TooMuchPainTheseDays's blog

By TooMuchPainTheseDays, history, 3 years ago, In English

(Give honest opinion)

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

Your time is up. My time is now.

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3 years ago, # |
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It's cheatchef not codechef

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3 years ago, # |
Rev. 2   Vote: I like it +72 Vote: I do not like it

Be Me
a newbie in cp
saw and got excited for snackdown
passed online qualifier by doing 3 problems
saw round 1A
solved 2 easily, solved 3rd in an hour
saw the cutoff Rank less than 1500 will qualify
saw current rank 6xx. 2 days left, gotta try 4th.
only 120 submission for 4th
spent 2 days grinding down 4th.
Scared, current rank is 4xxx.
finally solved 4th with 3 hours remaining.
calm that will qualify, current rank 1xxx
closed the system.
Now seeing current rank 4xxx
FML

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    3 years ago, # ^ |
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    lol same thing happened solved 4/6 and last hour seeing my rank 4k+

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    3 years ago, # ^ |
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    **The qualification criteria for Round 1A has been loosened from "Participants with rank <= 1500" to "Participants who have solved at least 4 problems".

    So all participants who have solved >= 4 problems are tentatively qualified for the Pre-Elimination round (pending plagiarism checks), and others can participate in Round 1B.

    The ranklist will be finalized after plagiarism checks.**

    --- by Codechef_admin

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    3 years ago, # ^ |
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    This is worse than Failed System tests....At least FSTs don't disqualify you....

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      3 years ago, # ^ |
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      Getting DQ'd for cheating is worse than FST? Well I mean... first, I'd argue that it's super easy to avoid getting DQ'd; just don't copy others' code. Second, the punishment for cheating should indeed be much much more strict than for being wrong but giving it an honest try, otherwise it doesn't sufficiently discourage cheating.

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        3 years ago, # ^ |
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        No, you misunderstood. I said inspite of his honest work, he was disqualified because other cheated. He saw himself qualifying until the last hours and then everything changed. This is obviously bad for him more than FST.

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          3 years ago, # ^ |
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          Ah, indeed, but you're not really DQ'd here, right? Just participate in round 1B, and you're good?

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            3 years ago, # ^ |
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            No I am through to round 2.Was feeling sad for the guy.

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              3 years ago, # ^ |
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              No, I mean the guy we are talking about here isn't DQ'd, he is just mildly inconvenienced.

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3 years ago, # |
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Thanks to some youtube bhaiyya/didi selling the idea that getting 1500 rank by cheating in an online contest will somehow help them get a good job.

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3 years ago, # |
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CodeChef really is helping people improve so much. Just look at the level of some people. Solving 4th and 5th in the last hour without any Wrong Answer. ORZ.

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    3 years ago, # ^ |
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    And that too in a variety of programming languages.Codechef OP.

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3 years ago, # |
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Bruh a lot of my friends failed to qualify from such incident :C! And a lot of them solved 4 problems in the first day and thought this was safe enough to qualify. But sadly, cheating grew exponentially! A lot of new EOIers(Egyptian Olympiad in Informatics participants) who are still newcomers got hyped when they solved 3/4 problems at first then the disaster occurred!

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    3 years ago, # ^ |
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    Codechef's moto:- Solve everything or retire

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    3 years ago, # ^ |
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    If you want to rip off CodeJam, atleast do it properly. Anyways, I solved the problems knowing this was going to happen. What was the expected solution of Flipping problem? I tried for every $$$i$$$, if $$$2^{i+1}-1-k$$$ is even and apply operations accordingly.

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      3 years ago, # ^ |
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      Every odd number or 0 had a solution. I solved it as if we are creating from all zeros to first k 1's because it was a xor operation hence order doesn't matter. let's say ith bit is high in k and i+1 th bit is low. You can turn on 2^i places on by turning on 2^(i+1) places on and then turning off last 2^i places off.

      Proof that only odd numbers are possible -

      There are two possibilities — Either don't apply any operation (if k==0) or apply few operations If we are applying few operation, 1st operation will always be used. So there will be 1 high bit after first operation. Now later operations can never change the parity of number of high bits. If you are turning odd number of places on then you will be turning odd number of places off too or if you are turning even number of places on then you will be turning even number of places off.

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      3 years ago, # ^ |
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      Well, it was a constructive algorithm. So it can be hard to get to the constructive part.

      Claim 1:

      If $$$k$$$ is an even number other than 0, then the answer is always $$$\text{NO}$$$.

