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Hi, what's the bug?

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We initially solved this problem for odd $$$n$$$ and extended it for even, so it might have been that for the setters it was easier to extend the logical line of reasoning for the solution. I think more testers could have been beneficial (at least higher rated ones) to get a better sense of G, we received some comments about hard implementation but nothing that made us worried about the difficulty. We’ll try to get a more wide sense of difficulties next time.

Yes, I think G was a bit too hard for its position (or maybe F was too easy). We were considering having to solve F in n log n (as a subtask or just on its own), maybe this would have helped -- but it would have also increased implementation.

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Maximum independent set (a set containing the maximal amount of vertices where no two are adjacent). The abbreviation is explained in the editorial now.

This was a really common question asked, "bbaaaabbb" also works.

No need to say sorry, just pointing it out :)

This is also explained as Solution 2 in the editorial!

There are only 2 things the MEX can do: increase or stay the same (it can never decrease since we are looking at larger and larger prefixes).

In the case of a positive difference, if the MEX stayed the same then it would have to be greater than the current value, which is impossible because that value had to appear earlier on in the prefix. Since the MEX is calculated on a permutation that can’t happen. So it has to increase.

In the case of a negative value, the MEX has to be less than the current value. But if it increased that means the current value changed its MEX, meaning its new MEX is at least (current value + 1) and it is actually greater. So it has to stay the same.

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