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: Wrong Answer)

from decimal import Decimal, getcontext
getcontext().prec = 30

def five(x):
    cnt = 0
    while x%5==0:
        cnt+= 1
        x = x//5
    return cnt

def two(x):
    cnt = 0
    while x%2==0:
        cnt+= 1
        x = x//2
    return cnt

N = int(input())

F = []
for i in range(N):
    A = Decimal(input())
    A = int(A*(10**9))
    F.append((five(A), two(A)))

F.sort()
# print(F)
j = 0
temp = 0
dp = [0]*19
k = 0
for i in range(N-1):
    if N-1-j < i:
        dp[min(F[i][1], 18)] -= 1
        # print(i, N - j - 1, dp)
    while N-1-j> i and F[i][0]+F[N-1-j][0] >= 18:
        dp[min(F[N-1-j][1],18)] += 1
        # print(i, N - j - 1, dp)
        j += 1
    k = sum(dp[:18-F[i][1]])
    temp += min(j, N-i-1) - k
    # print(j, N-i-1, k)

print(temp)

"""
3
0.1
0.1
10
"""