Any ideas for F?

Nothing unfortunate about it, logic trumps everything

No, that's not what you did. To compute the answer via the method described above, you would have to compute how many elements are lesser than the current element at any given iteration and add them and also keep and their count using a Fenwick tree/ BIT. That's the incorrect approach though because sorting would be more optimal.

I suppose 1600/1700.

Didn't think of that at all! Can you try to hack my submission if possible?

Well, did you take care of the fact that ordered_sets only keeps the unique elements in ascending order? And if you did, could you share me a link to your submission? I found a link on GfG that takes care of that using Fenwick tree but turned out it was an incorrect implementation!

No need to do that. Simple recursive dp solution works just fine. The states would be dp[index][team_num][if_last_ele_was_included].

It clashes with Atcoder Beginner Contest 203. Can we prepone it?!

Does anyone have the working link to the rating predictor of Codechef? The links that I find don't work

Let’s suppose if we start at a units, and we get g units in d+1days. Hence, a+(a+1)+(a+2)+...+(a+d)=g. Hence, (d+1)*(2a+d)==2g. Now, the rest of the proof is pretty straightforward !

It is. It's 2h 45m! Check the contests section

The contest was really brilliant in my opinion. Kudos to the author! Even after having not performed that well, you still made my day! :)

Since we have to make the sum as large as possible while keeping their bitwise AND to be 0, we observe that we can do this by setting each bit from 0 to k-1 in exactly (N-1) of the numbers. It's easy to see that the bitwise AND of this comes out to be exactly 0, while the sum remains as large as possible(can be proved by contradiction). Now, we have to assign these k bits to (N-1) of these numbers so, the total number of ways for each bit will be N choose N-1 which is N. Hence, for k bits, it would be N^k.

If a number is not coprime with n, then that doesn't necessarily make the product%n 0

Did anyone else solve B by figuring the minimum and maximum possible positions of each character 'M'? I thought that was really interesting! Also, the problems were fun, although I felt that the gap b/w C and D was a bit too much. Nevertheless, kudos to the authors!

Somehow, I didn't think of a dynamic programming solution for C. I just ended up counting the number of particles with decay ages from i=1 to i=k and obtained that by using prefix sums.

https://discuss.codechef.com/t/lunchtim-editorial/87198

LUNCHTIM had a nice intended solution. Although it could also be done trivially by using segment trees. Over and all, I really liked the problems. Kudos to the authors!

The rating predictor doesn't show anything for me. Strange!

It's a multisource BFS problem.

The idea is once you get the original matrix, there is only one optimal matrix that can be obtained. So, you find that matrix using BFS => First, store the indices of all matrix elements that have the maximum possible number in the matrix(note that it is always less than 2e6), and then move to it's neighbours and keep doing the same thing!

Think of brute force, and show that it'll work using the pigeonhole principle

My self worth took a big hit during this contest.. The round was really great though! Kudos to the authors..

I guess so, I'll be tumbling in infinite loops in my code to check its' strength!

Anyway, admirable work Codechef! :)

The problems were really interesting!

Kudos to the authors, and the coordinators involved! :)

Your inability to solve a problem doesn't make it rude!

Thanks for the contest, even though I couldn't perform well, I got to learn a bunch of things!

Kudos to the authors

You say that you don't, and yet you care enough to comment!

Don't know about others, but coming from a math background, this contest was really enjoyable. Kudos to all the authors! :)

liouzhou_101, I guess you must know that a considerable amount of accepted solutions for B are wrong. The fault being that they're considering that if the hero has his health points zero before the last attack, he will still be able to win. Now, these accepted solutions are churning out WA now. So, I just want to know that since the solutions are not going to be re-judged again, why are you folks choosing to make this decision?

Any idea why my code fails for Test Case 55?

One of the best contests I have seen in a while and I am saying this because this is one of the few contests where I was stuck to my laptop for solving D1, D2 until the very end. And, they didn't seem unsolvable which they usually do.

Thanks for these brilliant problems, authors

Much kudos to you liouzhou_101

which their spineless admins never take

If you were an admin, what would you have done?

Am I the only one who doesn't understand what this means? Correct me if I am wrong, but is it that the two subtasks will be about the same problem, but perhaps, they would ask us to compute different things?

Despite having done horribly in this contest, I'd like to say that this was a nice contest!

Can anyone share their approach for the fifth problem of the second division?

Could you share your approach for Div 2 E?

Could you tell me your approach?

between city*

Make a left_c and a right_c array which represents the maximum number of cities city i can traverse that lie strictly on it's left and it's right respectively.

Then, notice that if there lies a path b/w city i and city i-2, then city i can traverse all those different cities that city i-2 can because at time t and time t+2, all paths are exactly the same. If there just exists a path b/w city city i, and city i-1, then increment one to the left_c array for the city i. And, do similar things for right_c.

Answer for the i_th city is left_c[i]+right_c[i]+1

You can check my submission

There might be a greedy algorithm to it, but I solved it by trying all possibilities using dp

Yeah, but I thought C and D made up for it!

Was I the only one who went into C and D thinking that they are graph problems, but both of them turned out to be dp ones? :P

I thought that the problems were really nice. Kudos to all the authors! :)

Hey,Thallium54 I did the correct thing, and my time complexity is right, can you tell me why I am getting a TLE? Submission — 104815636

I solved it using prefix and suffix arrays.

