ISeeDS should be thanked for not participating in the contest

DoNtTellHimWhatToDo

Pretests_Passed, YaAAbu, SuccessfulHackingAttempt and FailedSystemTest can also hold sway the contest. And MemoryLimitExceeded, we should not forget. :)

Maybe should be accepted?

That's _Wrong_Answer

The one "for those who don't want to click" is good for me. I'm so lazy.

tourist blesses me!

seems like he will not come

[Error] no match for 'operator+' (operand types are 'div.' and 'div.')

Global round = max(div1, div2, div3) = div1; for divisions we use max;

I think that div1 > div2 > div3. My estimation is that div1 + div2 + div3 = div0.9

Let's downvote Errichto to make his contribution lower than Radewoosh

Generally, educational rounds have 7 problems and 120 minutes of time. So comparatively it is enough.

No, of course not. We need 300 minutes.

But there isn't a "Codeforces Global Round 1" in your contest history.

Most rounds is a bit late for Chinese,hope the contest can be earlier,and we won't stay up late and feel sleepy the next day.

accepted?

upd?

so every body going to forget " system_test " ,because he is unrated ...(read last line of editorial. )

me too

What is IOI brother?

Anybody prepare an IOI synchronized contest in Codeforces? That will be great.

I wish if there's something similar to this for the last rated round for Div1 before IOI2019.

That's bad.

What's the meaning of "contes"? I have never learnt a word called "contes".

:3 Meanwhile, the problem will be like:

maybe the contest doesnt have interactive problems xd

And MLE is for majk limit exceed (only joking :D)

why not ? because of just one mistake he did before ? i'm afraid for you ;

[DELETED]

you can do some contests generally -- pratice more. There're so many basic problems for newbies, such as Div2A/B

yashvardhan29 will become Foobar Elite today!

Added now.

Now that the round is over, I feel I should comment on that. We considered both D and E to be tricky problems, even some very good testers had problems with them. Perhaps C and D turned out to be a bit easier that anticipaded.

F on the other hand, was a different beast. I originally proposed a completely different problem — with in my opinion a cute observation. However, it was geometry and testers almost went on a riot because of the implementation difficulty.

About $$$10$$$ days before the contest I came up with the problem that is now $$$F1$$$, but the testers saw it as too easy, even easier than D or E — and the gap between $$$F$$$ and $$$G$$$ was afwul.

So I realised then that we can increase $$$m$$$ and it adds to the problem. Suddenly, it was too difficult and it was the gap between E and F that was brutal. With the contest fast approaching and the tester's ranks dwindling, we realised that the safest way to mitigate was to split F.

Some small gap still remained though, but I'd say that the ratio of ACs on adjacent problems around $$$4$$$ is fine (after all, it has to reduce from couple thousand to only a few in around $$$6$$$ problems). Please try to understand that it is sometimes quite difficult to judge the difficulty based on a small sample of testers. Additionally, if you find out two days before contest it is going to be speedforces, the best solution is typically to keep it that way rather than to perform some ninja fixes.

arsijo curse

My friend locked his problem E but he can't change his code after rejudged and already lost his rating

Why unrated after rejudged? I can't understand.

The round is always UNRATED or at the edge of UNRATED if arsijo is the coordinator.

Actually, I didn't get pretest passed in this problem, but I have an idea.

Solution 1
You can replace the problem to a new problem as follows by using euler tour method:

• Initially, there is an array $$$a = $$${$$$a_1, a_2, ..., a_N$$$} and $$$b = $$${$$$b_1, b_2, ..., b_N$$$}.
• Query type 1: Add $$$a_l$$$, $$$a_{l+1}$$$, $$$a_{l+2}$$$, ..., $$$a{r}$$$ by $$$x$$$.
• Query type 2: Get the maximum value among {$$$|a_l| * |b_l|, |a_{l+1} * b_{l+1}|, ..., |a_r * b_r|$$$}.

• You can divide at most $$$2 * B + 2$$$ subsegments, that the value of $$$a_i$$$ that increased in add query is the same in subsegment.
• Add query: Just add $$$x$$$ in proper subsegments. Complexity $$$O(B)$$$ for each query.
• Get-maximum query: Use Convex-Hull Trick and binary search. The convex-hull can calculate with precount. Complexity: $$$O(N)$$$ for precount, $$$O(B log B)$$$ for each query.

If you consider two first and two last characters of the string, you are able to find one in first two and one in last two that match. Add them to your palindrome.

Hint: Since no two characters are adjacent, and there are only 3 characters in the alphabet, there exists a character in the first two characters of the string that equals a character in the last two characters of the string.

Since no consecutive letters are the same, you can make any even-length palindrome into a longer odd-length palindrome. So let's fix the middle of the palindrome, how about s[n/2]. Then you can show that from the midpoint, you never have to go more than two spaces to the left, or two spaces to the right to find two letters that are the same.

So at worst, it will take 4 spaces to add 2 letters to your palindrome.

Be careful of n=4 case (may need to try other midpoint).

Can you share your cute solution :)

really kyot solution

i overcomplicated the problem :\

Am I the only one to try solving problem without that "no repeated letters" thing for 15-20 minutes before realizing that problem is totally different?

Does that make me gay? Oh, man :(

you guy are so funny, hahaha

How do you see this? Is it a browser extension?

But they are strong! :)

You can make the contest unrated for these 10 participants.

Well, After 1.5 hour I noticed that my ac submission has wa. It's kinda unfair.

I saw you hacked A. Can you tell me please what was the hack for it ?

Isn't problem F nice as well? Or I guess all the fun with it was too short for you to feel the excitement?

2 1 2 is also correct. You don't have to minimize number of parties.

I feel that mate. I got that twice

Both outputs are correct. Multiple solutions possible.

That's a good point, my solution also gives "2 1 2" but it passed system test... Maybe the intended meaning of "she can invite no parties" is "it is allowed that she doesn't invite any parties".

Take this one: aabbcc

please enlighten us with the solution!