TimeWarp101's blog

Codeforces Round #845 (Div. 2) and ByteRace 2023
By TimeWarp101, 23 months ago, In English

Hello, Codeforces!

NJACK — the Computer Science Club of IIT Patna is excited to invite you to ByteRace 2023 — Codeforces Round 845 (Div. 2) and ByteRace 2023 under Celesta — the annual Techno-Management Fest of IIT Patna.

The contest will take place on Jan/21/2023 17:35 (Moscow time). This round will be rated for participants with rating lower than 2100.

Many thanks to all the people who made this round possible:

You will have 2 hours to solve 6 problems.

The scoring distribution will be updated later.

$$$\color{white}{\text{I love Kaori}}$$$

UPD: Scoring distribution: $$$500-1000-1500-2000-2250-2750$$$

UPD: Editorial

UPD: Congratulations to the winners!

Official winners:

  1. jiangly_fan_fan_fan_fan
  2. ducati
  3. xiachong
  4. FasterThanLight
  5. Remask_588_handles

Unofficial winners:

  1. noimi
  2. jiangly_fan_fan_fan_fan
  3. Nyaan
  4. ducati
  5. neal

First solves:

A: noimi at 00:00
B: neal at 00:02
C: noimi at 00:06
D: noimi at 00:09
E: noimi at 00:15
F: sjc061031 at 00:13

PRIZES: 30 hoodies (customizable with name) will be given to:

  • Top 20 Indian participants
  • Random 10 from top 100 (rank 21-100) Indian participants

Note: we will identify Indian participants through their flags and they may be asked for address proofs later.

See you all in the standings!

UPD: Here is the list of people who won hoodies. We will contact you all soon. Congrats!

Top 20 Indian participants

Random 10 from top 100 (rank 21 — 100) Indian participants

About Celesta

Celesta is the annual Techno-Management Fest of IIT Patna. Celesta conducts a variety of events in various technical domains. Some of these are open and free for all, with exciting prizes and goodies for the winners!

You can head over to our website and check it out for yourself!

Good luck!

  • Vote: I like it
  • +273
  • Vote: I do not like it

Comments Comments (174)
| Write comment?
»
23 months ago, # |
  Vote: I like it -12 Vote: I do not like it

Auto comment: topic has been updated by TimeWarp101 (previous revision, new revision, compare).

→ Reply
»
23 months ago, # |
  Vote: I like it +23 Vote: I do not like it

As a VIP tester, I VIP-tested. Hope you enjoy the round!

→ Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    Thanks Celeste for the contest! I appreciate it.

    → Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Hello I am having problem with submission of Problem B.My Code runs fine on my local machine but its giving wrong answer when I tried to submit it here.. It seems to me the issue with compiler pls look into this..

    → Reply
    • »
      »
      »
      23 months ago, # ^ |
      ← Rev. 3   Vote: I like it 0 Vote: I do not like it

      There is no issue with the compiler. Check your code for inconsistencies that may be leading to undefined behaviour. Try custom invocation to find the bug.

      → Reply
»
23 months ago, # |
  Vote: I like it +36 Vote: I do not like it

As a problem setter I set problems.

शुभकामनाएं

→ Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

I hope the competition is not too difficult.

→ Reply
»
23 months ago, # |
  Vote: I like it +56 Vote: I do not like it

omg Spring Festival Eve round

→ Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it
  • Q1: Why only for the Indian?
  • Q2: Why hoodie? Why not any expensive prize?
→ Reply
»
23 months ago, # |
  Vote: I like it +1 Vote: I do not like it

omg new year eve round

→ Reply
»
23 months ago, # |
  Vote: I like it +10 Vote: I do not like it

Omg the round i tested is tomorrow. Hope you guys have fun!

→ Reply
»
23 months ago, # |
  Vote: I like it -105 Vote: I do not like it

Wtf with this indian rasism? Why all indian contests give prizes only to indian participats?

→ Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it +37 Vote: I do not like it

    They are Indian setters. They can't send prizes to other countries because it costs too much.

    → Reply
    • »
      »
      »
      23 months ago, # ^ |
        Vote: I like it -91 Vote: I do not like it

      Bla bla bla. You (indians) are the only one in this world, who are doing this. What a shame.

      → Reply
      • »
        »
        »
        »
        23 months ago, # ^ |
          Vote: I like it +30 Vote: I do not like it

        I am not Indian and I'm not saying whether they are right or wrong. I'm just saying what setters say in their comments usually.

        → Reply
        • »
          »
          »
          »
          »
          23 months ago, # ^ |
            Vote: I like it -30 Vote: I do not like it

          Well code ton also gives prizes here and there.
          Its only for indians that it cost so much.
          stop talking Abito

          → Reply
          • »
            »
            »
            »
            »
            »
            23 months ago, # ^ |
              Vote: I like it +13 Vote: I do not like it

            Ton rounds give crypto currencies. This is different. I am not defending them nor am I attacking them. I am only stating what setters say when someone asks them about this. Please stop.

            → Reply
            • »
              »
              »
              »
              »
              »
              »
              23 months ago, # ^ |
                Vote: I like it +23 Vote: I do not like it

              An Indian round without an Indian-rascism-shitposting thread is like a river without water. Sun Tzu | The Art of War

              → Reply
      • »
        »
        »
        »
        23 months ago, # ^ |
          Vote: I like it +14 Vote: I do not like it

        I am not sure why you are so invigorated by this. In this round, 2 packages were given to two participants in Vietnam and the contest organiser was Vietnamese.

        Please understand that this contest (including most of the Indian contests I assume you were referring about and the Vietnamese contest I mentioned above) are not sponsored, so they are giving prizes from their own money. Shipping costs are a lot, so the prizes are confined to the contest's country.

        Also, Codeforces is meant to be a rich and diverse platform for coding, not a way to get prizes. If that is your goal, I have a list of excellent gift stores I am sure you would like.

        → Reply
  • »
    »
    23 months ago, # ^ |
    ← Rev. 2   Vote: I like it +40 Vote: I do not like it

    I was part of organising similar events in previous years, generally sending prizes requires approval from college professors and they denied to approve the transactions for overseas participants since it was "costing too much". Too much pain to convince them for anything. Later on we did send amazon gift cards to overseas winners but couldn't get approval for some prizes :(

    → Reply
»
23 months ago, # |
  Vote: I like it -7 Vote: I do not like it

I would have been top1 in this contest but unfortunately wont be able to participate.

→ Reply
»
23 months ago, # |
  Vote: I like it +31 Vote: I do not like it

As a problem setter, I discovered that, upon giving this contest, your IQ will increase $$$555$$$ times.

PS: The problems are super fun. Hope you all will enjoy solving them! Don't forget to read them all :)

→ Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it +49 Vote: I do not like it

    My current IQ is ZERO, Do something for me as well

    → Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    My IQ did increase thanks to this contest. (Hopefully the same happens for rating too XD)

    → Reply
»
23 months ago, # |
  Vote: I like it -44 Vote: I do not like it

Clashing with Leetcode round :/

→ Reply
»
23 months ago, # |
  Vote: I like it +23 Vote: I do not like it

Setters are from my college, so excited. Best of luck people!

→ Reply
»
23 months ago, # |
  Vote: I like it +5 Vote: I do not like it

I need pants!

→ Reply
»
23 months ago, # |
  Vote: I like it +1 Vote: I do not like it

Hype overload!!

→ Reply
»
23 months ago, # |
  Vote: I like it +8 Vote: I do not like it

Good luck to everyone participating! Hope you enjoy the problems.

→ Reply
»
23 months ago, # |
  Vote: I like it +6 Vote: I do not like it

Nobody wants to watch the boring Spring Festival Gala.A CF round is much better compared to the bad shows.

→ Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

all the best everyone . have a good and learning contest for all.

→ Reply
»
23 months ago, # |
← Rev. 2   Vote: I like it +13 Vote: I do not like it

As floormate of the problem setters, I can confirm that this round will be really interesting.

→ Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

What prizes do u have for Belarus people?

→ Reply
»
23 months ago, # |
← Rev. 2   Vote: I like it +1 Vote: I do not like it

I need expert back.

Edit: It's done

→ Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

omg Spring Festival Eve round

→ Reply
»
23 months ago, # |
  Vote: I like it +12 Vote: I do not like it

Want to hear AwakeAnay side story.

→ Reply
»
23 months ago, # |
← Rev. 2   Vote: I like it +34 Vote: I do not like it

#845 is about to begin and there's no official editorial for #844

→ Reply
»
23 months ago, # |
  Vote: I like it +6 Vote: I do not like it

I have a bad feeling about this round

→ Reply
»
23 months ago, # |
  Vote: I like it +34 Vote: I do not like it

→ Reply
»
23 months ago, # |
  Vote: I like it +3 Vote: I do not like it

What's the scoring distribution? It's still not updated.

→ Reply
»
23 months ago, # |
  Vote: I like it +12 Vote: I do not like it

I sincerely wish people all over the world a happy Spring Festival! In the new year, I hope you can achieve everything you want, be healthy and have good luck.

→ Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Great to take part in a Codeforces Round while watching the boring Spring Gala! Wish everyone and I have a great positive rating delta as well as a great new year!

→ Reply
»
23 months ago, # |
  Vote: I like it +2 Vote: I do not like it

Congratulations to participants for participating on Chinese New Year Eve.

→ Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

I am having issue with submission of problem B. When running it on my local machine it is giving right answer but when I submitted the code it is printing different output. Please look into this..

→ Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    You should try it in custom test and check if you are having the same output as your local machine or not.

    → Reply
»
23 months ago, # |
  Vote: I like it +58 Vote: I do not like it

Congratulations to yzc2005 for making submission 190000000 (link won't work until after contest).

→ Reply
»
23 months ago, # |
  Vote: I like it -33 Vote: I do not like it

Trash Indian rounds with very standard problems. Bad E and F.

→ Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it +41 Vote: I do not like it
    • Makes Codeforces account
    • Solves all 6 problems in a little over an hour, getting 5th official place
    • Calls it a trash indian round with standard problems
    • Refuses to elaborate
    • Leaves
    → Reply
  • »
    »
    23 months ago, # ^ |
    ← Rev. 3   Vote: I like it 0 Vote: I do not like it

    i do not see issue with standard problems, earlier codeforces used to give standard problem variations only if you pickup few age old problems. It is just since recent 4-6 years that more constructive and math problems are coming and i do not see any issue in both standard/adhoc as they both enhance problem solving/approaching skill.

    → Reply
»
23 months ago, # |
← Rev. 2   Vote: I like it +3 Vote: I do not like it

Problem C is interesting.

And how to approach D?

→ Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    tree dp

    → Reply
    • »
      »
      »
      23 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      How exactly?

      → Reply
      • »
        »
        »
        »
        23 months ago, # ^ |
          Vote: I like it +5 Vote: I do not like it

        The answer is 2^(n-1)*sum(the max depth of the subtree with root i) where i iterate among all vertexs. (the depth of leaves is 1, the depth of parents of 1-depth vertexs is 2, and so on)

        → Reply
        • »
          »
          »
          »
          »
          23 months ago, # ^ |
          ← Rev. 2   Vote: I like it 0 Vote: I do not like it

          I guessed this solution after an hour of doing random stuff on the sample case :/

          → Reply
        • »
          »
          »
          »
          »
          23 months ago, # ^ |
          ← Rev. 2   Vote: I like it +8 Vote: I do not like it

          I explained it like this: If at least of the children has a possibility to be a 1, that possibility has to be 50%.

