vec![vec![0i64; 37]; 37];
    for &x in b.iter() {
        let mut two = 0;
        let mut five = 0;
        let mut y = x;
        while y % 2 == 0 {
            y /= 2;
            two += 1;
        }
        while y % 5 == 0 {
            y /= 5;
            five += 1;
        }

        if two >= 18 {
            two = 18;
        }
        if five >= 18 {
            five = 18;
        }
        c.push((two, five));
        tfs[two][five] += 1i64;
    }
    let mut ans = 0i64;
    for &(two, five) in c.iter() {
        for i in 0..19 {
            for j in 0..19 {
                if two + i >= 18 && five + j >= 18 {
                    ans += tfs[i][j];
                }
            }
        }
        if two + two >= 18 && five + five >= 18 {
            ans -= 1;
        }
    }
    println!("{}", ans / 2);
}

const eps: f64 = 1e-10;