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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <bitset>
using namespace std;

#ifdef LOCAL
	#define eprintf(...) fprintf(stderr, __VA_ARGS__);fflush(stderr);
#else
	#define eprintf(...) 42
#endif

using ll = long long;
using ld = long double;
using uint = unsigned int;
using ull = unsigned long long;
template<typename T>
using pair2 = pair<T, T>;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
ll myRand(ll B) {
	return (ull)rng() % B;
}

#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second

clock_t startTime;
double getCurrentTime() {
	return (double)(clock() - startTime) / CLOCKS_PER_SEC;
}

const int N = 18;
ll a[N + 1][N + 1];
char s[33];

pii read() {
	scanf("%s", s);
	ll x = 0;
	int L = strlen(s);
	int p = L;
	for (int i = 0; i < L; i++) {
		if (s[i] == '.') {
			p = i;
			continue;
		}
		x = 10 * x + (int)(s[i] - '0');
	}
	if (p == L)
		p = 9;
	else
		p = 9 - (L - p - 1);
	while(p--) x *= 10;
	if (x == 0) return mp(N, N);
	int y = 0, z = 0;
	while(x % 2 == 0) {
		y++;
		x /= 2;
	}
	while(x % 5 == 0) {
		z++;
		x /= 5;
	}
	return mp(y, z);
}

int main()
{
	startTime = clock();
//	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);

	int n;
	scanf("%d", &n);
	while(n--) {
		pii t = read();
		a[min(N, t.first)][min(N, t.second)]++;
	}
	ll ans = 0;
	for (int i = 0; i < N; i++)
		for (int j = 0; j < N; j++)
			for (int p = 0; p < N; p++)
				for (int q = 0; q < N; q++) {
					if (i + p < N || j + q < N) continue;
					if (mp(i, j) > mp(p, q)) continue;
					if (mp(i, j) == mp(p, q))
						ans += a[i][j] * (a[i][j] - 1) / 2;
					else
						ans += a[i][j] * a[p][q];
				}
	printf("%lld\n", ans);

	return 0;
}