      Proof: Lets consider a sequence of operations of $$$1,2,4,8,...,2^x$$$. Then we are gonna flip $$$2^{x+1}-1$$$ elements. Since there is an even number of $$$1$$$ and $$$2^{x+1}-1$$$ is odd, then an odd number of $$$1$$$ must remain.

      Construction for any odd $$$k$$$:

      First, find the smallest $$$2^x$$$ such that $$$1+2+...+2^x$$$(lets call such sum $$$S$$$) is bigger than or equal to $$$k$$$. Notice if we changed any element from that sum from positive to negative(lets call the element $$$a$$$), then it will contribute to a difference of $$$2a$$$. For example, if $$$a=16$$$, and we made it $$$a=-16$$$, this will make $$$S-k$$$ to decrease by 32. This gives us an insight to get $$$C$$$ such that $$$C=\frac{S-k}{2}$$$. We can then know when we need to remove $$$a$$$ ones or add $$$a$$$ ones by seeing binary representation of $$$C$$$. So we have constructed a sequence such that is made up of:

      $$$1(x)2(x)4(x)8(x)...2^x$$$

      where $$$x$$$ is either $$$+$$$ or $$$-$$$.

      Applying construction to the problem:

      This is by far, the hardest part with a lot of corner cases due to such cases like $$$K=N$$$ or close to it (at least for my construction). Notice that, if we have something like($$$N=K=9$$$) $$$1+2-4-8$$$, we can remove 1+2 one and then flip it with 4 to make it like such and finally remove 8. For example:

      $$$111111111$$$

      will become:

      $$$111111101$$$

      $$$111110001$$$

      $$$111111110$$$

      $$$000000000$$$

      This works as you first do $$$-1-2$$$ first then flip it with the $$$4$$$ to make it $$$4-(1+2)$$$ Now, you can do that with any sequence of — and +. If you have some + operations then -, you flip at the — operation. If you have — operation without any + operations before it, you simply remove the last few 1's. The only cornercase is when $$$K=2^x-1$$$, as the last operation will be + so you won't be able to flip. You can handle this case alone.

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        3 years ago, # ^ |
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        Cheaters looking at this solution: "No thanks I saved myself the trouble"

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        3 years ago, # ^ |
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        I think that it's easier to look at the problem not as the process of removal of bits from K to 0, but as growing up the sequence of bits from zero length to K (the signs are just inverted compared to your math).

        With this interpretation, we know that the last operation with the largest $$$i$$$ is always adding a big chunk of bits rather than is removing them (otherwise the length of the bits sequence will become negative). The ordering strategy becomes simple: first we add the biggest chunk of bits, then we handle all the bits removal operations (their sum is always smaller than the single biggest positive chunk) and finally handle everything else in any order.

        Here's the code and examples for K=9 and K=15:

        Spoiler
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        3 years ago, # ^ |
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        My solution was very similar to this. I most probably had some implementation issues or corner cases.

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      3 years ago, # ^ |
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      I had a completely different approach, by making observations from the brute force solution.

      Even doesn't work due to parity. For the odd case, it turns out that the direct observation that can be made is by thinking it backward since the operations are invertible, considering an initial array of $$$k$$$ zeroes, iteratively picking the maximum sized island of zeroes, and making the operation such that it fills the highest power of $$$2$$$ number of ones at the end in this island while performing this operation at most one new island of zeroes will be created due to size of the flip, we can just push them into the priority queue.


      Example
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    3 years ago, # ^ |
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    From CodeChef official announcement, they changed the qualification bar from "rank <= 1500" to "solving at least 4 problems". So I think getting 4 is safe :)

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3 years ago, # |
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For some cheaters eat sleep cheat repeat

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3 years ago, # |
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3 years ago, # |
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It's been a long time time since I wrote a cheating blog . Should I write another one??

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3 years ago, # |
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I knew I don't have many chances in the upcoming round, but I still spent hours on the 4th problem (because I found it interesting) and finally solved it which made me rank in the 400s. But to my utter dismay, I found that I am an idiot and my practice of 1.5 years is just me wasting my time. Also stuck between 1500 and 1600 here, I can't even reach Expert. I hope I don't regret doing CP in the future :(

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Life’s not fair, it never was, it isn’t now and it won’t ever be. Do not fall into the trap, the entitlement trap, a feeling like you’re a victim. You are not; get over it and get on with it.

