zwezdinv's blog

By zwezdinv, 2 years ago, In English

Hello, Codeforces! ☀️☀️☀️

We are glad to invite you to our second Div. 3 round, which will take place on Nov/17/2023 17:35 (Moscow time). The round will be rated for all participants with rating less than 1600.

You will be given 7 tasks and 2 hours 15 minutes to solve it.

The tasks were prepared by zwezdinv, meowcneil, EJIC_B_KEDAX, Sokol080808 and molney.

Remember, that only trusted Div. 3 participants will be included in official standings. To qualify as trusted participant you need:

  1. Participate in at least five rounds (and solve at least one problem in each of them)
  2. Do not to have a point 1900 or greater in the rating

We would like to thank:

Good luck everyone!

UPD: Editorial

  • Vote: I like it
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| Write comment?
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2 years ago, hide # |
 
Vote: I like it +58 Vote: I do not like it

As a tester I recommend you to participate in round, and don't forget to read all problems statements

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Vote: I like it +32 Vote: I do not like it

As a tester, I recommend you to take part in this round!

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Rev. 3  
Vote: I like it +32 Vote: I do not like it

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Vote: I like it +51 Vote: I do not like it

as a tester, this round is awesome

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Vote: I like it +73 Vote: I do not like it

As a tester, this round is very rofl.

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Vote: I like it +16 Vote: I do not like it

as a not tester, I hope this round will be great

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Does participating in at least five rounds count this round? Because I have been in exactly 4 contest before this :D

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petition to give Gornak40 black handle (as he is, in fact, huge male gorilla) :spinning_gorilla_emoji:

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Score Distribution?

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I hope everyone enjoys this contest ^_^

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Vote: I like it -38 Vote: I do not like it

Good thing that Gornak40 didn't create the tasks too, his problems are special

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Rev. 2  
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Green testing crying in the corner

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In the contest page shows 2 hrs but in the blog page shows 2 hrs 15 minutes how is it possible?

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Rev. 2  
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CR7 > MESSI

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Vote: I like it +1 Vote: I do not like it

Coming back to give contests after long time, feels good to be back!!

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I hope this round will help me reach 1200+ rating. As a participant, I wish you good luck!

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hope this round will help me reach cyan

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2 years ago, hide # |
 
Vote: I like it +21 Vote: I do not like it

I really enjoyed your first div 3 round (Codeforces Round 881 (Div. 3)). Very much looking forward to this one =)

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Wish I could be a pupil tonight!!!

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As a tester I recommend you to participate in round, and don't forget to read all problems statements, it can be usefully)))

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m1, m2, m3 codeforces sites are not working.

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Thx. Very nice round. Enjoyed problem G alot!

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    2 years ago, hide # ^ |
     
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    What is so special about it?

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    2 years ago, hide # ^ |
     
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    can you tell me ur idea to G ?

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      contest running bruh :|

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      2 years ago, hide # ^ |
      Rev. 2  
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      Well, lets count $$$tin$$$ and $$$tout$$$ ($$$tin_v$$$ = first time we entered vertex $$$v$$$, $$$tout_v$$$ = time we leaved vertex $$$v$$$ completely). Vertex $$$u$$$ is an ancestor of $$$v$$$ if and only if $$$tin_u \le tin_v$$$ and $$$tout_u \ge tout_v$$$, but $$$tout_v \gt tin_v$$$, so it brings us down to this condition: $$$tin_u \le tin_v \le tout_u$$$. Now lets make every $$$p_i = tin_{p_i}$$$. Every query we need to see if there is a number from $$$tin_x$$$ to $$$tout_x$$$, that is included in segment $$$[l; r]$$$. This can be done using segment tree in $$$O(log^2n)$$$ per query.

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    2 years ago, hide # ^ |
     
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    But why constraints in F are so low? My solution is linear with 10 lines of code...

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      2 years ago, hide # ^ |
      Rev. 2  
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      How do you verify that the answer is correct faster than $$$O(n^2q)$$$?

