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Everyone Should Virtual Participate In Former Chinese Official Rounds, Their Problem Of Very Good Quality
For Example, In The Provincial Team Selection Contest That Recently Ended, It Got Excellent Problems. I Especially recommened the T2 in Day 1. Its Statement Is Clear And Inspiring And The Testcases Are Well Designed.
Also T1 in Day 2. The Time Limit Is Well Designed.
EDIT: UP
You've stopped acting?
Agree. You can try to VP the excellent round.
First the contest is in a NO-FEEDBACK mode, so you can only test samples locally while you don't know the actual speed of the grader. The grader may slower of faster than your current computer, but there are $$$2$$$ problems have great time limit (Day1 B, Day2 A) so that you will easily receive the verdict TLE when the score has announced.
Second The the problem B have great time limit. The algorithm of $$$\frac{nq}{8}$$$ received a score between $$$44$$$ and $$$88$$$, and $$$\frac{nq}{w}$$$(Official Solution's time complexity) received a score between $$$88$$$ and $$$100$$$, all because of TLE.
Third The algorithm of $$$nq$$$ easily passed all of the testcases of B even it's easy to hack. A number of participates used this solution to pass the problem and successfully became a member of the provincial team.
You can try to VP it to feel the great quality of Chinese OI.
Also, Masters and Grandmasters can't enter the provincial team while some people with the rating less than $$$2000$$$ can (Same province). Guess why!
ok i failed to enter the provincial team
Ok I became the 'Grandmasters' in your words... /sad /sad
That's because the contest type has too much randomness. Sometimes we call it 'lottery contest'.
I didn't enter the team because of the two problems either :(. The conductors could have prevented the accidents. But apparantly they chose the opposite :(.
This is absolutely not fair. I've wrote a email to CCF about this. But no one knows if this will work.
It's definitely not going to work. Changing the test data after contest has no precedence.
(I'm the main of the author)
Is there a precedent that trivial brute force got a high score in official contest?
I don't know. I highly doubt the answer is "yes".
For example, in NOIP 2021, the problem
variancehas a simulated annealing solution which passes all tests with good parameters. I don't know if you think the SA solution in that problem can be called "trivial BF" or not, but at least its difficulty is also significantly lower than the official DP solution.SA in that problem is hard to hack or even impossible to hack. And if you want get a high score using SA, you still need to discover the key property of that problem, which is also the hardest part of the problem.
bitset = good problem
Give link to the website to participate from (if possible give an English website cuz I don't know chinese)
emm... Seems that this blog is just a rant written by CHN CPers, as Day1 B and Day2 A are controversial problems :)
Problems here:
day1A
day1B
day1C
day2A
day2B
day2C
The problems are in Chinese, so you need to use a translator.
up
Season 2 has come.
Last Saturday's problem C is a very good one. It reminds us of the importance of reading the constraints, and then tells us that it's not actually that crucial. Meanwhile, the solutions are very diverse, fully demonstrating the charm of problems with partial scores. Even solutions that got wrong answer on samples will still be rewarded with additional points for their efforts.
Now everyone should solve CSP-S2025 Problem B, This problem does a "good job" of distinguishing between different levels of ability. The person who solved it using $$$O(\alpha n2^k)$$$ / $$$O(\alpha nk2^k)$$$ algo got 92/88/80 points in this problem, but who solved it using $$$O(\alpha nk2^k\log n)$$$ algo even if $$$O(\alpha m2^k\log n)$$$ algo got 100 points.
CSP-S2025 Problem C is also a good problem. This problem effectively identifies those who didn't review string theory. And the excellent limitation about $$$t_{j,1}$$$ and $$$t_{j,2}$$$ is also very heart stirring (Note: this problem didn't guarantee that $$$|t_{j,1}|=|t_{j,2}|$$$)