Блог пользователя windva

Автор windva, история, 14 месяцев назад, По-английски

Hello, Codeforces!

I am pleased to invite you to participate in Codeforces Round 899 (Div. 2), which will be held on 25.09.2023 17:35 (Московское время).

All problems were written and prepared by me.

You will be given 5 problems, and one of them is divided into two subtasks. You will have 2 hours to solve them.

I would like to thank:

Score distribution: $$$500 - 1000 - 1500 - 2000 - (2000 + 2000)$$$

Good luck and enjoy the round!

UPD: Editorial is out.

UPD2: Congratulations to the winners!

Official Contestants

  1. Lycorisqwq
  2. zepto
  3. vjudge39
  4. Beelzebub_blue
  5. _t_W_

Official + Unofficial Contestants

  1. ksun48
  2. jiangly
  3. antontrygubO_o
  4. Sugar_fan
  5. KumaTachiRen
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14 месяцев назад, # |
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Auto comment: topic has been updated by windva (previous revision, new revision, compare).

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14 месяцев назад, # |
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As a tester,problems are awesome.Hope you enjoy it :)

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    14 месяцев назад, # ^ |
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    We hope that the contestants will not spend a lot of time just reading the meaning of the questions.

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14 месяцев назад, # |
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As a tester, I did not know that the contest would be scheduled this early. Well, that was quite early, I guess.

Anyways, the problemset was enjoyable. Everyone, good luck on positive $$$\Delta$$$, and I hope you enjoy the contest as well (regardless of the $$$\Delta$$$).

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14 месяцев назад, # |
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As a tester, the last problem is the best problem I've seen recently. Good Luck and Have Fun!

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14 месяцев назад, # |
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As a tester, windva is handsome, so please participate.

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    14 месяцев назад, # ^ |
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    Will there be a problem based on cuteness of windva ? ( i'm not sure just guessing)

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14 месяцев назад, # |
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As a tester, I personally did like problem D and E. Hope you do not give up early and have a good result

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14 месяцев назад, # |
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As a participant, windva is god. :blobaww:

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14 месяцев назад, # |
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As a tester, I really enjoyed testing this contest.

The problems are creative, and solving these problems require a variety of approaches. Each problem has its unique style, and will broaden your perspective to the next level. Plus, they are fun to solve!

I saw windva putting a tremendous amount of effort and time to run this round. Moreover, windva is really cute, so please do consider participating actively.

Big thanks to windva, for creating this beautiful set, allowing me to test it, and for being cute.

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14 месяцев назад, # |
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This should be hackable, but I'm too lazy to write a testcase against it.
224983900

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14 месяцев назад, # |
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Wish me luck I hope this contest is another step into turning gren

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14 месяцев назад, # |
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We are ready

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14 месяцев назад, # |
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Good luck everyone! Hope you get your color-up!

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14 месяцев назад, # |
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I hope this will be an unrated contest for me :)

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14 месяцев назад, # |
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As a participant, windva is 大. very good at 三行詩. :blobaww:

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14 месяцев назад, # |
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2 out of 3!

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14 месяцев назад, # |
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Hope to solve at least one problem :)

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14 месяцев назад, # |
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Can anyone please explain why when I read input this way it is not accepted : 225037625

but when I read it this way it is accepted : 225042785

I thought both of them the same and I even runned debug with the first test case for the first submission and it just works like how I expected

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    14 месяцев назад, # ^ |
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    in first case you are breaking the for loop and not taking the complete input which gets carried on to the next tc

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14 месяцев назад, # |
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I'm kinda new here, is this round rated?

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    14 месяцев назад, # ^ |
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    All codeforces rounds are rated, unless said otherwise in the announcement.

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14 месяцев назад, # |
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Hope that I can solve problem D :)

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14 месяцев назад, # |
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rated for the participants with rating lower than?

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    14 месяцев назад, # ^ |
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    The round is rated for participants of rating lower than 2100(Master).

