Hi everyone!

I would like to invite you to my second Codeforces round, which I have created with my friends Ashishgup and Vivek1998299.

With that said, I bring to your attention our new Codeforces Round #548 (Div. 2) that will take place on Mar/21/2019 18:35 (Moscow time). If your rating is less than 2100, this round will be rated for you; otherwise, you can participate out of competition.

I would like to thank Ashishgup and Vivek1998299 for their help with preparing problems, V--gLaSsH0ldEr593--V, mohammedehab2002 and Um_nik for testing the problems and _kun_ for coordinating our round. I would also like to thank MikeMirzayanov for the awesome Codeforces and Polygon platforms.

You will be given 6 problems and 2 hours to solve them. Scoring distribution will be announced later.

Good luck! :D

**UPD1:** Shortly after the contest, we'll be on the community Discord server to discuss the tasks.

**UPD2:** Score distribution is — 500-1000-1750-2250-2500-3000

**UPD3:** Editorial

Is there something wrong with the registration system before? I noticed that it said the contest will be rated for people whose rating is under 1900 on the registration page when I registered for this contest.The Candidate Masters who registered that time have the sign of out of competition. What should we do now? Should we register again or the administrator will help us fix the bug? Please fix it, thanks and wish all the participants have a good contest and get a good rating.

You can write comment in THIS BLOG, MikeMirzayanov has noticed the issue

thx

I was registered out of competition and had to re-register. I am not sure if everybody noticed that — is there a way to automatically transfer all out-of-comp candidate masters to normal registration?

Mike said here that he was going to fix it later. He probably hasn’t gotten around to it yet.

oh god

PLEASEno useless math so we can participate .... i know we don't matter for you anymore but make an effort for us and don't bullshit us.but i don't have much hope in this one as the problem setter is from india . oh well

keep faith in bhai ka swag , Ashishgup he will not disappoint u .

nah don't have anything goin for

The problem setter's handle has letter '_' in it. This won't be a good contests.

Codeforces round by an Indian coder. I am really exciting for this contest.

why so ?

I hope someone with the handle name "System_test_failed" reply to this comment

Let's hope your code will pass in system testing too. :)

We know what happens in this comment

Any categorization of people according to their respective nationalities is potentially provocative.

excited*

learn to speak English before trying to look smart online.

Let's hope for a contest with awesome questions with strong pretests !!!

Squad is Back on the occasion of

Holi.Super Excited.

Contest on

Holi.Confused whether to be happy or to be sad.

Anyway it doesn't matter, You won't be busy hitting colours at 9:05pm (IST) in the night.

Is there a separate discussion forum for Codeforces or do people have to write blog entries?

Does anyone know if Topcoder has the same functionality as Codeforces like being able to solve past problems, submit solutions for testing, viewing other people's solutions filtered by language sorted by executions time, etc.? Topcoder website seems like a disaster. I would ask this in the Topcoder forum except I can't find a way to make a post there and there customer support seems unable to help. Thanks.

Yes, it has most of it, but you need to download and launch Topcoder Arena

Thanks. Is there a separate discussion forum on codeforces or do I ask questions on Round blogs like this?

You should ask questions in blog entry, or comment on the relevant post

where?

.

Your own comment: https://codeforces.com/blog/entry/66067?#comment-500584 Contradicts the above comment!

Memes are just for fun, not to analyze much :P

Oh thanks. You opened my eyes.

he has anime profile picture so he's an irrelevant individual

So many girls in your expectations.

I appreciate your ability to notice the girls, despite of those colors xD

Thanks. xD

HAHAHHHHAHHAHAHHHHAHHAHAHHHHAHHAHAHHHHAHHAHAHHHHAH=)))))))))))))))))))))))))))))))))))))))))))))))))))

it is too late for our Chinese students (￣_￣ )

Because it needs to consider the time in Russia and the time in the author's country. It is late for Chinese students sometimes, but whether you participate in it or not still depends on you. You can balance your study, sleep and the training and the contests. If you really want to join the contest, just do it. If you are worried about the time and will not join it, you still can virtual participate or just practice the problems when you are free.