          Any number of XORing 50% possibility of 1 still gives 50%, so as long as one child could have a 1 in the previous round, we got a 50% chance of getting a 1.

          For leaves it's just 50% one time, then they become 0. For a node one above that, once all children are 0, it also becomes 0 the next round. So if e.g. the longest child chain has 2 children, we have to take 50% chance 3 times, 1 for the start, and 2 for the 0 to trickle up through the length 2 children line.

          → Reply
      • »
        »
        »
        »
        23 months ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Try to find the solution for every pair of nodes and times and then you'll soon realize that they are all the same before becoming zero.

        → Reply
  • »
    »
    23 months ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it
    Hint
    Hint 2
    Solution
    → Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Did you solve C? What was your approach?

    → Reply
    • »
      »
      »
      23 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it
      Spoiler
      → Reply
      • »
        »
        »
        »
        23 months ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        IS there any good article on two pointers and sliding window method?

        → Reply
        • »
          »
          »
          »
          »
          23 months ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          Competitive programmers handbook

          → Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    Thank you for the appreciation, it means alot. 🧡

    → Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Lol, problem F seems to be standard

→ Reply
»
23 months ago, # |
  Vote: I like it +16 Vote: I do not like it

I didn't do as well as I wanted but I enjoyed the contest, thanks setters ! :)

→ Reply
»
23 months ago, # |
← Rev. 9   Vote: I like it -24 Vote: I do not like it

Problem D can be solved by tree dfs(we need to find the maximal path from every vertex to it's childs), and we can solve E with any template for finding strong components (we just need to add an negative weight edge v--(-w)-->u for every u--(w)-->v, representing the edge can be reversed if cost>=w, and binary search for cost), but I could not solve C.

→ Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it +7 Vote: I do not like it

    Problem C is just two pointers :)

    → Reply
    • »
      »
      »
      23 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      any hints of problems c ?

      → Reply
      • »
        »
        »
        »
        23 months ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Check out the editorial, we have provided hints as well!

        → Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I don't think D was that trivial as it was pretty hard for me to see the idea. Or maybe it's just a skill issue on my part.

    → Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    same... C took me 1 hour and a half ( I didn't see that I WA'd B cuz I didn't check submissions) ☠️ but D took me like 3 minutes to think of the solution and 10 minutes to code (though this was after the contest already ended so I can't submit.)

    sha256 of my sol
    → Reply
»
23 months ago, # |
  Vote: I like it +1 Vote: I do not like it

How to prove Problem B solution?

→ Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    n!*n*(n-1)

    → Reply
  • »
    »
    23 months ago, # ^ |
    ← Rev. 3   Vote: I like it 0 Vote: I do not like it

    see below for my attempt to prove

    → Reply
    • »
      »
      »
      23 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      I did the same thing. I saw that each permutation gave the same answer.

      → Reply
    • »
      »
      »
      23 months ago, # ^ |
        Vote: I like it +1 Vote: I do not like it

      For the proof, 1 2 2 1 consider the permutation, if you put 3 in to anywhere
      1 2 3 3 2 1
      1 3 2 2 3 1
      ...
      it will increase the inversion count by (3-1) * 2

      Because if for the left half its inversion count is j, for the right half it will be 3-1-j, plus all elements on the right half have inversion with the left 3, overall it will be j+3-1-j + (3-1) inversion. For m, it will m-1+m-1 = (m-1)*2 inversions. Summing this for all m <= n gives n * (n-1)

      → Reply
  • »
    »
    23 months ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    Suppose $$$p_i,p_j$$$ such that $$$1\le i<j\le n$$$ are distinct elements of the permutation. Their reflections are $$$p_i',p_j'$$$, respectively. If $$$p_i<p_j$$$, then they contribute only $$$2$$$ inversions because $$$p_j>p_i'$$$ and $$$p_j'>p_i'$$$. If $$$p_i>p_j$$$, then they also contribute only $$$2$$$ inversions because $$$p_i>p_j$$$ and $$$p_i>p_j'$$$. Thus, for every pair of distinct indices $$$i,j$$$, they contribute $$$2$$$ inversions. There are $$$n\choose2$$$ ways to choose pairs of indexes. So, the answer is $$$n!\cdot{n\choose2}\cdot2$$$.

