AmShZ's blog

By AmShZ, 3 weeks ago, In English

Hi Codeforces!

Keshi, alireza_kaviani and I are delighted to invite you to participate in Codeforces Round #800 (Div. 1) and Codeforces Round #800 (Div. 2).

  • Start time: Jun/16/2022 17:35 (Moscow time)
  • Duration: $$$120$$$ minutes
  • Number of Tasks: $$$6$$$ for both divisions
  • Rated range: ($$$-\infty$$$,$$$1899$$$] for Div2, [$$$1900$$$, $$$\infty$$$) for Div1

We are honored to have set the 800th Codeforces round. We wish this wonderful platform all the best along with many other exciting rounds.

Huge thanks to the following people:

Thanks to NEAR for supporting this round, details can be found in this post.

We have worked hard to keep the statements clean and the pretests strong!

Please read all of the problems and their notes, enjoy your time and solve as many as you can! Good luck have fun to everyone!

The scoring distributions will be announced later.

UPD: Here are the scoring distributions:

  • Div. 1: 750 1000 1500 2250 2500 3000

  • Div. 2: 500 1000 1500 1750 2250 3000

UPD: Editorial is out!

 
 
 
 
  • Vote: I like it
  • +796
  • Vote: I do not like it

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3 weeks ago, # |
  Vote: I like it +61 Vote: I do not like it

Great contest! you will enjoy it.

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    3 weeks ago, # ^ |
      Vote: I like it -14 Vote: I do not like it

    BitmaskForces ?

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    3 weeks ago, # ^ |
    Rev. 2   Vote: I like it -37 Vote: I do not like it

    Good luck everyone...Hope everyone performs well :)

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    3 weeks ago, # ^ |
    Rev. 2   Vote: I like it -35 Vote: I do not like it

    lol .. He said Huge thanks to testers for providing invaluable feedback.

    If you dont believe me scroll and check.

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

      invaluable != not valuable

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      3 weeks ago, # ^ |
        Vote: I like it +14 Vote: I do not like it

      invaluable is actually of stronger positive meaning than valuable, its easier when u understand it as "unable to be valued as it is too precious to be"

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3 weeks ago, # |
  Vote: I like it -9 Vote: I do not like it

Orz

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

Yet another great Iranian round :) orz

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3 weeks ago, # |
  Vote: I like it +29 Vote: I do not like it

Hope everyone can be red in this contest (of course include me).

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

btw congrats to alireza_kaviani for qualifying in Iran's IOI 2022 team

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3 weeks ago, # |
  Vote: I like it +23 Vote: I do not like it

Codeforces orz
MikeMirzayanov orz
Polygon orz

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

Orz

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

Paranoid creepy tasks?

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    3 weeks ago, # ^ |
      Vote: I like it +36 Vote: I do not like it

    Don't worry, no surprises.

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      3 weeks ago, # ^ |
        Vote: I like it -21 Vote: I do not like it

      Don't worry, there will be no more than $$$10^9+7$$$ suprises.

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3 weeks ago, # |
  Vote: I like it +64 Vote: I do not like it

As a tester, problems are great and statements are short.

I encourage you to participate in this round!

You will definitely enjoy this round!

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

    Haha certainly....but sometimes short ones are the real fire ones...looking forward for this round....

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3 weeks ago, # |
Rev. 3   Vote: I like it +7 Vote: I do not like it

Where is translation into russian?

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

Another div1 with trygub support? It will be fine.

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

Hope all the best for everyone

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3 weeks ago, # |
  Vote: I like it +50 Vote: I do not like it

Finally, my first Div 1 contest. Yayyyy. Every time I used to fall back to expert just before Div 1 contests.

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

Good luck for everyone!!!

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

good luck for everyone

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3 weeks ago, # |
Rev. 4   Vote: I like it -29 Vote: I do not like it

good luck man!

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3 weeks ago, # |
Rev. 4   Vote: I like it 0 Vote: I do not like it

Good luck !?

Naah,it won't help :(

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

800th round! A great number

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

WOW! We have crossed 700. ٩ (◕‿◕。) ۶

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3 weeks ago, # |
Rev. 4   Vote: I like it -17 Vote: I do not like it

No hacking ??

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3 weeks ago, # |
  Vote: I like it -13 Vote: I do not like it

All the best everyone!

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

My first time participating in Div. 1, I am so nervous.

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

Wow, great!! 800 contests, its huge!! Thank you Codeforces❤️

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

We hope that 800th Codeforces round will be more interesting for all of us.

