Hello! Codeforces Round 820 (Div. 3) will start at Sep/12/2022 17:35 (Moscow time). You will be offered 6-8 problems with expected difficulties to compose an interesting competition for participants with ratings up to 1600. However, all of you who wish to take part and have a rating of 1600 or higher, can register for the round unofficially.

The round will be hosted by rules of educational rounds (extended ICPC). Thus, solutions will be judged on preliminary tests during the round, and after the round, it will be a 12-hour phase of **open hacks**.

You will be given **6-8 problems** and **2 hours and 15 minutes** to solve them. One of the problems in this round is **interactive**. Don't forget to read the guide on interactive problems.

Note that the **penalty** for the wrong submission in this round is **10 minutes**.

Remember that only the trusted participants of the third division will be included in the official standings table. As it is written by link, this is a compulsory measure for combating unsporting behavior. To qualify as a *trusted participant of the third division*, you must:

- take part in at least five rated rounds (and solve at least one problem in each of them)
- do not have a point of 1900 or higher in the rating.

**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.**

Thanks to MikeMirzayanov for the platform, help with ideas for problems and for coordination of our work. Problems have been created and written by ITMO University team: MikeMirzayanov, myav, Gol_D, Aris, Gornak40, DmitriyOwlet and Vladosiya.

We would like to thank: Kniaz, BledDest, Svyat, Be_dos, Timur2006, FelixDzerzhinsky, alphabet321, FedShat, vsinitsynav, Jostic11, erankyun, _Roma_, KKT_89, ABalobanov, lightseba, powergee101, DafuQ_o and AshrafEzz for testing the contest and valuable feedback. List of testers will be updated.

Good luck!

**UPD 1:**

We are in the middle of the round (and missing Gornak40 and Gol_D).

**UPD 2:** Editorial

Auto comment: topic has been updated by Aris (previous revision, new revision, compare).Good luck for everyone . Do your best to be the best !!

Good Luck everyone! I hope everyone has fun!❤️ Divs 3 are sometimes moody.

Its the 1729th codeforces round! The Hardy-Ramanujan number which is very well known in Mathematics

Nice!

Great!

For those who are not familiar with interactive problems. This might help you! Link

Finally, my first out of comp-

wait this is div.3, not div.4

As a tester, give me a contribution, please

As a participant, I will downvote.

SpoilerRepeating a funny thing, makes it unfunny

then can you help me in being funny? seriously I think I need help

Ya, I would like to,

do not comment!Things will automatically be funny :)

As a tester, I will upvote.

SpoilerFinally found a kind-hearted man whose name is FelixDzerzhinsky! Thanks for such kindness (❁´◡`❁)

:D HAHA

Good luck to everyone and I hope the problems will be enjoyable!

First unrated DIV3 ❤️, hope to solve 6 or more problems this time!

i wish i could be expert!

Surely you will reach Expert one day!

Don't lose the hope, grind yourself and love Competitive Programming ❤️

thank you brothers...take love

why unrated ?

It is unrated for those with rating >= 1600

ohh

it means it is unreated for me ?

No it's not. The ratings will be updated soon.

its currently in the unrated contest list in my profile so will it be updated soon and moved to rated contest ? Ps. This is my first contest so technically 0 rated as of now

Yes unless there is a separate post about the contest being unrated. Check this post where the contest was unrated.

Normally, a contest takes around 24 hours to update the rating. So hopefully, the ratings will be distributed within a few hours.

Hope to be Expert this round!

Hope to be pupil again

Codeforces Language Picker -- chrome extension to fix codeforces language picker.

Hopefully I'm gonna be green this contest.

.

Guy's name tho

Div3 Ftw

I hope that I will be able to solve A!!

spoilerIt is really a very very big challenge for me!!!

Auto comment: topic has been updated by Aris (previous revision, new revision, compare).As a tester, I hope you enjoy the problem set!

Good luck and have fun!Hoping to get a positive delta!!

Give me some upvotes. I'm sad! :(

okay, i give you up vote..Good Luck for upcoming contest!!!

now the time of div 3 bruuh

Hello cheater,

let's see if your solutions again gets skipped or not.

Div. 2.5

Please don't talk about contests before they are finished.

oh sorry but i didn't tell anything consider a spoiler + easiness is a mysterious ( sorry my english is not good) term maybe its just easy for me right? + i see people say mathforces speedforcess etc.. in other rounds why this comment consider deferent from them

In my opinion ANY comment referencing the contest during it shouldn't be allowed, I'm not sure to what extent the codeforces staff want/can block this type of unfair comments, but in any case saying a problem is easy is a bit worse than saying all problems are easy (although I agree they should be removed as well)

yeah you are right i won't write anything during the contest again

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Whenever someone starts a line with, "i am not simp", there is a high chance that they are not simp. Seems legit.

I know...

no but her insta is @zachscud

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that's not her id lmao

deleted

D is much easier than C.