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        2 years ago, hide # ^ |
         
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        We build a bamboo tree and just hung $$$n$$$-th vertex to $$$d$$$-th. In this case path from $$$n$$$-th vertex to $$$1$$$-st vertex will be exactly $$$d$$$.

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          2 years ago, hide # ^ |
          Rev. 3  
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          You didn't understand my question. How does the contest system make sure that your output is correct faster than in $$$O(n^2q)$$$? They can't assume everyone will use the bamboo solution.

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Didn't notice that only people competed in 5+ rounds are rated......

Already spent 1+ hour on the 7 problems

Sad QAQ

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B,C >> D,E

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    2 years ago, hide # ^ |
     
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    pC has tedious implementations. I'd say E is quite easy as I solved it in 10 minutes (compared to 80 minutes for C).

    However, B and C are easier when it comes to finding ideas. However, proving your idea works and implementing it makes it a lot more difficult, compared to E (can't say anything about D).

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What's the intended solution to G? The only two solutions I could think of seem completely overkill:

  1. Merge Sort Tree on DFS order times of the inverse values of the permutation.

  2. SQRT Decomposition on whether each node is a parent of at least one node in the buckets $$$[p_1, p_{\sqrt{N}}], [p_{\sqrt{N} + 1}, p_{2 \cdot \sqrt{N}}], \ldots, [p_{k \cdot \sqrt{N} + 1}, p_n]$$$. For the upto $$$2 \cdot \sqrt{n}$$$ remaining nodes that don't lie completely in a bucket check them each in $$$O(1)$$$ using dfs order ancestor checking.

Is there an easier approach?

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E was easier than B wtf

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Problem F,on example,2nd test case, in the third query already exist a path with distances 3 between 2 leafs , 5-4-2-3, but the answer wasn't -1 -1 -1.I'm receiving WA veredict for that?

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Greedy-forces :)

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oh f i just solve G 1 minute after contest time woooooooow

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I finished the G at the last minute, and because I forgot to write "return 0" ,I got compilation error and I passed it only 1 minute after the contest.

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In D it's hard to prove that 1 and 2 the only correct combination without same numbers. Bad problem in my opinion.

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    Its not that hard to prove :

    Condition of the problem is equivalent to $$$a_i/a_j = 2^{a_i}/2^{a_j}$$$

    WLOG assume $$$a_i \gt a_j$$$, we get $$$a_i/a_j = 2^{a_i-a_j} \leq 1e9$$$ $$$\iff a_i-a_j \lt = 30 \iff a_i/a_j \leq 30/a_j+1 \leq 31$$$

    $$$2^{a_i-a_j} \leq 31 \iff a_i-a_j \leq 4 \iff a_i/a_j \leq 4/a_j + 1 \leq 5$$$

    $$$2^{a_i-a_j} \leq 5 \iff a_i-a_j \leq 2 \iff a_i/a_j \leq 2/a_j+1 \leq 3$$$

    $$$2^{a_i-a_j} \leq 3 \iff a_i-a_j \leq 1$$$

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come onnnnnnn if the time was just one minute moreeeee ohhhhhhhhhhhhhhhhh

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One of the easier contest I've had in a while

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Problem Statements are too hard to understand.

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low quality problems on Greedy_forces

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In G, if It is guaranteed that the sum of n and the sum of q over all test cases do not exceed 1e5. shouldn't Binary Lifting work?

I got TLE on TC 10. My Submission

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    2 years ago, hide # ^ |
    Rev. 2  
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    bruh the sum of $$$q$$$ over all test cases do not exceed $$$10^5$$$, not the sum of $$$r_i - l_i$$$

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D will appear in my nightmares

other than that, the contest was great

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Free hack : submission

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What was the solution to D? I observed something that only 1 and 2 matters.. but couldn't get to the solution

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    ai * 2^aj = aj * 2^ai

    let suppose (aj>ai)

    2^(aj-ai) = (aj/ai);

    now

    aj/ai can be max 1e9, so aj-ai can be a maximum of 30,

    so now for every ai, chek possibilities for numbers in ai+1 to ai+30;

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O(n log(n)) solution of G using persistent segment tree 233188158

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$$$O(nq)$$$ solution for G: 233198776

Limits $$$n, q \le 10^5$$$, TL=3sec should be non-existent in 2023 :)

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D could be solved without knowing the fact that only the combination of 1 and 2 is valid.

