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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <inttypes.h>

#define MEW UINT64_C(1000000000)

#define INF UINT64_C(101010101010101010)

uint64_t kugyu(void) {
	char buffer[10240];
	char* nyan;
	uint64_t ret;
	if (scanf("%10000s", buffer) != 1) exit(1);
	nyan = strchr(buffer, '.');
	if (nyan == NULL) {
		strcat(buffer, ".");
		nyan = strchr(buffer, '.');
	}
	strcat(buffer, "00000000000000000000");
	memmove(nyan, nyan + 1, 10);
	nyan[9] = '\0';
	if (sscanf(buffer, "%" SCNu64, &ret) != 1) exit(1);
	return ret;
}

int N;
uint64_t asumisu[222222];
uint64_t k[222222];
uint64_t left[222222];
int count[20][20];

int main(void) {
	int i;
	uint64_t answer = 0;
	if (scanf("%d", &N) != 1) return 1;
	for (i = 0; i < N; i++) {
		asumisu[i] = kugyu();
	}

#if 0
	for (i = 0; i < N; i++) {
		printf("%" PRIu64 "\n", asumisu[i]);
	}
#endif

	for (i = 0; i < N; i++) {
		int j, jj;
		int c2 = 0, c5 = 0;
		k[i] = asumisu[i];
		left[i] = 1;
		for (j = 0; j < 18; j++) {
			if (k[i] % 2 == 0) k[i] /= 2; else { left[i] *= 2; c2++; }
			if (k[i] % 5 == 0) k[i] /= 5; else { left[i] *= 5; c5++; }
			if (left[i] > INF) left[i] = INF;
		}
#if 0
		printf("%20" PRIu64 " %20" PRIu64 "\n", k[i], left[i]);
#endif
		answer += count[18 - c2][18 - c5];
		for (j = c2; j < 20; j++) {
			for (jj =c5; jj < 20; jj++) count[j][jj]++;
		}
	}

	printf("%" PRIu64 "\n", answer);
	return 0;
}

/*

掛け算して MEW*MEW の倍数になればおｋ
→ A_i * A_j = MEW*MEW*k (kは整数)
→ A_j = MEW*MEW/A_i * k

k → (MEW*MEW)/A_i を約分した時の分母の倍数
A_j → (MEW*MEW)/A_i を約分した時の分子の倍数

MEW*MEW は2と5だけからなる
→2と5がその数以上あればおｋ、を管理する

*/