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

Hope for the +ve delta in this round

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3 weeks ago, # |
  Vote: I like it -13 Vote: I do not like it

Will this contest be rated?

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

    Rated range: (−∞,1899] for Div2, [1900, ∞) for Div1

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

I know I won't be able to solve any one of em, still gonna participate..

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

GAP

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

Hope I don't go back to being a pupil :/

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3 weeks ago, # |
Rev. 2   Vote: I like it +69 Vote: I do not like it

When someone says orz as a comment
68e3682b065c5b4a66b64c2e78865293

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

tell newbie what is orz

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

is it contest?

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

orz

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3 weeks ago, # |
  Vote: I like it -32 Vote: I do not like it

Benq When are we seeing you in the action?

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3 weeks ago, # |
  Vote: I like it +33 Vote: I do not like it

I guarantee you will enjoy the contest.

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3 weeks ago, # |
Rev. 3   Vote: I like it -26 Vote: I do not like it

nice

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3 weeks ago, # |
  Vote: I like it +23 Vote: I do not like it

I hope I don't lose points in this game.

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3 weeks ago, # |
  Vote: I like it +33 Vote: I do not like it

No offence but why does authors tend to hide the score distribution and reveal it just before the contest starts?

strategy?
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3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Lmao, what does +inf & -inf rating mean !!

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

    infinity, quite literally. If you did not know, negative ratings exist and they are not rated in most rounds (most rounds' rated range begin from 0)

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3 weeks ago, # |
Rev. 2   Vote: I like it +6 Vote: I do not like it

[1900, tourist] for Div1* :v

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

I hope that this contest is nice for all.

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

I hope we get to the max rating after this round ^_^

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

Good luck for everyone!!!

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3 weeks ago, # |
  Vote: I like it -8 Vote: I do not like it

Why only two div2 testers?

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3 weeks ago, # |
Rev. 4   Vote: I like it 0 Vote: I do not like it

Eight 100!!!

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3 weeks ago, # |
  Vote: I like it +65 Vote: I do not like it

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

Why so many orz comments? Could someone explain, please?

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

RIP Score distributions.

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

When will be the score distribution update? It's just 1 hour 12 minute remaining.

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3 weeks ago, # |
  Vote: I like it +15 Vote: I do not like it
waiting for score distribution be like
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3 weeks ago, # |
  Vote: I like it +4 Vote: I do not like it

My first contest!!

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

Not a Doctor Still doing a lot of operations.

Won't say bad round. I liked the Problems though.

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3 weeks ago, # |
Rev. 2   Vote: I like it -54 Vote: I do not like it

I think that this contest isn't good. I can't understand the problems well.

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    3 weeks ago, # ^ |
      Vote: I like it -14 Vote: I do not like it

    Why so many downvotes !? It's my personal opinion !

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

      Because people didn't like your opinion. I mean that is how this system works. If people don't like your opinion they downvote you. if they like it then they upvote you.

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

    I agree with your opinion!

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

If this was a Div2 I am a refrigerator. tf.

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

I don't mind difficult problems but a,b,c are very observation based and took me a lot of time, hence we don't get enough time for interesting problems like D

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3 weeks ago, # |
  Vote: I like it +35 Vote: I do not like it

I dont like this round (my opinion)

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

    I respect the work of autors. Its very cool to do problems for codeforces round. I dont like this types problem. Thats what I meant.

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

I guess I am just dumb.

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

Round #800 CLASSIC GREEDY ONE !!!!!!

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

I didn't expect adhoc tasks :(

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

I liked problem C. Sample cases were very useful for solving.

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

    I hated it for that reason, but problem was very good . They should'n have provided the 5th sample it was easy to guess then.

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3 weeks ago, # |
  Vote: I like it -30 Vote: I do not like it

The comment is hidden because of too negative feedback, click here to view it

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3 weeks ago, # |
  Vote: I like it +37 Vote: I do not like it

How to solve Div 1C?

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    3 weeks ago, # ^ |
    Rev. 4   Vote: I like it +25 Vote: I do not like it

    Reverse the graph and then do something very similar to dijstrak starting from n-1. Notice that at some node, you have a bunch of nodes you can go to and if you know their distances, you can kill some in descending order(so like their effective distance is +0 +1, +2 +3....) and find the optimum of that. Thus, what you can do is to store the current estimates incoming (after reversing) for each node and pick the best (similar to dijstrak) after adding some weights. Then, we process the nodes by current estimate similar to dijstrak. But this may potentially force you to reupdate the whole set everytime you process a node. But, you realise that you only add numbers greater than numbers you add before (since you process nodes in ascending estimates). So, you can just store the best estimate.