There is no doubt that I will participate in it,because I really enjoy it.However it just goes against my original intention of wanting a healthy life. :D thanks for your reply.

Should the pretest be strong enough or not ??

Ashishgup's contests run smoothly.

Hope for this contest too.

i think pretests should be strong but they should not be too strong otherwise hacking phase will have no meaning !!!

There is no hacking phase...

Happy Holi to everyone.

I think this contest will be harder than div 3

You're a racist.

Its not racism, its nazism!

Who know, who know)

Are you from another dimension, where div.2 easier than div.3?

it was wrong look one more again

I want to add some scores ^_^

Jeel_Vaishnav thanks for contest

Раунд от индуса..........

Codechef

Hope see some worthy problems this time...

Why "fidofido" is the bad word?

Woow this round, I will fall) but its doesn't matter, because it will "FUN"!

Hi guys! Unfortunately CF-Predictor doesn't work in last version of Chrome. They changed Cross-Origin Read Blocking policy and my extension can't send requests to the backend anymore. But the actual problem is that I can't update code and fix issue ASAP. I recently started working in Google and they have pretty strict policy about open source projects. I'll try to come up with some solution, but sorry terms are unknown.

Current solution is to use website version.

What about Firefox? Is the extension working there?

Yes, I suppose so. It even can works in Chrome if it doesn't updated yet.

You can just open any past contest and if you see additional column with rating change, extension is working.

Complexity distribution is so balanced

lol xx its hard to set a bc

haha , thats why i think why cf doesnt hold div1 rounds frquenttly :

reason : div2 problems are more likely equal to div 1 c , d .

then why create a seperate rounnd just for div1 e .

Nice Problem D!

it just prank bro

What is wrong with the pretest 8 on problem C?

I think you have not taken care of adding 10^9+7 while subtracting and then taking mod.

you might be doing (total-calculated_value)%mod, which gave WA to me , but using (((total-calcualted_value)%mod )+mod)%mod which gave AC,hope it may help.

D is cool problem

Liked Problem C a lot thanks!!

+1

During the contest, I got alot of "codeforces is temporarily unavailable" page and the problem was taking a lot in order to show anyone face the same issue like me?

Yeah, I also faced a couple of times. But it didn't bother me much.

You can always try m1.codeforces.com

zh_zhumash xD

LOL

when you don't want to solve graphs problems but cf has another opinion

A lot of people were saved by the authors from being hacked on A due to integer overflow.

PROBLEM

CIMPOSSIBLE WHY WA??I think your solution was wrong.

your opinion is wrong. my answer was correct.

who is the setter of problem d ? Ashishgup ?

Million dollar question-How to solve D?

for each number g you have to find 3 things:

a = how many numbers are relatively prime to g within range [1,m]

b = how many numbers will have gcd g again when paired with g within range [1,m]

c = how many numbers have gcd<g when paired with g. (This was the hardest part for me in this problem)

The rest is calculating E(g) with a bit of knowledge of probability

My code: 51649595

Is there O(n*k) solution for C? Or the constraints were there just to bamboozle innocent people like me? :)

No need to count connectivity of red edges. Merge all nodes with red edges to components and multiply each member count of adjacent components

Just a linear time solution. O(n)

Count the number of bad strings, and subtract them from total (n^k) to get the final answer.

Think how you can form a bad string, which doesn't traverse through any of the black edges. For simplicity remove the black edges from the graph and then, you'll get some connected components.

To Form a Bad string of length K, all the nodes should be from the same component. If you take nodes from different components, they must have to traverse through a black edge and hence it will not remain a "bad string".

