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)

#![allow(
  non_snake_case,
  unused_variables,
  unused_assignments,
  unused_mut,
  unused_imports,
  unused_macros,
  dead_code
)]
//use proconio::fastout;
use proconio::{input, marker::*};
use std::cmp::*;
use std::collections::*;
macro_rules! debug {
  ($($a:expr),* $(,)*) => {
      #[cfg(debug_assertions)]
      eprintln!(concat!($("| ", stringify!($a), "={:?} "),*, "|"), $(&$a),*);
  };
}
fn main() {
  input! {
    n:usize, A:[String;n]
  }
  // ip:整数部分, dp:小数部分
  // 1/512で0.001_953_125 で下9桁に届いてしまう。
  let mut cnt = vec![vec![0; 50]; 50];
  let mut p2 = vec![0; n];
  let mut p5 = vec![0; n];
  let A: Vec<String> = A;
  for i in 0..n {
    let s = A[i].clone();
    let mut ip = 0usize;
    let mut dp = 0;
    if let Some(i) = s.find('.') {
      // 小数の場合
      ip = s[0..i].parse().unwrap();
      let ss = s[(i + 1)..].to_string() + "0000000000";
      dp = ss[0..9].parse::<usize>().unwrap()
    } else {
      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;
    }
    cnt[p2[i]][p5[i]] += 1;
  }

  let mut ans = 0usize;
  for i in 0..n {
    let remain2 = max(18 - p2[i], 0);
    let remain5 = max(18 - p5[i], 0);
    for j in remain2..50 {
      for k in remain5..50 {
        ans += cnt[j][k];
      }
    }
    if p2[i] >= remain2 && p5[i] >= remain5 {
      ans -= 1;
    }
  }
  println!("{}", ans / 2);
}