No it is not. Like C took me 7 minutes at most, D took 15-20.

C took 20 minutes to understand the statement, and another 7 to implement it.

it took me 20 min to solve D and 20 min to solve C but another 30 min to find the mistake in my C

A took 10 min and B took 40 min

How did you come up with intuition that only 2 person will go together to have maximal groups and not even a group of 3 will give us the maximal answer?Is it some common idea or what?

If 3 people can form a group, then there must be 2 people in that group that can form a group. As we only want to maximize the number of groups, if 2 people can already form a group, why add more members to the group? I think basic observation is enough the come up with this idea.

any idea for problem no E?? binary search is one solution but it requires 60 quarries.

you can just ask 25 pairs (a, b) and (b, a), and you are able to print the answer once you've got different paths for an arbitrary pair. The probability of getting same path for each pair is 2^(-25).

u can just try every pair {1, i} and {i, 1} for i from 2 to 27. In each two query ({1, i} and {i, 1}) u have a 50% chance that u get distance in different directions. If u do, then the answer is ans1 + ans2, where ans1 and ans2 is results for queries. Also, if u get -1 at some point, it means, that the previous i was the greatest vertex in graph, so it should be the answer. So print i — 1 and return

Instead we can just take the pair (1,2) and query it 50 times. There is a probability of (1-2^-50) that we will get two different numbers. Just summing them ( with a +1 ofcourse) will give the result.

No, I too missed that part in the statement at first: after an answer is determined for a specific query, it wont change, so it will keep outputting the already computed value

Well, there is a possibility that all pairs are equal, right?! So why is the solution correct!! Could he give a wrong answer in my opinion.

yes, but that possibility is really really low, so it will almost certainly pass (and if it miraculously doesn't, submit a second time)

Thx

Yes, its possible, but the probability of it happening is equal to 1/2^25, which is veeeery small, so u can hope that it wont happen)

Please, can you explain where did the number come from(1/2^25)?

there are 25 pairs, the probability of each of them to equalize is 1/2 and also the probabilities are unrelated, so the probability for all of them to fail at the same time is (1/2)^25

Moreover, if you want your solution to fail then the average number of times you need to submit to get first WA is 2^25 ~ 10^7.5

Actually a bit less, remember there are multiple test cases. But in general yes, that's what I said, the solution won't fail

Gosh, I thought about that, but I was like "hey, this won't pass for whatever test case, there must be another way"...

Was E ternary search?

I think no. Try basic ternary give me WA TC 1x

I try every pair {1, i} and {i, 1} for i from 2 to 27 ^^

I thought so for a while but no, it's much simpler.

About problem E: what is the probability that out of 25 pairs of queries at least one pair of query will give distinct answers, e.g. query(a, b) != query(b, a)?

Roughly 1-2^-25 when n is big enough.

It is said that it may be equal but how you are so sure that it will be equal with 25 queries

The interactor chooses one of the two paths with equal probability.

But the probability of WA is not completely 0. Is there any deterministic solution?

Most likely not. It's the same as the Polynomial Hashes, there is a very small chance of failure, so nothing wrong probably

I thought that codeforces round always has some deterministic solution.

Well, again, Polynomial Hashes are not deterministic, yet still it's a popular way to solve problems on strings. There is nothing wrong about it. You just need to make sure the chance of failing is very small

I get it but usually there is always a deterministic solution. Polynomial hashes has deterministic alternatives.

Not always

"usually there is always a deterministic solution" well, there

, it just needs more (than 50) queries. People often confuseISa problemwitha problem statement. You should not consider a problem statement in isolation, rather all of statement, input format, input constraints as a package. For example, for any NP Hard problem, we all know that there are no polynomial solutions. That does not mean they are not solvable. You can set small constraints and expect an exponential solution with some optimization here and there. In other words, this is the difference between math olympiad and programming contest.I hope that satisfies your itchy feelings about not having a 100% guaranteed solution. This problem was designed and the constraint was set specifically for a probabilistic solution. If more than 60 queries allowed, one can find an exact solution using binary search.

I see, thank you.

same bro how you can even guess that number .

I think that problem E is more suitable for div 2,it's really hard !!

No, the idea, especially with No hacks, 50 test cases, really says it uses random. Also, as there is no binary search, you can only rely on distances, but they could all be simply distances on a part of a graph, so you cannot guarantee the answer is correct. The only way is to minimise the chance of fail

We rarely have "easy" problems with randomized solutions. This is a good way to introduce randomized solutions to beginners

Yup. First time I have ever solved a question with randomized solutions

Most unbalanced round ever, downvote.

