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;
int main()
{
  int n;
  cin>>n;
  vector<ll>a(n);
  for(int i=0;i<n;i++)
  {
    double d;
    cin>>d;
    a[i]=(ll)(round(d*1e9));
  }
  map<pair<int, int>, int>m;
  for(auto x : a)
  {
    int t=0;
    while(x%2==0 && x>0)
    {
     	x/=2;
      	t++;
    }
    int f=0;
    while(x%5==0 && x>0)
    {
      x/=5;
      f++;
    }
    m[{t,f}]++;
  }
  ll ans=0;
  for(auto x : m)
  {
    for(auto y : m)
    {
      if((x.first.first + y.first.first)>=18 && (x.first.second + y.first.second)>=18)
      {
        if(x==y)ans+=(long long)(x.second*(x.second-1)/2);
        else if(y>x)ans+=(x.second*y.second);
      }
    }
  }
  cout<<ans<<endl;
  return 0;
}