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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.Arrays;
import java.util.Scanner;

class Main {
	public static void main(String[] args) {
		new Main().run();
	}
	
	long gcd(long a,long b) {
		return a==0?b:gcd(b%a,a);
	}
	
	int[] f(long a) {
		int[] ret=new int[2];
		while (a%2==0 && ret[0]<9) {
			ret[0]++;
			a/=2;
		}
		while (a%5==0 && ret[1]<9) {
			ret[1]++;
			a/=5;
		}
		return ret;
	}

	void run() {
		Scanner sc = new Scanner(System.in);
		int N=sc.nextInt();
		long[][] cnt=new long[20][20];
		long ans=0;
		for (int i=0;i<N;++i) {
			double a=sc.nextDouble();
			long den=(long)1e9;
			long num=(long)(a*den);
			long gcd=gcd(num,den);
			num/=gcd;
			den/=gcd;
			int[] p=f(num);
			int[] q=f(den);
			for (int u=-10;u<10;++u) {
				for (int v=-10;v<10;++v) {
					if (u+p[0]-q[0]>=0&&v+p[1]-q[1]>=0) 
					ans+=cnt[u+10][v+10];
				}
			}
			cnt[p[0]-q[0]+10][p[0]-q[0]+10]++;
		}		
		System.out.println(ans);
		
	}	
	
	

	void tr(Object... o) {
		System.out.println(Arrays.deepToString(o));
	}

}