I guess you've never heard of mr Green Grape

Can anyone help me with my solution for Problem E? why is it not working and taking more iterations than 50? although my solution should run within log(1e18) < 50 (correct me if i m wrong here).

submission

log(1e18) > 50

Oops! I calculated log with base e (that is ~ 42). that is why i got confused.

thanks mate

I calculated the log with base 2 LOL and used binary search

cool E

Greattt

In question E how you can be so sure that ans will be find by just finding it upto 25

The chance of failure is 2^(-25), which is very small

According to some cursory analysis, my solution for E seems to have a

99.391488785%chance of passing a set of 100 test cases. Don't think this is the intended solution.i guess it is (i did the same btw), because hacks are disabled and there are 50 tests only. Moreover, the problemsetters did even tell us that amount of tests is 50, which is kinda strange.

The problem emphasized "with equal probability" in bold letters, and also made sure to state that "? a b" will not necessarily yield the same result as "? b a". They also mentioned the number of tests in the jury (which I've never seen mentioned before in the problem).

So yes, I am pretty sure the intended solution is to try "? a b" and "? b a" until you get something different. You do need to make sure you try distinct pairs each time, but there are easy ways of doing that (including ways that guarantee the correct answer if the result is ever -1). My submission: 171950029

Great round <3 <3 <3

How to solve C?

Find all the characters in string that lie between letters s[0] and s[n — 1] and use all of them for jumps.

sorry can you elaborate further how this achieve the minimum cost ps : sorry i meant the maximum jump in this case for example: abchhhhhhhhhhhhhhd shouldn't your solution fail?

abchhhhhhhhd it will be as a b c d (b-a)+(c-b)+(d-c)=4 and also as d-a=4

oh you are right i was thinking in a scenario where taking a character which is not between s[0] and s[n — 1] can lead to minimum cost too but this example doesn't satisfy it thanks alot

How solve E

I try every pair {1, i} and {i, 1} for i from 2 to 27 ^^

why till 27?? can you please explain?

Because there is a total of 50 queries, and for each vertex you need to use 2 queries

Thank you.

Hi Noobish_Monk. I have a doubt. If I print all my queries as 1 and 2 for 50 times, then if I get 2 different sets in my answer then it's just the sum . What's wrong with this approach. And the probability that I always get the same answer is 1/power(2,50) which should be approximately 0. Why do we need to print for 27 queries ??

The answer for the same query is the same However queries ? a b and ? b a are different So you can't just ask 1 2 and 2 1

I should have read that statement :( . Thanks mate :)

You can also do it from 4 to 29, because the problem states that n ≥ 3.

Thank you.

Cause we have 50 queries bro

Solution for E was probably meant to be different, But here is how I AC it:

IdeaI basically just checked 1-[2, 26] vertices in a format:

? 1 v

? v 1

And if the results differ, then the answer is their sum. If you got -1, it means the last valid run was the previous one, so you can output v-1. It worked for some reasons, probably not intended.

same thing for me too!

I never thought it would work but some how it worked

I share the same idea with you ^^

IT'S IS THE INTENDED SOLUTION

Can you explain how it is the intended solution like very first thing that comes to mind is Binary search but it didn't work.

Because of how many times the statement highlighted the idea of equal probability i think this is the intended solution.

There is no deterministic solution that guarantees 100% correctness. Suppose I was an adversary that doesn't have a specific graph in mind, or even a value of $$$n$$$, but I make answers to screw you over while always ensuring that there exists some graph that is consistent with all my answers. Then I can always ensure that each answer I give will be either -1 or some distance that is below the established lower bound from the program.

With 50 queries, you can only gain enough information to distinguish between $$$2^{50}$$$ possible answers, However, the range for $$$n$$$ goes up to $$$10^{18}$$$ (which is much larger than $$$2^{50}$$$), so you cannot guarantee a correct solution.

The problem dropped a lot of hints suggesting the randomized approach ("equal probability" in bold, clarifying that "? a b" and "? b a" are processed independently, specifying the number of jury tests, etc), so I'm quite sure this is the intended solution.

If you want your solution to fail then the average number of times you need to submit to get first WA is 2^25 ~ 10^7.5

Why can't we just use pair 1,2 and 2,1 25 times ? What is the probability that use these 2 pairs wont give me the answer in 25 queries ?

It is stated that the same pair will give the same result for all requests.

The problem states that repeating the same query will yield the exact same result every time. There is only a 1/2 chance that "? 1 2" and "? 2 1" will yield different results. If they both produce the same distance, then you will always get the same result for "? 1 2" or "? 2 1", so your probability of success will remain as 1/2. This is only for one jury test, so the probability of getting all 50 jury tests would be $$$\left(\frac{1}{2}\right)^{50} \approx 0.00000000000008817842\%$$$, which is not worth trying.