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14 месяцев назад, # |
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how do I become a tester for some codeforces round?

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    14 месяцев назад, # ^ |
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    I invited some of my acquaintances to test the round, and the coordinator invited in more testers. I don't know exactly from what route he invited them.

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14 месяцев назад, # |
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Guys I won't be able to participate if I just didn't will that effect anything after I registered?

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    14 месяцев назад, # ^ |
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    If you don't submit anything during the contest, then your rating won't change. You can cancel your registration anytime before the contest starts tho

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14 месяцев назад, # |
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I hope everyone reaches their goal <3

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14 месяцев назад, # |
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I couldn't solve C, should I commit suicide?

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14 месяцев назад, # |
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Why dfs in D is so slow. I got TL int $$$O(30 \cdot n)$$$, changed $$$30$$$ to $$$21$$$ and it's still $$$2.7$$$ seconds...

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14 месяцев назад, # |
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I am going to break my 1666 rating :))

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14 месяцев назад, # |
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Any hints for C ?

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    14 месяцев назад, # ^ |
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    If you fix the leftmost element which you will remove, then what would be the answer?

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      14 месяцев назад, # ^ |
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      what an elegant solution you've coded!

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      14 месяцев назад, # ^ |
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      Hmm, but can't you then proceed to remove even more elements down the line? Or would that not be beneficial since there's no use in flipping the parity twice?

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    14 месяцев назад, # ^ |
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    Work backwards. Start from $$$i=n-1, ..., 0$$$. There are 4 cases: The $$$i$$$ location is odd or negative, and for each, it can be $$$\geq 0$$$ and $$$<0$$$. For each case, you can do a greedy decision.

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    14 месяцев назад, # ^ |
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    you can do DP. Check my solution.

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    14 месяцев назад, # ^ |
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    You can get answer by checking first two element of array(obviously n>=2).

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14 месяцев назад, # |
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Why the skill difference between D and E so high everytime. btw how to approach E anyone?

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14 месяцев назад, # |
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This second question that came, well in the third test case why was the output 6 and not 7? You could include 1 too in the set.

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    14 месяцев назад, # ^ |
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    But then the set would contain everything, which isn't allowed.

    "Find the maximum number of elements in an attainable $$$S$$$ such that $$$S \ne S_1 \cup S_2 \cup \dots \cup S_n$$$."

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14 месяцев назад, # |
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What is the logic for B? Thanks.

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    14 месяцев назад, # ^ |
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    Store all element in map that represent union of all, after that for each element of map see which sets include that element and remove that set. which map element results in lowest removal is answer. here's my solution --> https://codeforces.net/contest/1882/submission/225124593

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      14 месяцев назад, # ^ |
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      Can you please explain in more detail?

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        14 месяцев назад, # ^ |
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        mp is storing union of all(you can use set too) after that i am seeing that how many element will remain if i want to remove all sets which contain that element. which element removal gives most element remain is answer , temp map(m) is storing the current union.

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    14 месяцев назад, # ^ |
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    You have to exclude at least one Element from the answer, so fix this element and add all the sets that doesn't contains it.

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14 месяцев назад, # |
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How do you solve D? (Hints)

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    14 месяцев назад, # ^ |
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    tree dp / rerooting

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      14 месяцев назад, # ^ |
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      I can solve it with tree DP easily for one root, but what is this rerooting trick you mention? Any links?

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        14 месяцев назад, # ^ |
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        If you know $$$dp[u]$$$ and there is an edge $$$(u,v)$$$, you can find $$$dp[v]$$$ in $$$O(1)$$$ https://usaco.guide/gold/all-roots?lang=cpp

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        14 месяцев назад, # ^ |
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        Think about how dp values change if you change the root to an adjacent vertex (old and new roots are connected by an edge). You'll notice that only the dp values of the old and new roots change, and this change can be calculated in $$$O(1)$$$. Now, you can just do a simple dfs traversal of the tree and always move the root when going up or down. All vertices will be the root at least once during this, and you can store those answers in a separate array.

