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: Accepted)

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <utility>
#include <algorithm>
#include <cstdio>
#include <iomanip>
#include <queue>
#include <deque>
#include <stack>
#include <fstream>
#include <cmath>
#include <random>
#include <complex>
#include <functional>
#include <bitset>
#include <sstream>
#include <cassert>
#include <time.h>

#define ll int64_t
#define Rep(i, n) for (ll i = 0; i < n; i++)
using namespace std;
typedef vector<ll> vec;
typedef vector<vec> mat;

const ll inf = 1LL << 60;

template<class T> inline void chmin(T& a, T b) {
    if (a > b) {
        a = b;
    }
}
template<class T> inline void chmax(T& a, T b) {
    if (a < b) {
        a = b;
    }
}
void printvec (vec &v) {
    Rep (i, (ll)v.size()) {
        cout << v[i] << " \n"[i+1 == (ll)v.size()];
    }
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    ll N;
    cin >> N;
    vec cnts(400, 0);
    vector<pair<ll, ll>> nums(N);
    Rep (i, N) {
        long double a;
        cin >> a;
        a += 1e-12;
        ll b = a * 1e9;
        ll cnt2 = -9, cnt5 = -9;
        while (b % 2 == 0) {
            cnt2++;
            b /= 2;
        }
        chmin(cnt2, (ll)9);
        while (b % 5 == 0) {
            cnt5++;
            b /= 5;
        }
        chmin(cnt5, (ll)9);
        cnts[cnt2*20+cnt5+200]++;
        nums[i] = {cnt2, cnt5};
    }

    // Rep (i, N) {
    //     cout << nums[i].first << " " << nums[i].second << "\n";
    // }

    ll ans = 0;
    Rep (i, N) {
        ll cnt2 = nums[i].first;
        ll cnt5 = nums[i].second;
        cnts[cnt2*20+cnt5+200]--;

        for (ll j = -cnt2; j <= 9; j++) {
            for (ll k = -cnt5; k <= 9; k++) {
                ans += cnts[j*20+k+200];
            }
        }
    }

    cout << ans << endl;    
}