zeus_iitg's blog

Invitation to CODEJUNK 2020
By zeus_iitg, history, 4 years ago, In English

Greetings Codeforces,

I would like to invite you all to CODEJUNK 2020, the competitive coding competition of Techniche, IIT Guwahati's annual Technical Fest, and which is being held in collaboration with The Coding Club of IIT Guwahati.

The contest will be held on CodeChef and will start at 20th July 2020, 20:00 hrs IST. You will have 5 problems to solve in 2hours. The ranklist will be ICPC style with the penalty set to 10 minutes.

The problems have been prepared and tested by me (zeus_iitg), Nishchay Manwani (EnEm) and Tejas Khairnar (BrownieTK).

You can also register for prizes and laddus worth Rs. 50,000 along with assured coupons worth Rs. 200 with a registration fee of Rs. 100.

For more information visit techniche.org/techolympics

Good luck. :)

NOTE: Registration is only for prizes. You can participate without registering as well.

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Comments Comments (35)
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4 years ago, # |
← Rev. 3   Vote: I like it +15 Vote: I do not like it

.

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4 years ago, # |
← Rev. 2   Vote: I like it +37 Vote: I do not like it

why competition is not free? You should make it free if you want more number of participants to take part.

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    4 years ago, # ^ |
    ← Rev. 2   Vote: I like it -16 Vote: I do not like it

    Deleted

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    4 years ago, # ^ |
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    The contest is conducted under the banner of Techniche, IIT Guwahati and sponsored by CodeChef. We had no jurisdiction over the contest. Also, the registration is only for prizes. You can take part without registering as well. As one of the problem setters, I would be glad if you participate. Hope you enjoy solving the problems.

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

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

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

Reminder. Contest starts at 20:00 hrs IST

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4 years ago, # |
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It will be a perfectly unbalanced problemset like every codechef contest is. Especially when we are having setters with such high ratings.

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4 years ago, # |
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coupons of worth 200 of what?

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    4 years ago, # ^ |
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    coupons worth Rs. 200 Rs. denotes Indian currency,i.e. INR

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4 years ago, # |
  Vote: I like it +15 Vote: I do not like it

The queue's such a pain, my submission hasn't been judged even after almost 8 mins now :(

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4 years ago, # |
← Rev. 2   Vote: I like it +6 Vote: I do not like it

Patience is the key to get accepted on Codechef

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4 years ago, # |
  Vote: I like it +32 Vote: I do not like it

Can international guys register for prizes?

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4 years ago, # |
  Vote: I like it +5 Vote: I do not like it

why illuminati as logo

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4 years ago, # |
  Vote: I like it +22 Vote: I do not like it

The test cases for The king of snakes weren't strong for $$$K = 3$$$. Checking the pair of integers about the mean gives AC which is wrong for a large set of test cases ..one of them being:

3 3
1 1 50

where $$$21$$$ is the correct answer but checking the pair of integers about the mean gives $$$18$$$ as the answer.

Since my ternary search code also passed I'm assuming that the test cases aren't wrong but just not strong enough.

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    4 years ago, # ^ |
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    Wtf .. How this solution passed!!! .. this observation works well with linear and quadratic terms but in cubic it's not always true .. These aren't week but extremely weak test cases

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4 years ago, # |
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please, anyone, explain the approach for problem C, D?

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4 years ago, # |
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For King of snakes when k = 3, we cannot determine x value based on mean and mean + 1, but many solutions got accepted, weak test cases I guess. How to solve for k = 3 ?

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    4 years ago, # ^ |
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    Not just for k=3 but for all the values of k given in the problem you can just do ternary search over the answer.

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    4 years ago, # ^ |
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    U can even brute force over all values by just maintaining prefix sums .

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4 years ago, # |
  Vote: I like it +7 Vote: I do not like it

Waiting for the editorial to see C and D graph theory problems! zeus_iitg

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    4 years ago, # ^ |
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    XOR of all odd distance node values from the node 1 will be the state… a non zero value is winning and 0 is losing.

    We can observe that it’s always possible to move from winning to losing for our opponent similar to how u prove a nim game… u just move from the odd distance to even distance by choosing an appropriate node… it’s similar to removing stones in the nim game because the even distance values aren’t considered in our XOR which means an operation on the odd distance node is like removing a stone.

    When in the losing state we can prove that we only move to a winning state. If we choose any node we will only obtain a non zero XOR since the current XOR is 0.

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4 years ago, # |
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This is the editorial for the 5th problem of the contest. Only 2 people managed to solve it during the contest. I tried my best to make the explanation clear. Feel free to ask for clarification if any part isn't clear. Here is the link Venkatesh is bored — editorial

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    4 years ago, # ^ |
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    any link for editorial of all questions??

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    4 years ago, # ^ |
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    I think you should also mention why calculating it bruteforcely will be enough, because of the result of phi(phi(m)) <= m / 2 for m > 1.

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    4 years ago, # ^ |
    ← Rev. 3   Vote: I like it 0 Vote: I do not like it

    I think the test data is weak or the constraints were not mentioned correctly because in worst case you should give all 'm' as prime in range [10^8, 10^9] and make them distinct too(you can easily find 10^5 primes in this range). Then the complexity is at-least O(T*sqrt(m)) since the complexity of calculating phi(m) would be O(sqrt(m)) [Although phi(m) = m-1 for prime but you need to check the primality of 'm']. Now T*sqrt(m) = 10^9.5 which should not pass the given TL....And I think that's the reason that many people didn't try this approach......any comments?

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4 years ago, # |
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Please publish the editorials as soon as you can zeus_iitg.

The problems were very good and I want to up solve them and I have tried all I've got.

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