_"Can I use prewritten code / templates?

No."_

it's clearly mentioned

phlap will ak gold in 15 minutes and take plat! Also, gggg0 and nyctivoe will ak platinum!

Man, you were so close! The ideia was to add random elements from the even array, so then the memory would be small. Here is my code, it passed after 1048576 submissions.

https://pastebin.com/U3yUCLRU

alright guys, I just want to share my deterministic solution, with 11*10+3 submission in worst case. firstly, we found N <= 1e9, but the digit of N <= 10, that's nice, so, we can read the N as string, and submit something that, if s[1] == sth(like '0'~'9', or 0 that the N's length may differ), and if so, do a infinite malloc so that you are surely getting MLE verdict, to differ this from others.

now, we get the input testcases. now, we output YES or NO, for output testcases. then we got all of these, then the code will be like: switch(n) {case 12831728: case 2347238: puts("YES"); break; default: puts("NO");}

I hope this will work.

123gjweq2

nope, stack overflow isn't documentation. documentation is stuff like cppreference.com or the official Python docs

Yes, you can.

may I ask what actually is this "12:00 ET" in +8 time? I searched it but no answer(I don't need to sleep early bruh)

and for Silver as well

You aren't allowed to use pre-written code or templates (or any online resource other than a reference for your language). Read the "Contest Conduct and Academic Integrity" section here: https://usaco.org/index.php?page=instructions

I didn’t type code this contest, but here’s an outline of a P3 idea. I think it works, but I’m hoping intended is easier to code?

If you wrote a brute force you probably had that when N is even there are two permutations, and the hard case is when N is odd. I had ugly casework on the position of N/2 (first, last, even in the middle, odd in the middle). N/2 has to be adjacent to 0 and/or N. Placing N/2 and 0 and/or N adjacent to it tells us which cells must have low numbers <N/2 and which must have high numbers >N/2. From there, for each case there's a way to place pairs in a certain order such that we have exactly two choices for each pair.

For example, if N/2 appears first and the second number is 0 (the case where the second number is N is symmetric), we know N/2 + 1 can only be adjacent to 0 and 1, so it must either be in the third position or the last, with 1 adjacent to it. But then N/2 + 2 can only be adjacent to 0, 1, and 2, so we have to pick 0 or 1 to put it next to and then place 2 adjacent to it. And we can keep going until we've placed all the numbers.

From here, for each case, each constraint implies two options of the form "once we place the i'th pair, it must be the x'th pair we placed on the left and there must be y pairs on the right", and we can compute the number of ways to satisfy all constraints using an O(N) dp (taking care to avoid inserting a log by sorting). This gives an O(N^2) sol.

The annoying part is translating the constraints on the input into constraints on the way we place the pairs into the arrangement--this looks substantially different for all four cases and while it works, it seems painful to implement.

Can you teach me? I just know the $$$O(3^k)$$$ solution.

I didn’t type code for this contest, but I think something like this should work?

1. Count the contribution of each bit separately (i.e, count sum of 1/union over each possible bit in the intersection). Fix a bit and consider only input strings with that bit set.
2. Run SOS to get the number of input bitstrings that are submasks of each possible bitstring.
3. Multiply each value in the SOS output by a magic number depending on the number of set bits.
4. Invert the SOS (i.e, run sum over supersets). For the right choice of magic numbers (constants which can be found through a PIE-like computation), this will get you the answer for each input bitstring.

Well, this is funny, I also only got up to k<=18 but mine is $$$O(k^22^k)$$$ XD

As a silver participant, I would like to say Silver P1 was nice, due to how easy it was :)