0ne8's blog

Invitation to IIT Hyderabad Code Clash 2026
By 0ne8, 4 months ago, In English

Hello Everyone!

Infero, the competitive programming club of IIT Hyderabad is glad to invite you to Code Clash, our programming contest conducted as part of our annual techno-cultural fest, Elan & ηVision. The scoring will be based on the ICPC format. It is an individual event, no teams will be allowed.The contest is open for all to participate.

Contest details:

The Winner receives prizes worth ₹7500!
The 1st Runner Up and 2nd Runner Up receives prizes worth ₹5000 and ₹2500 respectively and others will get Participation Certificates. To be eligible to receive the prizes, register here.

UPD: Contest Link : Code Clash 2026

The contest contains atleast 1 Interactive Problem. Those who don't know how to solve Interactive problems read here.

Note: Only Indians are eligible to receive prizes. Best of Luck to all Participants.

UPD: We are sorry about AI submissions and blatant cheating, we will go through the submissions and remove cheaters and announce the final results. Thanks for participating and we hope you liked problem set.

UPD2: The Editorial is available now, click here.

UPD3: The Top-3 participants are
1.hitman_py
2.ompatel45672
3..BOSS.
Congratulations to them and thanks for participating everyone!

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

CF BLog post at 08 Jan 26, 11:00 PM IST

Meanwhile Registration Deadline: 07 Jan 26, 11:59 PM IST

Now how to register?

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4 months ago, hide # |
 
Vote: I like it +11 Vote: I do not like it

As a tester, the problems are cool. Will be a fun way to spend a sunday afternoon :)

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

Few questions:

1: Is template usage permitted?

2: What languages are accepted?

3: When you say ¨Only Indians are elligible to recieve prizes¨ does this include certificates? For example, if someone from outside of India won 1st, The first place from India would be considered the official ¨#1¨ but the outsider still get a certificate stating #1? Just seems a bit confusing imo.

Seems like a fun round though! GL to all （＾ｖ＾）

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    4 months ago, hide # ^ |
     
    Vote: I like it +3 Vote: I do not like it

    ¨Only Indians are elligible to recieve prizes¨

    it's about prize money. Only participants from india are eligible for the prize money, rest is for everyone. if you(non-indian) get #1 and any indian gets #2 so the indian will get the #1 prize moeny.

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4 months ago, hide # |
 
Vote: I like it +13 Vote: I do not like it

Excited to participate (and get cooked T_T)

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4 months ago, hide # |
 
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do we have to be present in the campus or we can attend it from anywhere ?

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4 months ago, hide # |
 
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hope i learn something new from the contest

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4 months ago, hide # |
 
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How to attend this ??? In which platform ????

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4 months ago, hide # |
 
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hey 0ne8 i am not able to register as deadline is closed..can you open for a while?

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4 months ago, hide # |
 
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thanks 0ne8 for opening the registration

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4 months ago, hide # |
 
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Excited!!

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4 months ago, hide # |
 
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Contest lowkey fire

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

Editorial? till then please share solution ideas.

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4 months ago, hide # |
 
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any hints for d? i tried with two pointers with two separate pointers for left end to count all possibilities

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    4 months ago, hide # ^ |
    ← Rev. 2  
    Vote: I like it +8 Vote: I do not like it

    For each 1 <= l <= n, find the smallest and largest r, which satisfy l <= r <= n and MEX(a_l,..a_r) = k.

    Both of them can be done using a 2-pointer approach, each requiring 2 * n queries, thus a total of 4 * n queries

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4 months ago, hide # |
 
Vote: I like it +17 Vote: I do not like it

Problem G can actually be solved using just 2 queries. Choose two primes p1, p2 < 1e6 which satisfy p1 * p2 > 1e8.

Any 1 <= x <= 1e8 must be co-prime to atleast one of them. Additionally if x is co-prime to some prime p then lcm(x, x + p) would be co-prime to p.

So if x is co-prime to one of those primes(say p1)

lcm(x, x + p1) = x * (x + p1)

Once we know p1 and x * (x + p1), x can be easily found using binary search.

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    4 months ago, hide # ^ |
     
    Vote: I like it +12 Vote: I do not like it

    We can solve this even in a single query by choosing l=0,r=31; Let p = lcm(x, x+31). Since gcd(x, x+31) = gcd(x, 31), we get p = x * (x + 31) / gcd(x, 31).

    Case 1: 31 divides x

    gcd(x, 31) = 31, hence p = x * (x + 31) / 31

    and therefore p % 31 = 0.

    Case 2: 31 does not divide x gcd(x, 31) = 1, hence p = x * (x + 31) and therefore p % 31 != 0.

    Thus, by checking p % 31, we can determine whether x is divisible by 31. Now solve quadratic

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4 months ago, hide # |
 
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Hey when will the participation certificates be available and how to collect them?

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    4 months ago, hide # ^ |
     
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    They'll be automatically sent to you via Unstop later. It might take some time to finalize the ranklist.

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      4 months ago, hide # ^ |
       
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      When will the results be declared? contest ended almost 7 hours ago....

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4 months ago, hide # |
 
Vote: I like it +3 Vote: I do not like it

Had a good experience thank you for such contest

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