I'm currently participating, will we be able to upsolve once it finishes?

Some people are still coding. I'll post a link to the results here when the contest is over.

Thanks! This was a really fun competition.

For the record, my 51 points solution to the last task is as follows. Take any function that can translate ID to dishes (I assumed simply taking modulo 3 would have counter-examples, so I took a random number from rand() after fixing the seed with srand(ID)), Then with a high chance you're going to have small groups (consecutive table segments with same dish order) and you're going to see entirety of the group (including boundaries) in the data given. Give every odd person in the group his original order, but replace every even person's dish with the one that is not the dish of the next group on the right. That way no matter what the group size is, you avoid clashes within the group and with the neighboring groups. Unfortunately, you sometimes won't see even one side of the group in the final subtask because while the chance of seeing one boundary is , you have 50000 queries overall (I haven't realized that for a while and just assumed my function was not random enough, creating quite hilarious random function complications throughout the contest). I'm really interested to hear the full solution to this problem.

You can maintain a set of nonnegative values (up to 10^5) that are not in the current subarray. The smallest of them is the mex value of the subarray.

Solved it using map and some optimization,my approach was kind of naive.. Simply take the first subset 0 -> k,calculate the current mex in the map then 1 -> k+1,(you wont have to loop over the array anymore, just loop over the map to find the first positive integer that didn't appear) 2 -> k+2, And so on..

you will have to calculate the mex each time, but there are few times where the mex doesn't change

I thought Finland is best known for things like sauna, Sibelius, etc., but it seems that the reality is very different :D

My solution for larger N was: make a random solution; while some row/s/column/s have equal sums, pick two and swap two elements from them. Since there are O(N²) possible sums and only O(N) are used, you can guess that this will give a solution quicker for large N.

Note that it's quite easy to make every sums "equal", which is a complete reverse of given problem : You just pick any arbitrary permutation, and write down all of it's cyclic shifts in a rows.

After finding out this fact, I tried to pad some numbers to make those sums different. I first wrote down {2, 3, 4, .... N-1}, {3, 4, 5, .. N-1, 2} ... in rectangle [2, N-1] x [2, N-1]. Now I have 2 numbers of {2, ... N-1} and N numbers of {1, N}. I wrote 2 ~ N-1 to each row and column, and placed other 2N numbers appropriately to make all sums different. In result, it looks like this in N = 7 :

1 2 3 4 5 6 7
2[2 3 4 5 6]7
3[3 4 5 6 2]7
4[4 5 6 2 3]7
5[5 6 2 3 4]7
6[6 2 3 4 5]7
1 1 1 1 1 1 7

It's easy to show that it gives answer correctly from N >= 5. (code : https://cses.fi/95/result/40780)