But by trying different pairs each time, there is a 1/2 probability of success for each distinct pair. The probability that all 25 pairs fail is $$$\left(\frac{1}{2}\right)^{25}$$$, so the probability of being successful at least once is $$$1 - \left(\frac{1}{2}\right)^{25}$$$. A single success is enough to guarantee the correct answer. We need to pass 50 jury tests, so the probability of ACCEPTED verdict becomes $$$\left(1 - \left(\frac{1}{2}\right)^{25}\right)^{50} \approx 99.999850988\%$$$. It would be extremely unlikely for such a submission to fail, and I would be very interested to see if there is anybody who met this unfortunate fate.

Thanks for your explanation. This is helpful.

I read this statement of same result produced for same query but didn't realise how important this was.

Even after reading this, somehow I thought that using 1,2 & 2,1 still is random. But it seems like that the interactor is caching the result of the query.

Similar to these lines, IMAGINE if 1,2 & 2,1 produced different results with 1/2 prob but they can produce different result every time, then we can just query 1,2 & 2,1 50 times right ?

If duplicate queries were rolled independently, then yes, you can just query "? 1 2" 50 times. You can mix in "? 2 1" as well, but it is completely identical to "? 1 2" in this case, so it doesn't matter. In this scenario, the probability of failure becomes $$$\left(\frac{1}{2}\right)^{49}$$$ (all 49 queries after the first one match with the first query).

Probability of AC becomes much higher then (though it was already high to begin with), and the implementation becomes much easier too (though it was already easy to begin with).

Even without luck being involved, it can very well be the case that 1 and 2 are on opposite side of the cycle so longer and shorter both lengths are INDEED equal. So you must take different pairs to keep a backdoor in case your queried vertices fall on opposite sides.

Ah, you're right; I forgot about that case. We can deal with this by simply performing 25 queries of "? 1 2" and 25 queries of "? 1 3". This slightly hurts the probability of success, but it's still very high.

25 queries of "? 1 2" gives the same answer each time. So you are really just wasting 24 queries in this case. Same goes for "? 1 3".

onebit1024 This was part of a discussion on the hypothetical scenario that each query is rolled independently, even if the same query was made in the past. In such a scenario, it would be optimal to repeat the same query (e.g., "? 1 2") 50 times instead of trying to find new pairs. If all 50 queries yield the same result, then the guess should be on twice this value (since it's likely that both paths have the same length).

Obviously, for the original problem E, repeating the same query would be a waste, so you have to keep trying distinct pairs (so the possibility that both paths have the same length would not be an issue at all).

Oh ok, sorry, I didn't realize that you were discussing a hypothetical scenario. Well in this scenario, this approach will work.

actually, during contest I tried the same thing because I missed the part where (a,b) would give the same result each time.

"Suppose I was an adversary" I don't need to read anything else. The solution will fail against an adversary. That's the whole point, that it was not an adversary, and it will honestly return any of the shorter or longer path from a to be with equal probability, without ever trying to make you fail. The moment an adversary comes into play, it will need binary search i.e. 60 queries.

The fact that an adversary will always be able to screw you over is definite proof that there does not exist a deterministic solution that guarantees correctness, which further suggests that the randomized approach is, indeed, the intended solution. Some people were unsure if that was the case, but I don't actually think there should be any doubt.

Can anybody explain how they came up with the number 25 that checking this many pairs will have a different result?? really didn't get this part.

It's probability theory. We have a chance of success 1/2, same for failure. As we ask 25 times, we have a chance of failure equal to 2^(-25), which is a VERY small number

Since we have at most 50 queries, we can check 25 pairs in both ways. Assume we got a pair (a, b). Checking it both ways means that you chack a pair (a, b) and (b, a) since it can give different results. Since the output path length is randomized between the 2 paths(the short and the long versions, since it's a cycle), it's 50% chance for each to output.

Asking 25 queries in this format and checking for -1 or different results for the (a, b) & (b, a) query. If you get -1, it means the last vertice(call it v; 2<=v<=26) is the correct answer, since the current one is out of bounds and the last one wasn't. So we output the last valid run. If the results differ, you can output sum of the results, since you got both paths for (1, v) pair and their sum is a complete cycle.

This one failing has quite a low chance, since it's 1/2 for each path to be given for a query, it's 2^-25% chance of this idea failing, if I am not mistaken.

I hope this makes sense.

You're allowed 50 queries. You make a pair of queries "? a b" and "? b a", and hope that you get different results for these two. If the results are different, then you can add the two results to get the number of vertices.

I'm not sure if the guess counts as a query, but if it doesn't, then you have 25 pairs of queries to get this. If it does count, then you only get 24 pairs of queries, plus an extra query, and one for the answer.

It's annoying that this is a randomized solution, but the probability of this failing is less than 1%, and there are many hints in the problem statement itself to suggest that this is the intended approach (equal probability in bold letters, mentioning that "? a b" and "? b a" are rolled independently, specifying the number of jury tests, etc).

Also if you ask why the numbers of allowed querry is not greater it's because with 60 querry (which is log2(1e18)) you can just get the result with a binary search

Very interesting problems, especially F!I have really enjoyed it!

