maroonrk's blog

By maroonrk, history, 3 years ago,

We will hold AtCoder Grand Contest 053. This contest counts for GP30 scores.

The point values will be 400-700-900-1000-1400-2400.

We are looking forward to your participation!

• +78

 » 3 years ago, # |   -109 Is It Rated?
•  » » 3 years ago, # ^ |   +6 Yes!
•  » » » 3 years ago, # ^ | ← Rev. 2 →   +33 Correction: yes, for the range $[1200, \infty)$.
 » 3 years ago, # |   +63 Protip: don't start with A!
 » 3 years ago, # |   +23 Does atcoder do anything about people submitting their code in an alt to avoid penalty? like plag checks? I have like 6 penalty rn and when looking at standings, I see many accounts with submissions towards the end with no comp history or 1 comp. Is this normal or am I imagining things
 » 3 years ago, # |   +41 My notepad after trying to D:https://i.imgur.com/WEUuoTx.jpg(geranazavr555 fix images in spoilers please)
 » 3 years ago, # |   +18 Any race ranking? Do AtCoder admins or clist.by admins plan to create such one?
•  » » 3 years ago, # ^ |   0 aropan created the list on the clist. Thanks a lot.
 » 3 years ago, # |   +28 Where does the formula from the second-to-last line from editorial of C come from? During the contest I came up with the rest of the solution in like 15 minutes and spent some decent amount of time trying to calculate this, coming to the conclusion that it's impossible :P. So is this somehow obvious?
•  » » 3 years ago, # ^ |   +71 I also didn't manage to get the formula during the contest, but that's how I understand it. Let $X_i$ be the event where there is a greater value than $A_i$ among $B_1, \dots, B_{i+d}$. The probability we are looking for is $P(X_1 \cap \dots \cap X_n)$, which is equivalent to $P(X_1) \cdot P(X_2 | X_1) \cdot \dots \cdot P(X_n | (X_1 \cap \dots \cap X_{n-1}))$. Now it's easy to see that the mysterious $\frac{2i+d-1}{2i+d}$ is just $P(X_i | (X_1 \cap \dots \cap X_{i-1}))$ because $X_i$ won't hold only if $A_i$ is the greatest value among $A_1, \dots, A_i, B_1, \dots, B_{i+d}$.
•  » » » 3 years ago, # ^ |   +8 Thanks a lot, got it.
•  » » » 3 years ago, # ^ |   +5 what about (n + i-1) / (n + i) ? Can you explain more, pleaes?
•  » » » » 3 years ago, # ^ |   +14 Maybe it's still the same as 2i + d — 1 / 2i + d but for the case when i + d exceed N
 » 3 years ago, # |   +36 For problem D, I don't understand the proof in the editorial that $t_{i,j}>=T_j$ for $i  » 22 months ago, # | 0 Is there any detailed proof for problem D "don't need to check$i