Not sure if this is still needed, but putting it here for others like me who come here looking for an asnwer. There are 3 icons at the very right of the problem. Hover your mouse pointer over each of them. The one that says, "Edit", is the one you want. If your problem names are now out of order, e.g. A, B, D, then there is a similar icon above the problem-wise buttons that lets you edit problem labels.

Cheating is abhorrent and is ruining the competitive programming spirit on Codeforces. It is a pain for everyone. So it is great that you keep a keen eye on possible cheaters and call them out.

This particular user is a student in my class (my own profile is not that great, I know, we can discuss that separately if you like), has demonstrated consistent improvement in class, and practices diligently on Codeforces and other platforms that I indicate in class.

His performance in this contest is in line with what I see in class. He discussed the matter with me. His submissions for all problems in this contest except H use a shorter version of his usual template. Due to the way marks are assigned in the class, he attempted H first, just like in the recent Div 3 contest last week. In the meanwhile he got access to his old files, one of which he mistakenly submitted, and attempted other problems with his template.

I was not with him so cannot guarantee that cheating did not happen, but I am satisfied with his explanation especially because it is in line with the performance that I have seen, the evidence implying cheating is weak, and the alternate explanation is simple. Others, of course, are free to form their own opinions.

That he responded aggressively and disrespectfully to you in the heat of the moment is regrettable. I talked with him about it and he acknowledges his error. Hopefully, he will be more respectful in the future.

AI detectors are famously unreliable. I am not sure which one you used and will be grateful if you can share a link.

You are right. This leads to an $$$O(\log x)$$$ solution.

The editorial tells us that $$$y$$$ must have at least 2 on bits — one matching $$$x$$$ and one not matching $$$x$$$. Also, $$$y \lt x$$$.

1. If all bits in $$$x$$$ are on, i.e. $$$x=2^k-1$$$, then y cannot exist.
2. If only one bit in $$$x$$$ is on, it must be the MSB and $$$x=2^k$$$. Any $$$y$$$ will end up greater than $$$x$$$, so $$$y$$$ cannot exist.

If you discount these cases, then $$$x$$$ has an on bit in MSB and some on bits and some off bits afterward.

One easy solution for $$$y$$$ is to copy $$$x$$$, turn off the MSB (so now $$$y \lt x$$$), and turn on all the other bits, i.e. $$$y=2^{k-1}-1$$$ where $$$k$$$ is the bit length of $$$x$$$.

The time complexity is determined by counting the bit length of $$$x$$$.

MikeMirzayanov maybe you have some suggestions.

Any particular reason for the downvotes?

I understand that this blog is not suitable for public display — it is for a mashup in a private group. I tried posting blogs in the private group, but then Codeforces treated the links to the original problems as group links, thus making them invalid. Posting publicly is my solution to preserve the links. If anyone has a better solution, please let me know.

What does it mean to have no chill?

From Prohibited AI Use,

"You may not input the problem statement, its summary, any excerpt, or a sub-problem into an AI-based system to receive ready-made code"

how is using AI to retrieve code for, say, Dijkstra's algorithm, different from copying such code from a ready-made template file on my machine? I know that many regular contestants do the latter.

This is pretty neat. Thanks for sharing.

I figured it out. Because of my confusion with the range of x, the code decrements x. That maps x=0 to x=-1, which does not result in an error but in undesired behavior when used as a list index.

I spotted one mistake but that does not fix the Wrong Answer above. The mistake is that each array should have size n+1 instead of n because each 0 <= x <= n. I also fixed the ranges accordingly. But, as mentioned, the Wrong Answer stays the same.

I have implemented the same logic as the solution above for 2009D - Satyam and Counting, but keep getting

Wrong answer on test 2: wrong answer 7th numbers differ - expected: '100', found: '99'

I cannot see the corresponding input.

Can anyone please tell me what I am missing? https://codeforces.me/contest/2009/submission/279654100

Thank you for explaining patiently.

Consider the initial array (1, 1). As per the rules, the following is possible:

(1, 1) -> (1, 2) -> (1, 3) -> (1, 4) -> (1, 5) -> (1, 6) -> (1, 7) -> ...

Every positive number can thus be obtained.

Sorry if I am overlooking something trivial.

Can you please clarify some of the samples, copied below, from 2008G - Задание Сакурако?

Input:
2 10
1 1
Output:
11

Query: Can we not generate every positive number from 1 and 1? How can there be any mex other than 0?

Input:
3 1
1 2 3
Output:
3

Query: Can we not generate every positive number from 1 and 2? How can there be any mex other than 0? Also, 3 is part of the input array. We can choose to operate only on the 1 and 2, always leaving the 3 in the input array. How can it be a mex?

Input:
3 2
1 2 4
Output:
4

Query: Similar to previous case.

For Problem F, sample 6, reproduced below, can someone explain how the required 25 points can be obtained in 80 operations, as given in the sample solution?

3 25
9 2
4 3
8 10

For problem E, the image in the Notes section loaded for me without any gorillas and I lost a lot of time trying to make sense of the image.

For problem G sample 1, can someone share the route which corresponds to a starting time of 0? From what I understand of the problem, there should be no solution and the answer should be -1. Maybe I am overlooking something. I am copying the sample below for convenience.

5 5
100 20 80
1 5 30 100
1 2 20 50
2 3 20 50
3 4 20 50
4 5 20 50

Indeed, here is a solution. https://codeforces.me/contest/1950/submission/253859143

OP is 4 yeas old by now, but in case anyone is about to use it, please know that it does work on 29 March, 2024.

One straightforward optimization can be that once i crosses n/2, you can be sure that the answer is n.

For Problem E, the tutorial says we need the "smallest" value of r. Shouldn't it say the "largest" value of r instead?

Is there a way to subscribe to the calendar?