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)

#include<stdio.h>
#include<string.h>
#include<math.h>

#define DECIMAL_POINT 8
#define UPPER_DECIMAL 0
#define UNDER_DECIMAL -1

#define OFFSET (long long int)pow(10,9)

long long int array[200001][2];

int main()
{
    int n = 0;
    int d_p, d_c, t;
    long long int mul = 0;
    long long int answer = 0;
    char num_arrray[20];

    scanf("%d", &n);

    for (int i = 0; i<n; i++)
    {
        scanf("%s", num_arrray);

        d_p = DECIMAL_POINT;
        d_c = UPPER_DECIMAL;
        t = 0;

        for (int j = 0; j<strlen(num_arrray); j++)
        {
            int current_num = 0;

            if (num_arrray[j] == '.')
            {
                array[i][0] = t;
                d_c = UNDER_DECIMAL;
                t = 0;
                continue;
            }

            current_num = num_arrray[j] - '0';

            if (d_c == UPPER_DECIMAL)
            {
                t = t*10 + current_num;
            } else {
                t += (long long int)(pow(10, d_p)*current_num);
                d_p += UNDER_DECIMAL;
            }

        }

        if (d_c == UPPER_DECIMAL)
        {
            array[i][0] = t;
            array[i][1] = 0;
        } else {
            array[i][1] = t;
        }

    }

    for (int i = 0; i<(n - 1); i++)
    {
        for (int j = (i + 1); j<n; j++)
        {
            mul = array[i][0]*array[j][1]*OFFSET + array[i][1]*array[j][0]*OFFSET + array[i][1]*array[j][1];
            
            if ((mul%(OFFSET*OFFSET)) == 0)
            {
                answer ++;
            }
        }
    }

    printf("%lld", answer);

    return 0;
}