I see. Thanks!

In the editorial of E:

Then everything to the left of is is non-increasing. Everything to the right of it is non-increasing.

Shouldn't it be "Then everything to the left of it is non-decreasing"?

The length of the string $$$s$$$ is even.

The hardest Div.2 round I've seen.

There are $$$q$$$ statements, but your solution reads $$$n$$$ statements only.

Great contest! I like A,C and F personally.

en_translator is still working on translating today's editorial for H. Great thanks!

Will there be any editorial for ABC216 in English? I believe there are many people who need it.

P.S. There have been editorials in Japanese already.

Obviously, the order of the pairs $$$(A,B)$$$ makes no difference to the answer, so first sort the pairs so that $$$A$$$ is monotonously increasing.

Let $$$f(x)$$$ be the number of the subsets $$$S$$$ that satisfy the condition in the problem statement, and the maximum element in $$$S$$$ is $$$x$$$.

To calculate $$$f$$$, let $$$g(x,v)$$$ be the number of the subsets $$$S$$$ that the sum of $$$B_i$$$ for $$$i \in S$$$ is not greater than $$$v$$$, and the maximum element in $$$S$$$ is not greater than $$$x$$$.

Through the definitions of $$$f$$$ and $$$g$$$, we can see that $$$f(x)=g(x,A_x)-g(x-1,A_x)$$$.

We can do simple DP to calculate $$$g(1,\cdots),g(2,\cdots),\cdots,g(N,\cdots)$$$ in $$$O(N\max\{A\})$$$ time and the answer to this problem is $$$f(1)+f(2)+\cdots+f(N)$$$.

So we can solve this problem in $$$O(N\log{N}+N\max\{A\})$$$ time.

Let $$$cnt_i$$$ be the number of candies with color $$$i$$$. Consider all $$$i$$$ such that $$$cnt_i \gt 0$$$. There are $$$\binom{N-cnt_i}{K}$$$ ways to choose $$$K$$$ candies so that no chosen candy has color $$$i$$$. To choose $$$K$$$ candies, the sum of the number of different colors are $$$\sum_{i}\binom{N}{K}-\binom{N-cnt_i}{K}$$$. There are at most $$$O(\sqrt{N})$$$ different elements in $$$cnt$$$, because the sum of the elements in $$$cnt$$$ are $$$N$$$. For colors with the same $$$cnt$$$ values, we can deal with them together. We can preprocess factorials and the inverses of those in $$$O(N)$$$ time to calculate each binomial coefficient in $$$O(1)$$$ time. So we can solve this problem in $$$O(N\log{N}+N\sqrt{N})$$$ time, which is fast enough.

In fact, the second solution is equivalent to the solution used xor basis. Why? Consider how xor basis work: The new number will be put in xor basis if and only if it cannot be represented as xor value of some subset of current xor basis. On the other hand, $$$0$$$ and $$$x$$$ are connected if and only if $$$x$$$ can be represented as xor value of some subset of the numbers considered as edge values before. Therefore, the time complexity of the second solution is $$$O(N\log{N})$$$.

I'm not sure whether the first solution is correct, but I know that it is almost the same as the second one. The reason is that j and j^i are not connected is equivalent to i cannot be represented as xor value of some subset of the numbers considered as edge values before. In other words, if it breaks, it will break at the very beginning. However, my doubt of the first solution is that, why it is still right, even though if it adds one edge between $$$a$$$ and $$$b$$$ ($$$a \lt b$$$), it will break later when $$$j=b$$$.

The Final Round of Technocup 2019 starts this Sunday, on the March 21, 2021 at 13:00 MSK (10:00 UTC)!

Shouldn't it be Technocup 2021?

P.S. If I misunderstand this, can someone please explain it to me?

I have realized how powerful the AtCoder library it is... Look at this AC submission of E if you want to know reasons.

How to solve E and F?

This is what I want! Thanks so much!

FST is equal to "Failed to pass System Tests".

Although I didn't count how many people failed to pass the system test on some problem, I found that my rank changed from 1238 to 764.

This round was really Fst forces!

Due to the official competition source codes of other participants will not be available for an hour after the end of the round.

So maybe we shouldn't discuss anything about the contest especially the solutions and the hacks.