---
description: Math
---

# LeetCode 1808. Maximize Number of Nice Divisors



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** `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 <= 10^9`

## Solution

[English Version in Youtube](https://youtu.be/kYm4bYrktkw)

[中文版解答Youtube Link](https://youtu.be/xkqcceTrGJw)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV1Si4y1P7wF/)

```
class Solution {
    
    long mod = 1e9+7;
    
public:
    int maxNiceDivisors(int primeFactors) {
        int groups_of_3 = primeFactors/3;
        int groups_of_2 = 0;
        int remain = primeFactors - groups_of_3 * 3;
        if (remain == 1) {
            if (groups_of_3 > 0) {
                groups_of_3--;
                groups_of_2 = 2;
            }
        } else if (remain == 2) {
            groups_of_2 = 1;
        }
        
        long ans = 1;
        
        int num_3_used = 0;
        int limit = sqrt(primeFactors);
        int temp = 1;
        while (groups_of_3 > 0) {
            ans *= 3;
            ans %= mod;
            groups_of_3--;
            num_3_used++;
            if (num_3_used == limit) {
                temp = ans;
                break;
            }
        }
        while (groups_of_3 >= limit) {
            ans *= temp;
            ans %= mod;
            groups_of_3 -= limit;
        }
        while (groups_of_3 > 0) {
            ans *= 3;
            ans %= mod;
            groups_of_3--;
            num_3_used++;
        }
        
        while (groups_of_2 > 0) {
            ans *= 2;
            ans %= mod;
            groups_of_2--;
        }
        
        return ans;
    }
};
```
