Finding the divisors of a number?

Question

Python

Hello! There is a factorization algorithm (I heard that you can find divisors of a number through it):

def Factor(n):
    Ans = []
    d = 2
    while d * d <= n:
        if n % d == 0:
            Ans.append(d)
            n //= d
        else:
            d += 1
    if n > 1:
        Ans.append(n)
    return Ans

In short, we need an efficient algorithm for finding the number of divisors of a number. tell me how to do it?

Answer 1

Hello! There are ideas for such an algorithm, you can go to the root of the number and check whether the number is divisible, if it is divisible, then immediately add 2 (because if 24 is a multiple of 4, then it is a multiple of 24/4), you just need to check if the divisor * divisor is not a number.
Here is the
while implementation

def fact(a):
  i = 1
  o = 0
  while i * i <= a:
     if a % i == 0 and i * i != a: o += 2
     elif a % i == 0 and i * i == a: o += 1
     i += 1
  return o

Answer 2

n=int(input("n="))
flag=True
m=[]
i=2

while flag:
     if n%i==0:
            m.append(i)
            m.append(n//i)
     
     if n//i==i+1:
             flag=False
     i+=1
print(m)

If we consider the number 58, then we immediately find two divisors 2 and 28, etc.