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        14 месяцев назад, # ^ |
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        If you have the formulae to calculate dp for any parent from it's children then you can also use the same formulae to reroot the tree to the child from it's parent. Keep doing it recursively in dfs and calculate answer for every node

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14 месяцев назад, # |
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A ok

B Unimaginably bad

C Unimaginably bad

D Unimaginably bad

E1 Unimaginably bad

E2 out of my skill no comments

total: 6.9/100 (fail)

abort_dodger_3000

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    14 месяцев назад, # ^ |
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    For me, it's like

    A ok

    B out of my skill no comments

    C out of my skill no comments

    D out of my skill no comments

    E1 out of my skill no comments

    E2 out of my skill no comments

    total: 93.39/100

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14 месяцев назад, # |
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Quite similar problem to problem D : 1187E - Tree Painting

It's basically the same idea

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14 месяцев назад, # |
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B's pretest were pretty weak, probably gonna fst a lot

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14 месяцев назад, # |
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Why TLE? It is $$$O(n)$$$

Submission: 225143030

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    14 месяцев назад, # ^ |
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    It seems like O(20n). We allowed fast O(20n) solution, but the intended solution is actually O(n) and not guarantee slow O(20n) solution to get AC

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      14 месяцев назад, # ^ |
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      Are $$$3$$$ seconds not enough when $$$n = 2 \cdot 10^5$$$? It is C++!

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      14 месяцев назад, # ^ |
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      I don't think it really makes sense to allow a slower solution only if it's implemented efficiently. I think you should've either allowed the slower $$$O(20n)$$$ to pass easily or changed to like $$$a_i < 2^{60}$$$ and a 2 second TL to only allow the fast $$$O(n)$$$.

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        14 месяцев назад, # ^ |
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        I guess it was quite hard because diverse languages (ex. Kotlin, python) have to be considered. We had to guarantee O(n) solution implemented by Kotlin to get AC

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        14 месяцев назад, # ^ |
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        I agree with your point.

        However, we had to consider slow languages, so had to set TL at least 3s. And due to this, we intended to pass O(20n) instead, because we thought it would be impossible to block O(nloga) in TL 3s even if we increase the constraints. In fact, there were O(20n) codes which worked in nearly 1 sec so I thought it wouldn't depend on the subtle difference in code's efficiency.

        Thanks for your feedback :)

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          14 месяцев назад, # ^ |
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          Usually, a constant of $$$20$$$ is too small to distinguish between two solutions. So it was the right decision not to block $$$20n$$$. But maybe it would be better to increase the TL? I don't think it's easy to squeeze $$$O(n^2)$$$ into this problem in any reasonable time, even on C++.

          By the way, thanks for problem C, I really enjoyed it!

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    14 месяцев назад, # ^ |
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    You can check my O(20n) solution. which works at 1.3-1.4 seconds. 225156423

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14 месяцев назад, # |
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A: We can solve by greedy: Let b[i] be the minimum it can be, which means, if a[i]==b[i-1]+1 then b[i]=b[i-1]+2, otherwise b[i]=b[i-1]+1.

B: For each $$$t \in \bigcup_{i=1}^{n} S_i$$$, assume $$$t \notin S$$$, we can let $$$S = \bigcup_{t \notin S_i}^{} S_i$$$, and take it as a candidate of answer.

C: Observe that if we removed the i-th card, then for j>i we can get max(0, a[j]) score by appropriate order of operations. Assume the most top card we removed is i0-th card, then the answer will be a[i0]*[i0%2==1] + sum(j>i0)(max(0, a[j])).

details

D: When u is the root, it's useless to do operation on u, and for each other node v, we need to cost (a[parent[v]]^a[v])*size[v] to make a[parent[v]]==a[v]. Therefore, if we let 1 be the initial root of the tree, and u is the parent of v when root is 1, then node v will contribute (a[u]^a[v])*(n-size[v]) to the answer of nodes in subtree of v, and contribute (a[u]^a[v])*size[v] to the answer of other nodes.

