Editorial for LCC '22 Contest 2 S5 - Nutty Necklaces

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Author: Trent

In general, "stateful" transitions are hard to calculate in a loop. Hence, we begin by "unravelling" the necklace into a sequence of N integers from 1 to M such that no two adjacent elements are equal. Additionally, the first and last element cannot be the same, and we must also account for overcounting (since in a necklace ABA is equivalent to AAB as rotations are considered the same).

We start by calculating the number of such sequences. Because of the additional constraint of the last element not being equal to the first, we will handle the two cases separately. Let D_i be the number of sequences of length i such that no two adjacent elements are the same and the first element is different from the last element. Let S_i by the number of sequences of length i such that no two adjacent elements are the same but the first is the same as the last element.

Then, D_i=S_{i-1}\cdot (M-1)+D_{i-1}\cdot (M-2) (If the first and last elements are the same, just take one that isn't the last element; if the first and last elements are different, we can't choose the first or last element)

and S_i=D_{i-1} (since no two elements are adjacent, we must have a sequence where the last element isn't the same as the first add a bead of the first color to the sequence).

Now, we go back to our original problem. We must convert our sequences to a necklace. However, we must now consider rotations. Let's say the shortest period is of length P. Then, there are P overcounted sequences and we must divide by P.

For example, if P is 3 and N is 9, then the sequences ABCABCABC, BCABCABCA, and CABCABCAB are all equivalent (they're rotations of each other, but each counted as a separate sequence).

Now, we must solve the problem separately once again. How many sequences of length N are there such that the shortest period is of length P? Consider this: Take a sequence of length P, and clone it \frac NP times to get a sequence of length N. Obviously, this sequence must have a period of length P, so the number of sequences of length N with a period of P is equal to D_P. However, it might not be the shortest, but if the shortest period isn't of length P, then it must be a divisor of P. So, we can subtract all all sequences with a length equal to a divisor of P.

More formally, if Q_P is the number of sequences of length N with a shortest period of P, then Q_P=D_P-\sum (Q_i\text{, where }i|P)

Now finally, there are \frac{Q_P}{P} possible necklaces (since each necklace has P "rotations" that are counted in the Q_P sequences).

So, to reiterate, we wish to find the sum of \frac{Q_P}{P}, we can actually optimize this by only considering P such that P|N, since P is the period. Additionally, Q_P can be calculated if we know D_P and Q_i for all i|P. Notice that if i|P, then i|N as well. So, Q_P really only relies on other values of Q_P. So, we only need to calculate the value of P such that P|N. In general, there really aren't that many divisors of a number - in fact it works out to be roughly 7000 divisors at most. So, we can run an O(D^2) algorithm, where D is the number of divisors. For each value of Q_P, we will first calculate D_P and iterate through all of the remaining divisors of N, subtracting Q_i if i|P. Then, once Q_P is calculated, we add \frac {Q_P}{P} to our total.

Now, firstly, to get the list of divisors, we can iterate from 1 to \sqrt N. If j|N, then we add j and \frac Nj to our list of divisors.

Next, to calculate D_P, we must use matrix exponentiation. To begin, we define the transition:

D_P=S_{P-1}\cdot (M-1)+D_{P-1}\cdot (M-2) and S_P=D_{P-1}

So, \begin{pmatrix}D_P\\S_P\end{pmatrix}=\begin{bmatrix}M-2&M-1\\1&0\end{bmatrix}\begin{pmatrix}D_{P-1}\\S_{P-1}\end{pmatrix}. We then take \begin{bmatrix}M-2&M-1\\1&0\end{bmatrix}^{P-1} and multiply that by \begin{pmatrix}D_1\\S_1\end{pmatrix} to get D_P. And we are done.


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