I think the editorial for Round#810 is not out yet, and I found a solution for 810C: XOR Triangle which I thought was interesting, even if my solution is probably more complicated the intended solution, so I wanted to share it here.
Let's say the binary string in the test case is $$$b_{N-1} \dots b_0$$$ (yes, rightmost bit is labelled $$$b_0$$$, because it is the least significant bit). The first idea behind my solution is that you can solve it using DP, where you loop from right to left in the binary string and compute the answer for $$$n=b_i\dots b_0$$$ for all $$$i=0$$$ to $$$N-1$$$, where $$$N$$$ is the length of the given binary string. That is, you compute the answer for every $$$n$$$ which is a suffix of the binary string in the given test case.
Now, let's say we are trying to find the answer for $$$n=b_i\cdots b_0$$$. If $$$b_i=0$$$, then the answer is just whatever the last answer for $$$n=b_{i-1}\cdots b_0$$$ was, which we already know from the previous DP answer. Otherwise if $$$b_i=1$$$, we have to do some work.
If $$$b_i=1$$$, I split the answer into four cases:
- Number of triples where exactly 0 of $$$a,b,c$$$ have their $$$i$$$ th bit set to 1
- Number of triples where exactly 1 of $$$a,b,c$$$ have their $$$i$$$ th bit set to 1
- Number of triples where exactly 2 of $$$a,b,c$$$ have their $$$i$$$ th bit set to 1
- Number of triples where exactly 3 of $$$a,b,c$$$ have their $$$i$$$ th bit set to 1
For the rest of the post, let $$$lastVal$$$ be the value of $$$b_{i-1}\cdots b_0$$$, mod $$$998244353$$$. It is easy to compute $$$lastVal$$$ for all $$$i$$$ using a simple for loop, adding $$$2^i$$$ wherever $$$b_i$$$ is set in the binary string.
Case where 0 numbers have $$$i$$$ th bit set to 1
In this case, since the $$$i$$$th bit of all of $$$a,b,c$$$ are $$$0$$$, it is obvious that $$$a,b,c < n$$$, so $$$a,b,c$$$ are just arbitrary $$$i$$$-bit numbers. Let $$$d=a\oplus b$$$ and $$$e=a\oplus c$$$. Then, the legs of the triangle have lengths $$$d, e, d\oplus e$$$.
Case where 1 numbers have $$$i$$$ th bit set to 1
Without loss of generality, assume $$$a$$$ has $$$i$$$ th bit set to 1 and $$$b$$$ and $$$c$$$ have $$$i$$$ th bit set to 0. (We'll multiply the answer from this case by $$$3$$$ at the end to account for this arbitrary choice.)
Case where 2 numbers have $$$i$$$ th bit set to 1
Case where 3 numbers have $$$i$$$ th bit set to 1
