This contest is for div3 only. I do not think there will be a relationship between div2 and div3. I mean can something be easier than Div2A? Maybe Div3A+Div3B+DivC are so close to Div2A. We will see after the round :)

This contest is only Div3 , so i don't think there's any relationship with Div1 or 2.

"We plan to include in these rounds simple training problems that will help beginner participants to gain skills and to get new knowledge in a real contest."

Судя по тому, что контест будет "по правилам образовательных раундов", — нет.

Thank you, it is fixed now!

:P

We will experiment with formats of Div. 3 rounds, but there are some reasons to start with rules of Educational Rounds:

• I prefer Div. 3 participants to solve problems during a round instead of spending time trying to hack
• for non-trivial problems tests in Educational rounds much more complete than pretests in regular Codeforces round.
• my plan is that Div. 3 rounds will not affect regular Div. 1/Div. 2 rounds in terms of schedule. It means that Div. 3 rounds (as well as educational) can't use time of KAN to coordinate them. They completely prepared by writers with my little help. It is much easier to write a round by rules of Educational Rounds than Codeforces: issues in writer solutions/validator/checkers affect a round less, less requirements on completeness of tests, etc.

You might want to take a look at this!

Here

As of 5/5/2018, there is no such thing as a Div. 4 user. The lowest is Div. 3.

Probably, participants from the first division will not be at all interested by this problems. And for 1600-1899 the problems will be too easy.

So, I think, there will be no issues with Div3 problems, as at some points, we will figure out that it is eventually too easy, regardless of which problem in a contest (A, B, C, D, E, or even F).

In other words, soon we will realize none of any Div3 problems will be any tough to be obliged to be distinguished.

ACM-ICPC scoring is used in many contests.

If you don't like the format of educational round, you don't have to participate.

If you don't want to be rated because of these rounds, you can always do virtual participation of solve problems regardless.

once you, too, was thus noob

Hey bro cyan isn't too hot either. We're all noobs, just some less nooby than others.

I like books better than the TV show, but the TV show is also wonderful =) Especially Natalie Dormer =)

За что минусы то? Я просто спросил.

Да не получит он +300 рейтинга, я гарантирую это

One can only hope^_^

don't worry, they'll open div 0 for him

He does not even need to look to the problems to solve them. :)

Obviously...

You will be offered 6 problems

for a 2-hour contest, indeed. 20 minutes seems better with 3~5 hours contest.

20 mins of penalty is standard ACM rules. It encouages coding correctly instead of coding fast, which I think is ok. Also, the contest is targeted at people who can't solve all problems in 19 mins :)

Penalty is calculated as the sum of times each problem solved (+ 20mins penalties), not as the time of the last solution. So it's not doubled because of that, don't worry :)

What exactly? Please, give me some examples of such handles.

I agree. In Div 2, very few experts (the highest eligible rating) solve the set. In this one nearly every specialist and their mother solved it (not really but you get the idea). You never see 200+ eligible participants solve the set in a round.

.

I think Div3B is too simple for Div2B

Sort the elements, and the k'th element (1-indexed) is your answer. If k = 0, then just subtract 1 from the first element.

The no solution cases are when the k'th element equals the k+1'th element (why?) and when k=0 and the first element is 1, because subtracting 1 would be outside of your range (you may be tempted to try to be clever and subtract 2 so you don't have to handle that case separately, but that fails if the smallest element is 2).

if you mean E,E is simple. Just for each connected component see that every node is connected to exactly 2 other nodes.

Полный, но еще взломать могут.

Ratings will be changed after whet Open Hacks Phase ends (in 12 hours approx)

Thanks, but next time please send it via private message.

vovuh explained here: http://codeforces.net/blog/entry/59281?#comment-428862

"6-th test is anti-qsort test especially for Java solutions. Replace int with Integer and it will be ok."

my idea, maybe wrong.

dsu = new DisjointSetUnionWithRollbacks
for component in componets:
    flag = 1     
    for edge in component.edges:
        dsu.add(edge)
        for edge in component.edges:
            dsu.erase(edge)
            if dsu.numIgnoredEdges > 0:
                flag = 0
            dsu.add(edge)
    ans += flag

UPD: solution 2

dsu = new DisjointSetUnion

for component in components:
    [head: tail] = components.edges
    flag = 1    
    for edge in tail:
        if dsu.add(edge) == EDGE_IGNORED:
           flag = 0
    ans += flag

In fact, we are to check, if after each edge deletion, component becomes tree.

However, this solution is very complicated.

Simpler ones:

1. component is good if each node touches 2 edges

2. component is good if it contains k nodes and k edges (sorry, it turned out to be wrong)

Вроде не зря.

Если их на qsort взломают, пока они в div 3, они это увидят, запомнят и такое больше не повторится. Вреда это особо не принесёт, так как у них не такой большой рейтинг, чтобы о нем беспокоиться. Но если они сейчас не поймут, что qsort — штука опасная, то они могут на нем проколоться в будущем, когда это будет более болезненно.

Could you please explain the test you used to hack so many F solutions? I tried to find something in common in the solutions you hacked but failed to see the pattern. Your test seems to be tricky.

I think it's because some of the author/tester solution fail on the same test.

Testers' solutions receive time limit exceed, because of this your hack gave an "Unexpected verdict". I hope I have fixed it now. Sorry for long delay.

It's already posted, here

Yeh same question,i wonder how can i know when the sys test will start ? the contest just said "Finished"

Afaik Unordered_map is slower than normal map.

unordered_map is based on Hash table. Normally a search operation of hash table works in amortized(1) but worst is O(N) (caused by hash-collision). std::map, on the other hand, is based on red-black tree, which has O(logN) complexity in search operation on average. Your solution was hacked by anti-hash test.

Regardless of whether you are a trusted participant of the third division or not, if your rating is less than 1600, then the round will be rated for you. So although unrated participants are not trusted, they are still rated.

You ignored MAP[A[i]] that was originally stored.

test : 9 10 11 12 8 9 10 11 10 11

ans = 4 yours = 3