Use this for anything related to plagiarism - Reporting cheating, appeals, etc

Hopefully 1B will be fair

if plag not checked this time i will leave CODECHEF

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guys lets show codechef the power of unity . just dont participate from next contest as a protest against their plag checking . i would like to request all the people out there advancing to next round plz dont participate to stop it right here .

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But how are you sure that there will be no cheaters in 1B :rofl: :rofl:

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I have been watching guys who are newbies in Codeforces solving 5 problems … Ahhh only codechef things :woozy_face:

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There will not be enough cheaters on round 1B so congrats to all you guys will qualify in next round.

Cheaters are lazy too :slight_smile:

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The best part is that 40% of solutions would give a 90% and above match rate on a simple plagiarism checker which they obviously won’t run.

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~2000 out of ~20000 people cheating! That’s not really the cp community that CodeChef advertises.

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Best think is some 2 star guy who himself did plagiarism is saying in threads that he mailed the leaked solution to codechef when his c++ solution of equal beauty matched with 60 percent people.

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cause after cheating 4 solutions they thought they will qualify.LOL

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Lol.:rofl:

MY Snackdown’s rank is second most disappointing thing in my life first one is being getting ignored by my crush.

Solved four questions still haven’t qualified maybe codechef has done swap(me, 2 star(guy))

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If you don’t mind can you please explain your solution for Equal Beauty?

oh brother this question has really made me suffer a lot I can’t think any simpler solution for this

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Hi,can you explain the approach for the 4th problem?

Fantastic

Actually how people get the correct solution in last 1hour that too 1000 people in one are correct thats insane!!!
i think green ticks appear more Fastly at end.

the first loop is for dividing the whole array in two subsegment and making all the element in both the subsegment equal (optimal way to make equal is to change all element to the middle value)
the second loop isn’t complex its self explanatory
the last loop for see if I take the ith element and minimum element then what is the most optimal value I need to take to match with maximum element this is done by binary search first I take difference as ith element - min then I need to get a value equal or close to max - diff

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i don’t know if you get the solution but I make the solution look like a nightmare

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