Geeksforgeeks Solution For " Count digits in a factorial "

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Problem :- Count digits in a factorial

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Solution :- 

#include <bits/stdc++.h>
using namespace std;
long long findDigits(int n)
{
if (n < 0)
return 0;
if (n <= 1)
return 1;
double x = ((n*log10(n/M_E)+log10(2*M_PI*n)/2.0));
return floor(x)+1;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        cout<<findDigits(n)<<endl;
}
return 0;
}

Output:-