The statement equivalents to: Count the number of pair (i, j) such that $$$a_i * 2^{a_j} = a_j * 2^{a_i}$$$. Suppose $$$a_i \gt a_j$$$, then we have: $$$a_i = 2^{a_i - a_j} * a_j$$$. Knowing this, we can loop for each $$$a_j$$$, how many $$$a_i$$$ could satisfy this equality, and add it to the answer. The case when $$$a_i = a_j$$$ should be handled separately. My solution: submission

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First contest I could fullsolve, wow!

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Really great round! In my opinion, the condition of problem D was too confusing, but the problem is still good

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What will i receive when i hack a problem successfully?

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Problem D is fun to understand the statement and enjoyable to solve, I really like it!

Thank you guys for the contest!

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Problem B was difficult to understand

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can anyone tell me why people having fewer solve count and people having same solve count and high penalty have better rank than me. Thanks!

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Which data structure do I need to know to solve G?

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.

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It was a nice round. Hats off to you guys (editorial panel). Really enjoyed as a newbie coder.

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    i was good but very bad luck because i had a little mistke in B which made me late and when i decide to leave it i went to solve C and D i had oubt that in D the only combination is 1 and 2 but kept lookign for solution until idecide to try it and got AC and then got back to B and found the mistake anyways i solved 4 problems for first time

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Found a cheater Salman_Johir. This 233189111 is his submission. He copied this code from this video's code. Look at the both code. He just has changed the variables' name.

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    2 years ago, hide # ^ |
     
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    come on problem C is just an application of a known algorithm

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      2 years ago, hide # ^ |
       
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      Problem C is an application of a known problem, but if you check the code of the video and the code of the cheater Salman_Johir you will see: Their code is exactly same, code with 10-12 lines can be coincident due the algorithm,but if you give it a check you will see every bit of the code is similar.

      And this guy's contest number 904 was totally copied and his both solutions were skipped!

      Check here

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very week pretest in problem C

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Thank you for the problems!

It's nice there is a short and witty solution to each one (except G).

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i got an O(n) solution that was hacked for problem D (TLE). The official solution is O(nlogn). Anyone throw some light on this? https://mirror.codeforces.com/contest/1899/submission/233176732

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    Unordered maps can be hacked because of how collisions are handled. You can use custom hashing functions as parameters to the unordered map to solve that problem.

    struct custom_hash {
        static uint64_t splitmix64(int64_t x) {
            // http://xorshift.di.unimi.it/splitmix64.c
            x += 0x9e3779b97f4a7c15;
            x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
            x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
            return x ^ (x >> 31);
        }
    
        size_t operator()(int64_t x) const {
            static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
            return splitmix64(x + FIXED_RANDOM);
        }
    };
    typedef unordered_map<long long, int, custom_hash> safe_map;
    
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My submission for D got hacked https://mirror.codeforces.com/contest/1899/submission/233134841. Is it cause I used unordered_map instead of map? :)

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    Nope, but rather because you are using int to store the number of pairs.

    Think about worst case scenario, if I put all the $$$N$$$ values of array $$$a$$$ equal to $$$2$$$, then the number of pairs would be $$$N*(N-1)/2$$$, ain't it?

    And $$$N$$$ can be upto $$$10^5$$$, can u store this value in int? or do u need to use another data type?

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      I have defined the macro #define int long long Btw hacked by TLE not WA :)

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        Oh! I did not see that!!!

        Then it must've been the notorious unordered_map collisions resulting in $$$O(N^2)$$$ time complexity from $$$O(NlogN)$$$

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          2 years ago, hide # ^ |
           
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          Here 233255221 it works by using custom hash for the map.

          Never ever use raw unordered_map in contests, always use the above custom_hashed_map.

          You can read more about it in this blog.

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You didn't write that there would be hacks...

