will see :):

Usual — 150 min

And also notice the unusual duration of this contest (usual time + 15min)

Ahh Thank you so much YPK.

If you'd just added this to your original comment everyone would know what you are talking about without going back through the revision history.

How do you know they are new? Have you checked all the problems that have ever been published?

Not very new after all.

Exact same problem as D

I won't believe your words anymore

As a Participant I'll tell my feedback after the contest!

Well, let's hope all the problem statements are not quatrains like this one.

That's right.

Your graph indicates that you're a hard-worker. I wish you the best of luck.

I wish this round will be my last round in Green :)

))

Meanwhile problem-setters:

Bhest aaf lack

Are you planning to retire after this round? XD

Fixed, thanks for pointing out.

I just want to add that "this round will be my last round in grey" is more ambitious than "I will become pupil after this round". :)

Why are all of your memes light-hearted?

Both setters are from Cyprus. And as they mentioned, this is the first ever fully Cypriot round.

May be it will be postponed

No it's not? The system testing is excellent in terms of speed.

That's because of your bad profile picture.

You should register before the contest begins next time or 10 minutes after it starts, sadly you can't compete now but you can solve the problems after the contest ends!

4) don't tell spoilers.

If all the characters in the given string are c then the answer is 0. Else just use brute force whether the answer can be 1. If the answer cannot be 1 then the answer will be 2 as we can always select n and then n-1.

And here I was, shocked, bamboozled, confused whether I am crazy, hallucinating, or just sleepy when I saw blue Indian kids solve D in like 5 mins or so. To whoever authored this problem, if it was a weird coincidence, then oh well, shit happens. Otherwise, fuck you.

We are really sorry for this, we were not aware that this problem already existed and the idea -we thought- was ours and came from scratch inpsired by the game "among us".

What was the intended solution? I was trying DSU's. I had the idea, but couldn't come up with an implementation.

I replaced unordere_map/unordered_set with map/set and reduced the running time from TLE to 124ms.. not sure if some of the test cases were deliberately designed to make c++ unordere_map/unordered_set fail.

I really liked how authors were lenient on time and constraints as well.

I used DP. But I think basically it's a math formula.

It was neither. It was basically just computing $$$3 \cdot 2^{2^{k+1}-3} \bmod {10^9+7}$$$ fast.

It was simple combinatorics. You can color the root in any one of the 6 colors, all the subsequent nodes can be colored in any 4 of the remaining colors. So the answer is simply 6 * 4 ^ (number of nodes — 2).

i solved it using a basic loop
ans *= 4^(2^i) where i from 2 to k then multiply the answer by 6

We did not copy the problem nor were we aware that it already existed. We are sorry for this, our only inpsiration was the game "Among us" and the rest were our ideas.

could u link this CSES problem?

I rage quit when I saw my "friends" solve D in 5-7 mins, and I went on scratching my head on that problem for like an hour or so, without any solutions.

I'm not sure about the CSES problem you're referring to. If it's https://cses.fi/problemset/task/1712, then I guess you missed the fact that $$$2^{60}$$$ can easily be stored in a long long, and you can do straightforward modular exponentation to that value. The main part was to find the formula (although short).

So you can only solve problem you've known before? Sound like you got the delta you deserve.

Yes and union find set is used to solve D

It was simple combinatorics. You can color the root in any one of the 6 colors, all the subsequent nodes can be colored in any 4 of the remaining colors. So the answer is simply 6 * 4 ^ (number of nodes — 2).

the minimum number of operations $$$m$$$ can be only 0, 1, or 2.

the upper bound is 2 because we can always choose $$$x_1=n$$$ and $$$x_2=n-1$$$ to make all characters in $$$s$$$ equal to $$$c$$$. $$$n-1$$$ can't divide $$$n$$$ since the problem guarantees $$$n\geq 3$$$

if originally each character in $$$s$$$ equals $$$c$$$, $$$m=0$$$

if originally $$$s_n = c$$$, $$$m=1$$$ (choose $$$x_1=n$$$)

if originally $$$s_n \neq c$$$, we find the last position $$$p\in [1, n]$$$ s.t. $$$s_p=c$$$, then if $$$p>\lfloor {\frac{n}{2}} \rfloor$$$, $$$m=1$$$ (choose $$$x_1=p$$$), else $$$m=2$$$

We were not aware of that and are really sorry about it.

We will make sure that nothing similar happens again.

Thank you for your kind words, we tried our best and are glad you liked them!

Same :pensiveclownwobble:

Consider the number base n. Each good number is a number that has all digits equal to 0 or 1 in base n. Well, to find the kth of those, realize you can generate all numbers with only 0 or 1 value digits by just counting in base 2 and looking at binary representation, since binary is different masks of 0 and 1. This means you can just look at the number k in binary, and for each 1 in the binary corresponding to 2^x, you add to the result n^x.

I was also doing the same but turns out that's not required and neither correct, u just need to check for i from (n/2 + 1) to n if there is any character equals to c then ans {i} otherwise m has to be atleast 2 and that can be done using {n,n-1}

7 a abbbbab you can choose x = 6

The optimal index is not necessarily prime.. Any index i such that for j in 1 to N, if j is divisible by i, then s[i] is equal to C will give an answer of 1. If such an index does not exist, simply use the last 2 indices, and ans will be 2. Can be brute forced ( NlogN )

one counterexample is:

7 a

aabbbab

in this case you should choose $$$x=6$$$, which is not a prime

I am not sure, but from the judge, there is a wrong answer on the case:

3 2
1 3 imposter
2 3 crewmate

where you should print "2" instead of "1".

One of my solutions still shows pretest passed after system testing. On resubmitting the solution after system tests it is accepted. link is attached below

Solution : My solution

Link to blog : Blog entry

Hi Mike, can you help me with this? my submission on problem B was considered plagiarism, but our submissions coincided just because the solution to pB is quite simple, and contains only one simple bit manipulation procedure.

This is actually true about long long xD

Strange verdict was fixed. Rating will be updated in some time.