Formatted question description: https://leetcode.ca/all/1808.html

# 1808. Maximize Number of Nice Divisors

Hard

## Description

You are given a positive integer primeFactors. You are asked to construct a positive integer n that satisfies the following conditions:

• The number of prime factors of n (not necessarily distinct) is at most primeFactors.
• The number of nice divisors of n is maximized. Note that a divisor of n is nice if it is divisible by every prime factor of n. For example, if n = 12, then its prime factors are [2,2,3], then 6 and 12 are nice divisors, while 3 and 4 are not.

Return the number of nice divisors of n. Since that number can be too large, return it modulo 10^9 + 7.

Note that a prime number is a natural number greater than 1 that is not a product of two smaller natural numbers. The prime factors of a number n is a list of prime numbers such that their product equals n.

Example 1:

Input: primeFactors = 5

Output: 6

Explanation: 200 is a valid value of n.

It has 5 prime factors: [2,2,2,5,5], and it has 6 nice divisors: [10,20,40,50,100,200].

There is not other value of n that has at most 5 prime factors and more nice divisors.

Example 2:

Input: primeFactors = 8

Output: 18

Constraints:

• 1 <= primeFactors <= 10^9

## Solution

If primeFactors <= 3, simply return primeFactors.

For greater values, consider the prime factors of n. The smallest nice divisor of n is the product of all distinct prime factors of n. For example, the prime factors of 12 are 2,2,3, so the smmalest nice divisor of 12 is 2*3=6. If n = p1^a1 * p2^a2 * ... * pk*ak, where a1 + a2 + ... + ak <= primeFactors, then n has a1 * a2 * ... * ak nice divisors. The problem is to find the maximum value of a1 * a2 * ... * ak.

To get the maximum value, there should be as many numbers 3 as possible, and the remaining numbers should be 2 (at most two numbers 2). Calculate the product using pow function and deal with large values accordingly.

class Solution {
static final int MODULO = 1000000007;

public int maxNiceDivisors(int primeFactors) {
if (primeFactors <= 3)
return primeFactors;
int quotient = primeFactors / 3;
int remainder = primeFactors % 3;
if (remainder == 0)
return (int) (pow(3, quotient) % MODULO);
else if (remainder == 1)
return (int) (pow(3, quotient - 1) * 4 % MODULO);
else
return (int) (pow(3, quotient) * 2 % MODULO);
}

public long pow(long x, int n) {
long power = 1;
while (n > 0) {
if (n % 2 == 1)
power = power * x % MODULO;
x = x * x % MODULO;
n /= 2;
}
return power;
}
}