Sarthak product subgraph

[sg60692]

The idea is to first create a $NPln(P)$ solution and then optimise it

Root the tree arbitrarily
For each connected subgraph we calculate it using the highest node in the subgraph, (must be unique)
For each node $u$ and each value $x$ we calculate the number of connect subgraphs such that u is the highest node of the subgraph and the product is equal to $x$

let $f_{u}(x)$ denote the above quantity
Clearly
$$ans = \sum_{u} {\sum_{x=1}^{P} f_{u}(x)}$$

let
$$g_{u}(x) = \sum_{i=1}^{x}f_{u}(x)$$
Therefore,
$$ans = \sum_{u} g_{u}(P)$$

lets look at following case,
We have a tree rooted as $u$ and another tree rooted at $v$
We want to combine these tree such that $u$ is parent of $v$,
We want to calculate the new $f_u$
Lets first find the new subgraphs formed by highest node $u$ that also contain $v$, denoted by$h$
$$h(x) = \sum_{i | x}f_{u}(i)*f_{v}(x/i)$$

The above mention is a Dirichlet Convolution
Clearly the updated $f_u$ be denoted by ${f'}$

$$f'{(x)} = f_{u}(x)+h(x)$$

Using this we can find and update the f values in dfs fashion

Instead of performing the above Drichlet Convolution naively we use the trick - Drichlet Convolution with prefix sums

Time Complexity:
$$O(N*P^{\frac{2}{3}})$$
We perform combine $N-1$ times, each time for one edge, each combine take $O(P^{\frac{2}{3}})$ operations

Space Complexity:
$$O(N*P^{\frac{2}{3}})$$
We have to save $O(P^{2,3})$ and the recursion depth can go upto $N$