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: Wrong Answer)

import numpy as np

N=int(input())
A=[]
PRECISION_DIGITS=9
NO_ENTRY = 0
cnt=np.full((PRECISION_DIGITS*2+1,PRECISION_DIGITS*2+1),NO_ENTRY,int)
# cnt=np.zeros((PRECISION_DIGITS*2,PRECISION_DIGITS*2),int)

# print(cnt)

for n in range(N):
	tmpA = float(input())
	tmpA = round(tmpA*10**PRECISION_DIGITS)
	#print(tmpA)

	five = 0
	two = 0
	# (10**9) -> (two,five)=(9,9)
	# min = 10**-9, ignore < 10**9 ???
	while (tmpA % 5) == 0 and five < PRECISION_DIGITS*2:
		five += 1
		tmpA //= 5
	while (tmpA % 2) == 0 and two < PRECISION_DIGITS*2:
		two += 1
		tmpA //= 2

	if cnt[two][five] == 0:
		cnt[two][five]=1
	else:
		cnt[two][five]+=1

# print(cnt)
ans = 0

for x1 in range(PRECISION_DIGITS*2):
	for y1 in range(PRECISION_DIGITS*2):
		if cnt[x1][y1] == NO_ENTRY:
			continue

		for x2 in range(PRECISION_DIGITS*2):
			for y2 in range(PRECISION_DIGITS*2):
				if cnt[x2][y2] == NO_ENTRY:
					continue
				# 2**a(a>=0) and 5**b(b>=0) => integer
				if 0 <= x1+x2 - PRECISION_DIGITS*2 and 0 <= y1+y2 -PRECISION_DIGITS*2:
					if x1 == x2 and y1 == y2:
						# same entry
						# Total number of combinations choose 2 items
						# nC2 = n*(n-1)/2
						ans += int((cnt[x1][y1]*(cnt[x2][y2]-1))//2)
					else:
						ans += cnt[x1][y1]*cnt[x2][y2]

		# count up completed entry
		cnt[x1][y1] = NO_ENTRY

print(ans)