For those from 2017, you can find them on the site (they're on the right as you get to the tasks section of the competition (to get to the right year, choose the year from the past editions section, at the right of the top menu). For 2018, only the solution to the hardest problem is posted:( . None of us has had time to upload them, there are some sketches though in the comments section of last year's post. If you find anything unclear, I can try to answer some questions about them. I hope we'll eventually post the solutions to last year's problems as well as to this year's ones.

For the online mirror? Not yet, it will be soon. I'll update the post with the link to the online mirror, but you can count on it starting just after the contest. As for official participation, your country has already registered, we're open to new countries generally, although we're already getting pretty close to the contest itself, so it'd be a bit difficult to sign up a new country for the official version in such short notice (yet doable).

Nope, that's also unfortunately the reason for which we couldn't afford accepting several unjustified teams from a country. We only have one server and want the feedback system to go smoothly so we had to constraint it. The limits are usually around 20 submissions per problem and we've increased them in the beginning in the past when we made sure that it wouldn't negatively affect the judging queue. We're trying to maximize it without the expense of a not-working system. The maximum number of submissions appears on CMS on the submissions tab of a problem and may vary from a problem to another (mainly based on its TL)

Sorry for the late reply (you definitely got answered back by now), but yes it will. Something's wrong for some reason. We'll make it work for tomorrow. Sorry for the inconvenience, I hope you can still partially enjoy the round.

Surely, the contest was scheduled for 9 AM Romanian time initially and at some point they moved it to 1 PM Romanian time. So they must have forgotten to shift the results, too :D

me too

Same here

Same here.

I guess everyone has the same problem :(

They will probably extend the contest.

It worked for a second. Error is back again :(

Ah so the server wouldn't load for 30 minutes but they still keep the tasks there so that the lucky somebody could download them and have some advantage. Thanks for sharing :)

Yes, they will (your last submission was a 0, the one before it a 13, and all the 3 before that also 0s)

I gave my soul for my 100-71.5-0-34 and then two of my teammates hit me with 100-100-100-33 and 100-100-100-87 :D Still haven't heard from the third one and I think I don't want to.

Didn't even try the third problem and thought about the last one almost all the contest :D That strategy produces noob looking 162 xD

My prediction
Gold : 273 and above
Silver : 200 to 204.59
Bronze : 140.5 to 200

I love interactive problems. Most of the times they are out-of-the-box and they are really cool to think about, even sitting somewhere in the comfortable chair.

Why you don't like constructive problems?

Hi! The online mirror is still running so we'll post the solutions afterwards. Hopefully (although I'd say unlikely) we'll have them written nicely and published on our site. If not, then you'll find them in comments explained by either us, or some other participants

B

C Cat: (without proof)

• Make the permutation zero-based.
• If there is some i with p[i] + p[n - 1 - i] ≠ n - 1, then the answer is  - 1. The multiset of values {p[i] + p[n - 1 - 1]} is invariant under the double-swap operation and at the end it should be all n - 1.
• After this, we only need to worry about indices , the other values will be automatically correct as p[n - 1 - i] = n - 1 - p[i].
• While there is some i with p[i] ≠ i and p[n] ≠ n - 1 - i, do a double-swap with i and p[i]. (Similar to what you would to to sort a permutation with the minimal number of single swaps.)
• Now for every index i, either p[i] = i or p[i] = n - 1 - i.
• If p[i] = n - 1 - i and p[j] = n - 1 - j and i ≠ j, i ≠ n - 1 - j, then we can do a double-swap with i and j followed by a double swap with i and n - 1 - j to get p[i] = i and p[j] = j.
• If three is a single pair (i, n - 1 - i) with p[i] = n - 1 - i, then we print -1. Otherwise print all double-swaps made.

D Mouse: (I don't know how you get 100, but you can get 90 with the following randomized idea).

• We keep a n × n table C that keeps track if p[i] can be j for all pairs (i, j).
• If at any point we have n ones in our table, then we know the answer.
• If we query q and get 0 as a response, then we can set C[i][q[i]]) to zero for all i. We will ignore all non-zero responses.
• Just asking random queries already scores some points, but we can do better if we weight queries based on our current information, try 50 - 300 queries and pick the one with the highest weight.
• In my weight function, I included an approximate probability for the query to give me a zero and how many ones I can cross off if that happens (with a higher weight for ones in a row with few ones).
• As a final optimization, if we determine that p[i] = A, then we cross off (j, A) for all j ≠ i. To make this happen more often, we do the following: If there is some i with at most 6 values of i, then we try to do a query to narrow down the possible values of p[i] and only weight those queries with the probability to get zero as a response.

Another solution to B ( https://ideone.com/4EZfS7 ), which I think is easier:

The grid

XXd
XXd
rr.

the number at the end is 6 times the number in the top-left.

We can "stack" these "blocks" to easily create powers of 6

XXd....
XXd....
rrXXd..
..XXd..
..rrXXd
....XXd
....rr.

We can change the 'r's and the 'd's into 'X's to change the coefficients of the powers of 6. There's also a separate case for the last "block".

This will have a maximum n of ceil(log_6(10^8))+1, which just happens to be 49.

They should be uploaded on the site pretty soon. We also hope to send them to oj.uz in the near future