Link of the problem ?

Lets calculate $$$dp_i$$$ = maximum possible amazingness of $$$A_{1 \ldots i}$$$.

$$$dp_i = \max_{j=1}^{i-k}(dp_{j-1} + beauty(A_{j \ldots i}))$$$ (maybe I messed up some indexes by $$$\pm 1$$$, let it be an exercise for readers).

The idea is that for a fixed $$$i$$$, $$$beauty(A_{j \ldots i})$$$ changes at most $$$O(\log(C))$$$ times where $$$C=\max(A)$$$.

To find beauty of A, we can use xor basis. To quickly be able to find beauty of subarray of A, we can build sparse table over A where each table element will contain xor basis of a corresponding subsegment. Then to find a beauty of subsegment we split it into $$$2$$$ smaller segments for which we already know xor basis and merge them into one xor basis. Since merge of two xor basises takes $$$O(\log^2(C))$$$, total complexity for finding beauty of subsegment is $$$O(\log^2(C))$$$.

Then for each $$$i$$$ we can do binary search to find $$$O(\log(C))$$$ positions where beauty of $$$beauty(A_{j \ldots i})$$$ changes. Since each such position requires $$$O(\log(n))$$$ queries for beauty of subsegment, total complexity for calculating these indexes for all positions is $$$O(n\log(n) \log^3(C)) $$$. I'm pretty sure you can cut off at least one of four logs in this solution but the constant behind it is actually pretty good so it might be enough.

When for a fixed $$$i$$$ we found $$$O(\log(C))$$$ segments where $$$beauty(A_{j \ldots i})$$$ is constant all what's left to do is to query for a maximum $$$dp_j$$$ on each segment which is doable with segtree.