E1: We can use 3 operations to swap two numbers in the permutation: [I]*s*[II]t[III] --> [II]*t*[III]s[I] --> [III]*s*[I]t[II] --> [I]t[II]s[III], and do 2 opartions on 1, n will cause no change. Therefore, we can build operation sequences for p[i] and q[i] separately, and if the parity of lengths of them is the same, we can fill (1, n) to operation sequence of p[i] if it's shorter, or fill (1, m) to operation sequence of q[i]. Otherwise, if n is odd, we can add n operations on 1 for p[i], similar for m is odd. If both n and m are even, the answer doesn't exist.

Why answer doesn't exist
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    14 месяцев назад, # ^ |
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    How did you arrive at the answer for D?

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      14 месяцев назад, # ^ |
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      since doing xor on a whole subtree is kind of like doing suffix xor on an array, you can consider do something like difference array, that is, make every edge $$$(u, v)$$$ have a weight of $$$w[u] \text{^} w[v]$$$, then the weight of $$$v$$$ is indeed the xor of path from root to $$$v$$$(except the root, but it is not important), and the problem become "Make all edge have weight $$$0$$$, in one operation you can make an edge's weight xor with $$$c$$$", then it is obvious that xor edge with it's weight is optimal.

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    14 месяцев назад, # ^ |
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    When u is the root,it's useless to do operation on u

    Can you explain why?

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      14 месяцев назад, # ^ |
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      My understanding was as follows: If you XOR the root and a certain set of bits get flipped in root's value, then those bits will get flipped for the entire subtree (since operation is on the entire subtree, not just root). So, essentially the problem remains identical just with the flipped bits i.e. you want to match those bits in the root's value with values in its subtree using the XOR operation on the subtree.

      Consider an example with only 1-bit numbers: If root's value is 1. And its subtree has node1=0, node2=1, node3=0, node4=0, node5=1. Then, if you apply operation on the subtree rooted at the root itself, these numbers will become root=0 and subtree will have node1=1, node2=0, node3=1, node4=1, node5=0. Clearly, the nodes in subtree which had different value compared to root still remained the same before and after the operation.

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    14 месяцев назад, # ^ |
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    For D[problem:1882D], the conclusion that it is useless to do operation on the root (to minimise cost) seems obvious.

    However, in your statement/solution, are you assuming that for a node v to become same value as root, the value of node (v) has to (at some point) become equal to the original value of its immediate parent after a sequence of operations? If so, how can we prove it for optimal cost?

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14 месяцев назад, # |
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Tried DFS and Rerooting for each bit in D, got TLE, shouldn't 20 * n pass?

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    14 месяцев назад, # ^ |
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    No need to do it for each bit. If you want to make a whole subtree equal then you need to make all children of the root equal to root otherwise its not possible. There is no way we can change the root value to be equal to the value of the child without changing the child.

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    14 месяцев назад, # ^ |
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    Please take a look at this comment

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14 месяцев назад, # |
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The contest is maybe very good But it DID NOT FIT ME. :( problem D is DP. I am not good at DP at all:( especially on tree :(:( E1 WA#10, mistake, didn't pass pretest until contest over. :(

problem C is fun. feel good over A B C.

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    14 месяцев назад, # ^ |
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    How did you reach purple without a stronghold on dp?

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      14 месяцев назад, # ^ |
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      Dp on trees is more annoying than normal dp. I struggled with the code a bit too

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14 месяцев назад, # |
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For Problem C, DP was way straight forward than the greedy method for me. DP Solution

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    14 месяцев назад, # ^ |
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    Thanks for this, I really need to get better at dp lol-

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    14 месяцев назад, # ^ |
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    This is the method I used as well in testing. In fact I did not know there is a greedy solution until today.

