CoderAnshu's blog

Codeforces Round #793 (Div. 2)
By CoderAnshu, 3 years ago, In English

Hi Codeforces!

I am pleased to invite you to Codeforces Round #793 (Div. 2), which will take place on May/22/2022 17:35 (Moscow time). You will be given 6 problems and 2 hours to solve them.

The round will be rated for participants of Division 2 with a rating lower than 2100. Division 1 participants can participate unofficially in the round.

All problems in this round were prepared by me and rivalq. We have worked on this round for a long time and hope you find every problem interesting.

I would like to thank:

Score Distribution:

750 – 1000 – 1500 – 2000 – 2500 – 2750

Good luck and have fun! See you in the standings. Make sure to read the statements very carefully.

Thanks to ak2006, video editorials for some of the problems will be available on his channel after the round.

Congratulations to winners!!

Global Top 5:

  1. jiangly, superfast AK!

  2. Beduardo

  3. SSRS_

  4. neal

  5. turmax

Official Top 5:

  1. Beduardo

  2. Brewess

  3. Kizuna_AI

  4. Focalors

  5. csyakuoi

Special mention to AwakeAnay for the only Indian to AK the round!

UPD 1: Links to video editorials of Problem B and Problem C

UPD 2: Editorial for problems A-D is here.

UPD 3: Editorial for E is added.

UPD 4: Editorial for F is added.

  • Vote: I like it
  • +616
  • Vote: I do not like it

Comments Comments (177)
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3 years ago, # |
  Vote: I like it +49 Vote: I do not like it

AmShZ orz, The captain of Iranian testers

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3 years ago, # |
  Vote: I like it +63 Vote: I do not like it

CoderAnshu supremacy!

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3 years ago, # |
  Vote: I like it -60 Vote: I do not like it

I downvoted you. Then, your contribution was raised by 1

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3 years ago, # |
  Vote: I like it +5 Vote: I do not like it

Orz

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3 years ago, # |
  Vote: I like it +131 Vote: I do not like it

As a true cyan (tester), I would recommend you to try the next problem if you get stuck ( '◡' )

Also, if you wouldn't mind
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3 years ago, # |
  Vote: I like it +32 Vote: I do not like it

As a tester, the problems are fairly interesting and quite challenging.

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3 years ago, # |
  Vote: I like it +25 Vote: I do not like it

As a tester, I can confirm that the problems are high-quality and interesting, and cover a good range of topics, and would recommend participating in this contest.

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3 years ago, # |
  Vote: I like it +40 Vote: I do not like it

As a tester, I really did test the round this time.

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3 years ago, # |
  Vote: I like it +3 Vote: I do not like it

I hope it will be great #793 div.2

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3 years ago, # |
  Vote: I like it +6 Vote: I do not like it

CoderAnshu lord supremacy. :)

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3 years ago, # |
  Vote: I like it +13 Vote: I do not like it

A contest by Indian Coder after such a long time!!! Very Excited!! Hoping that this contest turns out to be the best in recent times.

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    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    How can you forget #782 div2 by my friend Newtech66 just a month ago?

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      3 years ago, # ^ |
        Vote: I like it +5 Vote: I do not like it

      Ohh yes! Now indian are making some interesting contest. And they do all this with hectic college schedule in India.(-_-)

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    3 years ago, # ^ |
      Vote: I like it +7 Vote: I do not like it

    The best duos of India. (CoderAnshu and rivalq) supremacy

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3 years ago, # |
  Vote: I like it +107 Vote: I do not like it

As a problem setter, I would really like all of you to participate.

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3 years ago, # |
  Vote: I like it -22 Vote: I do not like it

i hope the round will be a bit better than previous

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3 years ago, # |
← Rev. 2   Vote: I like it +18 Vote: I do not like it

For the last two contests I have been fsting on problem D due to stupid implementation mistakes. Hopefully this contest I will be able to break the streak.(Hopefully by solving problem D )

Edit: Broke the streak by not solving D this time.:clownglasses:

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3 years ago, # |
  Vote: I like it +18 Vote: I do not like it

As a tester, I get to write a comment for some contribution. hue hue hue.

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3 years ago, # |
  Vote: I like it +11 Vote: I do not like it

Notice in the last line

Make sure to read the statements very carefully.

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3 years ago, # |
  Vote: I like it +1 Vote: I do not like it

All the best Everyone !

