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Hello! Codeforces Round #582 (Div. 3) will start at Aug/30/2019 17:35 (Moscow time). You will be offered 6 or 7 problems (or 8) 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 rating 1600 or higher, can register for the round unofficially.

The round will be hosted by rules of educational rounds (extended ACM-ICPC). Thus, during the round, solutions will be judged on preliminary tests, and after the round it will be a 12-hour phase of open hacks. I tried to make strong tests — just like you will be upset if many solutions fail after the contest is over.

You will be given 6 or 7 (or 8) problems and 2 hours to solve them.

Note that **the penalty** for the wrong submission in this round (and the following Div. 3 rounds) 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 participants of the third division*, you must:

- take part in at least two 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 my work. Thanks to my good friends Mikhail awoo Piklyaev, Maksim Neon Mescheryakov and Ivan BledDest Androsov for help in round preparation and testing the round.

Good luck!

I also would like to say that participants who will submit wrong solutions on purpose and hack them afterwards (example) will not be shown in the hacking leaders table.

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**UPD**: Thanks to Artem Rox Plotkin for testing the round!

**UPD2**: Editorial is published!

[deleted]

I hope I will get some positive rating from the contest.

YOU WILL NOT!!!!!!!!!!!!!!! I AM SURE THAT AFTER 5 CONTESTS YOUR RAITING WILL BE BELOW 750!!!!!!!!!!!!!!!!!!!!!1

Why are you being mean?

You are not even blue yet LMAO

In my real account my raiting is 1700+

No one cares..

Just learn to be nice to people

I am just trolling in all my comments(not in posts).my small contribution can proof this. But this comment is not trolling and is true

First,div3 after becoming expert hope my rating remains constant.

your rating will remain constant because div.3 is unrated to experts and higher like you and me.

Raiting does not matter, matter only nice girls!

Now I can say "I won't lose my rating ANYWAY!" proudly :D

Now I can say "I won't gain any rating ANYWAY" sadly :(

This will be my first contest. I hope I will do well and good luck for everyone for the same....

Hope the ranking of problems be better from the last contest.

After this contest,I think I will be a expert.How excited I am!

Hope you will be an expert

Thanks!

A japanese friend told me that reaching blue in a div3 (and yellow in a div2) is dishonorable.

I thought it was orange :(

to my eyes: yellow = master, orange= international master

Vovuh...master of preparing div3 contests...

I'm still not a master unfortunately :(

How did you write "master" like that?

I'm sure you can see the html-code of my message to understand how I did it :)

master

will be master

It means that CF support html code.

master

My Final exam is going on... But I won't miss div-3..

Div-3 is love..Vovuh's rounds are always good!

Yours are also good.

Comment from Div3 #579

Now tell Codeforces isn't ratist :c

Div 3 rounds are better than "Educational rounds" in teaching the most valuable skill for Codeforces, namely SPEED.

Depending upon what your purpose. If you have high rating is seem not :)) But I like Div3 rounds too :3

Hopefully I will become back to expert soon. I have been very disappointing in myself recently.

Firstunratedcontestfor me. XDHello,Guys. This is my first contest after a lot of time. I studied Quantic Math in this period. I think that I can beat majority of the Div3 contestants even tough I am at 1361. :)

Letsssssss GOOOOOOOOOOOOOOOOOOOOOOOO

~~LoOooOoOLoOOooOoLoOOoOoOOooOooOOLoOooOOoOOoOOL. ROFL. LMAO. You think majority of contestants can only solve less than 3 problems?~~↑that's just bullshit

All the best. The time I solved 5 problems I didn't really expected that. it was pure luck. one lucky contest will be enough to bring you up to specialist or even 1500+. Good luck and have fun! :)

Note that this is vovuh's round. It's likely to have problems with multiple queries. So guys, a notice for you, remember to clear the data of the last query before answering the queries!

