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)

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author
 */

// Created by White Knife on 15/04/20.
// doot diddly donger cuckerino Hahahahahah

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("unroll-loops")


#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace __gnu_cxx;
using namespace std;

// @formatter:off
#define F               first
#define S               second
#define all(x)          (x).begin(), (x).end()
#define sz(x)           (long long) (x).size()
#define sim             template<typename T
#define ris             return *this
#define dor             > debug & operator <<
#define eni(x)          sim > typename enable_if<sizeof dud<T>(0) x 1, debug&>::type operator<<(T i)
sim> struct rge { T b, e; };
sim> rge<T> rang(T i, T j) { return rge<T>{i, j}; }
sim> auto dud(T *x) -> decltype(cerr << *x, 0);
sim> char dud(...);
struct debug {
#ifdef LOCAL
  ~debug() { cerr << endl; }
  eni(!=) { cerr << boolalpha << i; ris; }
  eni(==) { ris << rang(all(i)); }
  sim, typename U dor(pair<U, T> d) { ris << "(" << d.first << ", " << d.second << ")"; }
  sim dor(rge<T> d) { *this << "["; for (auto it = d.b; it != d.e; ++it) *this << ", " + 2 * (it == d.b) << *it; ris << "]"; }
#else
  sim dor(const T&) { ris; }
#endif
};
#define pr(...) " [" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "] "

using ll = long long;
using ull = unsigned long long;
using ld = long double;
//using li = __int128;
//using uli = unsigned __int128;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vb = vector<bool>;

using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vpii = vector<pii>;
using vpll = vector<pll>;
template <typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

const ld PI = acos(-1);

template<typename T> inline T mini(T& x, T y) { return x = min(x, y); }
template<typename T> inline T maxi(T& x, T y) { return x = max(x, y); }
template <typename T> T rand(T x, T y) { return rand() % (y - x + 1) + x; }
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());

// @formatter:on


class AIntegerProduct {
public:
  
  void solve(std::istream& cin, std::ostream& cout) {
    int n;
    cin >> n;
    map<pii, int> cnt;
    for (int i = 0; i < n; ++i) {
      double x;
      cin >> x;
      ll val = ll(x * 1e9 + 0.5);
      int a = 0, b = 0;
      while (val % 2 == 0) {
        a++;
        val /= 2;
      }
      while (val % 5 == 0) {
        b++;
        val /= 5;
      }
      cnt[{a, b}]++;
    }
    ll ans = 0;
    for (auto p1: cnt) {
      for (auto p2: cnt) {
        if (p1.F.F + p2.F.F >= 18 and p1.F.S + p2.F.S >= 18) {
          if (p1 < p2) {
            ans += p1.S * p2.S;
          } else {
            ans += p1.S * (p1.S - 1) / 2;
          }
        }
      }
    }
    cout << ans << '\n';
  }
};

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
	AIntegerProduct solver;
	std::istream& in(std::cin);
	std::ostream& out(std::cout);
	solver.solve(in, out);
	return 0;
}