1808. Maximize Number of Nice Divisors
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 mostprimeFactors
. - The number of nice divisors of
n
is maximized. Note that a divisor ofn
is nice if it is divisible by every prime factor ofn
. For example, ifn = 12
, then its prime factors are[2,2,3]
, then6
and12
are nice divisors, while3
and4
are not.
Return the number of nice divisors of n
. Since that number can be too large, return it modulo 109 + 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 <= 109
Solutions
Solution 1: Problem Transformation + Fast Power
We can factorize $n$ into prime factors, i.e., $n = a_1^{k_1} \times a_2^{k_2} \times\cdots \times a_m^{k_m}$, where $a_i$ is a prime factor and $k_i$ is the exponent of the prime factor $a_i$. Since the number of prime factors of $n$ does not exceed primeFactors
, we have $k_1 + k_2 + \cdots + k_m \leq primeFactors$.
According to the problem description, we know that a good factor of $n$ must be divisible by all prime factors, which means that a good factor of $n$ needs to include $a_1 \times a_2 \times \cdots \times a_m$ as a factor. Then the number of good factors $k= k_1 \times k_2 \times \cdots \times k_m$, i.e., $k$ is the product of $k_1, k_2, \cdots, k_m$. To maximize the number of good factors, we need to split primeFactors
into $k_1, k_2, \cdots, k_m$ to make $k_1 \times k_2 \times \cdots \times k_m$ the largest. Therefore, the problem is transformed into: split the integer primeFactors
into the product of several integers to maximize the product.
Next, we just need to discuss different cases.
- If $primeFactors \lt 4$, then directly return
primeFactors
. - If $primeFactors$ is a multiple of $3$, then we split
primeFactors
into multiples of $3$, i.e., $3^{\frac{primeFactors}{3}}$. - If $primeFactors$ modulo $3$ equals $1$, then we split
primeFactors
into $\frac{primeFactors}{3} - 1$ multiples of $3$, and then multiply by $4$, i.e., $3^{\frac{primeFactors}{3} - 1} \times 4$. - If $primeFactors$ modulo $3$ equals $2$, then we split
primeFactors
into $\frac{primeFactors}{3}$ multiples of $3$, and then multiply by $2$, i.e., $3^{\frac{primeFactors}{3}} \times 2$.
In the above process, we use fast power to calculate the modulus.
The time complexity is $O(\log n)$, and the space complexity is $O(1)$.
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