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;
using ll =long long;
typedef pair<int,int> P;
#define SORT(a) sort((a).begin(),(a).end())
#define REV(a) reverse((a).begin(),(a).end())
#define For(i, a, b)    for(int i = (a) ; i < (b) ; ++i)
#define rep(i, n)       For(i, 0, n)
#define debug(x)  cerr << #x << " = " << (x) << endl;
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

void coY() {cout <<"Yes"<<endl;}
void coN(){cout <<"No"<<endl;}
 
//Write From this Line

//const ll mod = 1e9+7;
//const ll mod = 998244353;
int main()
{
	//積としてありえるのは、1,2,3,...,10^8通り。
	// 全部10^9して考えても良いかも。そうすれば、2の数,5の数の和が18以上のやつだけ考えれば良い。1
	int n;
	cin >> n;
	cout << setprecision(15);
	vector<double> D(n);
	rep(i,n) cin >> D[i];
	vector<vector<int>> mas(19,vector<int>(19,0));
	ll ans = 0;
	rep(i,n){
		double d = D[i];
		ll tmp = d * 1e9;
		// tmp が2と5を何個ふくむか。
		int ni = 0, go = 0;
		while(tmp % 2 == 0){
			ni++;
			tmp /= 2;
		}
		while(tmp % 5 == 0){
			go++;
			tmp /= 5;
		}
		ni = min(18,ni);
		go = min(18,go);
		int needni = 18 - ni;
		int needgo = 18 - go;
		for(int N = needni; N <= 18; N++){
			for(int G = needgo; G <= 18; G++){
				ans += mas[N][G];
			}
		}
		mas[ni][go]++;
	}
	cout << ans << endl;
}