We will hold Polaris.AI Programming Contest 2026(AtCoder Beginner Contest 457).
- Contest URL: https://atcoder.jp/contests/abc457
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20260509T2100&p1=248
- Duration: 100 minutes
- Writer: sounansya, harurun4635, vwxyz0
- Tester: vwxyz0, cn449
- Rated range: ~ 1999
- The point values: 100-200-300-400-475-525-625
We are looking forward to your participation!









I hope I get high score
unsigned __int128_t score = -1;
Hope too.
(stO tty Orz)
excited
Hope to solve A to D , thanks
Hope too.
glhf
I'll try opening problems in order EDCBAFG this time. Hope I'll succeed!
What a buggy E!
solved upto $$$E$$$ in 40 min, stared at $$$F$$$ for the rest. good round.
So why is there a problem that was released just two months ago that is exactly the same as today's task G? Even the problem names are almost the same.
Simply copy and paste the code, delete the specific solution output, and you'll get AC.
and https://www.luogu.com.cn/problem/P11546
Finally I solved problem F
G is a Joke.
I couldn't understand E at first and I also tried to solve F, but I failed.Only ABCD……
Me too, I misunderstood what E meant
I have a very good solution to problem F. Consider working backwards. d_n-1 is always 1. Consider the newly introduced i. i has only two possibilities: maximum or second largest. If it's neither maximum nor second largest, let dp_k represent the distance k between the maximum value and i. If i is the maximum, dp_0 += dp_di-1. If i is the second largest, dp_di += dp[di — 1]. If neither is true, choose any value and sum them up: dp_k+1 = dp_k * (n-i-1). Then, by maintaining the offset, we can optimize the problem from O(n^2) to O(n).
Problem G is almost the same as the one from POI 1998.
joke
G https://www.luogu.com.cn/problem/P16380
Also https://www.luogu.com.cn/problem/P5939
Did anyone use a RMQ data structure for E?
Yes.
Joke contest.G is as same as [NordicOI 2026] Catching Apples and E is dirty
I hope I can solve E one day