nihal2001 posted a blog for asking the following question:

There are $$$n$$$ stones, two players take turn to choose $$$x$$$ stones, such that $$$1 \leq x \leq n$$$ and $$$x \And n = 0$$$. The one who cannot make choose lose.

Determine who is the winner, The first one or the second?

I proved that If $$$n$$$ belongs to https://oeis.org/A295897, the second player win(can be solved in $$$O(\log n)$$$).

My Question(Multiple piles version of above question):

There are $$$n_1, \cdots, n_m$$$ stones, two players take turn to choose an index $$$i$$$ and $$$x_i$$$ stones, such that $$$1 \leq i \leq m$$$, $$$1 \leq x_i \leq n_i$$$ and $$$x_i \And n_i = 0$$$. The one who cannot make choose lose.

Who will be the winner?

This turn out to compute Sprague-Grundy function for $$$n_1 \cdots, n_m$$$. The first player win if and only if $$$sg(n_1) \wedge \cdots \wedge sg(n_m) \neq 0$$$

I know how to compute $$$sg(n)$$$ in $$$O(n^2 \log n)$$$, but I can't give it a formula or compute efficiently like $$$O(\log n)$$$ or $$$O(\log^2 n)$$$

Thanks in advance.

UPD: $$$O(n^{log_2 3} \log n)$$$ implement based on wery0's comment.