Those of us who have been doing cp for a long time, have come to love the representation of numbers as an infinite prime power vector i.e.

Where $$$n$$$ is a number and $$$p_i$$$ is the $$$i$$$ th prime number. This has many useful properties for problems, such as mapping

All the above operations are done elementwise. However in some cases, the problem authors may forbid you from factorising the numbers, by perhaps increasing $$$n$$$ to around $$$10^{36}$$$ where there isn't any simple factorisation algorithm to do it in reasonable time. However using effective prime factorisation, we can use these properties freely on a set of numbers. If this set contained all integers, then it would require us to truly prime factorise the integer. However, we only need to use these operations on a subset of integers, making our task a lot simpler. We will make a finite array $$$q$$$ of size $$$t$$$ with the following properties.

$$$(1)$$$ Any element in $$$x \in S$$$ is representable by $$$q$$$, i.e. there exists $$$[f_1, f_2, \ldots f_t] = v'_n$$$ such that $$$x = \prod q_i^{f_i}$$$

$$$(2)$$$ For all the above operations, replacing $$$v_n$$$ with $$$v'_n$$$ should produce the same result.

Let us show, that if $$$q$$$ satisfies $$$(1)$$$ and $$$gcd(q_i, q_j) = 1$$$ for all $$$i < j$$$, then $$$q$$$ satisfies $$$(2)$$$.

Let us now consider how we can find $$$q$$$ given $$$S$$$. Let us proceed element wise, and assume we have a valid solution, and we want to add another element.

Now let us add the next element $$$x \in S$$$. Go through $$$q$$$ until we spot an element that has a common factor with $$$x$$$. Let that element be $$$q_i$$$.

Let $$$P(q)$$$ be the set of elements produced by some product of elements in $$$q$$$. Notice that if $$$q$$$ has 2 elements that divide each other $$$a \mid b$$$, then $$$a, b$$$ is replaceable by $$$a, \frac{b}{a}$$$, and $$$P$$$ will not lose any elements. Let us do this recursively until neither divides the other, or an element becomes 1. Let us call this procedure reduce_pair.

(It maybe useful to look at the example implementation in parallel from here)

We run the procedure reduce_pair on the elements $$$q_i, x$$$, if $$$x$$$ becomes 1, we're done. If $$$q_i$$$ becomes 1, we remove it from the set and continue.

We then run an iterative algorithm on a list $$$L$$$, initialised with $$$[q_i, x]$$$. Let the elements before $$$L[ptr]$$$ be all co-prime. Initialise $$$ptr$$$ with 1. Run reduce_pair on all elements before $$$ptr$$$ with it. Then add their gcd to the end of the list if its not one, and divide both elements by it. This will make all elements upto $$$ptr + 1$$$ coprime. There can only be as many elements in this list as there are true primes in $$$q_i, x$$$. Therefore we can deduce that the size of the list will be at most $$$O(\log{\log{max(S)}})$$$.

Since this algorithm will be run $$$O(\log{x})$$$ times, the total algorithm will run in $$$O(n^2\log{A} + n\log{A} \times (\log{\log{A}})^2)$$$ Where $$$A = max(S)$$$, and we will get our array $$$q$$$.

Example implementation : 159590539