Ah okay. That's a relief.

Me too. Then I noticed that there are no updates in the chat window. Then I suspected that my Internet connection is down, so I tried to open Codeforces. Then, I restarted arena. It couldn't log in.

Haha, I also wonder how many points I would get by solving nothing

I've managed to read Hard, and then got the same problems.

Also is this safe to discuss the problems now?

Confirmed. (At least, so far as official TCO advancement and ratings are concerned.)

I think that's a sign:))However, if you are on a plane, then how could you write here? Or it hasn't departed yet?

The contest appeared again also.

Did you mean Sep 9th, which is a Saturday?

My initial plan was to go to Poland to hunt a slot for wildcard in case I fail in 3B.

Now the decision is more difficult. Should I go to Poland to increase the probability of qualification?

TCO10-12 (at least) had the following rule: In the event that a round must be cancelled for any reason, the round will start the following day at the same time. I am rather unhappy that this rule no longer exists -- Sept 9th is not very good for me, while reserving the next day is usually easier :(

What happens if someone qualifies both to the Finals and the Wildcard Round? Is the Wildcard Round spot given to the next competitor?

L = 2 is easy (brute force all ratios from 2 to b - 1).

I think L = 3 is the main part.

By the way, if you find a short solution, you can just append zeroes to get longer solutions.

You can add 0 at the end of your answer. So you only need to find the shortest one.

For L = 2, the answer should be . If b ≠ 3 and b + 1 is not a prime number, The shortest answer is 2.

For L = 3, there are 4 cases.

pb² + qb + r|qb² + pb + r. pb² + qb + r|(qb² + pb + r) - (pb² + qb + r) = (q - p)(b - 1)b. So you can enumerate the value of q - p and find the divisors of (q - p)b(b - 1).

pb² + qb + r|rb² + qb + p. pb² + qb + r|(r - p)(b² - 1). It is similar to the first case.

pb² + qb + r|qb² + rb + p. Let x = qb + r, y = p. k(yb² + x) = xb + y. So x / y = (kb² - 1) / (b - k). You only need to enumerate k, and find gcd of kb² - 1 and b - k.

pb² + qb + r|rb² + pb + q. It is similar to the third case.

Can you explain your flow solution?

The following is not a formal proof, but one that seems good enough for me :) Suppose that we need to remember three cells, let's name them a, b and c. Then the path must first visit a for the first time, then b for the first time, then c for the first time, then a for the second time, then b for the second time and finally c for the second time (if the path was for example abcbac then we could take a 'shortcut' between the two a's, costing at most two instead of one, but saving the cost of b). If we need to start in one corner and end in the opposite corner, then (for topological reasons) this structure (abcabc) is not possible without the path crossing itself (try it on paper), so we never need to remember three cells.

Below are two examples. The first example shows that we do need to remember two cells and the second example shows that if the destination (or source) is not on the border, then we need to remember more cells. The second example can be extended to any number of cells that need to be remembered (if the grid can be large enough). In the examples S stands for source, D for destination and lowercase letters for initially blocked cells that need to be removed in the optimal path.

SS..####
SS..####
##..####
....####
...a....
..#.....
.....#..
....b...
####....
####..##
####..DD
####..DD

SS#####.......
SS#####.......
..###....###..
..###...cDD#..
..#....#.DD#..
..#...b...##..
.....#....##..
....a...####..
####....####..
####..######..
####..........
####..........

EDIT: krijgertje was faster :)

My explanation doesn't look formal enough, but anyway, here's my attempt.

Usually, like Case 1, simple Dijkstra works. The cost to move the square from position p to position q is the number of cells that are covered by q but not by p.

It fails only when things like Case 2 happens. We shouldn't count the cell A twice.

This pattern can be nested in more complicated ways. In Case 3, we have to keep both A and B before reaching A for the second time.

In Case 4, do we have to remember three cells? No. Instead of using A+B+C, we can use A+C+D to achieve the same cost.

Let's call a cell "special" when the cell is visited twice. Suppose that in some case we have to remember three special cells. Call them A, B, C in the order of first visits. Then, the second visit to A must be after the first visits to B and C (otherwise we can remember B or C after we discard A). Possible cases are:

• (1) A1 -> B1 -> C1 -> A2 -> B2 -> C2
• (2) A1 -> B1 -> C1 -> A2 -> C2 -> B2
• (3) A1 -> B1 -> C1 -> B2 -> A2 -> C2
• (4) A1 -> B1 -> C1 -> B2 -> C2 -> A2
• (5) A1 -> B1 -> C1 -> C2 -> A2 -> B2
• (6) A1 -> B1 -> C1 -> C2 -> B2 -> A2

In (2) to (6), we can find a "shortcut path" like Case 4. (1) never happens. As in the picture on the bottom, the square will be trapped into a closed region.