Now suppose cnt = no. of nodes in a connected component. Then no. of bad strings formed by nodes of that particular connected component is = cnt^k

Total no. of bad strings = sum of bad strings of each component.

Note : If the difference is less than 0 after taking modulus with 1e9+7, Add 1e9+7 to make it positive.

One minute silence for the three Indians: alwaysGREEEN, Abhinaviitbhu and imhemant.

Who were the only people to get

hackedtoday, on the occasion of holi.and a minute for me for 3 fail hacks

How to solve C?

First, notice that this is an inclusion-exclusion type of problem. So the answer will be something like: all possible permutations — all impossible ones. After realizing that, you'll soon notice that all the impossible ones are the collection of nodes that don't have any black edge at all.

Hence, we can simply isolate collection of nodes by using disjoint set. Simply merge when the edge is red and skip the black edge. When you have got a bunch of disjoint sets, you can calculate how many permutations can be constructed from a disjoint set of size x. pow(x, k)

The total answer would be pow(k, n) — (for each disjoint set of size x, pow(x, k))

Solution

a

Can someone explain how to solve C

lets do tree with only RED . we have forest and answer is n ** k — a1 ** k — .. — af ** k, whee ai number of vertexes in i tree

Can u please elaborate. I mean how did you get this?

Count how many sequences are bad (not good) and subtract it from N^K. If we make the graph only with red edges, for each connected component with size X we will get X^K bad sequences from this component. Sum of bad sequences for each component will give total number of bad sequences. It's not possible to get any more bad sequence because merging any two component would need black edge which we can't use in bad sequences.

Got it. Thanks a ton!

How to solve E?

Spoiler1One can see that we are matching the clubs with a certain range of potentials. If there are no deletion, we can run any bipartite matching algorithm and meanwhie keep the maximum mex.

Spoiler2Now that there are deletions, a smart way to get around this is build the graph after all deletion and add edges to it in reverse order. In this way, we are only adding edges, and then we can check whether there are new augmenting paths in the new graph.

The amortized complexity will be $$$O(VE)$$$.

Thanks a lot!

Is this the right approach?

We start with a bipartite graph with 0 potential vertices on the right. We will first add all persons that were never removed to the clubs on the left. We will process the days in reverse order. We will add the person that was removed on this day to our bipartite graph after we have processed this day. As long as the bipartite matching is maximal, we add the first available potential (first 0, then 1, then 2, etc) to the bipartite graph and rerun the matching algorithm. When the matching is not maximal anymore, the final value we tried to add is our MEX value for this day. The MEX can only increase during this process, so we never have to remove added potentials anymore in the graph.

Edit: solved while processing the days in chronological order with a similar approach as described above (but instead of adding potentials, we remove them when the matching isn't maximal). https://codeforces.com/contest/1139/submission/51653759

RobeZH, how are you calculating mex using bipartite matching. Can you please explain. Thanks

First we build a graph to check that whether the mex can be 1 (, more specifically, say we put the potentials on the right hand side of the bipartite graph, we build it such that only potential 0 is connected to the sink, and see if there is a perfect matching). If yes, we incrementally build the graph to check that whether the mex can be 2, and so on. If no, we know that the previous number is the mex for the current set of people.

Thanks human!

Help me solve problem B, My solution O(n^2) :'(

You can solve it in a single iteration from n-1 to 0, making it O(n).

Check my simple solution.

thank you, I misunderstood the problem. I think it don't need to end at n-1,

Oh sad.. Better luck next time :)

Think greedy.

Use the maximum possible number of chocolates from

a_nand with considering the picks use the maximum possible number of chocolates froma_n-1and so on...Can some one help me with problem D?

I got wrong answer on test 13.

Here's the codeToday's problem

C, is the first tree problem I have solved till date without traversing (dfs/bfs) the tree.How so?

I used UFDS (Union Find Disjoint Set).

Check my solution.

Oh, yes, I forgot about DSU solution.. But that means that you are expert and never solved MST problem, wierd.