I think my A problem is brief enough.Is there any person can give me some advice to simplify my approach further?

Can anyone tell me what i am doing wrong in my solution of problem C ?

My submission

As the solution uses sorting I might suggest it's that the last index is not the last in your sequence. Or you could possibly go in the wrong order if p >= q or less than q. IDK, but that's most likely the mistake

Try "baba".

I fell into the same trap and got WA. Thankfully, I figured it out before the contest was over.

I didn't read your code (but I checked "baba" via custom invocation), but it's most likely due to your use of sorting. The assumption that your range covers all instances of first character and all instances of last character only applies if first character < last character. In the scenario where last character < first character, your range likely does not capture all instances of first and/or last character.

Here is the counter case for your solution: 1 yydia

correct ans: 24 5 1 2 4 3 5

Your ans: 24 4 1 4 3 5

Now it's Accepted!

Accepted

Thanks Noobish_Monk LightBrand99 Nikhil19259

In the p > q part, you should include p and q in your answer.

Yeah Now it's accepted.

https://codeforces.com/contest/1729/submission/171949266 Can anybody help to find a short test where my solution for C does not work?

one counter test case: 1 puvh correct ans: 8 2 1 4 your output: 8 3 1 2 4

WorstDiv 3Whats wrong with my solution to C? 171872810

Check 1 zzzzza

Output:

25 6

5 4 3 2 1 6

Edit: Thanks :D

To the author of problem E: how did you generate the tests? I'm curious to know. Btw, awesome contest!

Yeh same doubt regarding the E part of the contest, in all the test cases, input is in this form-

3 9195979e279504391c49d2f080c1d8c755d044ca what is this — 9195979e279504391c49d2f080c1d8c755d044ca ?

Shouldn't this be a long integer? Thanks.

I think you store the length of the permutation as well as it's index in sorted order. Not sure though...

i was trying to solve F with condition s(l1,l1+(w-1) ) != s(l2,l2+(w-1)) when i read l1 != l2 then when i realaize my mistake the contest was over :(

Can someone find a mistake in my code for 1729C - Jumping on Tiles?

Code: 171911705

Thanks!

Check 1 zzzzza

Try "baba".

I fell into the same trap and got WA. Thankfully, I figured it out before the contest was over.

I didn't read your code (but I checked "baba" via custom invocation), but it's most likely due to your use of sorting. The assumption that your range covers all instances of first character and all instances of last character only applies if first character < last character. In the scenario where last character < first character, your range likely does not capture all instances of first and/or last character.

Oh my god I solved E. I have never tried random algorithms in competitive programming ever. Too bad the contest is over by now, but holy fuck I would never believe that it would actually work. Holy Jesus. 171950813

Nice problem D! It makes me difficult even though it's only div 3. Can anyone give me a solution ?

My solutions problem c solution problem D solution

For C or D?

For C, jumping directly to the end yields the minimum cost, which is |last character — first character|. How to maximize jumps? Observe that if you're in $$$\alpha$$$ and you jump to $$$\beta$$$ and then to $$$\gamma$$$ such that $$$\alpha <= \beta <= \gamma$$$, then the cost is the same as jumping directly from $$$\alpha$$$ to $$$\gamma$$$. So you can jump to any character in the range [current character, final character] and it would not affect the final cost.

For example, with "logic", you can go from "l" to "i" to "g" to "c" because "l" > "i" > "g" > "c". To maximize jumps, you need to visit every character in the range [starting character, final character] in order (either non-increasing or non-decreasing order, depending on the relative order between first and last character).

For D, if you calculate $$$y - x$$$, you basically get the person's balance from paying for their meal. This can be negative if they don't have enough money. You can then sort the balances, and try to pair the one with the lowest (i.e., most negative) balance with the highest balance. If the result is non-negative, it's a valid pair, and you can move on. If not, then the lowest balance is hopeless and you move to the next lowest balance (try with highest balance), and so on.

very ez understand, thanks

Would you please review 171943515? Maybe it is wrong when $$$n$$$ is small, but I do not know how to fix it.

int isn't big enough to hold the values required.

Damn it you hacked me twice ^_^. Thank you very much!!!!!!!!!! When debugging in a limited time, it is easy to ignore some very basic facts...

In problem E, all the AC solutions I've checked till now are querying

$$$?$$$ $$$1$$$ $$$x$$$ and $$$?$$$ $$$y$$$ $$$1$$$ where ($$$2 \le y \le 26$$$).

But how is that acceptable when the interactor chooses with a certain probability? Yeah, the probability of getting a correct answer is very close to $$$1$$$ ($$$1 - 2^{-25}$$$) but certainly not $$$1$$$. I thought of such a solution but ignored thinking since it can create a situation where same solutions may get AC and WA which is unexpected. I don't think this was a good/standard problem for a contest.

