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 PyPy3 (7.3.0) (judge verdict: Runtime Error)

N = int(input())
A = [0]*N
B = [[0]*19 for x in range(19)]
C = [0]*N
D = [0]*N
a,b,c,d,e,ans = 0,['0']*9,0,0,0,0
for i in range(N):
    c = (input()+'.0').split('.')
    b = ['0']*9
    for k,v in enumerate(c[1]):
        b[k] = v
    A[i] = int(c[0]+''.join(b))

# print(*A)

for i in range(N):
    d,e = 0,0
    while (A[i] % 2 == 0) and (d <= 18):
        A[i] //= 2
        d += 1
    while (A[i] % 5 == 0) and (e <= 18):
        A[i] //= 5
        e += 1
    C[i] = d
    D[i] = e
    B[d][e] += 1

for i in range(N):
    c = C[i]
    d = D[i]
    B[c][d] -= 1
    for j in range(18-c,19):
        for k in range(18-d,19):
            ans += B[j][k]

print(ans)