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 C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <vector>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <iomanip>
#include <numeric>
#include <queue>
#include <cmath>
using namespace std;

int main() {
    
    int d = 20;
    int n;
    cin >> n;
    unordered_map<int, unordered_map<int, int>> um;
    long long int res = 0;
    for (int i = 0; i < n; i++ ) {
        string s;
        cin >> s;
        long long int t;
        int n1 = 0, n2 = 0;
        if (s.find('.') == string::npos) {
            t = stoll(s);
            while (t % 2 == 0) {
                t /= 2;
                n1++;
            }
            while (t % 5 == 0) {
                t /= 5;
                n2++;
            }
            for (int j = -n1; j < d; j++) {
                for (int k = -n2; k < d; k++) {
                    res += um[j][k];
                }
            }
            
            um[n1][n2]++;
        } else {
            int index = s.find('.');
            n1 = s.size() - 1 - index;
            n1 = -n1;
            n2 = n1;
            s = string(s.begin(), s.begin() + index) + string(s.begin() + index + 1, s.end());
            t = stoll(s);
            while (t % 2 == 0) {
                t /= 2;
                n1++;
            }
            while (t % 5 == 0) {
                t /= 5;
                n2++;
            }
            for (int j = -n1; j < d; j++) {
                for (int k = -n2; k < d; k++) {
                    res += um[j][k];
                }
            }
            um[n1][n2]++;
        }
        // cout << n1 << " " << n2 << endl;
        // cout << res << endl;
    }
    cout << res << endl;
}