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CristianoPenaldo's blog

By CristianoPenaldo, history, 8 months ago, In English

I was using the Berlekamp-Massey (BM) algo for yesterday's H, but my sol worked too slow. Here is my idea:

(1) Divide $$$[1, n]$$$ into scales. Let $$$S(scale)$$$ be the scale $$$scale$$$, which is $$$S(scale)$$$ is $$$\{x | x \times scale \geq n, x \times scale/2 < n\}$$$. For example, if $$$n=7$$$, scale $$$1$$$ is $$$[7, 7]$$$, scale $$$2$$$ is $$$[4, 6]$$$, scale $$$4$$$ is $$$[2, 3]$$$, scale $$$8$$$ is $$$[1, 1]$$$. It is guaranteed that $$$scale$$$ is a power of $$$2$$$ in my sol.

(2)Fetch $$$O(log scale)$$$ consecutive items for each scale using the matrix fast power algorithm. The matrix is constructed in step (3) from the last scale.

(3)Using the BM algorithm to build a recurrence of length $$$O(log\,scale)$$$ and a recurrence matrix $$$M \in \mathbb{Z}/p\mathbb{Z}^{O(log\,scale) \times O(log\,scale)}$$$.

I mean for each scale we fetch some items $$$I$$$, get a recurrence using the BM algorithm, and construct a recurrence matrix $$$M$$$ using that recurrence, for example $$$a_r = a_{r+1} + a_{r+2}$$$, then

The matrix $$$M$$$ and the items $$$I$$$, and the fast matrix power algorithm, are used to fetching items for the next scale. This is a coarse-to-fine algorithm. The problem is that, for each scale, I need to fetch $$$O(log\,scale)$$$ items and each item costs $$$O(log\,scale^3log\,n)$$$ using a fast matrix power algorithm (matrix multiplication is $$$O(log\,scale^3)$$$, and there are up to $$$O(n)$$$ elements for each scale. Therefore the complexity for each scale is $$$O(log\,scale ^ 4 log\,n)$$$ and the overall complexity is $$$\sum\limits_{scale} O(log\,scale ^ 4 log\,n) = O((logn)^6)$$$ for each test case which is too slow. Any idea to speed up?


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