I am talking about exclusive tree problems usually given in cf.

I believe MST is found on a given general graph.

Or maybe I just don't remember. Thanks for judging me :)

Can you talk a little about your solution using DSU?

Check my solution for more clarity.

Thank you!

In fact, it can be solved by just dfs.

How to write code to hack others? I use this code to hack problem B, why the return is "Invaild input"?

## include

using namespace std; int main() { int n=100000; printf("%d\n",n); for(int j=n;j>=1;j--){ int t=10000000-j+1; if(j==n){ printf("%d",t); } else printf(" %d",t); } return 0;

}

Dude, you gotta give input to the code they've submitted on which you think that expected outcome will be different from their's output, nor the code.

Hope you get it.

You need a newline at the end of the input.

Nice set of Problems. Three cheers to the authors :D

Already waiting for the Editorial. Please publish them soon :)

What is the intended solution for D? I managed to pass the pretests with inclusion-exclusion and some optimisations.

My solution was calculating the probability of obtaining an array of length

`n`

with the gcd of the array(except the last elemnent) as`x`

with ending element`y`

where $$$x,y \in [1,m]$$$, $$$ n \in [1,\inf]$$$ and`y`

is coprime to`x`

.Then, summed up the contributions with a bit of math.

Problem E: The mex of the multiset S is the smallest non-negative integer that is not present in S. For example, the mex of {1,2,3} is 0. Why the mex of {1, 2, 3} is not 4?

0 is non-negative and 0 is smaller than 4.

0 is a non-negative integer. The MEX of {1,2,3} is 0, because 0 is not in the set. It would be 4 if the set was {0,1,2,3}.

EDIT: Triple ninja-d :) Because 0 is nonnegative?

non-negative integermeans zero is included.{0,1,2,3,...}Oh thanks, I'm sleepy :D

Problem C: Count connected components of red nodes. If a connected component has p nodes. Impossible paths are p^k, sum similarly for all connected components. Finally subtract this value from n^k, also subtract no of nodes that have zero red edges. Is my idea anywhere close at least? I couldn't manage the time to implement though..

I think the answer is just $$$n^k-\sum p^k \pmod{1e9+7}$$$

What is the reasoning behind your formula?

There are n^k paths in total. If there are p nodes in a connected component(>=0 red edges, ==0 black edges), invalid paths are p^k. Sum all such p^k, subtracting this from n^k gives valid paths.

oh shiiiiit , i solved problem C but i forgot to Print the number of good sequences modulo 109+7.

E has an awesome solution)) Really nice, and didn't even get close to solving it

contest was good !! Happy Holi everyone :)

can E be solved by maximum matching?

Yes. I used Kuhn's Algorithm

can prob E solved by bipartite match?

Yes. But there is tricky part, it is not just reduction to maximal matching problem.

so sad...

When I recognised how to solve this problem, it was 10 minutes before the conteat ends.

It was a very unbalanced round. The jump in difficulty from B to C was to great. The problems were interesting but I felt like there should have been a problem between B and C.

yup! but petr can solve every question. he solved such hard problems of Atcoder world finals where gennady just solved one . so before listening to twice , listen to his stream

I have the same thoughts about C and D))

I made an idea for Problem D and couldn't implement it for 1 hour 30 minutes XD

why ?

My idea is very hard to implement(at least for me).

My solution idea for D:We can define the weighted/directed graph to solve this problem. Each node has distinct set of primes. This means the prime divisors of greatest common divisor made by array $$$a$$$ so far. For example, if the $$$a = [24, 30, 18]$$$ then the state of $$$a$$$ is $$$(2, 3)$$$. Of course there can be self-looping edge exist.