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got AC on Problem C during contest but got tle on system testing......?isn't its complexity O(n)!!!

Code : 233167945

UPDATE:: got TLE because of "auto" iterator!

TLE : 233167945

AC : 233276048

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    No, it's not; you have a nested loop when you iterate through vector v.

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    Rev. 6  
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    The AC code passed in 982 ms / 1000 ms, that's very very close to the time limit. Maybe the use of the range-based for-loop had something to do with your code getting TLE, maybe it didn't. We can't really know since if your code is that close to the time limit, even the same exact code can pass on one submission and get TLE on another.

    If you're wondering why your "$$$O(n)$$$" solution is on the verge of TLE, it's because you create an array int a[MAXN]; every test case. Creating and initializing an array of length $$$n$$$ is $$$O(n)$$$ and since you're initializing an array of length $$$MAXN$$$, the time complexity ends up being $$$O(t\cdot MAXN)$$$ which is quite slow.

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    Same... I guess Its just a system failure :\ 233286944

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      Even though I'm using an unordered map. Everything should work just fine.

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      And to be honest, I dont even understand why in the contests where the all problem tests are executed during the online sumbmission, the "aftertesting" is the case.

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Best contest

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it was interesting. but, i wasted time to search something good to solve problem B, i thought to bruteforce that 'k' at first, as it was.. but don't know why i felt that stupid and moved to C:). Now i saw it's E was easier than that, it was likely not arranged in increasing difficulty :) also, i felt C easier than B.

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Can someone explain what is wrong with my solution? My O(n) solution for D got hacked https://mirror.codeforces.com/contest/1899/submission/233160169

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    I'm not a python user so I can't get you an 100% precise answer but : - I found that dictionary using only string keys are way faster than other dictionaries, this would have been enough to get you an AC : https://mirror.codeforces.com/contest/1899/submission/233292415 - It's also faster to use pypy instead of python3 (there is even a warning when submitting a code)

    As for the reason of your code being too slow, i tried to rewrite your sum a couple times and couldn't find anything way faster than what you wrote so it's most probably the insertions in the dictionary that are too slow. I saw that the worst insert case is 0(n), which happen if there are tons of conflict in the hashmap. So maybe the hack that got you to be too slow is designed to create lots of conflicting keys... This last part is a guess but I thought the first part was worth sending this answer.

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Can anyone please tell why my O(n) solution got tle on problem D. https://mirror.codeforces.com/contest/1899/submission/233294769

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Please some help me. my rank in common stading is 2198 But in friend standing and profile contest page it is 2619 and and my rating change is calculated on 2619 rank

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    Common standings, when "unofficial participants" is off, doesn't show all rated participants. From the announcement:

    Remember, that only trusted Div. 3 participants will be included in official standings. To qualify as trusted participant you need:

    • Participate in at least five rounds (and solve at least one problem in each of them)
    • Do not to have a point 1900 or greater in the rating

    Regrdless of these extra restrictions, the rouns is still rated for everyone with < 1600 rating. Your rank among all rated participnts is the 2619, but your rank among trusted participants is 2198.

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      Thanks bro but i still have a doubt. Participants near 2619 rank have lower solve count than me and participants near 2198 rank same solve count and penalty as mine and still my rank is below to those who have solves less problem than me. Sorry for annoying question and terrible english

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        The people around 2619 in the trusted participant standings also have a lower rank among all rated participants:

        Your rank among trusted participants (link):

        Your rank among all rated participants (link):

        Near 2619 located random guy's rank among trusted participants (link):

        Same random guy's rank among all rated participants (link):

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Attention!