Matthew McConaughey

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3 years ago, # |
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not the first time actually, when I solved 4 I thot it's enuff but well cheaters had another plan(I'm not sure its due to cheater or wot but my rank goes from 800 to 2200+ in last 50 minutes), now will have to try hard in 1B :sadge:

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3 years ago, # |
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cheaters think cp stands for copy paste not for competitive programming

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3 years ago, # |
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Shame, shame.

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3 years ago, # |
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Sometimes I literally feel bad for CodeChef admin's and problem setters, they are trying hard to gift some awesome contest, but people are started to decrease interest on CodeChef for those ****ing cheaters ***h*l*s. Sorry, Can't control me from saying those.

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3 years ago, # |
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Wait, people are still competing in Codechef these days?

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    3 years ago, # ^ |
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    You should too

    We know about cheating. That doesn't mean the problems are bad.

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      3 years ago, # ^ |
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      Why was the time duration decided to be 2 days? It was pretty obvious this would entice cheating.

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        3 years ago, # ^ |
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        By your logic, women in short skirts entice rapists.

        There is only one way to stop cheating. Participants should stop cheating. That's it.

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          3 years ago, # ^ |
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          But atleast the government (in this case, codechef) can improve their security to prevent rapes in the first place. Having contest with a stricter time limit was a necessity.

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            3 years ago, # ^ |
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            Yes I agree. Cheaters get encouraged to cheat only because codechef doesn't do plag checking. With an usual time limit + plag checking after the competition, the experience will be smooth and everyone will be happy.

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              3 years ago, # ^ |
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              Look, I know Codechef sucks.

              But the fact that something this simple has not been done should be a strong indication that it is not as easy as you think.

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          3 years ago, # ^ |
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          An a-bit-off-topic question here: is cheating one of the main reasons for codechef to make long challenges only rated for div 3? Or would you mind sharing your other concerns? You know, long challenges are rare and rated ones are rarer...

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            3 years ago, # ^ |
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            It is one of the reasons, but not the only one. I can’t go in more detail, sorry.

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3 years ago, # |
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Grandma is very disappointed. Never giving another contest on codechef.

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3 years ago, # |
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How are you sure that cheating occured? E was like 2000 CF rating and given 3 days the number of submissions is not that high.

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    3 years ago, # ^ |
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    Well, it is a similar problem to an old Codejam problem. But for someone who didn't see that Codejam problem, it is actually not that easy. Fixing $$$N=K$$$ corner case took me a while. So I rate it at least 2100. Talking about ur theory of 3 days, then the number of solved of this problem after few hours should converge not diverge. According to your theory, if I tried in a 3000 problem, I will be able to get to it in few days as I will diverge to the solution. But this doesn't happen. I am sure if I tried in a 3000 problem, it is likely that I won't be able to get it at all...

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3 years ago, # |
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obvious leak (just tryna stick to the 2-word word limit)

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3 years ago, # |
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UPDATE: The qualification criteria for Round 1A has been loosened from "Participants with rank <= 1500" to "Participants who have solved at least 4 problems".

So all participants who have solved >= 4 problems are tentatively qualified for the Pre-Elimination round (pending plagiarism checks), and others can participate in Round 1B.

as if more than 1500 people solved 4+ problems without cheating.

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    3 years ago, # ^ |
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    Well that is still a bit plausible, seeing as the numbers for problem D gradually increased over the 2 days. But >1500 (my rank jumped from ~850 to 2400 during the leak) people solving a problem they've been stuck on since more than 46 hours on having a sudden epiphany is much more unlikely. I'd say problem 4 was a div 2 mid level prob maybe. Hacker Cup Round 1(which had similar long 1 day format) C was definitely tougher yet many solved it

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      3 years ago, # ^ |
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      The increase of problem D submissions near the end of the contest isn't completely unbelievable. There are at least a few legitimate explanations that don't involve cheating:

      • A lazy person could have foolishly assumed that solving 3 problems was just enough to advance to the next round. Then this person kept slacking off, until he/she realized that this won't work. And finally scrambled to attempt to actually solve the 4th problem in the remaining few hours.

      • Busy people may just have no free time to participate seriously until the weekend. So they wake up on Saturday morning and give problem D a try for the first time. Some of them succeed. Coincidentally there's just half a day left until the contest is over, so the timing of their submissions looks suspicious.

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    3 years ago, # ^ |
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    as if the fourth problem wasn't leaked yesterday....

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3 years ago, # |
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Delete Codechef

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3 years ago, # |
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its a secret collab between telegram and codechef

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Its cheateChef. I was loosing my confidence. So, I stop doing contest on codeChef. Its so frustrating for newbie contestant.

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Deleted