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3 years ago, # |
← Rev. 3   Vote: I like it +9 Vote: I do not like it

Hopefully the "best of 2022" which were replaced in the previous round will be seen here.

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3 years ago, # |
← Rev. 2   Vote: I like it +1 Vote: I do not like it

I hope I solve about 3 problems and become a specialist !!

Good luck for all participant!

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3 years ago, # |
← Rev. 12   Vote: I like it +2 Vote: I do not like it

All the best for all participants

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3 years ago, # |
  Vote: I like it +8 Vote: I do not like it

As not a tester,give me contribution!

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3 years ago, # |
  Vote: I like it +59 Vote: I do not like it

No offense though

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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Everything is so sorted, just look at the testers list! Efforts put everywhere.

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3 years ago, # |
  Vote: I like it +10 Vote: I do not like it

Much Awaited Contest

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    3 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    of cource! it seems like very very very great contest and also much awaited.

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3 years ago, # |
  Vote: I like it +1 Vote: I do not like it

I will be taking part in this contest.. I am getting back after nearly 1 year.. wish me good luck bois!!

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3 years ago, # |
  Vote: I like it +7 Vote: I do not like it

Btw, 793 is a semiprime number. Hope that means at least half of problems will be great!

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3 years ago, # |
  Vote: I like it -6 Vote: I do not like it

Good luck!

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3 years ago, # |
  Vote: I like it +15 Vote: I do not like it

Best of luck to everyone!!

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3 years ago, # |
  Vote: I like it +43 Vote: I do not like it

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3 years ago, # |
  Vote: I like it +54 Vote: I do not like it

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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I believe in CoderAnshu & rivalq supremacy!

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3 years ago, # |
  Vote: I like it +11 Vote: I do not like it

SwapOnPermutationForces.

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    3 years ago, # ^ |
      Vote: I like it -19 Vote: I do not like it

    This justifies your WA on B.

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      3 years ago, # ^ |
        Vote: I like it +11 Vote: I do not like it

      Well just because I didn't initialize the answer big enough :(

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3 years ago, # |
  Vote: I like it -36 Vote: I do not like it

Yet another unbalanced round... #SpeedForcesOP

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3 years ago, # |
  Vote: I like it +9 Vote: I do not like it

Now I can finally sit and watch my rank fall due to so many wrong submissions.

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3 years ago, # |
  Vote: I like it +6 Vote: I do not like it

Problems are tricy ig.

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3 years ago, # |
← Rev. 2   Vote: I like it +3 Vote: I do not like it

Would anyone please explain why the output of problem C is 7 for this input?

1

13

1 2 3 5 6 7 7 8 8 9 9 10 10

Thanks.

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    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Put 1 in the middle.

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    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    10 9 8 7 3 2 1 5 6 7 8 9 10

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    3 years ago, # ^ |
    ← Rev. 3   Vote: I like it 0 Vote: I do not like it

    Array: 10 9 8 7 1 5 3 6 2 7 8 9 10

    10 9 8 7 1 5 3 6 2 7 8 9 10

    In reverse: 10 9 8 7 2 6 3 5 1 7 8 9 10

    My Approach if u want to know
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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

FML, I think I have the solution for E. Rip. I think it involves traversing each tree in the forest and doing a tree DP.

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3 years ago, # |
  Vote: I like it +9 Vote: I do not like it

Guys!! Hope you liked the round. Please share your feedback on every problem. I will be really thankful.

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    3 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    Because i solved till C, it was very great round!)

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    3 years ago, # ^ |
      Vote: I like it -40 Vote: I do not like it

    unbalanced and stupid problems

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      3 years ago, # ^ |
        Vote: I like it +1 Vote: I do not like it

      Not fair. Frankly speaking, it happens not so often, but A was pretty. As for B, i am not a pro of bitwise operations, but even i didn't found this one too hard. And C... some stupid mistakes, but after about 30 min i did found the right solution by painting samples into notepad, so i do not think that this round was disbalanced

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        3 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Eeeew pupil, ofc it won't be unbalanced for you.

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          3 years ago, # ^ |
            Vote: I like it +4 Vote: I do not like it

          What, how did ya call me????????? I am a specialist now, baby!!!!!!!!!!!!!!!!!!!

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    3 years ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    I was able to solve till $$$C$$$, and it felt okay, I had almost got $$$D$$$, it was a nice constructive problem, and I don't know why I was getting WA on test 2.