Your solution 233122161 for the problem 1899C significantly coincides with solutions sk.dubey/233122161, 21H51A6655_/233126912, ayush_1217/233140420, sudhanshu_2306/233141014, mokamal2003/233141318, quantum_quill/233144312, Dineshsingh10001/233146424, huseyn_quliyev/233146946, fuck_23/233147330, mostakem123/233147919, Ritesh027/233148153, vaibhav0543/233148375, artemfad/233148803, akpalert/233148863, sihare_123/233149177, ayushverma150702/233149667, kanikapunia119/233149712, Anonymous-2002/233149818, Ooppa/233149911, Shreshth/233150012, zakeer_shaik/233150023, jaikiran1027/233150227, gaurang_07/233150492, MADARA1287/233150624, aashish1110/233150766, krishnaP/233150907, O-G/233150925, saurav_24/233151260, Crypto_Wiz/233151484, ajaykumar2004/233151824, ankitsingh7979092/233151869, saigowthamkumar011/233152117, QuickSilver/233152299, ashisharyan/233152315, shivamjha/233152449, Vivek_117/233152467, MK2001/233152516, Vishnu_Shukla/233152625, i_ayush23/233152627, vishalgoud_777/233152721, pratyushraghuvanshi73/233152789, am_2002/233153165, AdarshPandey_24/233153187, ArsheeyaTabassum/233153292, kalluboiiii/233153350, saideepu5692/233153507, 21b01a0526/233153853, Vedula_Aakanksha/233153939, hiteshkumar8642/233153988, Gismodos86/233154249, vishal22101/233154272, mani_gsn123/233154292, Krishnendu016/233154388, rijuoddusual1999/233154603, Hitzjack28/233154628, yashasrikommu/233154943, Gen_Raiden/233155200, 21H51A0557/233155230, priyanshu_kumar/233155254, naruto2022/233155484, ayush_agr0311/233155989, jaiwanth/233156185, FakeCheck/233156461, ashraful_ridoy/233156638, Sadanand26/233157829, Sandeep_Tripathi/233158057, cosmodivine/233158176, aniket_main_/233158715, devil_1627/233159058, nullifier24/233159145, pankaj001/233159970, mamalak08/233160008, akashmaity57/233160166, zddtuus/233160574, BethHarmon/233160824, Harish_74/233161058, Nullpinter/233161472, roytijesha_30/233161900, manoj_kumar__10/233162028, itzkamal22/233162197, Harshitsingh30/233162730, saurabh033/233162777, sameer693/233162943, ja_g_an/233163193, yoshith/233163562, smfaizalam786/233163821, gauravsingh9864/233164082, sauravmh/233164326, imyash_b/233164337, HayatKoujan/233165365, praghavendar/233165583, Aparna165/233165627, SilentSpecterCode/233166087, arifakhtar51/233166627, uncool/233166850, 228r5a6620/233167211, IndianPsycho/233167868, AnialwaysOP/233168083, nithin93/233168736, mynameone/233169012, xcoderdevesh/233169033, Sangeetha_87/233169375, rahul__04/233169658, harsh___mishra/233169692, GoD_GaMerS/233169748, YAGNAPRIYA_2004/233169775, rushikesh_kotkar/233169998, c_dola/233170299, Gaurav_17/233170303, Vaibhav__Prasad/233170399, kiran_2244/233170630, Addictiveboy/233171085, Vaibhavp22/233171185, mnish_0.9369/233171432, Vishwas23/233171627, madhurgupta12/233172083, bvrith_95/233172398, jaythakor/233172632, bdbhatt01/233172642, maharshi1712/233174750, harshitchopra09/233174763, sumanth43028/233174807, yousofkortam/233174877, Terminator2004/233175070, chuvya_24/233175081, gupta05/233175542, A.Hazem_2003/233175598, aadi900/233175898, lakshmi_sowmya/233175992, deepanshu.26.g/233176074, Cpp_Master/233176451, SSingh_248/233176571, isham_026/233176598, Priyank295/233177099, AhmedLaithy/233177384, sanskar.resoluteai/233177489, suraj9192/233177560, HardCoder404/233177647, pravallika1805/233178427, kishanthakor/233178717, vinit__jha/233179019, tesla_11/233179191, komalkbairwa/233179363, NITHINKRISHNA/233179422, Mahin273/233179551, Yuvraj_17/233179572, aakash_811/233179605, ritu_raj_07/233179950, Perhac/233179955, crack__/233180114, falcoln/233180457, sharmahimanshu/233180660, sadhukhansubham588/233180881, mdmahfuzahmed2223/233180900, 21h51a7326/233180993, ankit7890/233180996, gin_spirit/233181047, sidhart25/233181535, AGOLAPUAMULYA/233181573, aryankrsingh4/233181649, vigneshdvs/233181691, Medhavi_Sahgal/233182302, suka_gehu/233182574, tejateju69/233182741, amaanahmad52/233182742, sowmya_04/233183322, mohamed_eltaweel/233183364, gugajcvda/233183394, Aryan_a1996/233183486, uday03meh/233183491, hawdiatarun/233184376, sabbirkhanoni/233184675, manasraj007/233184991, -.........__/233185098, anuragthakurNitrr/233185294, comalonwizme/233185303, verma.shivam2605/233185657, drynor/233185824, giriramireddy/233186059, hexaoctalai/233186216, CMRIT25_21R01A7301/233186650, 777_Raju_Ranjan/233186919, beanomahto/233188428, utkarsh_tyagi437/233188469, sudoaptrana/233188779, anoosh_op/233189369, shivansh2611/233189616, Ahmed_Mahmoud2/233190016, A_mish/233190062, Snehal652/233190096, _Mridul/233190188, rishabh0903/233190516, captive_geek/233190897, subba_reddy13/233191688, kanishks281/233193052, Zakir060/233193086, vishalsaw1092003/233193258, pavanjadhav17/233194069, pyansu07/233194321, blitzzz15/233194411.

