Say we need to solve a recurrence of the form :

Here, *f* is an arbitrary function which can be computed in *O*(1). An example of such a problem is 553E - Kyoya and Train. The tutorial gives a solution in .

I recently saw a similar problem in codechef may long SERSUM, where I used another technique which I had thought of when I saw the prior problem. In this post I explain this very simple and intuitive sqrt decomposition solution, running in which works in almost the same time as solution because of a very low constant. Here goes:

Let us divide into blocks of size say *K*, giving us a total of ⌈ *N* / *K*⌉ blocks. Block *t* consists of [(*t* - 1)*K*, *tK* - 1]

Let .

Let *C*_{t} = *A*_{t}*B*.

When processing the block *t* + 1 we assume that we know *C*_{t}, which is computed when block *t* has been completely processed. So, we compute *C*_{t} a total of times using FFT in time .

Consider some *n* between [*tK*, (*t* + 1)*K* - 1]. Then,

.

Clearly, the right part can be calculated in *O*(*K*), and we have an extra time of *O*(*NK*) because of this part.

Overall complexity = , which is minimized taking , and overall complexity is .

An optimization:

Note that the right part can be computed with only one mod operation if we use m2 = m * m, and keep subtracting m2 when the total sum exceeds it. Eventually we take total sum modulo m. This is a very common technique, which we use in matrix multiplication as well.

This means that we should use a block size even greater than for better results, because of a huge constant difference in the two parts. (Computing *C*_{t} using FFT, and computing the remaining sum in *O*(*k*) using brute force iteration with the optimization above.) Just as an example, for *N* = 10^{5}, and *m* = 10^{9} + 7 (which is a bad mod value for FFT problems), the best performance comes when you take *K* ≈ 6 × 10^{3}. For this input size it runs in about 2 seconds, which is pretty good.