Maximize Number of Nice Divisors in Python

Suppose we have a number pf represents number of prime factors. We have to make a positive number n that satisfies the following conditions −

• The number of prime factors of n (may or may not be distinct) is at most pf.

• The number of nice divisors of n is maximized. As we know a divisor of n is nice when it is divisible by every prime factor of n.

We have to find the number of nice divisors of n. If the answer is too large then return result modulo 10^9 + 7.

So, if the input is like pf = 5, then the output will be 6 because for n = 200 we have prime factors [2,2,2,5,5] and its nice divisors are [10,20,40,50,100,200] so 6 divisors.

To solve this, we will follow these steps −

• if pf is same as 1, then

  • return 1

• m := 10^9 + 7

• q := quotient of pf/3, r := pf mod 3

• if r is same as 0, then

  • return 3^q mod m

• otherwise when r is same as 1, then

  • return (3^(q-1) mod m)*4 mod m

• otherwise,

  • return (3^q mod m)*2 mod m

Example

Let us see the following implementation to get better understanding

def solve(pf):
   if pf == 1:
      return 1
   m = 10** 9 + 7
   q, r = divmod(pf, 3)
   if r == 0:
      return pow(3, q, m)
   elif r == 1:
      return pow(3, q-1, m) * 4 % m
   else:
      return pow(3, q, m) * 2 % m

pf = 5
print(solve(pf))

Input

5

Output

6