Dude, he is making contests, actually making actual contribution to the community. What have you done to do the same? It's pathetic to see the entitlement which participants seem to have. Even I had a terrible contest, but I wouldn't take it out on the authors. Try to give constructive criticism, if possible or don't comment at all :)

Apparently, yes

Can anyone prove why brute force works in B? I calculated such numbers upto 10, 000 only to find a pattern!

Well, now notice the difficulty gap between B and C. There's a reason why I mentioned that.

Is it just me, or does the scoring distribution of Division 2 seems a bit scary?

Lol, any reason why it should be unrated?

Can someone please explain something? It seems to me that the people in the scoreboard have already been through the problem set are playing the role of testers in a way, and are not going to participate in the contest today! Am I right about this?

It's interesting to note that before I made this comment, -is-this-fft- had a lot of downvotes, but as soon as I made this comment, people started upvoting him like batshit crazy. Now, I don't care if y'all mofos like me or not, life's much bigger than that. I just taunted this dude because I didn't like the condescending tone that he put forth(which is obviously strictly my opinion) that undermines the work done by the setters(alteast) in creating nice contests from the past couple months, hopefully in the future

Nibba, you don't speak like this with that hair!

Well, I am unbiased in terms of this round being unrated for me regardless. I don't have anything against the setters. I thought the problems were still nice. My problem lied in the technical issues that transpired. Anyway, if you still feel like giving a nice sarcastic comment, feel free to do so, brother! :)

MikeMirzayanov, I started with problem B today, because A looked too tough for me. After getting the implementation intact, I submitted my solution, and naturally it didn't submit. I left the contest out of frustration, because I saw people had already submitted A by that time. Now, I definitely agree that the setters did a lot of hard work, but at the same time, the starting five minutes of the round are much more valuable for a newbie/pupil. So, you should make this round semi-rated. I think it's a good compensation for both the setters and the participants. And, note that-regardless, this round will be unrated for me. So, I am actually unbiased over here.

Agreed on that. But, we are wanting to make this contest reason due to a bigger reason other than having weak pretests. I started with problem B today, because A looked too tough for me. After getting the implementation intact, I submitted my solution, and naturally it didn't submit. I left the contest out of frustration, because I saw people had already submitted A by that time. Understand that we are talking about this issue. Don't trivialize it by talking shit about weak pretests. I understand that you might get to CM after this round, because you didn't FST. But, don't lose your conscience and start acting like an asshole. Apologies if I sounded too rough, but I am calling like I see it. And by the way, this round(rated or unrated) is the same for me either ways. So, I am actually unbiased over here, unlike you.

I think that the best way to clear this will be to have a lower rating bound for asking questions. It should be atleast 1600(or 1400 in the absolute worst case). As a pupil, I still think that this should happen

Says who?

I don't think it's racist to say 'average Indian contestant asks more bad questions / makes more bad requests than average contestant from most countries'.

Yeah ok, I get it. I guess you're right, then! By the way, your argument goes exactly like that of Ben Shapiro's. By the way, this logic can be applied to many things, we can also use this logic to condemn hiring of women in most fields, because on an average, women do tend to resort to household work, after a particular amount of time in comparison to a guy. So, I guess, the wage gap is justified because of this strawman argument.

I guess we know why this is becoming more common now. If you consider yourself witty enough to insinuate certain connotations, then be bold enough to accept it as well!

By former, I referred to the former among AtCoder and Educational Contests, i.e. Atcoder!. By this, I think he meant that he solved more AtCoder Contests than the Edu. rounds, xD

The very last one!

Also, could you try to make a more balanced contest? This one was worse than most Codechef Cook-offs

Weak test case for C

Just to be clear, is the point value system of ARC consistent with that of ABC in terms of difficulty of problems? As in will ARC B be as tough as ABC D, ARC C as tough as ABC F,... and so on?

Yo Um_nik, I've a feeling that you'll really like this round!

I honestly think that somebody should tag the GMs and above here, because the expected quality of the contest is pretty damn high, and most people think the contrary!

Uhmm, I don't understand! Everyone's cf rating is still the same now, do you mean to say that the cf rating of a person will go up if he/she performs well and vice versa? If no, then what exactly do you mean to say?

I wonder if Tourist, Petr, Errichto, neal and ksun are not submitting just because they know something that people who have completed the set don't? Like, they're thinking about something bigger

Also, ecnerwala made his final submission first, and yet he is shown to be behind Benq Can anyone clarify how are rankings calculated?

Why can't I see Petr, Gennady or Mikhail in the standings? Are they already advancing to the finals?

https://prismatic-age-179203.appspot.com/

Check it out yourself!

It was asked in the Google Online Challenge, conducted today!

Sorry if this is a stupid question, but can anyone tell me if my method of finding leaves is incorrect. I am just trying out all those elements whose adj[u].size()==1 Here is my function input, and check - vector<vector> adj(n+1); for(int i=0; i<n-1; i++) { int v, u, w; cin>>v>>u>>w; adj[v].pb(u); adj[u].pb(v); } for(int i=1; i<=n; i++) { if(i!=1&&adj[i].size()==1) { leaves.pb(i); } }

Hey, how did you guys know about this test? Are there more tests to be held, and can we still apply for the intern?

Wait, do we have subtasks in div2(asking because of the score distribution/and that it isn't mentioned)

During the contest, I misread the question B as the total number of shapes of length n one can get if for any block, there is at least one block that shares one side with that block, thereby removing the problem constraint of n consecutive days. I still haven't been able to come up with any solution! Could anyone help with that?

Just realized the solution for the equation in question A was 228, this guy is a frickkin legend!

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