Divisor count function
Return the number of positive divisors of n (the divisor-count function tau(n)), for n up to about 10^12. Counting by looping to n is far too slow; instead factorize n by trial division up to sqrt(n), and use that if n = product of p_i^e_i, then tau(n) = product of (e_i + 1). Return 0 for n <= 0.
Implement
count_divisors(n: int) → intExamples
in
[12]out6What a strong answer looks like
State your approach and its time/space complexity out loud before you optimize. Handle the edge cases (empty input, duplicates, overflow), and say why you chose this over the brute force. Green tests are the floor, not the grade.
0:00 of about 20 min
solution.py
InputExpectedGot
[12]6not run yetsample