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: Wrong Answer)

use text_io::*;
use std::process::exit;
use im_rc::HashMap;

fn main(){
    let mut n:usize = read!();
    let mut v = [[0;20];20];
    for i in 0..n {
        let mut aaa: f64 = read!();
        let mut aa: f64 = 10000000000.0 * aaa;
        let mut a: usize = aa as usize;
        let mut ni: usize = 0;
        if a != 0 {
            while a % 2 == 0 && ni < 19 {
                ni += 1;
                a /= 2;
            }
            let mut go: usize = 0;
            while a % 5 == 0 && go < 19 {
                go += 1;
                a /= 5;
            }
            *&mut v[ni][go] = *&mut v[ni][go] + 1;
        }
        else {
            *&mut v[19][19] = *&mut v[19][19] + 1;
        }
    }
    let mut sum:usize = 0;
    for i in 0..20 {
        for j in 0..20 {
            for k in 0..20 {
                for l in 0..20 {
                    if i+k>19 {
                        if j+l>19 {
                            if v[i][j]*v[k][l]!=0 {
                                println!("{} {} {} {} {} {}",i,j,k,l,v[i][j],v[k][l]);
                            }
                            if i == k && j == l && v[i][j] != 0 {
                                let aaaa: usize = v[i][j] - 1;
                                sum += (aaaa * (aaaa + 1));
                            } else {
                                sum += v[i][j] * (v[k][l]);
                            };
                        }
                    }
                }
            }
        }
    }
    println!("{}",sum/2);
}