Shayan's blog

Codeforces Round 978 (Div. 2) — Video Tutorial
By Shayan, 7 weeks ago, In English

Note: The text editorials will be provided by the authors of the round. This video tutorial acts as an additional resource for those who prefer video over text, not as a substitute for the text editorial.

2022A — Bus to Pénjamo

Video

2022B — Kar Salesman

Video

2022C — Gerrymandering

Video

2022D1 — Asesino (Easy Version)

Video

2022E1 — Billetes MX (Easy Version)

Video

2022E2 — Billetes MX (Hard Version)

Video
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7 weeks ago, # |
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Second comment XDDD

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7 weeks ago, # |
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Nice contest,but didn't have chance to take part in it.

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7 weeks ago, # |
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Oh! nice. Contest was at 1AM IST. Most Cheaters from India could not have participated.

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7 weeks ago, # |
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Hard D2, but really amazing.

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7 weeks ago, # |
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Is the n limited to 69 people in D1, because it says the limit of n<=10^5?

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7 weeks ago, # |
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hey,can anyone provide me the problems of same type of b,c and d here?i would be grateful

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7 weeks ago, # |
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B was a very bad problem.

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    6 weeks ago, # ^ |
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    So true. It seemed like only those who were already aware of the greedy algo could solve it.

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7 weeks ago, # |
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To all those who are downvoting this blog, please understand that this is not the official editorial and this guy had nothing to do with the preparation of the problems to be used for the round.

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7 weeks ago, # |
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any ideas for D2 anyone ?

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7 weeks ago, # |
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For solving D2, you gotta make three key observations:

Spoiler
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    7 weeks ago, # ^ |
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    Dang, I was so close to getting it :(. Thanks for the solution though!

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    7 weeks ago, # ^ |
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    How about n=3?

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      7 weeks ago, # ^ |
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      for n = 3, the minimum number of queries is actually 4. This is the only case where you have to ask > n queries. For every n > 3, you can solve it with atmost n queries.

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7 weeks ago, # |
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For B, I'm taking the max x elements from the array and choosing them (subtracting them by the minimum of this subarray), sorting the array again and then repeating until all elements are zero. It's failing at some 294th testcase, why?

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    7 weeks ago, # ^ |
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    Consider n = 3, x = 2 and the Elements {2, 2, 2}

    According to your approach you would remove (seconds, third), (seconds, third), (first), (first) giving you an answer of 4.

    Whereas you can pair up (first, second), (first, third) and (second, third) giving you ans answer of 3.

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      7 weeks ago, # ^ |
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      So what actually is the issue in the logic?

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        7 weeks ago, # ^ |
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        Issue is, you do not want to make a number 0 when you can decrement other numbers. This is done to ensure that you keep as many different models as possible to choose from. The brute force logic will be to keep sorting the array, taking largest x (or all if count is less than x) non-zero elements of the array and decrement them by 1. This needs to be repeated until all elements become 0. Obviously this will give TLE as array elements can go upto 10^9.

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        7 weeks ago, # ^ |
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        Issue is that the pairing logic its in the update rule you are taking maximum X elements and then removing minimum of these each step.

        You logic works like this for {2,2,2}

        Take {2,2} and make it {0, 0}

        Now we have {2, 0,0} then -> {1, 0, 0} then -> {0, 0, 0}

        But if we look at one decrement at a time take {2,2} in the first pairing decrease it by 1, {1,1} now the maximum 2 elements are {2,1} instead of {1,1} which will end up giving the right answer.

        Your logic implicitly assume that the maximum elements don't change during the min(max k elements) operations, but if you work this out as shown above then you can see that they do change. So you would have to change your algorithm to account for that.

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      6 weeks ago, # ^ |
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      got it

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7 weeks ago, # |
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I thought for a few hours and believe that I've got a perfect proof for problem B. But the comment section is too small and I can't write it down.

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    7 weeks ago, # ^ |
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    I also want know about your proof!.

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      7 weeks ago, # ^ |
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      Hell. I discovered that my proof is almost identical to the video tutorial. But, anyway, the proof framework for this problem is very advanced. You should try to play with it a bit.

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      6 weeks ago, # ^ |
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      ??

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    7 weeks ago, # ^ |
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    Bro thinks he's Pierre de Fermat lol.

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7 weeks ago, # |
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indeed a good contest and problem c was nice. cheater count was significanlty less, maybe because most of the cheaters were sleeping

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7 weeks ago, # |
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Got stuck at B, luckily I solved D1 very quickly so I didn't lose points. At least I learned something new, cool problems!

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7 weeks ago, # |
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Shayan Why do we need to remove edge in E2 ?

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7 weeks ago, # |
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People are saying that problem B is quite a famous and standard prob. Can someone suggest some sources where I can make myself familiar with more of these famous and standard problems? Thanks!

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    7 weeks ago, # ^ |
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    its just greedy, We just get the maximum as it should be different cars, then divide by the given limit X, add 1 if remainder. I dont think its even a standard problem. If so, i would also like to know how.

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      7 weeks ago, # ^ |
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      Watch Shayan's Stream. He says that it is a famous prob. Many in the comments were also saying the same.

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7 weeks ago, # |
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Where is text editorial :(

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7 weeks ago, # |
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nice problemset, thankyou

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7 weeks ago, # |
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So，how can I see the text version of the solution？

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7 weeks ago, # |
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did author forget about the text tutorial?

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7 weeks ago, # |
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D2 is really interesting I think though I didn't participate

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6 weeks ago, # |
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I am not unable to understand how the solution of question B works :(

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