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 Python (3.8.2) (judge verdict: Accepted)

N = int(input())
A = [input() for _ in [0]*N]

for i,a in enumerate(A):
    b = a.split(".")
    if len(b)==1:
        A[i] = int(a)*10**9
        continue
    c = b[1]
    n = len(c)
    c += "0"*(9-n)
    A[i] = int(b[0] + c)

cnt = [[0]*50 for _ in [0]*50]
B = [(0,0)]*N
for k,a in enumerate(A):
    i = 0
    j = 0
    while a%2==0:
        a//=2
        i+=1
    while a%5==0:
        a//=5
        j+=1
    B[k] = (i,j)
    cnt[i][j] += 1

for i in range(1,50):
    i = 49-i
    for j in range(50):
        j = 49-j
        cnt[i][j] += cnt[i+1][j]

for i in range(50):
    i = 49-i
    for j in range(1,50):
        j = 49-j
        cnt[i][j] += cnt[i][j+1]

ans = 0
for i,j in B:
    me = int(i>=9 and j>=9)
    ans += cnt[max(0,18-i)][max(0,18-j)]-me

print(ans//2)