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 contest will be held online on Codeforces.
- Start Time: 14:00 IST on 11th January, 2026
- Duration: 2 hours.
- All the problems were created by nimish.agarwal.2006, atigdng, HyperBoolean, Adamant18, FranXenoz, COdERBRO513 and me.
- Special thanks to Ceilings, Siddarth_MSR and Titan493ASST for testing the problem set and coordinating the event.
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!
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?
It will be re-opened soon.
How to attend this ??? In which platform ????
The contest will be held on Codeforces.
It is open now, register here
Can you please reopen the registration?
It is reopened, register before Noon.
Please re open it...........
As a tester, the problems are cool. Will be a fun way to spend a sunday afternoon :)
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＾）
¨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.
Excited to participate (and get cooked T_T)
So humble sir
how to be so humble and dedicated sir
do we have to be present in the campus or we can attend it from anywhere ?
No need to be present in the campus, it's an online event.
hope i learn something new from the contest
How to attend this ??? In which platform ????
hey 0ne8 i am not able to register as deadline is closed..can you open for a while?
thanks 0ne8 for opening the registration
Excited!!
Contest lowkey fire
How to attend it ???
Here's the link for registration https://mirror.codeforces.com/contestInvitation/d3cc2f5c1e31d04e048e2cf7ade7c6ac9012ccf9
Is it rated ???
No, it is unrated.
Editorial? till then please share solution ideas.
any hints for d? i tried with two pointers with two separate pointers for left end to count all possibilities
For each
1 <= l <= n, find the smallest and largestr, which satisfyl <= r <= nandMEX(a_l,..a_r) = k.Both of them can be done using a 2-pointer approach, each requiring
2 * nqueries, thus a total of4 * nqueriesFor E can you provide some hint ?
Problem G can actually be solved using just 2 queries. Choose two primes
p1, p2 < 1e6which satisfyp1 * p2 > 1e8.Any
1 <= x <= 1e8must be co-prime to atleast one of them. Additionally ifxis co-prime to some primepthenlcm(x, x + p)would be co-prime to p.So if
xis co-prime to one of those primes(sayp1)Once we know
p1andx * (x + p1), x can be easily found using binary search.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
Hey when will the participation certificates be available and how to collect them?
They'll be automatically sent to you via Unstop later. It might take some time to finalize the ranklist.
When will the results be declared? contest ended almost 7 hours ago....
Had a good experience thank you for such contest