I did feel bothered by the fact that the intended solution is randomized, especially since I was trying much more complicated approaches before the contest ended, and only got AC after it was over.

However, I think it's really a matter of opinion on whether this is appropriate. Randomized algorithms have numerous practical applications in the real world, so this kind of recognition that a randomized algorithm would work is a good skill to have, which can be argued as being among the type of skills that competitive programming should cover. Also, the problem did drop a lot of hints that this was the intended solution ("equal probability" in bold, mentioning that "? a b" and "? b a" are processed independently, mentioning the number of jury tests, etc), so I think it's justified in this case.

I personally think it was a great problem (maybe I have some bias since I solved it quite quickly in the contest). It makes you think outside of the box and consider non standard methods that you don't regularly see in CP, which ultimately is a great way to test/improve problem solving.

Can someone explain how does the rating work, I'm confused a little bit because I'm new to codeforces.

It depends on your rank in the contest and your current rating.

In problem E ,is it really fair to give such a probability based question in a contest because it's not like the probability of getting them all equal is zero right,just because of that many of us didn't go for that approach and the binary search one was exceeding the queries(which should actually be the intended solution).

The probability of success is over 99%, and the problem dropped a lot of hints to go for a randomized approach ("equal probability" in bold letters, clarifying that "? a b" and "? b a" are processed independently, mentioning the number of jury tests), so I think this was justified. I understand how one would assume there is a deterministic solution, but randomized algorithms have numerous practical applications in the real world, and recognizing when a randomized algorithm would be effective is a valuable skill for a programmer to have, so it can be argued that it should fall within the scope of competitive programming. I understand the concerns, but I think it's a matter of subjective opinion and cannot definitely be ruled as "unfair".

Is there a deterministic solution to the problem E, which

alwaysworks no matter how the generator is configured?No. An adversary can always ensure that the result for each query is either -1 or some distance that is less than the current lower bound that was established. Therefore, there are essentially two possible types of information to be gleaned from a single query. With 50 queries, this only allows identifying up to $$$2^{50}$$$ different values, which is less than $$$10^{18}$$$. So there is no 100% correct deterministic solution.

Are hacks rated?

No

Is square root decomposition the intended solution for problem F?

there is no need in sqrt decomposition, the only data structure we actually need is prefix sum array

In problem E, shouldn't it be "in C++ you should use function

fflush(stdout)"?Any hints for G?

Hint1If you picked a occurrence of $$$t$$$ starting at $$$i$$$ then the previous occurrence which needs to be picked should have $$$startingindex <= i - m$$$

Hint2If the rightmost index that satisfies the above condition starts at $$$j$$$ then you can't pick a $$$startingindex$$$ which has $$$endingindex$$$ before $$$j$$$.

Hint3The above problem can be solved using dynamic programming

Hint4For number of ways you need to consider only $$$index >= startingindex$$$ of last occurrence and $$$index <= endingindex$$$ of last occurrence which gives the minimum answer for $$$dp[i]$$$ where $$$dp[i]$$$ is the minimum answer if $$$ith$$$ occurrence of $$$t$$$ in $$$s$$$ is picked.

CodeAn interactive problem without Binary Search ,,,,,, surprising

There was one such 1713D - Tournament Countdown

how to solve problem c? I used priority queue but got wa

The maximum number of jumps must be on the characters which in the range between the first and the last character

ex) s =

"codewars"character 'w' doesn't belong to ['c' -> 's']

so you shouldn't jump on it

Take the rest of the characters, sort them, then print their indexes.

My submission

The minimum distance is simply the distance from jumping directly from the first to the last.

How to maximize jumps? If you go from $$$\alpha$$$ to $$$\beta$$$ to $$$\gamma$$$, where $$$\alpha \leq \beta \leq \gamma$$$, the total cost is the same as going straight from $$$\alpha$$$ to $$$\gamma$$$. So you can keep jumping to an intermediate index,

providedthat the character in it lies between the previous character you were in and the next character that you're going into. So if you need to jump from "b" to "f", then you can jump into all of the characters in the range [b, f], in non-decreasing order. By the same argument, if your final character is actually smaller than the first character, you can still jump into all characters in the range, but in non-increasing order this time.Anyway，I dont think E is a good problem.

Definitely because if you go to exact statement then it will always be a point of discussion whether (2^-25) = 0 or not .

Randomized algorithms are not that bad!

Randomized algorithms are the best thing in the whole universe.

Especially when they're not the intended solution

Based.

Btw, are u talking about today's E? Wasn't the randomized solution intended?

No, in this E the random solution is 99% intended I was talking about 1310B - Double Elimination, the random solution imo wasn't intended.

I agree. Giving such an easy problem at E is confusing. I got the idea in maybe 10-20 seconds but kept wondering whether there are any tricky scenarios or edge cases. It should've been in D.

Same too. Got the idea immediately but said to myself "it cant't be that stupid".