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    14 месяцев назад, # ^ |
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    I tried the same approach but didn't bass the example.

    It's much easier to think about Dp solution for problems like this than to think about a greedy solution -$$$ at $$$ $$$least$$$ $$$for$$$ $$$me$$$-.

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    14 месяцев назад, # ^ |
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    In your solution, can you explain the calc(v,i+2,1)+v[i] case when the parity is odd?. I was thinking on the exact same grounds but cannot understand why this is needed.

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      14 месяцев назад, # ^ |
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      Let's say we are on an odd position i now. If we take this value right now and if there exists another value at position i+2 it will become even after the operation.

      Example: [1,-5,2,-3,5]

      After taking the element at position 1: [1,-5,2,-3,5] > [-5,2,-3,5]

      Here you can see that the value 2's position has changed from being ODD to Even.

      But if we take the element at position 3 first: [1,-5,2,-3,5] > [1,-5,-3,5]

      As you can see the position of the element 1 has not changed from being ODD to Even. So we can take this element even after taking the 3rd element.

      So, for that it is optimal to take the value of (i+2)th element before taking ith element. So by calc(v,i+2,1), I am saying that "I will take the ith value after taking the (i+2)th value."

      I hope it clears up the idea behind it.

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        14 месяцев назад, # ^ |
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        Hi bro, still did not understand this line - ->So by calc(v,i+2,1), I am saying that "I will take the ith value after taking the (i+2)th value

        why we need to check this for only i and i+2th element. like example 1 -2 -2 -2 4 -2 -2. Here we can take 4 and then 1 as well which is not at position i and i+2, How is this case handled?

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          14 месяцев назад, # ^ |
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          It is a recursive function. And it has 3 branches for odd positions.

          • (1) Take the current value and make the next value ODD and calculate from there.

          • (2) Do not take the current value and calculate from the next position.

          • (3) First calculate from the i+2 th value, then take the current value. So, making no changes to the positions of the next elements.

          And it has 2 branches for the even positions:

          • (4) Remove the current value from the array and make the next position even. And calculate from there.

          • (5) Do nothing to the current value and calculate from the next position which is an ODD position.

          So for the case you have mentioned we follow these sequences:

          3->2->5->1

          The numbers represent the type of operation at each time stamp.

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14 месяцев назад, # |
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it was a very good contest and I do learn a new thing from D problem thank you Windva for writing these Masterpiece

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14 месяцев назад, # |
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Why multitest in D???

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14 месяцев назад, # |
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The system tests for E1 are weak. A test with two decreasing permutations of length 2500 causes my solution to TLE. I'm not sure why some tests were included for E2 but excluded for E1, since I'm pretty sure test 87 from E2 would have caused the same TLE on E1 (the code that causes the TLE is unchanged between my submissions). This also seems like the most intuitively "hard" case, so I'm surprised it wasn't included early on among the pretests for both versions of the problem.

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14 месяцев назад, # |
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D is too similar with Leetcode 834. It is possible to "transform" 834's solution to D's solution by just adding weight of the egdes by a[u]^a[v]

Leetcode 834 solution (in Chinese version) :https://leetcode.cn/problems/sum-of-distances-in-tree/solutions/

My solution: https://codeforces.net/contest/1882/submission/225158821

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    14 месяцев назад, # ^ |
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    To have that transformation you need the main idea anyway, so I don't think this is a major issue.

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14 месяцев назад, # |
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C was too much ad hoc for me, fuc*** my whole contest :(, D was straight kinda straight forward, submitted just after the contest was over :(

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14 месяцев назад, # |
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Problem B Is too much for div2B

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14 месяцев назад, # |
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Video Editorial For Problem A,B,C,D

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14 месяцев назад, # |
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My DP solution for C is not working. Can anyone plzz check?

Code
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14 месяцев назад, # |
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Super Fast Delta Delivery!

