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 <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using VI = vector<int>;
using VL = vector<ll>;
using VS = vector<string>;
template<class T> using PQ = priority_queue<T, vector<T>, greater<T>>;
#define FOR(i,a,n) for(int i=(a);i<(n);++i)
#define eFOR(i,a,n) for(int i=(a);i<=(n);++i)
#define rFOR(i,a,n) for(int i=(n)-1;i>=(a);--i)
#define erFOR(i,a,n) for(int i=(n);i>=(a);--i)
#define each(i, a) for(auto &i : a)
#define SORT(a) sort(a.begin(),a.end())
#define rSORT(a) sort(a.rbegin(),a.rend())
#define fSORT(a,f) sort(a.begin(),a.end(),f)
#define all(a) a.begin(),a.end()
#define out(y,x) ((y)<0||h<=(y)||(x)<0||w<=(x))
#define tp(a,i) get<i>(a)
#define line cout << "-----------------------------\n" 
#define ENDL(i,n) ((i) == (n) - 1 ? "\n" : " ")
#define stop system("pause")
constexpr ll INF = 1000000000;
constexpr ll LLINF = 1LL << 60;
constexpr ll mod = 1000000007;
constexpr ll MOD = 998244353;
constexpr ld eps = 1e-10;
constexpr ld pi = 3.1415926535897932;
template<class T>inline bool chmax(T& a, const T& b) { if (a < b) { a = b; return true; }return false; }
template<class T>inline bool chmin(T& a, const T& b) { if (a > b) { a = b; return true; }return false; }
inline void init() { cin.tie(nullptr); cout.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15); }
template<class T>inline istream& operator>>(istream& is, vector<T>& v) { for (auto& a : v)is >> a; return is; }
template<class T>inline istream& operator>>(istream& is, deque<T>& v) { for (auto& a : v)is >> a; return is; }
template<class T, class U>inline istream& operator>>(istream& is, pair<T, U>& p) { is >> p.first >> p.second; return is; }
template<class T>inline vector<T> vec(size_t a) { return vector<T>(a); }
template<class T>inline vector<T> defvec(T def, size_t a) { return vector<T>(a, def); }
template<class T, class... Ts>inline auto vec(size_t a, Ts... ts) { return vector<decltype(vec<T>(ts...))>(a, vec<T>(ts...)); }
template<class T, class... Ts>inline auto defvec(T def, size_t a, Ts... ts) { return vector<decltype(defvec<T>(def, ts...))>(a, defvec<T>(def, ts...)); }
template<class T>inline void print(const T& a) { cout << a << "\n"; }
template<class T, class... Ts>inline void print(const T& a, const Ts&... ts) { cout << a << " "; print(ts...); }
template<class T>inline void print(const vector<T>& v) { for (int i = 0; i < v.size(); ++i)cout << v[i] << (i == v.size() - 1 ? "\n" : " "); }
template<class T>inline void print(const vector<vector<T>>& v) { for (auto& a : v)print(a); }
inline string reversed(const string& s) { string t = s; reverse(all(t)); return t; }

int main() {
    init();

    int n; cin >> n;
    ll ans = 0;
    map<int, map<int, int>> m;
    FOR(i, 0, n) {
        string s; cin >> s;
        int dot = -1;
        FOR(j, 0, s.length())if (s[j] == '.') {
            dot = j;
            s = s.substr(0, j) + s.substr(j + 1);
            break;
        }
        ll x = 1, y = stoll(s);
        if (dot != -1)FOR(j, 0, s.length() - dot) {
            x *= 10;
        }
        int xt = 0, yt = 0, xf = 0, yf = 0;
        for (; x % 2 == 0; x /= 2, ++xt);
        for (; y % 2 == 0; y /= 2, ++yt);
        for (; x % 5 == 0; x /= 5, ++xf);
        for (; y % 5 == 0; y /= 5, ++yf);
        for (int j = xt - yt; j < 14; ++j)
            for (int k = xf - yf; k < 14; ++k) {
                ans += m[j][k];
            }
        ++m[min(yt - xt, 13)][min(yf - xf,13)];
    }

    print(ans);

    return 0;
}