Isn't intentional code obfuscation forbidden?

Please check this: https://codeforces.com/contest/1729/submission/171937226

MikeMirzayanov

This submission too: https://codeforces.com/contest/1729/submission/171874872

MikeMirzayanov

So I was just casually looking at the hacks corner and found a few hacks for problem A which seem way too odd.

171954115, by: vaibhav_1710 to: chaudharyvaibhav184; 171952586, by: andreifilimon to: wavetome;171877008, by: mihir2808 to: suiiiii

Now how and

whyin the world did all these people put an a is equal to some random no. check? I don't know where to report these so I'm just posting it here. There are points for hacks so this should count as a violation right?those are probably made specialy so people who check a lot of submissions can find those stupid mistakes and take the points. I agree to you thats like a violation but as those are there why not to make a easy hack. (there are a lot of this mistakes in submissions) by the way a good way to hack smth i tried a lot on problem C was big tests...

There are some telegram channels which sell these solutions during the contest for as cheap as 20-30 cents per question.

That's not good. That should be reported.

Can those be Trojan Horses for cheaters? In edu rounds (Edu, Div 3, Div 4) there are no points given for hacks. Therefore, there's no incentive to plant deliberate mistakes for future hacks.

Thanks for being the only person to clear my doubts. I didn't know hacks didn't account for any points for div. 3 onwards. Now I feel stupid for posting this.

I like problems with randomized solutions. It would be great if there were more such problems as today's E.

Do you like randomized algorithms?

I don't know what to say about $$$E$$$, I had that idea early, but I doubted it would work, then i submitted it in the last minutes (because why not) and somehow it worked, but I got high penality.

I'm certain that many people (maybe hundreds) had this solution in their head but they didn't try because it depends on luck :D

So i don't like this random algorithm.

me not touching the yes/no button as it will disturb the perfect pattern

I had a doubt regarding the E part of the contest, in all the test cases, input is in this form-

what is this — 9195979e279504391c49d2f080c1d8c755d044ca ?

Shouldn't this be a long integer? Thanks.

Any hints for F? I'm guessing there's some number theory property related to mod9 as $$$w$$$ can overflow int64.

Hint: any number in decimal mod 9 = the sum of its digits mod 9. It would be a good exercise to prove this statement.

Hint about proving this statement & other similar statementsConsider each digit's contribution to the total modulo. What is $$$1$$$, $$$10$$$, $$$100$$$, $$$1000$$$, etc modulo $$$M$$$? It is possible to prove statements about any arbitrary modulo using this method.

Why wa on tc 2. problem c. https://codeforces.com/contest/1729/submission/1719150491.

Try this input:

The map yields values from 1 to 26. When adding elements to the 2D vector, the first index is given by the map, so the first vector index also ranges from 1 to 26 (1 for 'a', 26 for 'z'). However, when clearing the vector at the end of the test case, only indices 0 to 25 are cleared. So any instances of 'z' saved from earlier test cases (at v[26]) will continue to remain in future test cases.

(note that v[26] is technically out of bounds, since v was declared to have size 26; this will not necessarily be detected as an error, and the program may continue to use v[26] as the unallocated memory location that sits 26 steps from the start of the vector, until an issue arises with this location later)

thanks man.

In problem E, hacks are disabled. Is system testing also disabled or can it be a FST?

The problem statement says that there will be 50 test cases and there were 50 pretests. So I'm presuming that there will not be any system testing for E

Curious to know if rating changes have been applied or not for this round. It is showing me as a unrated contest atleast for me when I tab the "all contests" dropbox in rating graph. Or this is a normal case when rating is not updated till now.

It shows it as an unrated contest if rating changes haven't been applied. As of now, the hacking stage has just gotten over and the system testing still has to commence.

How to do Question E

If

`n<10^15`

or`number of queries < 60`

my solution of ternary/binary search would have worked. My naïve idea= ask query as`"? 1 m"`

. if this returns -1 decrease value of m or else increasebut this uses 60 queries. How to approach this types of question

The issue with binary search is that you can only eliminate at most 1/2 of the search space consistently; in fact with question E I think there is no way to get a solution that you can be absolutely 100% sure is correct.

There is a solution though that has about a 99.99% chance of being correct. It's only possible because of the problem emphasizing that one of two paths will be chosen

with equal probability.MikeMirzayanov is the only one without laptop in the picture.

Also I don't know the others. Does anyone else know all of them??

It's the ITMO university team. Just read earlier in the post. There are seven in the team, but two are not in the picture (specified in the caption). I'm pretty sure the seven of them all know each other, and there are probably other people (possibly from ITMO University as well) that would also know them, but I'm not sure why you're so interested in them.

I was just curious, since they were all with Mike and probably at the headquarters from where this brilliant site is maintained.