    It would be really helpful if anyone can explain what does this means ? I got this on $$$D$$$

    My submission: https://codeforces.cc/contest/1682/submission/158075759

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      3 years ago, # ^ |
        Vote: I like it +6 Vote: I do not like it

      I guess there is a problem in amount of your edges, the compiler is trying to say you have to print another edge, but your code started to answer on the next test case, so you probably don't have enough edges printed. Sorry if I'm wrong

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        3 years ago, # ^ |
        ← Rev. 2   Vote: I like it +3 Vote: I do not like it

        Yeah, you were right, thanks for your help, I got one test case failing "100100100100", and now it got accepted :)

        It's a very sad thing that I wasn't able to find this basic test case during the contest..

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    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    strong pretests for most of the problems and very clarifying statements .thanks for the round.

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    3 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    Questions are very interesting. From B to D all of them involved a decent amount of thinking. Loved the round.

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    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I got TLE for using unordered_map instead of map in C... Bad contest :(

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    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    sorry @CoderAnshu, I was upvoting but my mouse sucks, I accidentally clicked on downvote, I am sorry for it. hope you don't mind it

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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

LinearForces :-)

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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

C felt kinda tricky.

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3 years ago, # |
  Vote: I like it +3 Vote: I do not like it

Why does the number of solvers of problem C go so high?

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3 years ago, # |
  Vote: I like it +16 Vote: I do not like it

I can solve D quickly, but I can't handle A,B,C.It's weird.

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    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    how to solve D?

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    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    god damn stupid greedy thought always messes up my first three problems. Well it sometimes works, but the moments it doesn't work really affect my mind.

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3 years ago, # |
  Vote: I like it +32 Vote: I do not like it

amazing samples guys

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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

What's the solution to D?

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    3 years ago, # ^ |
    ← Rev. 2   Vote: I like it +7 Vote: I do not like it

    I create egdes as a circle, find a edge to remove, then for each two node (x, y) I need to change the degree, I remove the edge of (y, y + 1) and add the edge (x, y + 1) 158057894

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      3 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      amazing simple solution! I have done a garbage code, by using link list and removing subgraphs of size 3 from boundary. In the end size 3 or 4 graph remains which is handled separately.

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    3 years ago, # ^ |
    ← Rev. 2   Vote: I like it +12 Vote: I do not like it

    If number of $$$1$$$'s is odd or there are no $$$1$$$'s at all the answer is "no", because the sum of degrees should be even (each edge is considered in $$$2$$$ degrees) and we should have at least $$$2$$$ leaves.

    To construct a solution, if the string is all $$$1$$$'s, just choose anyone of them to be the root and connect it to all the other nodes.

    Otherwise, connect each $$$0$$$ to the node before it ($$$0$$$ at node $$$1$$$ is connected to node $$$n$$$). Now we have chains starting with $$$1$$$ and ending with $$$0$$$ (call such endpoints "tails"). To make all the nodes in these chains valid, we have to connect odd number of edges to the tails.

    To do so, if all the $$$1$$$'s are consumed in the chains, just connect the tail of any chain to the tails of all the other chains. Otherwise, choose any unconsumed $$$1$$$ and connect it to all the chain tails and to any other unconsumed $$$1$$$.

    Such construction guarantees that no intersections occur, as we first add some edges between adjacent nodes (can't be intersected) and at the end we choose only one node and connect it to some other nodes.

    Submission

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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

can anyone explain that how to solce c :*

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    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I just binary searched the answer. You can in linear time check whether an answer X is possible. Although I have to say it was tricky to implement, two WA submissions and like 30 minutes debugging for edge cases

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    3 years ago, # ^ |
      Vote: I like it +7 Vote: I do not like it

    All numbers that appeared two or more times can appear in LIS of both original and reversed array. Then we distribute all numbers that appeared only once evenly in LIS of original array and reversed array. Note that we can share one number in both LISs, so the answer is x + (y + 1) / 2 where x is the number of numbers that appeared two or more times, and y is number of numbers that appeared once.

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    3 years ago, # ^ |
    ← Rev. 3   Vote: I like it +4 Vote: I do not like it

    I just count how many number appears 2 times or more as m2, 1 time as m1 and the answer is m2 + (m1 + 1) / 2.