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14 месяцев назад, # |
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In problem C i got AC but i don't know if my idea is good or just lucky

practically i maintain a multiset where i keep the values that i took

a priority queue where i keep the value i didnt take but are on odd positions

a cnt where i mantain the values i didnt take on even positions

i have a value tura that means if i have to look on even or odd positions , because i deleted a even number or odd number of numbers before

so practically when i am on a odd position i see if its > 0

if it is i take it and tura becomes 1 -tura

if i am on a even position i see if i took a position before if i did and this one is bigger than 0 i take this position as odd also

if i didnt i will see if i have cnt > 0 if yes it means i can take it if its bigger than 0

then if i cant do that either i see if v [ poz ] > min ( multiset.begin() , priority_queue.top())

if yes i take it and i give up a element to the left

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14 месяцев назад, # |
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Can someone pls explain this solution by Jiangly for problem B? https://codeforces.net/contest/1882/submission/225097723

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    14 месяцев назад, # ^ |
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    I hope this commented code helps:

    225190010

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    14 месяцев назад, # ^ |
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    The variable Or is used to store the union of all the sets. We enumerate the number $$$x$$$ to erase for the answer. Then it's clear that all sets containing $$$x$$$ should not be selected, and we can select all other sets . So we can know the largest union set to get which doesn't contains $$$x$$$.

    For code implement, uses bitmask to record each set, and some bit operators to check the conditions above.

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14 месяцев назад, # |
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Since October 2022, I have participated in 3 CF Rounds, all in May, this year. After that time this one was the first. This year I participated in the IMO and IOI. Hence, during the last few months, I was training for both IMO and IOI. I started training for the IOI after the IMO. During that period I have not participated in any CF Round. I was virtually participating in different 5-hour IOI-style contests, mostly IOI. To understand the problem statements clearly, I read them slowly. And now I think I am not in my best form of CF-style contests. Fortunately, I managed to solve A, B, C, D and to write the code of E. After I finished coding, then finding and fixing bugs, around a minute was remaining for the contest to end, so I quickly tested my code on the test cases available in the statement and sent the code. But… I got WA on TEST 1. What?… and after a few seconds of checking my code, I found the following:

BUG

I wrote this to locally debug my code to find bugs, and forgot to change it into a comment… I quickly fixed that, but it was already late for a minute… The contest was over… During the system testing I hoped that my idea of the problem was wrong, or that there was another bug in the code (so that the lateness of 1 minute wouldn't change anything), but, fortunately or unfortunately, after the end of the system testing, I sent the fixed code and got AC… WOW… I thought I would be more relaxed should I get WA.

P. S. By the way, the Honourable Mention from the IMO and Bronze medal from the IOI, which I earned, made me feel happy that a story like this happened during the regular round but not the official contest.

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14 месяцев назад, # |
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Editorial please!

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14 месяцев назад, # |
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In problem D, how to prove that in the optimal cost solution, all values in the subtree of a node v should become equal to the original value of the node v itself after certain operations?

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    14 месяцев назад, # ^ |
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    Assume you would make an operation on the node $$$v$$$. Then all the nodes in the whole tree will be $$$xor$$$-ed. If you only look at one bit position, that means either we $$$xor$$$-ed with $$$0$$$, then nothing changes. Or we $$$xor$$$-ed with $$$1$$$, then all bits $$$1$$$ changed to $$$0$$$-s and vice versa, so the differences are still the same. We did not get closer to our goal. So using an operation on the root is futile. This means all numbers will be equal to the number in the root.

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14 месяцев назад, # |
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I'm getting WA on test 5 E1, it says:

Checker Log

wrong answer incorrect A

what does that mean?

