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;
#define ll long long
#define fr(i,j,k) for(int i=j;i<k;i++)
#define f(n) fr(i,0,n)
#define f1(n) fr(i,1,n+1)
#define pb push_back
#define F first
#define S second
#define all(x) x.begin(), x.end()
const int mod = 1e9 + 7;
const int maxn = 1000005;

void solve() {
	int n;
	cin >> n;
	vector<pair<int,string>>v;
	f(n) {
		string s;
		cin >> s;
		v.pb({s.size(), s});
	}
	sort(all(v));
	ll ans = 0;
	map<ll,int>mp;
	for (auto &i : v) {
		int n = i.F;
		string s = i.S;
		s = " " + s;
		int pre[n + 5] = {};
		for (int i = 1 ; i <= n ; i++) {
			pre[i] = (pre[i - 1] | (1<<(s[i]-'a')));
		}
		ll h = 0;
		for (int i = n ; i > 1 ; i--) {
			if (mp.count(h)) {
				ans += __builtin_popcount(pre[i] & mp[h]);
			}
			h *= 37;
			h += (s[i] - 'a');
		}
		mp[h] |= (1<<(s[1]-'a'));
	}
	cout << ans << '\n';
}
int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int c = 0;
    int t;
    if (!c) {
        t = 1;
    }
    else {
        cin >> t;
    }
    while (t--) {
        solve();
    }
}