There is a special node called

all, this node contains all prime numbers in $$$[1, m]$$$. For all non-special nodes there is an edge exists directed fromallto that node.Calculate weight of all each edges, then you can solve the equation such like; $$$EX(node) = 1 + \sum_{\text{all neighbors}}^{} EX(neighbor) * weight$$$ where $$$EX$$$ means the expectation of length of sequence started from that state. There is an exception; $$$EX(NULL) = 0$$$.

Example for $$$m=10$$$

all->null: 1 (1), [2]: 3 (2, 4, 8), [3]: 2 (3, 9), [2, 3]: 1 (6), [2, 5]: 1 (10), [7]: 1 (7), $$$EX(ALL) = 1 + 0.1 EX(NULL) + 0.3 EX([2]) + 0.2 EX([3]) + 0.1 EX([2, 3]) + 0.1 EX([2, 5]) + 0.1 EX([7])$$$null: 5 (1, 3, 5, 7, 9), $$$EX([2]) = 1 + 0.5 EX([2]) + 0.5 EX(NULL)$$$null: 3 (1, 5, 7), $$$EX([2, 3]) = 1 + 0.4 EX([2]) + 0.1 EX([2, 3]) + 0.2 EX([3]) + 0.3 EX(NULL)$$$You can find weight of all edges with fast factorization.

More simple example $$$m=4$$$

all->null: 1 (1), [2]: 2 (2, 4), [3]: 1 (3), $$$EX(ALL) = 1 + 0.25 EX(NULL) + 0.5 EX([2]) + 0.25 EX([3])$$$null: 2 (1, 3), [2]: 2 (2, 4), $$$EX([2]) = 1 + 0.5 EX(NULL) + 0.5 EX([2]) = 2$$$null: 3 (1, 2, 4), [3]: 1 (3), $$$EX([3]) = 1 + 0.75 EX(NULL) + 0.25 EX([3]) = \frac{4}{3}$$$Thanks for awesome problemset and super fast system testing !

I have enjoyed the contest a lot.

if a problem submit twice, all accept, which is the score in the problem?

The first one will be ignored and you will get penalty of -50 for re-submission.

thanks

Why is this invalid input for hacking A? Why is this N^2 solution passing for A?? Please help.

Because servers are now upgraded and with compiler optimizations almost around O(10^9) passes in one sec.

Probably compiler optimization.

Thanks!

The compiler optimizes the increments in the for loop into an addition. After optimization, the complexity is $$$O(n)$$$.

Thanks!

I was kind of confused by problem B, it implies you can decided to buy a chocolate or not, but in reality you always buy a certain chocolate, even if 0 times...

Can anybody tell me how to solve problem D? I was thinking to solve it like E[len]=E[number of 1's]+E[number of 2's]+E[number of 3's]+------E[number of n's] in the sequence but could not able to solve it.

See my solution: https://codeforces.com/blog/entry/66067?#comment-500828

No. of N-length arrays: Sum(C[i] ^ N) / m ^ N. ( 1 <= N <= ...)

C[i]: Count of multiples of C[i] <= m. (Make sure that we do not re-count multiples, C[x] -= C[y])

If you re-arrange 1st statement, you can get sum of infinite GPs for each C[i].

Sum the answer for all C[i].

You are taking C[i]^N for length N but it contains all the elements that have GCD>1 so length will no longer be N and it will be more than N

Good contest

In problem D, can someone explain the solution with mobius function?

My solution is $$$\sum_{i=1}^{\infty} i*q^i = q / (q-1)^2$$$, and the answer is $$$\sum_{i=1}^{m}mu[i] * q / (q-1)^2$$$, $$$q=[m/i]/m(i=1\dots m)$$$, but it is wrong :(

Because you count something twice or more. Let call the first expression is f(q), then you want f(i) is the answer with some first element have gcd() = i, but you also count the other with gcd = 2i, 3i, ... So you need to subtract f(2i), f(3i) to get the right answer.

