1808. Maximize Number of Nice Divisors: RETRY, credit: https://leetco…

…de.com/problems/maximize-number-of-nice-divisors/solutions/3220691/just-a-runnable-solution/

src/solution/mod.rs

@@ -1363,3 +1363,4 @@ mod s1803_count_pairs_with_xor_in_a_range;
 mod s1805_number_of_different_integers_in_a_string;
 mod s1806_minimum_number_of_operations_to_reinitialize_a_permutation;
 mod s1807_evaluate_the_bracket_pairs_of_a_string;
+mod s1808_maximize_number_of_nice_divisors;

src/solution/s1808_maximize_number_of_nice_divisors.rs

@@ -0,0 +1,96 @@
+/**
+ * [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 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
+ *
+ */
+pub struct Solution {}
+
+// problem: https://leetcode.com/problems/maximize-number-of-nice-divisors/
+// discuss: https://leetcode.com/problems/maximize-number-of-nice-divisors/discuss/?currentPage=1&orderBy=most_votes&query=
+
+// submission codes start here
+const MOD: i64 = 1_000_000_007;
+
+impl Solution {
+    // Credit: https://leetcode.com/problems/maximize-number-of-nice-divisors/solutions/3220691/just-a-runnable-solution/
+    pub fn max_nice_divisors(prime_factors: i32) -> i32 {
+        let prime_factors = prime_factors as i64;
+
+        let st = [0_i64, 1, 2, 3, 4, 6];
+        if prime_factors < 6 {
+            return st[prime_factors as usize] as _;
+        }
+
+        let mut result = st[3 + prime_factors as usize % 3];
+        let mut base = 3;
+        let mut exp = prime_factors / 3 - 1;
+        while exp > 0 {
+            if exp & 1 == 1 {
+                result = result * base % MOD;
+            }
+            base = base * base % MOD;
+            exp >>= 1;
+        }
+
+        result as _
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    fn test_1808_example_1() {
+        let prime_factors = 5;
+
+        let result = 6;
+
+        assert_eq!(Solution::max_nice_divisors(prime_factors), result);
+    }
+
+    #[test]
+    fn test_1808_example_2() {
+        let prime_factors = 8;
+
+        let result = 18;
+
+        assert_eq!(Solution::max_nice_divisors(prime_factors), result);
+    }
+}