    Essentially its just do dijstrak with the following modification: dist[v] = min(dist[v], dist[u] + indeg[v]) indeg[v]--

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

For problem D: if the sequence is decinc, there are only at most $$$3$$$ representations of it, that are of interest. For maximum increasing subsequence, you must use all elements except maybe one and if you don't use an element in the increasing sequence, it's always either first or last element.

Note that if the sequence is decinc, then so is all of its subsequences, hence you could kind of make a binary search for each $$$L$$$ on the largest $$$R$$$ such that $$$[L; R)$$$ is decinc.

But my solution failed pretests because it uses binary lifting and needs $$$O(n \log n)$$$ memory. I'm pissed off it doesn't pass.

Now that I think of it, you can probably also do it in $$$O(n)$$$ with two pointers?

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3 weeks ago, # |
  Vote: I like it +58 Vote: I do not like it

I've seen a version of Div1D with answering for whole sequence in a few Polish competitions, so the key observation (that we can build the answer greedily) was known to me (I'm not saying that the rest is correct, last rounds weren't generous for me).

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    3 weeks ago, # ^ |
      Vote: I like it +65 Vote: I do not like it

    I can write a bit more: assume that we've split the prefix into two sequences (decreasing and increasing) and the next two elements are $$$x$$$ and $$$y$$$, where $$$x$$$ can be appended both to increasing and to decreasing sequence. It turns out, that if $$$x<y$$$, it's optimal to append $$$x$$$ to increasing one. If $$$x>y$$$, it's optimal to append $$$x$$$ to decreasing one.

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    3 weeks ago, # ^ |
      Vote: I like it -219 Vote: I do not like it

    No need to show off commenting every time you get first place, big head. Hopefully you FST again and get another -100 :)

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      3 weeks ago, # ^ |
        Vote: I like it +46 Vote: I do not like it

      Yeah, nobody was interested in the information that subproblem has already appeared somewhere, mb

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

Approach for Div2C??

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

    Beginning from last non-zero element, iterating backwards, keep adding arr[i], sum must never be >=0, except at Oth element. That was the opeartion about.

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

      i was iterating from backwards and decreasing the current element say a[j](initial array) by k such that it gets equal to m[j](given array) and adding k-1 to a[j-1] lastly i check that array is equal or not alongwith my pointer is on the first element or not this approach is giving wrong answer

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

    let j = position of last non-zero number (1-indexed)

    There must be no prefix sum of length i(0<i<j) with sum <= 0

    A[j] must be negative and total sum of array A = 0

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3 weeks ago, # |
  Vote: I like it +24 Vote: I do not like it

Div2B was hard -_-

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

Can anyone make me understand the div2 problem A statement , please, I am really confused how the define prefix here and minimum possible creepiness

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

    Yeah problem statement was not clear But prefix means abs(count(1s) — count(0s)) till that index eg: 0 1 0 0 1 pref 0 -> has 1 0s and 0 1s so its abs(1-0) — 0 Similarly till 1 to 4 i.e index 0 to 3 more formally 1's count = 1 and 0s count =3 so absdiff = (1-3) = 2 But since we need to minimize creepiness we need to do alternate 1s and 0s so they can cancel each other at certain prefix and abs diff = 0

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

    for every prefix there will be a creepiness. out of all the creepiness of every prefix, we will have to take the maximum creepiness. now this maximum creepiness should be minimum.

    form the testcase a=3 and b=7. this means there are 3 0's and 7 1's

    there can be multiple answer to this, but lets take the testcase answer: 0001111

    now the prefixes for there are: (beside them i will write the creepiness(c)) [c= abs(a-b)]

    0 : a=1, b=0, c=1 00 : a=2, b=0, c=2 000 : a=3, b=0, c=3 0001 : a=3, b=1, c=2 00011 : a=3, b=2, c=1 000111 : a=3, b=3, c=0 0001111 : a=3, b=4, c=1

    out of all the prefixes, the max creepiness is 3.

    in this example, no matter how you arrange the 0's and 1's, you maximum creepiness will never be less than 3.

    but this not how you should arrange your 0 and 1. You should first alternatively arrange them and then if there are some left from either just 1 or 0, just print all the remaining of them.

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

I'll suffer from PTSD from now on for string problems :(

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

what's the solution for div1A? I have no idea why I got AC

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

    Hey, Can I ask you, As a div 1 participant how was Div1B?

