As a tester, I encourage you to participate in the round. Nice problems.

I think problem E has a simpler explanation.

After doing the math and finding out that going from ($$$x_0$$$, $$$y_0$$$) to ($$$x_1$$$, $$$y_1$$$) changes your overall parity by $$$x_0 \oplus y_0 \oplus x_1 \oplus y_1$$$. You can do some more math and find out that going from ($$$x_0$$$, $$$y_0$$$) to ($$$x_1$$$, $$$y_1$$$) to ($$$x_2$$$, $$$y_2$$$) changes your overall parity by $$$(x_0 \oplus y_0 \oplus x_1 \oplus y_1) \oplus (x_1 \oplus y_1 \oplus x_2 \oplus y_2) = x_0 \oplus y_0 \oplus x_2 \oplus y_2$$$.

Then we can easily see that this extends as follows: going from ($$$x_0$$$, $$$y_0$$$) to ($$$x_1$$$, $$$y_1$$$) to ($$$x_2$$$, $$$y_2$$$) to ... to ($$$x_n$$$, $$$y_n$$$) changes your overall parity by $$$x_0 \oplus y_0 \oplus x_n \oplus y_n$$$. This means that only the parity of the first and last points matter, all other points get cancelled out by the XOR. Additionally, the first point is fixed, so we only care about the last point.

Let's call points of the same parity as the starting point good, and the other points bad. If you are the first player, you want to make the last point be good, you can do this if the number good points are not less than the the number of bad points. You can do this by repeated choosing the bad points to deplete them until your last turn, then choosing (or forcing the other player to choose) a good point. Similarly, if you are the second player and the number of good points is greater than the number of bad points, you can just keep choosing the good points until the last turn, then choose (or force the other player to choose) a bad point.

I think there are some small typos.

"Let's reorder the columns of this matrix by bringing the even-numbered columns to the left and the odd-numbered columns to the left.". I think it should be bringing the odd-numbered columns to the right.

"Each query consists of two integers 1≤x<y≤n. For each query, calculate modulo 998244353 the number of permutations p of length n such that p[y]=maxyi=1p[i] and 2⋅p[x]<p[x]." I think the statement says 2.p[x] < p[y].

I don't know the completely trivial solution :(

Can you explain it?

Is this Div2 rated for participants with rating < 2100 ?

In the tutorial of D, when you write $$$sz_i$$$, do you actually mean the size of the subtree rooted at $$$i$$$ or do you just mean $$$s_i$$$?

I think you forgot to link the editorial in the contest materials section.

I have a similar solution but with a segment tree instead of BS.

https://mirror.codeforces.com/contest/1808/submission/201087741