We can use a recursive backtracking algorithm to generate all possible values of \(f_{1..k}\). At each step, we consider every possible candidate of \(f_i \in [2, 41]\). Since \(f_i\) has to be a divisor of \(P\) within \([2,41]\), we can just precompute these. We start the backtracking with \(P\), and at each step, divide by each choice of \(f_i\) until either \(P\) becomes \(1\), or the sum of factors used so far exceeds \(41\).

Will this backtracking be efficient enough? First, note that the deepest we can go is \(k=20\) by choosing each \(f_i=2\) (since minimizing each factor maximizes the array length). This mean \(P \le 2^k \approx 10^6\). It's somewhat difficult to prove an upper bound on this recurrence, but intuitively, the larger the factor we pick at each step, the fewer remaining branching factors there will be. Emprically, the model solution passes the full test set in under \(10\) seconds on a modest system, taking no more than \(20\) million recursive steps on any test case.

Another approach is to enumerate all [partitions](https://en.wikipedia.org/wiki/Partition\_(number\_theory)) of \(41\), of which there are [under 45000](https://www.wolframalpha.com/input?i=number+of+partitions+of+41). There are many known [algorithms](https://code.activestate.com/recipes/218332/) to do so, taking linear or constant steps per partition. Even exponential ones will likely work in our submission window. After generating each partition of size \(k\), we can check whether the product equals \(P\) in \(\mathcal{O}(k)\), for an overall time complexity of \(\mathcal{O}(45000*k)\).