I am writing this post in response to the accusation of plagiarism during Codeforces Round #909 (Div 3) for Problem C. I want to express my genuine concern and assure everyone that I did not cheat during this contest.

Firstly, I want to highlight that as an Expert-rated participant, the contest was unrated for me. There was no incentive for me to cheat in a contest that wouldn't contribute to my rating. Moreover, I have consistently participated in Codeforces contests for over 1.5 years, and my commitment to fair play remains the priority.

Regarding the code similarity issue, I understand that there are resemblances between my code and that of others and the number is really huge (I'm shocked too seeing that). However, I want to clarify that I used a portion of the maxSubSum function from GeeksforGeeks. I am unsure whether this is permitted on Codeforces, but I take responsibility for not explicitly mentioning the source.

I urge you to please review my coding style, variable names, and previous submissions. You will find that while I might have used a common algorithm or function, the overall structure and approach are uniquely mine. I also noticed similar solutions from high-rated coders in the contest, which adds weight to my stance.

I want to emphasize that I have learned from my past mistakes during my initial days on Codeforces. I was involved in plagiarism in two contests when I was a Newbie, but I have since rectified my approach and have strictly adhered to honest coding practices. My friends and family know how dedicated I am to problem-solving, I miss family gatherings just to give CF contests. It is my hobby. I

I genuinely fear the possibility of my account getting permanently banned for an unintentional mistake. I fear because I have been working hard to achieve milestones on my account. I feel happy whenever I look at my graph which shows my commitment to CP and fear losing motivation and strength. I can get back to the same rating by creating a new account but this profile shows my struggle and journey.

I urge the Codeforces community and administrators to thoroughly investigate the matter. And I apologize for using third-party code. Please!

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

Hi. I got a plag which I think is wrong, the links to both the solutions are —

his — 233160268

mine — 233148751

The question itself was very basic and majority of the solutions had same approach but even then, both of our solutions are very different and only thing that match is the mantra at the top which is a very famous mantra. Please remove this plagiarism and make this round rated for me once again.

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

Lmao, guess I gotta write about it here. I got a system message about coinsiding solutions with kiyomaru28 (233110440233185978 and 233115355233171404). The solutions literally have nothing in common other than a giant ASCII art, wich I use in all my codes, that are available on codeforces for everyone to see (one of the first uses: 228909025), except that for some reason they credited a different person for the original art.

No idea how codeforces system desided that it's cheating while a lot of people probably use giant headers, like AtCoder one or something from top competitors here and have no problems with it

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

    yea, the exact same thing happened with me as well... I guess they only plag check headers now .. and not bother to look for what is inside.