How do you realize when system testing is over in edu rounds and div3 rounds? At the end of hacking stage, it says "Final standings, open hacking phase finished". But don't system tests take place after this, in that case how is it final?

System testing in these rounds usually happen 4-5 hours after hacking phase

I am aware that system testing takes place a few hours after hacking. I am just curious as to how you realise whether sytem testing has finished or not. Like for round 820, what indication do I get to understand that system testing is done or not?

All it says is "final standings" which isn't true I think because system testing hasn't started.

when will the ratings be assigned?

it's already updated

I am not able to see yesterdays contest in the list of contests i have taken part in .

i was joking, rating will update after 12 more hours

ok thanq

Why are ratings still not published, is this round rated or unrated

This round had 12 hours of hacking phase, which will be used to re-judge all of the solutions as far as I know. The hacking round finished a few hours ago, now hacks are being analyzed most likely and after re-testing, ratings shall be published, so a few more hours probably.

I am having trouble with this, please share your insights.

Cool update. Would love to have these in future rounds as well :)

I think if this problem was placed in the A place,there will be more people solved it.Many people had the corret idea,but they think this solution is not fit to E problem.This kind of problems may be good,but where they should be placed is still a problem.

Well, A through D are easier than E, so everything's fine, imo. And if it had been set as A, more participants would have been determined to dodge this round.

I think we are forgetting, that this is a Div3 round. This E is even harder, than usual E Div3 problems, imo again.

When will the rating change? my rating has not changed yet

Same stuff, pal. I am so looking forward it :D

Where's the editorial

Problem F video editorial.

when can I see the results? 12 hours-fade is already up...

they look awesome!

Finally ratings are updated :D

Hey but it didn't reflected on my profile still now.

Waiting for expert :D

Who knows why my raiting is unrated ? why?

why is taking so long to update ratings? :/

System testing ended few minutes ago, so you may expect rating change soon!

Ok cool

my rating is yet not changed.neither the contest appears in the list of contests i have taken part in

G can be solved in $$$\mathcal{O}(m + n\cdot \log(n))$$$ time and $$$\mathcal{O}(n + m)$$$ memory, by using dp + segment tree.

First compute starting positions of occurences of $$$t$$$ in $$$s$$$. This can be done in linear time in many ways (I used z function). It's easy to notice that we can apply operations in order from right to left. Then, we will compute $$$dp_i$$$ = answer if the rightmost operation was applied to range $$$[i,i+m)$$$.

To compute $$$dp_i$$$, we could iterate to the left of $$$i$$$ on the starting index of the next operation (let's call it $$$j$$$). Two things need to happen:

This two conditions define a valid range of values for $$$j$$$. We just need to be able to combine the $$$dp$$$ values of this range. We can use segment tree to get this merged value.

Overall, we compute occurences of $$$t$$$ in $$$s$$$ in $$$\mathcal{O}(m + n)$$$ and then compute $$$dp$$$ values in $$$\mathcal{O}(n\cdot \log(n))$$$, so final complexity is $$$\mathcal{O}(m + n\cdot \log(n))$$$ with $$$\mathcal{O}(n + m)$$$ memory.

Here is a link to my submission :)

so cool

That's what I thought! This task seems similar to a (slightly harder) problem from POI 2018, stage 2 problem "Konduktor". Link: https://szkopul.edu.pl/p/default/problemset_eng/oi/25 You change the language to English in the top right corner. Also, forgot to say, nice explaination of your solution

Since the query range is fixed, you could overkill this further by replacing the Segment Tree with a Minimum Queue for a $$$O(n+m)$$$ solution. submission

I solved 2 problems in this round but still, the rating is not updated. what is the estimated time it takes to update the ranking?

It will be updated soon.

System Testing has ended way back.... then why so long in rating update?

It takes time (probably around 24 hrs from contest end) for educational round rating to be updated.

But this isn't an Educational round...

Read the second paragraph of this blog

Ohh ok... my bad

This Round is showing under Unrated Category now!! Why?

Oh yes :/

how ?

Aris Is it unrated round?

Editorial section still not uploaded

Ratings also not updated :(

Let's see I think it may be Unrated thats whay:(

I like E. Open my mind.

I didn't. Even though the probablity is 50-50 still there's a chance that the ans for "? a b" is same as that of "? b a" for the first 50 queries. Change my mind :(

Probability of all 50 queries failing and none giving an useful information is 2^-25, that's basically somewhat like 99.9998% chance of a solution passing.

I don't get this. It fails for an edge case right. What if I'm the testcase author and I wantedly set such a case?

Testcases are random as well, it's not a preset. The answers are generated randomly while testing.

it's not about case work. In statement it is said, that The interactor chooses one of the two paths with equal probability

I get it now. Thanks ;)

You're welcome!

Probability 0.00000003 is small enough to never happen, that's it

See I get this but they've mentioned that the answers for the queries are pre-determined right so there is a chance that the solution fails for a case.