We will hold AtCoder Regular Contest 210.
- Contest URL: https://atcoder.jp/contests/arc210
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20251116T2100&p1=248
- Duration: 120 minutes
- Number of Tasks: 6
- Writer: maspy
- Tester: Nyaan
- Rated range: 1200 ~ 2799
The point values will be 300-600-700-800-1000-1100.
We are looking forward to your participation!
I was automatically registered as unrated, how to change it to rated?
Is it div.1 or div.2?
Is div1+2.
wow 6 prob arc!
So hard!!!
A,B,D very easy!!
D ???
D ez?I don't think B is very sz.
only four contestants, hos lyrics and three other 1dan+ contestants, passed D. are you one of them?
Compared to problem D, which is simple, I prefer to believe that your code was generated by AI
Isn't that a bit much for you to say?
It is true
I'm not as capable as you.I hold you in high regard.
ARC_T[1,2,3]=ABC_T[3,5,5]
Then why did I solve abc432_c but get WA in arc210_a?
Then why did I fail to solve both B and C? I submitted C seven times :(
Losing rating again.
I think this ARC's C is much harder than previous ABC's E.
How to solve problem C? I don't know at all greedy (as an atcoder maxrating>=2100 user)
Can't believe another nonsensical ARC appears.
I love you
love you too.
That's like flipping a coin and being shocked that you got heads.
Why nonsensical? I think D,E are good.
Why is my code wrong? link
I generated 5 million cases at random but all of the answers are correct.
I have the same problem, what troubles me is why the errors are concentrated in the test points of 03_mid.
It's possible that the data in 03_mid has been specially constructed, reflecting some potentially flawed practices in our code.
Coincidentally, I also generated many sets of data, but none of them had any issues.
Thx.
Look down, plz. The same issue might exist in your approach.
It seems there's an issue with your approach. The process of simulating the combination of smaller coins into larger ones should start from the smallest denomination and merge upwards, rather than only considering values 20 less than the current coin.
A direct implementation of the above solution would be $$$O(n^2)$$$. Therefore, you need to merge the coins from the smallest to the largest denomination in one go, and then apply a greedy approach from the largest to the smallest denomination, breaking down the previously simulated merged coins. This would bring the complexity down to $$$O(n)$$$. Referring to my code might be helpful.
That's very helpful. Thx.
Thx.
Just curious, when will rating get updated on kenkoooo. Most of the time it's pretty fast.