Congrats!

Guess is really not easy for me xD.

But D just needs a little observation and it's a trivial ds problem :)

Yes

Same here.

I think it's pretty easy to see that $$$a_n(y)$$$ is proportional to $$$y^{(A+1)n}$$$ (assuming the definition of $$$a_n(y)$$$ is as I mentioned below maroonrk's post).

Recall that $$$a_n(y)$$$ is an integral of a homogeneous function $$$e(x_1,\dots,x_n,y)$$$ of degree $$$nA$$$ over the $$$n$$$-dimensional box $$$(x_1,x_2,\dots,x_n,y)\in [0,y]^n\times \{y\}$$$. In simpler terms, if we scale all of $$$x_1\dots x_n,y$$$ by a factor of $$$r$$$ then the value of $$$e$$$ scales by $$$r^{nA}$$$.

Suppose we have evaluated $$$a_n(y_1)$$$ for some real $$$y_1$$$ and want to relate it to $$$a_n(y_1r)$$$ for some positive real $$$r$$$. Consider the one-to-one mapping from points in the original box $$$[0,y_1]^n\times \{y_1\}$$$ to the new box $$$[0,y_1r]^n\times \{y_1r\}$$$ that scales element-wise by $$$r$$$. The value of $$$e$$$ scales by $$$r^{nA}$$$ after applying the one-to-one mapping. Also, the volume of the new box is a factor of $$$r^n$$$ larger than the volume of the old box. Multiplying these two factors gives that $$$a_n(y_1r)=r^{n(A+1)}a_n(y_1)$$$, from which it follows that $$$a_n(y)$$$ is a polynomial in $$$y$$$ of degree $$$n(A+1)$$$.

But noticing that the 1 in the formula is actually the upper bound of $$$x_i$$$ is also hard

I think you can still solve the problem even if $$$1$$$ in the formula was some other constant $$$c$$$. It would just add one extra step. Compute $$$h(z)$$$ in the same way as before, then replace $$$[z^N] \exp h(z)$$$ in the answer with $$$[z^N]\exp (ch(z))$$$.

https://atcoder.jp/contests/arc180/editorial/10317

I think the definition of $$$a_n(y)$$$ in the editorial is off by a factor of $$$y^n$$$. From the way it is currently defined it is proportional to $$$y^{An}$$$, not $$$y^{(A+1)n}$$$. Instead of the expected value, it should be the $$$n$$$-dimensional integral.

Here's how I make the observation for problem A.

Consider string ABAB. There are 2 different operations we can perform, which are reduce ABA B and A BAB. Notice that either way, the resulting string would be AB. So, there are 2 different string you can make for ABAB --> ABAB and AB.

Try with longer string, ABABA. Different operations you can perform are ABA BA , A BAB A , and AB ABA. Whichever you choose, the resulting string is the same, ABA. You can choose either to perform the operation again or not. Finally, there are 3 different strings you can get, ABABA, ABA, and A.

Up to this point, you can infer the pattern if the strings are alternating (ABABABAB... or BABABAB....).

What happened when we find 2 consecutive characters, such as ABAB AAAA. You will notice that you can reduce ABABA to (ABABA, ABA, A) with the above methods but you can't change the last substrings (AAAA).

Therefore, the number of possible strings you can have depends on the alternating characters.

For example: BBABABAABABAAAABA. You can separate this string into several alternating substrings. B | BABABA | ABABA | AA | ABA. Ignore non alternating substrings such as B and AA. Find the number of different substrings you can create for each alternating substrings and multiply it to get the result.