Hello everyone!
We are happy to invite you to participate in the International Coding League, 2025 (ICL'25), conducted by the students of BITS Pilani, one of the flagship events of our technical fest APOGEE.
This will be a team programming contest and the duration of the contest will be 2 hours.
Contest Link: International Coding League — Round 1
Registration Form: Click Here
Both rounds of the contest will be held on Codeforces, and would be in teams of 1 or 2. The prize pool for the online round is INR 12,000 (5K + 3K + 2K + 2 * 1K). To be eligible for these prizes (and to participate in the next round), teams must fill the registration form.
The Top 25 teams in this round along with 15 teams from BITS Pilani would qualify to the second round, which would be held offline at BITS Pilani, Pilani campus on 30th March, 2025. The prize pool for the offline round is INR 18,000 (10K + 6K + 2K).
The problem setters and testers are: D3VC, DaBest, Quick-One, lefteryx, lycogno, nannu, ritviksingh791
We hope you enjoy the contest and have fun!
Happy Coding! :)
UPD1: Registrations for the contest have started on Codeforces! You can now form your team and register. Ensure that your team name matches the one filled in the Google Form to be eligible for prizes and qualifications.
UPD2: The contest has started.
UPD3: There was an issue with setter's solution for problem E. We will not be considering the problem for the final ranklist.
Quick-One orz
Quick-One ORZ
Quick-One orz
Quick-One orz
When is the deadline for registration?
26th March 2025, 20:00 IST
Is it for college students only or open to all?
It's open to all
is this only for Indian people?
No, it’s open for everybody.
Are cross-college teams allowed?
Yes
Why clash with codechef;)
I'm trying to register for the contest but i don't see any team names in the drop down menu
Might be because you are not in a cf team. go to your profile > teams > create new team
Are the problems sorted in difficulty order?
Could you allow viewing others' submissions?
How to solve G ?
You can solve using dynamic programming, $$$dp$$$ [ $$$i$$$ ][ $$$j$$$ ] = minimum cost to make $$$F$$$ [ $$$i$$$ ]= $$$j$$$.
The transitions are; $$$dp$$$ [ $$$i$$$ ][ $$$j$$$ ] = min( $$$dp$$$ [ $$$i$$$ ][ $$$j$$$ ], $$$dp$$$ [ $$$i-1$$$ ][ $$$k$$$ ] + max( $$$j-k$$$ , $$$0$$$ ) ).
Note that this transition can be done only |$$$k-j$$$| $$$\ge$$$ $$$T$$$ [ $$$i-1$$$ ]
Final answer is minimum of $$$dp$$$ [ $$$n-1$$$ ][ $$$j$$$ ] over all $$$j$$$
Time complexity. $$$O$$$ ( $$$n$$$ . $$$B²$$$ ) where B can be around $$$3000$$$
Nice solution, thanks :) I should have been able to do it
How to solve F?
https://math.stackexchange.com/questions/4910001/formula-for-number-of-graphs-with-n-vertices-and-k-connected-components
Test data of C is weak actually
How to solve J ?
Can you please open the solutions of other to see? Also when will the editorial be released. I really need to know the approach for C.