Get number of digits in factorial result without compute it.

factorial_digit_count.cc

#include <iostream>
#include <cmath>
using namespace std;

typedef long long ll;
typedef long double ld;

ld factorial_digit_count(ld n)
{
    // log(ab) = log(a) + log(b)
    ld digits = 0;
    for (ll k = 2; k <= n; ++k) {
        digits += log10((ld)k);
    }
    return floor(digits) + 1;
}

int main()
{
    // log10(n^k) = k*log10(n)
    ld n = 19941994;
    ld n_factorial_square_digits = 2*factorial_digit_count(n); // digits of (19941994!)^2
    ld n_to_n_digits = n*log10(n); // digits of 19941994^19941994
    cout << (ll)n_factorial_square_digits << " " << (ll)n_to_n_digits << endl;
}