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

      actually pretty simple problem on dp on trees

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

        Can you please explain your solution, or suggest some similar dp on trees problems.

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

    I don't know if i am eligible to explain something to CM, but this is how I solved.

    • First I observed that if the first element = x, then we go right x times from first element.
    • If we go x times right from first then we must go left x times from second element.
    • since the second element = a[1] = ( number of rights from second — number of lefts from second). So we get right_times[1] = a[1] + x.
    • Similary construct for all indices. If any of them is negative then its not possible since we cant go negative number of times in some direction.
    • This is my code

    PS: We also need to check if at some point right[i] becomes zero then all the array elements from that index must be zero, since we dont move towards right anymore.

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

      oh, I understood, thanks! not bad problem though

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

tox1c_kid got trolled

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

It's funny that I solved C faster than B and solved B faster than A :D

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3 weeks ago, # |
Rev. 4   Vote: I like it 0 Vote: I do not like it

first time solving A,B,C and D, even though I toke 14 mins in problem A, I hope I get specialist.

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

Do you have a small test case that fails this simple and wrong solution for 1E?

  • For each element, count how distinct values on the left that the smaller than the element
  • For each element, count how distinct values on the right that the smaller than the element
  • Take the sum of the minimum of the two counts for each element
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    3 weeks ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    I also had the same idea, but here is the problem: It might be better to take the right maximum and then the left maximum afterwards.

    test case:

    5

    4 3 5 1 2

    The min for 4, 3, 1, 2 is clearly 1. But the min for the 5 would be 3. However, we can take right maximum and then left: 5 -> 2 -> 0. So answer is actually 1+1+2+1+1 = 6 instead of 7.

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

My C shouldn't have passed... caught the countercase too late though, so FST likely.

B... doable, but I forgot I was counting bad spans which can just extend to the beginning because they're already bad.

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

    edit: facepalm for not knowing how my solve was correct...

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3 weeks ago, # |
  Vote: I like it +172 Vote: I do not like it

1693C - Keshi in Search of AmShZ is the problem of the year.

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    3 weeks ago, # ^ |
      Vote: I like it +39 Vote: I do not like it

    Disagree, C was standard imo, for example 102341H - Hypno is very similar, but a bit harder.

    D, E and F were rather nice, a bit of AtCoderForces spirit, but still way better than most AGCs.

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      3 weeks ago, # ^ |
        Vote: I like it +29 Vote: I do not like it

      My bad, I had never seen anything similar to C.

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        3 weeks ago, # ^ |
          Vote: I like it +49 Vote: I do not like it

        Nope, not your bad, it's just your opinion versus mine. The fact that for you it was good and interesting can only be a plus for the problem. :P

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      3 weeks ago, # ^ |
        Vote: I like it +26 Vote: I do not like it

      wow, it's an honor.

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

Problems were interesting.

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3 weeks ago, # |
Rev. 4   Vote: I like it 0 Vote: I do not like it

So hard..... I can not solve (Div 1. C)1693C - Keshi in Search of AmShZ, I thought topological sorting + DP may work, but what if the graph has a loop and is not a DAG?`

`

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    3 weeks ago, # ^ |
    Rev. 4   Vote: I like it 0 Vote: I do not like it

    you run a djikstra-like algorithm on the reversed graph. In each iteration, my solution has an invariant that the nodes that have been popped from the set have the smallest values of $$$dp[u]$$$, where $$$dp[u]$$$ is the minimal cost required if one starts from node $$$u$$$ and goes to node $$$n$$$. We remove edges only when we are at the node from which the edge emanates (goes out). Then you can see how a djikstra-like algorithm always maintains this invariant, and hence answer is $$$dp[1]$$$.

    details: 160892338

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

      I appreciate it, I think I do not understand the idea of djikstra totally, it is time to take my DS book up again.

      Sadly, I may get a down rank.

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3 weeks ago, # |
Rev. 2   Vote: I like it +62 Vote: I do not like it

D1C is amazing and suitable for commemorable 800th round, thanks!

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

I hope to reach expert next round :( :(

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3 weeks ago, # |
  Vote: I like it +37 Vote: I do not like it

rainboy orz on astonishing participation! He won't become a GM after this contest but he really deserved it. This strategy of solving the problems starting with the harder ones is really not that bad, especially for educational purposes.

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

The best round recently I think! Thank the author!

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

small feedback: A is kinda cool, B&C both too standart IMO

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

Not sure about the complexity of "dive" function in my solution to D: 160848374. If anybody wants to try to create the test were the complexity is worse, you're welcome (don't know if it's correct or not).