    → Reply
  • »
    »
    23 months ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    Consider a specific permutation of size $$$n$$$. For any two different indices $$$i$$$ and $$$j$$$, either they form an inversion in the original array OR they form an inversion in the reverse array. Each such pair will contribute exactly one such inversion (note that there are no duplicates in the original array, since it is a permutation). There are $$$\frac{n(n - 1)}{2}$$$ pairs of indices.

    This will already count all inversions that are between indices that are either both in the original array or both in the reverse array. All that's left is to count inversions where one index is in the original array and the other index is in the reverse array. Because both the original and reverse arrays are permutations, it's easy to count them: the value $$$k + 1$$$ in the original array will have $$$k$$$ values smaller than it which are present in the reverse array. Thus, the total number of such indices is $$$\sum_{k = 1}^{n - 1} k = \frac{n(n - 1)}{2}$$$.

    Add them up and we have exactly $$$n (n - 1)$$$ such inversions. This is for a single specific permutation, without actually depending on what the permutation is. Since there are $$$n!$$$ permutations, the required answer is $$$n! n (n - 1) \bmod (10^9 + 6)$$$

    → Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Consider any pair of numbers $$$i$$$ and $$$j$$$, such that $$$i$$$ is to the left of $$$j$$$.

    In the final array you will have ... $$$i$$$ ... $$$j$$$ ... $$$j$$$ ... $$$i$$$ ...

    You can see that no matter which value is higher, each pair will give 2 inversions.

    There are $$$ \frac{n(n - 1)}{2} $$$ pairs, each will contribute $$$2$$$ inversions per permutation and there are $$$n!$$$ permutations

    → Reply
»
23 months ago, # |
  Vote: I like it +3 Vote: I do not like it

how to solve problem C?

→ Reply
  • »
    »
    23 months ago, # ^ |
    ← Rev. 3   Vote: I like it +5 Vote: I do not like it

    Sort the input array $$$v$$$. Keep track of 2 pointers $$$left$$$ and $$$right$$$, initially both at the 1st position of the array, and keep a map that stores the frequency of all the divisors from $$$v[left]$$$ to $$$v[right]$$$. If the size of the map is not equal to $$$m$$$, then increase $$$right$$$ and update the map, otherwise increase $$$left$$$ and update the map. Keep track of the minimum difference between max and min while iterating.

    → Reply
    • »
      »
      »
      23 months ago, # ^ |
      ← Rev. 2   Vote: I like it 0 Vote: I do not like it

      This approach just kept giving me TLE :( 190022533 What is the complexity of it? N*sqrt(N) should pass those limits or am I wrong?

      → Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

how to solve b

→ Reply
  • »
    »
    23 months ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    Consider three cases separately,

    case 1: where the inversion is completely in the first half (i.e. the original permutation),

    case 2: where the inversion is completely in the second half (i.e. the reversed permutation),

    case 3: where one end of the inversion is in first half and the other in the second half.

    The answers for these cases will be t+(nC2-t)+(1+2+3+...+(n-1)) where t will depend on the permutation. But as you can see, the sum is simply 2(nC2) which is same for each permutation of length n. There are a total n! such permutations. So the answer is n!*2*nC2 which is n!n(n-1).

    → Reply
»
23 months ago, # |
  Vote: I like it +15 Vote: I do not like it

Due to Codeforces not loading properly near the start of the contest, some participants may have accidentally submitted the same code multiple times. In my case, for example, I used the main site to submit and got a 502 Bad Gateway, so I switched to one of the lightweight versions (m1, m2, or m3, not sure exactly which one) and submitted over there instead, not realizing that my first submission actually went through. This was treated as a resubmission and I lost 50 points because of it.

