Hi everyone,
The second Indian ICPC Amritapuri onsite regional round will be held tomorrow in Bengaluru, Coimbatore and Kollam.
We plan to host a stream covering the round. The stream links will be posted shortly. The problems will be posted in this blog when the contest starts. The editorial will be posted after the contest ends.
Hope you enjoy the contest!
Contest Details
- Public ranklist Link — TBD
- Date & Time — 29 December 2024, Sunday, 9:00 IST (Tentative start time)
- Duration — 5 Hours
- There will be 8 to 20 problems to solve
- Problem statements — Google Drive
- Will follow standard ICPC scoring system (20 minutes penalty and 1 point per problem)
- The questions may not be sorted by difficulty
- Stream link — Codechef Channel AryanC Channel
- Mirror streams on — CodeChef and aryanc403 channel
UPD : The stream has started and the problem statements have been released on this Google Drive
8-20 problems why do I have a feeling there's gonna be 14 problems tomorrow O-O
Why not 20 :)
Would be shits and giggles if we kept only 8. :P
ranklist please
Could you please share the ranklist
Please share Problem Statements as well.
Is there a public ranklist for this round? The link in the stream (https://www.codechef.com/ICPCAMROS24) is inaccessible to the public(?).IIRC we could view ranks on codedrills last time.
Solutions for L and K?
In K, i was able to get a diophantine equation with some inequalities and a variable to minimize.
In L, i don't have a proper proof but i believe we can calculate answer for different levels independently and maximize the sum at each level.
We will post thr complete editorial soon (after the closing ceremony), give me some time
When will the final ranklist be published ?
It will be published today.
Still not released or ami missing something ?
Daldo sir :)
Since there is no editorial yet, i might as well elaborate on what i think the solution to L is, I mentioned before that the levels are independent which is not completely true and it was the mistake we were making in contest.
Basically the idea is that, for any level $$$i$$$, we have some players to pick such that they allow us to form a valid tree i.e. a valid pairing at the start that leads to this result on level $$$i$$$ and the sum of player's supporters is maximum. So, we start from last round, for each player $$$j$$$ on some level $$$i$$$, we will have a left threshold above which we can choose any player. We can proof that if we choose a player on some level $$$i$$$ to maximize sum then those players will also be chosen on the next level. We observed this and thought the answer for different levels is independent.
But there is one more condition, let's say we pick some players $$$i_1,...,i_k$$$ then we can pick at most player between any two already picked players on the next level. It's easy to modify the code for this condition, basically we maintain the players picked on the last level and then make sure we only choose at most one player between any two players.
UPD: Found my mistake
dp[i7][i6][i5][i4][i3][i2][i1]whereijis no of yet to be decided losing teams injthround.ij <= 2^{7-j}$$$O(2^{21})$$$ states per testcase, 7 transitions for each state.
We start with
dp[1][0][0][0][0][0]Transitions looks like following
dp[i7][i6][i5][i4][i3][i2][i1]->dp[i7][i6][i5-1][i4+1][i3+1][i2+1][i1+1]+(1+2+3+4+5)*aj, notably we do not need index in dp state.Thanks
Would be great if you could create a gym from the problem statements and the actual tests!
Yeah that would be great if someone does so
when will final RankList be Published? Can u upload RankList now?
Nice contest, waiting for the ranklist :)
Couldnt make it to the regionals this time but next year for sure!
Can we please have the final ranklist ?
Can we have the questions in codeforces pls :)
Can someone explain how to solve question D ?
It was discussed on stream
Final rank list : here