We gladly invite you to participate in the Three Subregionals Cup 2017 – a team-based online programming competition parallel to the ACM ICPC Subregionals in Moscow, Minsk and St. Petersburg. The tournament will be organized by the team of Yandex.Contest and the jury of the Subregionals. The online rounds will be held on the Yandex.Contest system.
Each online contest will be available for solving for three days starting the day of the QF.
Start times are: Moscow – October 22, 12:00 MSK, Minsk – November 2, 11:00 MSK, Saint Petersburg – November 4, 13:00 MSK.
Teams that participate in the onsite version of one of the three quarterfinals will be able to participate in the online versions of the other quarterfinals via Yandex.Contest. Teams that don’t participate in any of the three quarterfinals onsite will be able to participate in all of them online via Yandex.Contest. Cumulated results of three subregional contests will be calculated based on Grand Prix 30 scoring system.
You can register for the online rounds at any time. You can represent a single person or create a team using the teams link. Each invited team member has to confirm his participation in the team. You can choose participation type and team members upon registering for the contest.
The rules of the online rounds correspond to the onsite quarterfinal's rules.
For your convenience a test contest is available.
UPD Welcome to the Moscow QF
Starting of the contest postponed by one hour. ETA of starting 12:00 MSK. Sorry for inconvenience.
It seems we cannot participate the contest as team.
fixed by Oleg
Do you have some editorial for Western Subregional problems? I'm interested in problem J, I've accepted some solution in upsolving, but don't know whether/why it is correct — sort values by first coordinate and then assume that Y is some cyclic segment of this array.
I have the same question. Anyone have some ideas?
Unfortunately, tests are quite weak.
Try to split indices in a such way that the first inequality satisfies in both ways to choose X and Y (it's always possible). Then trivially determine correct one to satisty the second inequality.
So how to do the first step properly?
Consider sequence a1, a2, ..., an a1 >= a2 >= ... >= an | sum_X = 0, sum_Y = 0
Let's consider elements one by one. Move i into X iff sum_X <= sum_Y else move into Y.
Now one can prove that sumX ≥ sumY - minY and sumY ≥ sumX - minX
I got AC with a even more weak strategy: randomly permute the given goods in a line, then try to split it into first i/second n-i group (for i=1..n). Check if there's one solution. If not, do random again. With no more than 50 times randomization I found a solution for all test cases.
The virtual contest for the Western Subregional will be opened at 12:30 Moscow time.
Sorry for the delay, but we prefer to check the settings twice before contest will be opened for virtual participation.