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14 месяцев назад, # |
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Thank you to the author and helpers for the contest! I liked all the problems except D, which I felt was standard and boring, and E2, which I can't solve. C and E1 in particular I like a lot. Thanks again :)

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14 месяцев назад, # |
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Thnx

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14 месяцев назад, # |
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DP Solution for C, considering

dp[i][0][0] => not removing the ith index as an even index from the start,

dp[i][0][1] => not removing the ith index as an odd index from the start,

dp[i][1][0] => removing the ith index as an even index from the start,

dp[i][1][1] => removing the ith index as an odd index from the start,


dp[0][0][1] = max(0ll,a[0]);dp[0][1][1] = a[0];

for(i=1;i<n;i++){
        ll x,y ;
        x = dp[i-1][0][0]+max(0ll,a[i]);
        y = dp[i-1][1][1]+max(0ll,a[i]);
        dp[i][0][1] = max(x,y);
        x = dp[i-1][0][1];
        y = dp[i-1][1][0];
        dp[i][0][0] = max(x,y);
        x = dp[i-1][1][1]+a[i];
        y = dp[i-1][0][0]+a[i];
        dp[i][1][1] = max(x,y);
        x = dp[i-1][1][0];
        y = dp[i-1][0][1];
        dp[i][1][0] = max(x,y);
}

submission — 225203776

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14 месяцев назад, # |
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can anyone explain B to me??

I understood that we need to opt out the set with least no of unique elements wrt other sets,but how to implement this.

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    14 месяцев назад, # ^ |
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    For me I just used the fact that all elements <= 50 so I tried union of every set not containing X for every X in [1, 50], if that union if not the same as union of every set then it works.

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14 месяцев назад, # |
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Hi, can someone tell me why the following code gives me a runtime error (I ran it in an IDE outside of Codeforces and no RTE was picked up). Thanks.


int main(){ ll t; cin >> t; while(t--) { ll n; cin >> n; vector<vector<ll>> s(n); vector<ll> freq(50),k(n); ll maxi=0,m=0; for (ll i=1;i<=50;i++) { freq[i]=0; } for (ll i=0;i<n;i++) { cin >> k[i]; vector<ll> a(50); for (ll i=1;i<=50;i++) { a[i]=0; } for (ll j=0;j<k[i];j++) { ll b; cin >> b; a[b]=1,freq[b]+=1; if (freq[b]==1) { m+=1; } } s[i]=a; } for (ll i=1;i<=50;i++) { vector<ll> freqnew=freq; for (ll j=0;j<n;j++) { vector<ll> c=s[j]; if (c[i]==1) { for (ll l=1;l<=50;l++) { freqnew[l]-=c[l]; } } } ll num=0; for (ll i=1;i<=50;i++) { if (freqnew[i]>0) { num+=1; } } if (num<m) { maxi=max(num,maxi); } } cout << maxi << endl; } return 0; }
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14 месяцев назад, # |
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Really fun round ngl

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14 месяцев назад, # |
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I and wxy_is_a_dog/2255153050 had studyed the Replacing DP,and we studyed https://www.cnblogs.com/wlzhouzhuan/p/12643056.html this blog.So our solutions have something in same by accident.

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14 месяцев назад, # |
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I got a message stating that my solution to the problem 1882C - Карточная игра coincides with qilby's submission (both submissions: 225107761, 225130187). The submissions are very similar (and what's even funnier, our nicks are also pretty similar), but the coincidence happened only due to how short the solution to that problem was (all variable names used there are pretty standard like $$$s$$$ for sum and $$$ans$$$ for answer or $$$ll$$$ for long long type). Moreover, I've submitted the solution $$$\approx$$$ 30 mins after De_Coder so it would make no sense to take that long if I had been cheating. The solution didn't use any data structures/advanced algorithms hence I didn't use any code written before the contest.

MikeMirzayanov

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3 месяца назад, # |
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For Question B, Set and Union. Test Case 3.

s2 U s3 U s5 = {1,3,5,6,8,9,10} but the test case missed 1. i.e s2 U s3 U s5 = {3,5,6,8,9,10}. Why is 1 missing for test case 3?

The answer for test case 3 should be 7 right?