    Example: 2 2 3 4 5 6 6 -> 2 3 6 5 6 4 2

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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Spent near one hour on E and then realized I made the wrong observation. I think A-E are all great problems.

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3 years ago, # |
  Vote: I like it +17 Vote: I do not like it

Those who solved problem C with unordered_map, depending on the compiler version, fall on test 26 or 28. Test 28 is a my hack test :)

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3 years ago, # |
  Vote: I like it +8 Vote: I do not like it

I think Problem B was slightly confusing in my opinion.I am not able to understand B problem clearly.

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3 years ago, # |
  Vote: I like it +18 Vote: I do not like it

how to solve E?

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    3 years ago, # ^ |
    ← Rev. 2   Vote: I like it +3 Vote: I do not like it

    Build a graph with $$$(x_i, y_i)$$$ as edges. For each i, find path from $$$i$$$ to $$$a_i$$$. A pair $$$(x, y)$$$ can be swapped if the edge $$$(x, y)$$$ is the end of two paths. Edit: add code 158085317

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      3 years ago, # ^ |
        Vote: I like it +3 Vote: I do not like it

      By "...the end pf two paths" do you mean the path where all of it's subtree nodes are located in their correct position?

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        3 years ago, # ^ |
          Vote: I like it +4 Vote: I do not like it

        Sorry I don't get what you mean but I just realized that it may be beneficial to name an edge with its index rather than its two end points since vertices are moving. Let's say the path from $$$i$$$ to $$$a_i$$$ consists of edge $$$e_{b_1}, e_{b_2}, \dots, e_{b_n}$$$, and path from $$$j$$$ to $$$a_j$$$ consists of edge $$$e_{c_1}, e_{c_2}, \dots, e_{c_m}$$$. If $$$e_{b_n} = e_{c_m}$$$, we can swap that edge. Then remove that edge from the end of two two paths and look for another edge that is the end of some two paths.

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          3 years ago, # ^ |
            Vote: I like it +3 Vote: I do not like it

          I see. But how do you do it in linear time for checking i and j?

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3 years ago, # |
  Vote: I like it +52 Vote: I do not like it

Great round! Especially enjoyed E and F.

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3 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

i literally did c in 8 minutes by using same approach as in editorial but was not confident enough and waited for it to fail main tests. but it passed :) edit: people made it hard just because it is problem "C". conclusion: don't judge a book by its cover

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    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    that's so true, i suppose C can be easier than B for those who have no idea what bitwise is

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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Great problems, although didn’t solve E.

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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Why unordered map gives tle on codeforces?

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    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Hash collision resulting in O(n) per access?

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3 years ago, # |
← Rev. 3   Vote: I like it +12 Vote: I do not like it

is CF getting harder or people are getting so much better? Feels like an 1800 rate problem a few months ago would be just 1600 now. So many solved C

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3 years ago, # |
  Vote: I like it -10 Vote: I do not like it

bullshit pretest for C

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    3 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    Why would you use an unordered map when it is quite easy to exploit the hash collisions for it lol

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    3 years ago, # ^ |
      Vote: I like it +12 Vote: I do not like it

    Congratulations!! You've reached the time which every coder reaches where he switches from hash map to map

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3 years ago, # |
  Vote: I like it +2 Vote: I do not like it

Ratings updated preliminarily. We will remove cheaters and update the ratings again soon!

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3 years ago, # |
  Vote: I like it +20 Vote: I do not like it

Here is some feedback, loved the round overall! I think BCD in particular are excellent.

A
B
C
D
E
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3 years ago, # |
← Rev. 5   Vote: I like it -7 Vote: I do not like it

Is it just me or do these submissions look so similar? - https://codeforces.me/contest/1682/submission/158041174 - https://codeforces.me/contest/1682/submission/158063005 - https://codeforces.me/contest/1682/submission/158043928 - https://codeforces.me/contest/1682/submission/158045408

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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

158034396

why TLE?

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    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Because you use unordered_map

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      3 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Doesn't it satisfy time constraints(nlogn)? I have used it many times. What else should we use then here? (sorting would have also taken nlogn)

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        3 years ago, # ^ |
          Vote: I like it +5 Vote: I do not like it

        unordered_map for int can degenerate to O(n) for lookup/insert (map is guaranteed O(log n)).

        But yes, sorting and then linear pass will work too.

