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# (.NET Core 3.1.201) (judge verdict: Wrong Answer)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using System.Text;

class Util{
	static void Main(){
		Sol mySol =new Sol();
		mySol.Solve();
	}
}

class Sol{
	public void Solve(){
		int N = ri();
        List<decimal> AArray = new List<decimal>();
        List<bool> isInt = new List<bool>();
        
        int ans = 0;
        for (int i = 0; i < N; i++)
        {
            decimal tmp = decimal.Parse(Console.ReadLine());

            int seisu = (int)tmp;

            if(seisu == tmp) {
                AArray.Add(tmp);
                isInt.Add(true);
            } else {
                decimal syosutenIka = (tmp - seisu) * 100000;
                if((int)syosutenIka != syosutenIka) continue;
                else {
                    AArray.Add(tmp);
                    isInt.Add(false);
                }
            }
            
            for (int j = 0; j < isInt.Count - 1; j++)
            {
                if(isInt.Last() && isInt[j]) {
                    ans++;
                } else {
                    decimal seki = AArray.Last() * AArray[j];
                    if((int)seki == seki) ans++;
                }
            }
        }

		Console.WriteLine(ans);
		Console.ReadLine();
	}

	static String rs(){return Console.ReadLine();}
	static int ri(){return int.Parse(Console.ReadLine());}
	static long rl(){return long.Parse(Console.ReadLine());}
	static double rd(){return double.Parse(Console.ReadLine());}
	static String[] rsa(char sep=' '){return Console.ReadLine().Split(sep);}
	static int[] ria(char sep=' '){return Array.ConvertAll(Console.ReadLine().Split(sep),e=>int.Parse(e));}
	static long[] rla(char sep=' '){return Array.ConvertAll(Console.ReadLine().Split(sep),e=>long.Parse(e));}
	static double[] rda(char sep=' '){return Array.ConvertAll(Console.ReadLine().Split(sep),e=>double.Parse(e));}
    static int[] generateNums(int num, int N){return Enumerable.Repeat(num, N).ToArray();}
    static int[] generateSuretsu(int N){return Enumerable.Range(0, N).ToArray();}
}

public static class Combination {
	//使い方:int[][] Is = Combination.Enumerate(nums, k, withRepetition:false).ToArray();
    public static IEnumerable<T[]> Enumerate<T>(IEnumerable<T> items, int k, bool withRepetition) {
        if (k == 1) {
            foreach (var item in items)
                yield return new T[] { item };
            yield break;
        }
        foreach (var item in items) {
            var leftside = new T[] { item };

            // item よりも前のものを除く （順列と組み合わせの違い)
            // 重複を許さないので、unusedから item そのものも取り除く
            var unused = withRepetition ? items : items.SkipWhile(e => !e.Equals(item)).Skip(1).ToList();

            foreach (var rightside in Enumerate(unused, k - 1, withRepetition)) {
                yield return leftside.Concat(rightside).ToArray();
            }
        }
    }
}