Is it possible to make any adjustments such that these kinds of duplicate resubmissions are not penalized? Note that the main site normally blocks the user from submitting an identical code twice, so I think it would be perfectly justified to remove such identical codes that did manage to get through (due to using the lightweight site, which implies technical issues that the participant should not be punished for experiencing).

Here are my submissions: 189973808 189973881.

→ Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    Tagging MikeMirzayanov because I think this should be applied universally across Codeforces contests, if possible.

    → Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    this issue happened to me too and It made my submission for problem A later about 1 minute:(

    → Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Was stuck in problem C for a long time, any intuition on how to solve it?

→ Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it
    Hint
    → Reply
»
23 months ago, # |
  Vote: I like it -14 Vote: I do not like it

i swear there is like no other solution for problem C other than annoying brute force and also to make the execution somehow fast

→ Reply
»
23 months ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

Read the problem statement for E as the minimum weight of edge reversal weight so that every node is reachable from one another. IMO this version seems cooler (and harder) than the original one.

→ Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    It's as simple as the original problem (both versions need to check strong components). The original is checking "whether there is only one component with 0 in-degree", and your version is checking "whether there is only one strong component".

    → Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Why did 189996915 not pass? I thought since ind and fac[n] are modded M already, ind, fac[n] <= 1e9+6. and (1e9+6)^2 <= max ll??

→ Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    But subtraction modulo M may give a negative number, so you also need to add M and take modulo again.

    → Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Is d tree dp? I couldn't really think of a solution, how do you solve D?

→ Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it +2 Vote: I do not like it

    No, it's math. For each node to be equal to 1, calculate total no. of arrays for which it is possible. Also find its relation with its height.

    → Reply
»
23 months ago, # |
  Vote: I like it +27 Vote: I do not like it

The TL on F is stupidly high, a simple $$$O(n^2)$$$ passed: 190000292

→ Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    Maybe there's larger test in system testing…

    Also there's possibility that the problem setter forgot to disable O(n^2) solutions for Div2F.

    → Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it +11 Vote: I do not like it

    I hacked it, the system tests are too weak

    → Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Anyone knows why simple dfs from first vertex in topological sort works in E? (for checking the condition)

As far as I know, topological sort only works for graphs without cycles

→ Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    If you think about it, the first vertex after the topological sort is the one which will always belong to the first SCC (i.e. a SCC with 0 indegree). So, just checking whether all the other nodes are reachable from this vertex is enough for the problem.

    → Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

I didn't see the constraint on n (problem D) carefully and set 100000 for the arrays. However, instead of getting RE, it got TLE ??? Just changing it from 100005 to 200005 made the program pass...

→ Reply
»
23 months ago, # |
  Vote: I like it +4 Vote: I do not like it

I liked problem D. It had such a simple solution (after thinking it through). Missed the submission by a few seconds tho ^^

→ Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Yeah, D was quite frankly hard to think about.

    → Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Can someone please look into my profile and suggest something, I don't know but cannot perform well for the last 7-8 contests. Please do recommend anything. It would be a great help. Thank you!!

→ Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    it could be that you took a break and you need a little bit of time to de-rust..

    → Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Can't believe I screwed up the contest by not precomputing the divisor array globally as I did the question by finding out the divisor of elements for every array as I thought what's the harm, finding divisor is sqrt(a[i]) time complexity anyway --> worst mistake. What a sad feeling it gives when you could've solved the question but lack of knowledge or stupidity comes in the way. PS: 1. Learned something new 2. Won't be forgetting it ever again

→ Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Problem E is a template of Tarjan.

→ Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

How to solve C?

→ Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Essentially what you want to find first is a subarray of numbers such that the number of factors of all numbers in that subarray is exactly $$$m$$$. Once you have done that, all we want to do now is to minimize the difference between the maximum and the minimum numbers in that subarray. You can do this by having $$$2$$$ pointers, and incrementing the $$$1st$$$ pointer towards the right while making sure that the number of factors remain $$$m$$$

    → Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Can someone help me find, what might be the issue with this submission for problem D?

190024130

Idea is to find the depth of each node and sum them. Finally multiple that by pow(2, n-1).

→ Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Hints for D please?

→ Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Check out the editorial, we have provided hints as well!

    → Reply
»
23 months ago, # |
  Vote: I like it +14 Vote: I do not like it

system test when ?

→ Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it +7 Vote: I do not like it

    Waiting for systests too, bit odd its so late. It annoys my I can't check all the pretests before system testing finishes...

    → Reply
»
23 months ago, # |
← Rev. 2   Vote: I like it +7 Vote: I do not like it

Here goes how solved A-F today in brief.

Upd: Add all problems now.

→ Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

for those who need help with C,D you can find the video editorial here — https://www.youtube.com/@grindcoding. Happy coding!

→ Reply
»
23 months ago, # |
  Vote: I like it +18 Vote: I do not like it

When will the System testing start? I have to cry myself to sleep also.

→ Reply
»
23 months ago, # |
  Vote: I like it +12 Vote: I do not like it

why is it taking so long for system testing

→ Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Main test? MikeMirzayanov

→ Reply
»
23 months ago, # |
  Vote: I like it +14 Vote: I do not like it

Speedforces

→ Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

can't wait to upsolve.System testing :(.

→ Reply
»
23 months ago, # |
  Vote: I like it +1 Vote: I do not like it

anyone solved C with binary search

→ Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    can you explain your solution with binary search? I got WA with binary search.

    → Reply
    • »
      »
      »
      23 months ago, # ^ |
      ← Rev. 2   Vote: I like it +1 Vote: I do not like it

      First I used seive to get the factors of all the numbers and used them in a map and I observed that the number of factors would not be more 125.

      Then I sorted the array.

      Then I binary searched on 0 to 1e18 (I took the upper bound randomly high) where for every mid I used sliding window to check if the given mid is possible or not .

      Then shifted the window accordingly.

      → Reply
      • »
        »
        »
        »
        23 months ago, # ^ |
        ← Rev. 3   Vote: I like it 0 Vote: I do not like it

        thanks. I did a binary search for the length, but it turned out to be wrong.

        → Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

why am i getting WA on pretest 2 for problem c : 190021791

→ Reply
»
23 months ago, # |
← Rev. 3   Vote: I like it -22 Vote: I do not like it

Nice Contest

→ Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

my submission is not tested on main tests please look into it 190002478 TimeWarp101

→ Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Why this solution:190038867 for problem A is giving WA. Thanks.

→ Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    cout before cin

    → Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    In the case of N = 1 for some testcase, you do not cin >> A[i] for the vector A. Thus, you return before you have read input for the current testcase

    → Reply
»
23 months ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

In D, I was memsetting a bool array of size 2*10^5 every single test case, which could have been 10^5 times. I'm surprised it got accepted. Is memset really that fast? Is it doing something magic in the background? https://mirror.codeforces.com/contest/1777/submission/190020821

→ Reply
»
23 months ago, # |
  Vote: I like it +1 Vote: I do not like it

when will be ratings updated?

→ Reply
»
23 months ago, # |
  Vote: I like it +5 Vote: I do not like it

to me, C is too hard and D is too easy, I'm stupid

→ Reply
»
23 months ago, # |
  Vote: I like it +19 Vote: I do not like it

i don't want to offend you but the datamaker of F have no brain?

