Can you really count divisors?

• Time : $O(\sqrt{n})$
• Space : $O(1)$

long countDivisors(const long& n) {
    long long sum = 0;
    for (long long i = 1; i <= sqrt(n); i++) {
        sum += n / i;
    }
    return 2 * sum - (long long)sqrt(n) * (long long)sqrt(n);
}