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: Time Limit Exceeded)

#include <math.h>
#include <stdio.h>
#include <string.h>
typedef long long ll;
ll a[200000];
int d[200000];
int n2[200000];
int n5[200000];
int min(int x, int y) {
	return x > y ? y : x;
}
int max(int x, int y) {
	return x < y ? y : x;
}
int main() {
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		char s[1001];
		scanf("%s", s);
		int len = strlen(s);
		ll num = 0;
		for (int j = 0; j < len; j++) {
			if (s[j] == '.') {
				d[i] = len - j - 1;
			} else {
				num *= 10;
				num += s[j] - '0';
			}
		}
		a[i] = num;
		while (num % 2 == 0) {
			num /= 2;
			n2[i]++;
		}
		while (num % 5 == 0) {
			num /= 5;
			n5[i]++;
		}
	}
	// for (int i = 0; i < n; i++) {
	// 	printf("%lld %d %d %d\n", a[i], d[i], n2[i], n5[i]);
	// }
	ll ans = 0;
	for (int i = 0; i < n; i++) {
		for (int j = i + 1; j < n; j++) {
			if (d[i] + d[j] <= min(n2[i] + n2[j], n5[i] + n5[j])) {
				ans++;
			}
		}
	}
	printf("%lld\n", ans);
	return 0;
}