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++ (Clang 10.0.0) (judge verdict: Runtime Error)

#include<iostream>
#include<cstring>
using namespace std;
struct Trie{
    int next[26];
    int cnt;
    int type;
}T[1000005];
int dp[100005][26],val[1000005];
int le;
void Insert(string s){
    int p=0,len=s.size();
    for (int i=len-1;i>=0;i--){
        int r=s[i]-'a';
        if(T[p].next[r]<=0){
            T[le].type=r;
            T[p].next[r]=le++;
        }
        p=T[p].next[r];
        T[p].cnt++;  
    }
	val[p]++;
}
void dfs(int x){
	for(int i=0,v;i<26;i++){
		v=T[x].next[i];
		if(v<=0) continue;
		dfs(v);
		val[x]+=val[v];
		for(int j=0;j<26;j++){
			dp[x][j]+=dp[v][j];
		}
	}
	dp[x][T[x].type]=val[x];
}
long long xx;
long long query(string s){
    int p=0,len=s.size();
    xx=0;
    for (int i=len-1;i>=1;i--) {
        int r=s[i]-'a';
        if(T[p].next[r]<=0) {
            return 0;
        }
        p=T[p].next[r];
    }
   	for(int i=0,v;i<26;i++){
   		v=T[p].next[i];
   		if(v<=0) continue;
   		xx+=dp[v][s[0]-'a'];
	}
	return xx;
}
string s[1000005];
int n;
long long ans;
signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    le=1;
    cin>>n;
    for(int i=1;i<=n;i++){
    	cin>>s[i];
    	Insert(s[i]);
	}
	dfs(0);
    for(int i=1;i<=n;i++){
    	ans+=query(s[i]);
	}
	cout<<ans-n;
	return 0;
}