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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

const int MAXN = 1e6 + 5;
int n;
double a[MAXN];
ll tmp[MAXN], ans = 0;
int cnt2, cnt5;
ll sum[101][101];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i], a[i] *= 1e9;
        tmp[i] = a[i];
        cnt2 = 0;
        cnt5 = 0;
        while (tmp[i] % 2 == 0) {
            cnt2++;
            tmp[i] /= 2ll;
        }
        while (tmp[i] % 5 == 0) {
            cnt5++;
            tmp[i] /= 5ll;
        }
        ans += sum[max(0, 18 - cnt2)][max(0, 18 - cnt5)];
        for (int j = 0; j <= min(cnt2, 18); j++)
            for (int k = 0; k <= min(cnt5, 18); k++) sum[j][k]++;
    }
    cout << ans;
    return 0;
}