May not get 2 Dan, but the problems are very interesting!!! Thank you, writer!!!

Sympathetic

+1

I think the idea behind problem C is quite inspiring, although I still can not understand the editorial. I have realized that selecting some value s means a mirror symmetry, but can't go further. Waiting for some other hints too.

E is more of a statement parsing problem but in a good sense. I don't think it's an easy problem for most participants because you need good skills in constructing math models and understanding what is important.

B is quite easily solvable if you notice that if they need to pass each other twice, they might as well start at the point where they passed each other for the first time.

and then 1233 -> 1216

Go and do more ABC problems. They are much easier.

When you are likely to solve problem E~F in ABC, then back to ARC. I am sure you can enjoy the problems.

Ordering of the group matters like they will come in the order given there so you just need to let the group sit like upcoming can't seat so here only two kind of group make exists (i.e. size of 1 and 2) so group of size 1 can sit anywhere, ans can be "NO" only when a group of size 2 arrives! with some condition.

So we just need to let groups sit so that it can occupy one or two more space than their own sizes which will lead to not make a sit for upcomers.

Remember group of 2 only will face difficulty always!

My Code

My approach is to simulate the worst case that each arrived group will pick a position one unit away from previous group if possible. i.e. x group1 x group2 x group3 ... where x is empty seats.

Consider a sorted sequence and arbitrary two operations with $$$s_1$$$ and $$$s_2$$$. The sequence changes by $$$+2(s_1-s_2)$$$, the order of elements is unchanged. Here $$$s_1 = a_p$$$ and $$$s_2 = 2s_1-a_r$$$ for some $$$p,r$$$, so $$$2(s_1-s_2) = 2(a_r-a_p)$$$. Obviously then, if $$$g$$$ is the GCD of all elements, then their remainders modulo $$$2g$$$ can't change. In an even number of operations, it's clear that the smallest possible value of the first element is $$$a_1$$$ modulo $$$2g$$$.

The remainders modulo $$$2g$$$ won't change in one operation either, since $$$2a_p-a_i = 2(a_p-a_i)+a_i \equiv a_i$$$ modulo $$$2g$$$. If the number of operations is odd, all that matters is that the order of elements in the sorted sequence is reversed, so the smallest possible value of the first element could be $$$a_N$$$ modulo $$$2g$$$ instead. (Since $$$g$$$ is the GCD of differences, each $$$a_i = b_i g + r$$$, only parities of $$$b_1, b_N$$$ matter.)

Finally, it's always possible to construct a sequence in which the smallest element is $$$a_1 \% 2g$$$ by first adding and then subtracting some 2*differences, see Bezout's identity. The rule on non-negative elements isn't broken then.

Why are you --it2? I think you should just leave it. AC code Linkk