→ Reply
»
23 months ago, # |
  Vote: I like it +5 Vote: I do not like it

to me: D was easier than C

→ Reply
»
23 months ago, # |
  Vote: I like it -7 Vote: I do not like it

my friend got 8000th standing, and i got 6600th, yet he still got more rating, i got my rating reduced he had his 9th contest this round and this was my 7th :/

→ Reply
»
23 months ago, # |
← Rev. 5   Vote: I like it 0 Vote: I do not like it

I got WA in Problem-C

Can anyone help me please.
My approach for Problem C:
  1. Sort the array A.
  2. Count and store all the multiples(1 to m) for all n elements. If all multiples from 1 to m are not exist in the array then the answer will be -1.
  3. Take two pointer L=0, R=n-1
  4. If multiples of the element in index L can be removed(all multiples from 1 to m still exist in rest of the array), then remove element from index L and increase L. Stop if we can't remove the element(multiples from 1 to m is not exist in rest of the array).
  5. Do the same from index R and decrease R untill we can't remove the element(multiples from 1 to m is not exist in rest of the array).
  6. Answer should be A[R]-A[L]
Example:
5 7
6 4 3 5 7

Here,
n=5, m=7
A=3 4 5 6 7
Multiples of
3->1 3
4->1 2 4
5->1 5
6->1 2 3 6
7->1 7

Frequency of the multiple 1, f[1]=5
As well as f[2]=2, f[3]=2, f[4]=1, f[5]=1, f[6]=1, f[7]=1
Now, L=0, R=5-1=4
remove A[L]=3 and change the frequency of the multiple
f[1]=4, f[3]=1
L=L+1
Then we can't remove A[L]=4 because frequency of 4 that is f[4] will be 0. So, we stop here and start from index R,
We can't remove A[R]=7 because frequency of 7 that is f[7] will be 0. So we stop.
The answer is A[R]-A[L]=7-4=3

My submission: https://mirror.codeforces.com/contest/1777/submission/190090177

→ Reply
»
23 months ago, # |
  Vote: I like it +5 Vote: I do not like it

Hey I find this Round has become unrated.Does anyone know why?

→ Reply
  • »
    »
    23 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    Temporary rating rollbacks are common. Generally, they need to recalculate rating changes (e.g., after identifying a significant set of cheaters). It should be updated soon afterwards.

    It's also possible that this contest was decided to be unrated due to some serious issues, but there was no announcement about this, nor did I see anyone discuss any such issue, so I think this is extremely unlikely. The contest should be rated; please be patient.

    → Reply
    • »
      »
      »
      23 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Now it has been rated again.Thanks for your explanation!

      → Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

What the hell man...one more codeforces round that got unrated...i just noticed that the ratings have been withdrawn..this sucks man..it absolutely sucks

→ Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

why the rating of this contest rolled back?

→ Reply
»
23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

please update problem ratings

→ Reply
»
14 months ago, # |
  Vote: I like it 0 Vote: I do not like it

For the record, I was supposed to receive a hoodie but I didn't and the contest organizers refuse to reply.

→ Reply
  • »
    »
    14 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Hey, the hoodie should reach you soon, probably within 1-2 weeks. My apologies if we were unresponsive.

    → Reply
    • »
      »
      »
      13 months ago, # ^ |
        Vote: I like it +8 Vote: I do not like it

      It did not reach and I have not received any communication about the progress after your comment.

      → Reply
      • »
        »
        »
        »
        11 months ago, # ^ |
        ← Rev. 2   Vote: I like it 0 Vote: I do not like it

        TimeWarp101, still no updates.

        I will continue posting every few months if I don't receive any updates.

        → Reply
        • »
          »
          »
          »
          »
          3 months ago, # ^ |
            Vote: I like it +5 Vote: I do not like it

          Update: They reached out around a month ago to confirm my address. This is after more than 1.5 years since the contest.

          Let's see if they actually deliver it this time.

          → Reply
          • »
            »
            »
            »
            »
            »
            6 weeks ago, # ^ |
              Vote: I like it +1 Vote: I do not like it

            Since I have been updating this thread for quite a while now, I wanted to update that I have finally received the hoodie after around a quarter shy of two years.

            Uniquely enough, it is a pink hoodie with a butterfly printed in the front.

            → Reply