Actually, what you're looking for is $$$\sum_{i=1}^{\infty} (i+1)*q^i*(1-q)=\frac{q*(2-q)}{1-q}$$$(you need to factor out the $$$1-q$$$ for wolfram alpha to give you the result: https://www.wolframalpha.com/input/?i=(sum+of+i+from+1+to+inf+(i%2B1)*a%5Ei)*(1-a)*1 . Make sure to add $$$\frac{1}{m}$$$ at the end for the special case of length 1. AC code: 51651931

I get it, Thx!

It seems that I missed the case that the first element is 1.

When will tutorial be out?

[Problem B]I just realized I didn’t understand the task properly. But now I wonder if there is a nice algorithm to solve the version of this problem that I had in mind.So the condition is such that the number of chocolates we buy has to be a strictly increasing segment. In particular we are allowed to leave some types of chocolates on the right side. For example:

Should produce 6.

I had the same misconception in the beginning, couldn't thing of anything better than O(N^2).

Same happened to me. Lost a lot of time for not reading again the problem statement.

Couldn't solve it because I understood the same.. After almost 2 hours thinking i'm pretty sure there isn't anything better than N^2.

This problem can be solved in O(N).

Could you explain how?

You run the loop greedily from the last index. Then we take min(Ai,x-1) at the ith index where x is the current maximum. If it becomes 0, then we break the loop and print the final answer.

Here is the link to my submission for better understanding. https://codeforces.com/contest/1139/submission/51651782

This can be solved in O(nlog(n)) it is simply LIS dp problem in your variant

I don't know if we are understanding the same problem.

For input array = {1 1 4 3 5 1} result would be 11 (0 1 2 3 5 0).

How would you solve it with LIS?

I was stuck at B because of same understanding. I am also wondering how to solve it in O(n) even though it was not the original problem.

This should produce 1 not 6

Lol , looks like I was not the only one who understood the problem in this way during contest time. I was stuck at this testcase for B (ofcourse because of same assumption as yours)

5

8 6 5 3 4

here answer should be 12 by taking 3 from 8 , 4 from 6 , and 5 from 5.

can anyone suggest how to solve this problem , where order should be increasing and we have to maximize the sum just like given above testcase.

The answer for the above test case will be 10 and not 12 by taking 0 1 2 3 4.

Thank you , but now I am curious about solving this imaginary problem .

If we take your approach [3,4,5,0,0], then we see that for j=2 (here 5), and i=3 (here 0) , Aj is not less than Ai as 5>0, and neither it is 0. Hence, we are not satisfying any of the conditions.

I liked the C priblem a lot. At first I found it difficult to solve it, but after that i understood it was easy.

Please publish the editorial as early as possible. Can't wait to see the solution of problem D, E and F!

Thanx. Nice problems. It is good that in pretests exists test for TLE, at least for E. There is happen that seems like optimization do not need, but pretest got TLE and one saves time for testing and will not get disappointment after contest.

How did (n*n) solution got accepted in Q A? Link https://codeforces.com/contest/1139/submission/51625556

It's magic of GNU C++

Maybe compiler optimization

The above block is optimized to

`ans += i + 1;`

you can see that here.compiler optimization + server upgrades !!!

I know that E's Solution is Khun's algo with reversing queries, but I am interested to know was there any direct HopcraftKarp Solutions that passed it should be $$$O(N^2sqrtN)$$$, since the number of edges <= 5000 and the number of nodes <= 5000 that totals to $$$\cdot{2*10^9}$$$ operations approx.

but as I don't really understand how HopcraftKarp works I am not sure can it be optimized to pass?

I think if we use Hopcroft Karp we need to binary search on the answer and so we attain complexity of $$$O(N^2 \cdot sqrtN \cdot logN)$$$. Can you explain your approach of using Hopcroft Karp and attaining complexity $$$O(N^2 \cdot sqrtN)$$$?.

Sorry, apparently I really didn't understand Hopcroft Karp that well, a friend of mine explained it to me afterward and told me that a simple $$$O(N^2sqrtN)$$$ Hopcroft Karp is not doable.

