### lunchbox's blog

By lunchbox, 4 months ago,

This post contains the editorials for all tasks contained in the AtCoder DP Contest, since there is no official editorial. This is written by nwatx and me.

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## Z — Frog 3

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• +242

 » 4 months ago, # |   +3 LUNCHBOX OFZ
•  » » 4 months ago, # ^ |   -8 LUNCHBOX OFZ
•  » » » 4 months ago, # ^ |   -22 LUNCHBOX OFZ
•  » » 4 months ago, # ^ |   0 LUNCHBOX OFZ
 » 4 months ago, # |   +16 LUNCHBOFZ
 » 4 months ago, # | ← Rev. 2 →   +20 Someone finally wrote one!
 » 4 months ago, # |   +6 Thanks a lot! It will be really helpful :)
 » 4 months ago, # |   +8 Good Job,,, I was waiting for this from ages. Thanks a lot ... OFZ
 » 4 months ago, # |   +13 ​I’m telling you, lunchbox is as cracked as he is jacked. Saw him the other day benching 696lbs while winning IOI. I asked him what he was doing and he said "better than you" and walked out with a explosion.
 » 4 months ago, # |   +1 Thank you very much. I was solving these problems for past 2-3 days but couldn't find editorial. What a nice timing.
 » 4 months ago, # |   +3 For the problem T-Permutation, I came up with the following idea: We first create n-1 edges directed from small numbered index to large numbered index. i.e. if we have $i^{th}$ sign as $<$ then we create an edge from $i$ to $i+1$. Now, the answer is the number of topological orderings for this graph. But I am stuck here. Can anybody help me how can I find the answer? Thanks in advance.
 » 4 months ago, # |   +14 I can't find the Editorial for N — Slimes anywhere in this post... Am I missing something?
•  » » 4 months ago, # ^ |   0 Thanks, I just added it!
 » 4 months ago, # |   +3 This is one of the best list which covers all varients of Dp and gradient starting from absolute beginners. Very nicely written Analysis and Code! Thanks a lot:)
 » 3 months ago, # | ← Rev. 10 →   +10 For task U(Grouping), you can reduce finding cost[mask] from $O(2^n \cdot n^2)$ to $O(2^n\cdot n)$ by doing like this. Spoilerfor (int i = 1; i < 1 << n; i ++) { int first = (int) log2(i), last = (int) log(i & -i); int p_first = 1 << first, p_last = 1 << last; if (first == last) cost[i] = 0; else cost[i] = cost[i -p_first] + cost[i - p_last] - cost[i - p_first - p_last] + a[first][last]; } And yeah, you can reduce calculation time from log2 by some precalculation, so you can do it in $O(2^n)$
 » 3 months ago, # | ← Rev. 2 →   +5 I think in the analysis of W-Intervals the range is supposed to be [0,l-1] not [0,a-1]. lunchbox
•  » » 3 months ago, # ^ |   0 Thanks for catching that!
 » 2 months ago, # | ← Rev. 8 →   0 sed
 » 2 months ago, # |   0 Time complexity for K is wrong.
 » 5 weeks ago, # |   0 Can someone explain problem L more clearly? Why is the transitions like that, and why in the analysis dp[i][i] = 0 but in the code dp[i][i] = a[i]?
•  » » 4 weeks ago, # ^ |   0 Errichto explained it well on his stream about that contest. In case you didn't saw that https://youtu.be/FAQxdm0bTaw?t=6892The transition formula , can be rephrased in that way."Pick up the left bound , the opponent will pick up the best value on interval dp[L + 1][R]. The difference between your turn and opponents' next turn will be your answer for that interval.Pick up the right bound , the opponents will pick up the best value on interval dp[L][R — 1]. The difference betwene your turn and opponents' next turn will be your answer for that interval.Find max between these two choices."Can't explain why dp[i][i] = 0 in analysis , while in the actual implementation dp[i][i] = a[i].