Do I have to register to join this contest? (My current rating is under 1600)

Of course. You

have toregister if you want to take part in a contest. For those rating equal or more than $$$1600$$$, they will be registered "Out of competition" for Div.3 contests. That is, their rating will remain unchanged after the round and they won't be listed in the official standing board.Alright, thanks for the information

Yep

I have a great idea! Make div3 rated for (blues,) purples and yellows but update ratings only if the delta is negative.

Why don't do the same thing to blues? They are unofficial for div.3 as well :(

sorry, I thought blues are also div3.

Let's escape this division guys! :D

Finally, not a Mathforces Round.

P.S Here was bad info

I don't think you should write that during the round.

My fault

D2 solution accepted with 1900ms

Me: heavy breathing for 14 hours

Unofficial editorial by me.

A (Chips Moving)

ExplanationIt is easy to see that there is no way for even valued chips to move to odd positions without paying, and vice versa. Let's say there are E even valued chips and O odd valued chips. The answer is min(E, O).

CodeB (Bad Prices)

ExplanationSince we want to know the minimum price in the future, lets iterate in reverse and keep track of it. When we are considering the i-th price, we already know the minimum price we saw.

CodeC (Book Reading)

ExplanationWe write K = floor(N / M) numbers, the numbers are f(k) = k * M % 10 for k = 1, 2, ..., K. These numbers f(k) must repeat every 10 instances of k, that is f(k) = f(k+10). Let k = 10 * q + r. We write q numbers of f(0), f(1), ..., f(9); plus another r numbers f(1), f(2), ..., f(r).

CodeD (Equalizing by Division)

ExplanationFor each number i in reverse order, lets see if all numbers could equal i. To do this, we maintain at each number, an int[] count[i] of how many numbers reached here by 0, 1, 2, ..., 18 divisions of 2. Then, for a number i with some count, we can calculate if K numbers could equal i in a greedy way: count[0] of them got here in 0 divisions, count[1] of them got here in 1 division, etc. and see how many divisions we need to find K numbers.

Afterwards, we divide the number i by 2, which increases the counts at count[i//2] appropriately.

CodeE (Two Small Strings)

ExplanationIts always possible. Consider the graph with nodes 'a', 'b', 'c' and all possible directed edges (even self edges). There are three types of edges: forward edges 'ab', 'bc', 'ca'; backward edges 'ba', 'cb', 'ac'; and repeat edges 'aa', 'bb', 'cc'. Our goal is to walk around this graph without touching the banned edges S or T.

Evidently, if we don't ban forward edges, we will answer 'abc' * N. If we don't ban backwards edges, we will say 'cba' * N. So lets say we ban a forward edge and a backwards edge. Without loss of generality, we could say the forward edge banned is 'ab'. Then the backwards edge is 'ba' or 'cb' or 'ac', and we have an answer 'b' * N + 'c' * N + 'a' * N.

CodeF (Unstable String Sort)

Explanation (grindy)We have 2N — 2 inequalities S[P_i] <= S[P_{i+1}] and S[Q_i] <= S[Q_{i+1}]. For each inequality S[u] <= S[v], draw a directed edge from u to v. Notice if a path exists from u to v, it implies S[u] <= S[v] by transitivity. If a path exists from u to v and from v to u (ie., u and v are in the same strongly connected component), then S[u] <= S[v] and S[u] >= S[v], so the letters S[u] and S[v] must be equal.

We can use Kosaraju's algorithm to find the strongly connected components. If there are less than K strongly connected components, the task is impossible. Otherwise, we could greedily fill the answer.

Code (grindy)Explanation (nice)If S[u] <= S[v], draw a directed edge from u to v.

If x occurs before y in P, and x occurs after y in Q, then a path exists from both x to y and y to x, ie. x and y are in the same strongly connected component, so S[x] == S[y].

We can use the above fact to deduce when there may still be P[j] with j > i such that S[P[i]] == S[P[j]].

Code (nice)G (Path Queries)

ExplanationSort the queries. Every time q increases, we can only add edges to the graph under consideration. Therefore, we only need to maintain the correct answer every time we add an edge to the graph.