I am getting TLE on pretest 6 in Problem A. Please help.

Link to the submission.In this code I'm taking each substring and just checking the last digit is even or not. Is it not one of the right way to do so?Your complexity of solution is O(n^2) and I guess java is slower than c++ and the multiplier for java might not be enough for an O(n^2) to pass but in c++ it is, just bad luck i assume!

Thanks buddy. Bad luck today indeed.

Why did my Submission 51646792 fail? Does it have something to do with modular arithmetic?

I am getting WA on pretest 10 in Problem B. The longest testcase, I suppose.

Link to my submission.Please Help.Learn the range of different data types in your preferred language.

int can only handle values up to about 2 billion.

Using long passes. See https://codeforces.com/contest/1139/submission/51652989

Why do i need to make the ans[] in the above code as long too? I tried it with sum as long, but not with ans array as long.

You don’t need to. As long as sum is long, it will pass. See https://codeforces.com/contest/1139/submission/51653644

Thanks Redux. You came as a savior today. Getting the logic but failing at implementation tired me out today

Can someone check if my solution for E is correct? It's not maximum matching.

First, I find max mex for the given situation without considering updates. Here's how I do that: I go from $$$0$$$ to as far as I can go by randomly picking a club which has a member of the required potential. When I choose the $$$i$$$'th club for potential $$$j$$$, I make a directed edge from all other clubs who have a member with potential $$$j$$$ to $$$i$$$. Now, if I get stuck at some potential k, I will bfs from the unused clubs to find a club which has a member with potential $$$k$$$. Now, if the path to this club is like $$$p_{x}$$$ -> $$$p_{a_{1}}$$$ -> ... -> $$$p_{y}$$$, where $$$p_{x}$$$ is the unused club and $$$p_{y}$$$ has a member with potential $$$k$$$, I can replace $$$p_{a_{1}}$$$ with $$$p_{x}$$$, and $$$p_{a_{2}}$$$ with $$$p_{a_{1}}$$$ and so on. After this I update the edges of the graph as required. By the end, I will find the max mex for the given situation.

Now, for updates: First notice that the max mex can only decrease or remain same. Now, I maintain the reverse of the graph I mentioned above, if a member from club $$$i$$$ is chosen for potential $$$j$$$, then I make a edge from club $$$i$$$ to every other club which also has a member with potential $$$j$$$. Now, for each member which leaves, we see if we were using this member in our chosen max mex, if not, we do nothing. If yes, then we bfs from this club to either find a club which is unused right now (in which case max mex remains same) or we find a used club from which we have taken a member with the highest potential (in this case max mex becomes this new potential). In both cases we update the entire graph again as required.

My solution for C runs in O(nk) time and it still TLEs, does anyone know why it exceeds the time limit

https://codeforces.com/contest/1139/submission/51649583

edit: I found the reason: apparently running a DFS has a much larger constant than I thought. I ended up passing by precomputing the DFS order and then doing DP directly on that

https://codeforces.com/contest/1139/submission/51920785

In Problem D how to find no of possible ways to get an array of length k with gcd say xYou need to get

needed prime divisorsandavoided prime divisors. That's the key.benim neden degerlendirmem dusuyorki bir problem cozdum ve bir kez bile yalnis gondermedim puanym 844 ve 24 degerlendirmemi kaybetdim bunun nedenini bana soyliyebilirmisiniz lutfan

Hello, Div.2, my old friend

I've come to solve your tasks again...

I have a solution to Problem

Bbut cannot pass the pretest 10 (very large values).could anyone help me to find out flaws in my solution?

The basic idea is to find the most right tangent line of the curve of $$$a_i$$$, it treats the final solution as two parts:

We can get the answer easily by adding those two parts together.

here is my solution： https://codeforces.com/contest/1139/submission/51664626

In problem D. Why the example 3 has the output 3333333338?