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<bits/stdc++.h>
using namespace std;
const int N=2e5+10;
typedef long long ll;

int get(){
    int ch;
    do{
        ch=getchar();
    }while(isspace(ch));
    return ch;
}
void read(ll &x,int &k){
    x=k=0;
    x=get()-'0';
    bool flag=0;
    while(1){
        int ch=getchar();
        if(isspace(ch))break;
        if(ch=='.')flag=1;
        else{
            if(flag)++k;
            x=x*10+ch-'0';
        }
    }
    while(x%10==0){
        x/=10;
        --k;
    }
    while(k<0){
        ++k;
        x*=10;
    }
}
typedef pair<ll,ll> P;
map<P,int> mp;

int main(){
    int n,num=0;
    scanf("%d",&n);
    for(int i=0;i<n;++i){
        ll x;
        int k;
        read(x,k);
        ll tmp1=1,tmp2=1;
        while(x%2==0){
            x/=2;
            tmp1*=2;
        }
        while(x%5==0){
            x/=5;
            tmp1*=5;
        }
        for(int i=0;i<k;++i)tmp2*=10;
        ll gcd=__gcd(tmp1,tmp2);
        ++mp[P(tmp1/gcd,tmp2/gcd)];
    }
    ll ans=0;
    for(auto it=mp.begin();it!=mp.end();++it){
        if(it->first.second==1)ans-=it->second;
        for(auto it2=mp.begin();it2!=mp.end();++it2){
            if(it->first.first%it2->first.second==0&&
                it2->first.first%it->first.second==0)ans+=it->second*(ll)it2->second;
        }
    }
    printf("%lld",ans/2);
    return 0;
}