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    3 years ago, # ^ |
    ← Rev. 2   Vote: I like it +5 Vote: I do not like it

    you can read this blog for efficient using of unordered_map

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3 years ago, # |
  Vote: I like it +6 Vote: I do not like it

Nice round, loved the problems.

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3 years ago, # |
← Rev. 2   Vote: I like it +4 Vote: I do not like it

Why the guy's submission is so werid? https://codeforces.me/contest/1682/submission/158020409

update: also this guy's submission: https://codeforces.me/contest/1682/submission/158055059

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    3 years ago, # ^ |
    ← Rev. 3   Vote: I like it +16 Vote: I do not like it

    I think in the past several months, they have found a way to trick the CF cheat detectors. That's why there are so many ACs for easy to medium problems now.

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3 years ago, # |
  Vote: I like it +4 Vote: I do not like it

such a poor round wasted 2 hours

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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

A should have been little easier :( I wasted more than 20 mins thinking there should be easier solution for A by having right one in my mind. Lesson learnt: Solve what comes to your mind first for A atleast

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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

In 3rd Problem, If the array a = [1,2,2] , Then according to codeforces, correct answer is 2. Can anyone explain how is it possible. My Claim : Answer should be 1 and not 2 since it is asked "strictly increasing".

Thanks in Advance !!!

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    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    2 1 2

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    3 years ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    a = 2 1 2 a' = 2 1 2 (reversed)

    second and third element form LIS edit: Boy but you got it in the Contest? Hopefully it's only due to luck

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      3 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Hi,I also have similar doubt, we also need to check that the singleton element which we are considering to add in both the sequence should be either minimum or the maximum element otherwise we won't be able to add it in both the sequences.

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        3 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        a = 1 3 2 3 1 (LIS)

        a = 1 3 2 3 1 (LDS)

        2 is neither max nor min

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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I don´t understand the problem B, why is {0,2,1} = 0?

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    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    swaping 1 and 2 is sufficient ans = 1&2 = 0

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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Thanks for super fast editorial.

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3 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

Good Round, I loved thinking about problem C, it was simple but tricky. going to remember it always. A great Round from an Indian problem setter.

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3 years ago, # |
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for problem :c

test case :
5
1 1 2 3 3

the given answer is 3,could some one please explain how it can be 3.

i am getting answer as 2

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3 years ago, # |
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I have some doubt in problem C.

For this test case

14

150482826 299981842 846268930 299981842 16711396 299981842 160038184 87399809 36731648 410799926 150482826 16711396 410799926 355178268

answer is 7, but my solution is outputting 6, may I know the problem here?

(My approach is to build a sequence greedily by sorting the array and then taking minimum of them and then placing rest of the numbers on left and right of that minimum found (leftMost after sorting) (Counting them just), so that I get a strictly increasing sequence even if reverse, numbers that come twice are put in either of the sides.)

Then I output the final answer as min(leftSum, rightSum) + 1

I am counting these leftSum and rightSum in the above approach.

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    3 years ago, # ^ |
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    You are almost correct, but instead of always taking the minimum as the one being used by both the "leftSum" and the "rightSum", you can choose any of the number to be shared by both sequence. An example of this is "5 3 4 5 3" where both the left sequence and the right sequence used "4" in the LIS. Hope this helps :)

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3 years ago, # |
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Thanks for this wonderful round!

I am finally a CM.

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3 years ago, # |
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in problem B, with the input:

1

8

0 1 2 7 4 5 3 6

it seems like the answer is not correct, if im wrong, pls point out my mistakes.

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    3 years ago, # ^ |
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    in problem statement it is clearly mentioned that value of X will always exist so this kind of test case not possible.

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3 years ago, # |
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It's Indian territory. love IIT CoderAnshu

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3 years ago, # |
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In problem D:

for input of: 1 4 1100

my submission: [submission:https://codeforces.me/contest/1682/submission/158117585] gives output of: YES 3 4 1 3 2 4 but the accepted answer is: YES 2 3 3 4 4 1

but I think my printed solution should also get accepted.. can someone please help me with this one. Thanks in Adv!!

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    3 years ago, # ^ |
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    (1, 3) and (2, 4) intersect internally in the circle.

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3 years ago, # |
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Please help me

My solution for problem E got RTE in test 8, but I don't know why; please tell me why

Thank you!

My submission: https://codeforces.me/contest/1682/submission/158149795

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