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



Sample input:
5
7.5
2.4
17.000000001
17
16.000000000


Sample output:
3

Submission status: Runtime Error

Solution code:
input_n = input()
N = int(input_n)
a = []
for i in range(N):
  a[i] = int(input())

integer_counter = 0
for i in range(N):
  for j in range(N-i):
    if i != j+i:
      product = a[i]*a[j]
      if "." not in str(product):
        integer_counter += 1
print(integer_counter)

Submission status: Runtime Error

Solution code:
input_n = input()
N = int(input_n)
a = []
for i in range(N):
  a[i] = int(input())

integer_counter = 0
for i in range(N):
  for j in range(N-i):
    if i != j+i:
      product = a[i]*a[j]
      if "." in str(product):
        integer_counter += 1
print(integer_counter)

Submission status: Runtime Error

Solution code:
n = int(input())
arr= []
c=0
for i in range(n):
  s = float(input())
  arr.append(s)

for i in range(n):
  for j in range(n):
    if(type(arr[i]*arr[j]))=='int':
      c++

Submission status: Runtime Error

Solution code:
# -*- coding: utf-8 -*-
# 整数の入力
a = int(input())
 
# 浮動小数点の入力
num_list = []
for i in a:
  num_list.append(float(input()))
 
count = 0
for i in num_list:
  for j in num_list:
    if i != j and isinstance(i*j, float):
      count += 1
 
# 出力
print("{}".format(int(count/2)))

Submission status: Runtime Error

Solution code:
N = int(input(""))
a = []
count = 0

for i in range(N):
  a.append(float(input("")))
  
for i in range(N-1):
    for j in range(i+1, N):
        if a[i]*a[j] - int(a[i]*a[j]) == 0
            count += 1
          
print(count)

Submission status: Runtime Error

Solution code:
# -*- coding: utf-8 -*-
# 整数の入力
a = int(input())

# 浮動小数点の入力
num_list = []
for i in a:
	num_list.append(float(input()))

count = 0
for i in num_list:
  for j in num_list:
    if i != j and isinstance(i*j, float):
      count += 1

# 出力
print("{}".format(int(count/2)))

Submission status: Runtime Error

Solution code:
#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <cmath>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int mod = 1000000007;

int main() {
    ios::sync_with_stdio(false);
    int n;
    ll a[69][69] = {}, z = 0;
    double d;
    cin >> n;
    while (n--) {
        cin >> d;
        ll b = (ll)round(d * 1e9);
        int x = 0, y = 0;
        while (b % 2 == 0) b /= 2, x++;
        while (b % 5 == 0) b /= 5, y++;
        a[x][y]++;
        //cout << x << ' ' << y << '\n';
    }
    for (int i = 0; i < 69; i++) for (int j = 0; j < 69; j++) {
        for (int k = 0; k < 69; k++) for (int l = 0; l < 69; l++) {
            if (i + k >= 18 && j + l >= 18) {
                if (i == k && j == l) z += a[i][j] * (a[i][j] - 1) / 2;
                else if (i < k || (i == k && j < l)) z += a[i][j] * a[k][l];
            }
        }
    }
    cout << z << '\n';
}