| Note that using a \(1\) in the output array will contribute to the sum without affecting the product. The problem reduces to finding any factorization \(P = f_1 * ... * f_k\) such that every \(f_i \ge 2\) and \(\sum f_i \le 41\). We can simply pad out the list with \(1\)s until we reach \(41\). | |
| Let's consider the prime factorization of \(P = p_1^{k_1} * p_2^{k_2} ... p_M^{k_M}\). We claim that if an answer exists, it can always be attained by taking the prime factorization and padding out the rest with \(1\)s. | |
| To see why this is true, note that for any two integers \(a, b > 1\) it's always true that \(a*b \ge a+b\). Since composite factors can only be attained by merging prime factors, the overall sum can only stay the same or increase if we use composite factors. We might as well just use the prime factorization if it sums less than or equal to \(41\) (otherwise the answer does not exist), and then pad with \(1\)s. | |
| The prime factorization can be generated in \(\mathcal{O}(\sqrt P)\), and padding up to \(41\) takes constant time, so the overall running time is \(\mathcal{O}(\sqrt P)\). | |