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

fn main() {
    let mut n: usize = read!();
    let mut v = Vec::new();
    for i in 0..n {
        let mut a: String = read!();
        v.push(a);
    }
    let mut cnt = vec![vec![0; 19]; 19];
    let mut p2 = vec![0; n];
    let mut p5 = vec![0; n];
    for i in 0..n {
        let s = v[i].clone();
        let mut ip:usize = 0;
        let mut dp:usize = 0;
        if let Some(i) = s.find('.') {
            // 入力値が小数の場合
            ip = s[0..i].parse().unwrap();
            let ss:String = s[(i + 1)..].to_string() + "0000000000";
            dp = ss[0..9].parse::<usize>().unwrap()
        } else {
            // 入力値が整数の場合
            ip = s.parse::<usize>().unwrap();
        }
        let mut x = ip * 1_000_000_000 + dp;
        while x % 2 == 0 {
            x /= 2;
            p2[i] += 1;
        }
        while x % 5 == 0 {
            x /= 5;
            p5[i] += 1;
        }
        p2[i] = min(p2[i], 18);
        p5[i] = min(p5[i], 18);
        cnt[p2[i]][p5[i]] += 1;
    }

    let mut ans = 0usize;
    for i in 0..n {
        let remain2 = 18 - p2[i];
        let remain5 = 18 - p5[i];
        for j in remain2..19 {
            for k in remain5..19 {
                ans += cnt[j][k];
            }
        }
        if p2[i] >= remain2 && p5[i] >= remain5 {
            ans -= 1;
        }
    }
    println!("{}", ans / 2);
}