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)

//#pragma comment(linker, "/STACK:134217728")

#include <iostream>
#include <sstream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <memory>
#include <cctype>
#include <cstring>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <algorithm>
#include <numeric>
using namespace std;

typedef long long Int;
typedef pair<int,int> PII;
typedef vector<int> VInt;

#define FOR(i, a, b) for(i = (a); i < (b); ++i)
#define RFOR(i, a, b) for(i = (a) - 1; i >= (b); --i)
#define CLEAR(a, b) memset(a, b, sizeof(a))
#define SIZE(a) int((a).size())
#define ALL(a) (a).begin(),(a).end()
#define PB push_back
#define MP make_pair

int f(PII a, PII b)
{
    Int m = Int(1e9);
    Int mm = m*m;
    
    Int x = Int(a.first)*Int(b.second) % m;
    Int y = Int(b.first)*Int(a.second) % m;
    Int z = Int(a.second)*Int(b.second) % mm;
    return (x*m + y*m + z) % mm == 0 ? 1 : 0;
}

void solve(int test)
{
    int n;
    cin >> n;
    int i, j, k, l;
    int cnt[47][47];
    CLEAR(cnt, 0);
    FOR(i, 0, n)
    {
        double d;
        cin >> d;
        d *= 1e9;
        Int a = Int(d + 0.5);
        int cnt2 = 0;
        int cnt5 = 0;
        while(a % 2 == 0)
        {
            a /= 2;
            ++cnt2;
        }
        while(a % 5 == 0)
        {
            a /= 5;
            ++cnt5;
        }
        ++cnt[cnt2][cnt5];
    }
    
    Int res = 0;
    FOR(i, 0, 47)
        FOR(j, 0, 47)
            if(cnt[i][j] != 0)
            {
                if(i + i >= 18 && j + j >= 18) res += Int(cnt[i][j])*Int(cnt[i][j] - 1)/2;
                FOR(k, 0, 47)
                    FOR(l, 0, 47)
                        if(PII(i, j) > PII(k, l) && i + k >= 18 && j + l >= 18)
                            res += Int(cnt[i][j])*Int(cnt[k][l]);
            }

    cout << res << endl;
}

int main()
{
    int T = 1, t;
    //cin >> T;
    FOR(t, 0, T)
    {
        solve(t);
    }
    return 0;
}