Lets say the graph is a set of disjoint connected components, each component having s_i nodes (s for size). Then, the answer is sum_i (s_i choose 2), which we can write as (1/2)(sum_i (s_i)^2 — sum_i s_i) = (sum_i (s_i)^2 — N)/2. Thus, it is only necessary to maintain S_2 = sum_i (s_i)^2.

Whenever we add an edge, if it connects two disjoint components with sizes sx and sy, then S_2 gets increased by (sx + sy)^2 — sx^2 — sy^2 = 2 * sx * sy.

CodeIs this official editorial? I am confused why its not part of the post.

It’s unofficial

Your implementation of G has failed on a test in which function

findmakes $$$O(n)$$$ recursive steps. I believe it would be better to use rank heuristics here.thanks Vovuh, it was cool contest.

Hopefully I go back to blue after open hacking is done!

How does my D2 pass in time ? 59740108.

Isnt the complexity O(2*10^5 * K * log(K) ) while iterating the map since maximum numbers in map can be 2*10^5.

Am I missing something or are the test cases weak ?

Heh. I'm TLEing when I really shouldn't be.

No, your complexity analysis is wrong. The total number of elements across vectors in the map is at max Nlog(a[i]). So, you can say that your amortized time complexity is O(Nloga[i]logN)

Can anyone tell me why this Java solution for D2 is TLEing?: 59748534

Shouldn't my solution be NlogN?

Java Arrays.sort() uses quicksort which is O(n^2) in worst case

Thanks! I was unaware that Arrays.sort() was so bad :(

Yes! I was able to verify that using Arrays.sort was the problem. I implemented mergesort, and it passed in 202 ms.

Arrays.sort in Java has a worst case running time of O(N^2). Converting your nums array to arraylist and using Collections.sort will make your code pass comfortably

https://pastebin.com/7UtCkk9u

TLE before even beginning your algorithm: https://codeforces.com/contest/1213/submission/59763079

In B i accidently took array smaller than the limits so i tried to hack myself and i can't hack it can anyone help me?

Difficult to hack, only hackable if those memory locations are assigned next to OS memory locations. Keep trying :P

Can anyone tell me why my problem G TLE? I think it is O(N + M) (assuming that dsu operation is constant). Is there any infinite loop possibility?

59762107

return (pr[x] == x ? x : root(pr[x]));

this line needs to be

return (pr[x] == x ? x : pr[x] = root(pr[x]));

omg :( silly me

Im actually used to code it like this

thank you XD, i think im just too tired right now..

Sad for him :(((( be hacked 4 problems :<<

Will those hack tests be included in final tests? What if my code fails in one or more of them?

Yes, if you not pass all test, you will be hack. =)))))))))

He is trolling lol. if your code fails in one of them, you get WA!

Ah! makes sense.

Problem E :36 cases o.0 It's so afraid :<

1st solution is of Ashishgup right??

Solution for F

if d occurs at i position in a and j position in b then it can be said that

s[min(i,j)]=s[min(i,j)+1]=.......s[max(i,j)] make an array of intervals and add every [i,j] in it

sort the array

now merge the intervals which share common index

if length of the interval array is less than k then it is NO

else

if length of the interval is greater than 26 then merge all intervals from 26th index

assign letters from a to each interval

now sort intervals on the basis of their position in A.done AC

Why the following brute force solution for D1 gives WA in test 5. https://codeforces.com/contest/1213/submission/59764056. Can someone have a look.

русский Я очень люблю раунды vovuh. Спасибо! 1213E - Two Small Strings удивительно.

Ελληνικά Μου αρέσουν πολύ οι γύροι vovuh. Ευχαριστώ! 1213E - Two Small Strings είναι εκπληκτικό.

English I really like the rounds vovuh. Thanks! 1213E - Two Small Strings is amazing.

Español Realmente me gustan las rondas vovuh. ¡Gracias! 1213E - Two Small Strings es asombroso.

This summer vacation I'v joined lots of cf's contests. before the new term , I wish I could be an 'expert'.