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

    I think I did the same thing, guessing that it would work (somehow it did). I would also like to see a proof of why this type of memoization runs in linear time.

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

      Yeah. I’m doing more stuff, but I can believe that the transition is correct.

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

I am able to solve only problem A.

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

    Glad to see your persistence, try our best to prepare for the next round!

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

    try upsolving at least B before looking at solution, its not difficult. I always try to solve at least one more problem after the contest

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

      I am able to solve b but in contest unable to solve it.

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3 weeks ago, # |
Rev. 3   Vote: I like it -11 Vote: I do not like it

see https://codeforces.com/blog/entry/103952?#comment-923384

UPD: why downvotes ? I simply removed my original post

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3 weeks ago, # |
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div 2 B was little tricky compared to c

may be c should have been swapped with b

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3 weeks ago, # |
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is "wrong answer jury had better answer" different from "wrong answer expected ,found"?

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    3 weeks ago, # ^ |
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    It's 2 different types of problems: "jury" thingy appears where you have to calculate a minimum or maximum answer for example and at some point there is a possibility to have a better answer than yours, but "expected" simply stands for a wrong answer.

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3 weeks ago, # |
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Thanks for the round, Div1 pC was really interesting!

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3 weeks ago, # |
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What was the sol for Div2 C?

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    3 weeks ago, # ^ |
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    start from right most and think of total left and right moves to get desired, and compare with the first value

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    3 weeks ago, # ^ |
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    I solved this problem using RBS (Regular Bracket Sequence) pattern. If you consider the positive numbers as the number of open brackets and the negative numbers as the number of closing brackets, you will see "If the total sequence is regular and it's closed then the answer is yes".

    So you will get Yes for ( ()()() ) or ((())) or ( ( () ((())) ) ). But if the bracket is like this ()()() then it's no because in this case you can't get back to the first index.

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    3 weeks ago, # ^ |
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    If at any index < last index prefix sum is <= 0 || total sum != 0 then ans is no

    here last index is index of last non zero element.

    Take care of few edge cases such as all zero elements

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3 weeks ago, # |
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Great contest. The problemset was amazing and so fun to solve. Enjoyed this contest so much.

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3 weeks ago, # |
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I think the problem setter is a Radiohead fan. Some of the problem title is similar to Radiohead's song title . Or its just a coincidence .

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3 weeks ago, # |
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very good and balanced round + interesting problems. Though I FSTed on div2c but was a silly mistake.

Overall thanks for the amazing round

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3 weeks ago, # |
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Enjoyable contest <3 ,, thanks author Fastest problem rating distribution .........

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3 weeks ago, # |
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How to solve Div2 D problem?

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

    do DFS on the tree.

    if the node children returned a number which is greater than or equal lv, just return the min of the returned value of the children and the rv

    if it is less than the lv, you need to make another operation to fix this issue, so you will increase sol by one, then you will need to return rv.

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      3 weeks ago, # ^ |
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      But how do you prove that the configurations doesn't matter?

      In the sense , say a node has value 5.. 5 can be written as [1,1,3] , [1,2,2] and so on. But in your solution , you just see the min(sum, Ri).

      Why does your solution work? Proof?

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        3 weeks ago, # ^ |
        Rev. 3   Vote: I like it +1 Vote: I do not like it

        I can't understand what you mean by configuration and why there are 3 numbers in your examples.

        I am not good at writing formal mathematical proofs, but I will try to proof it logically.

        The current node doesn't care how many nodes under it, it only cares about the maximum value they all need, because it is the maximum value I can give to the current node, I can decrease it if I want, but I cannot increase it without an additional operation.

        But notice that the sum of all the child nodes values may be greater than Rv, so I take the min of them

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3 weeks ago, # |
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Can anyone tell the testcase for which my submission is failing (Div2C):- Submission

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3 weeks ago, # |
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luck maatters much in life

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3 weeks ago, # |
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Finally going to be pupil !!!

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3 weeks ago, # |
Rev. 3   Vote: I like it 0 Vote: I do not like it

I want to prove that other than the similarity in the code in question A Also I got Ac before him/her in question B

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3 weeks ago, # |
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can someone tell me why i got tle'd in this submission https://codeforces.com/contest/1694/submission/160921292 Thank you

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3 weeks ago, # |
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For Problem A — Directional Increase , Can someone please check what edge case am i missing — https://codeforces.com/contest/1693/submission/161166279