We hope you liked the problems! Unfortunately, problem C turned out to be harder than usual. Please read its editorial, we hope you'll find that the intended solution is not that hard.

1763A - Absolute Maximization

Idea: DreadArceus
Prepared by: DreadArceus

1763B - Incinerate

Idea: gupta6007
Prepared by: gupta6007

1763C - Another Array Problem

Idea: .utk.
Prepared by: .utk.

1763D - Valid Bitonic Permutations

Idea: warks
Prepared by: warks

1763E - Node Pairs

Idea: crimsonred
Prepared by: nobita1729, DreadArceus, crimsonred

1763F - Edge Queries

Idea: nobita1729 DreadArceus
Prepared by: nobita1729 DreadArceus

Will be posted soon.

• 23 hours ago, # ^ | ← Rev. 2 →   +19 from tourist's submission lets have dp[l][r] as number of ways we can construct our array from l to r, with elements [n-(r-l+1)+1, n] (basically (r-l+1) of highest distinct elements) regardless of K the next element we are gonna put is n-(r-l+1), so we try to extend out dp to the left and to the right to calculate dp[l-1][r] and dp[l][r+1]. Do not forget to check if the element we put next is valid to restrictons given wth (i, j, x, y)

  • 11 hours ago, # ^ | I understood the solution of D that is provided in the editorial. But trying to understand the dp approach for the same. I saw many solved it by dp but not able to get my head around. Please explain anyone.

  • 6 hours ago, # ^ | ← Rev. 2 →   0 Can you please tell why we always choose "(r-l+1) of highest distinct elements"?

• 22 hours ago, # ^ | Hey, thanks for pointing this out. The validator for problem F removes bridges and then ensures that all the subgraphs remaining have no articulation points. This is enough to fail all the graphs that will cause solutions to the problem to fail. The Petersen graph doesn't have a Hamiltonian cycle and thus fails the condition given in the statement but will not fail any solutions or the validator. Since we cant find Hamiltonian cycles in all the subgraphs efficiently, the validator is made like this. According to the statement, the Petersen graph should fail though, sorry for the confusion. Do you have any suggestions to improve the validator?

  • 22 hours ago, # ^ | Thanks for the explanation. The condition in the problem statement seems like it might be infeasible to check. Probably the correct thing to do would have been to use a different condition or disallow hacking.

    • 22 hours ago, # ^ | We had wanted to disable hacking but, in the end, thought that this validator would be enough. We were only thinking about the cases that cause solutions to fail and overlooked graphs like these. Sorry for the inconsistent validator. We will do better next time!

• 20 hours ago, # ^ | Did you find out why it happened? I'd really like those points back :(

  • 19 hours ago, # ^ | They are removing the cheaters ! :) they will come back

• 21 hour(s) ago, # ^ | I think that biconnected graphs are enough to satisfy my condition,and finding a hamilton cycle for such graphs is equivalent with finding one in any graph.

• 19 hours ago, # ^ | Thanks. Fixed it.

• 18 hours ago, # ^ | The problem statement states that . The power of each monster is at least . If there are still some monsters alive, will decrease by at least every iteration. If reaches zero before all monsters die, the answer is "NO". Thus, the loop will run for a maximum of iterations per test case before reaches zero, and since there are at most test cases, the loop will run for iterations in total at maximum, well within the time limit.

  • 18 hours ago, # ^ | Thanks, makes sense.

    • 17 hours ago, # ^ | I like your solution, although more difficult to implement. Could you explain how you come up with it ?

      • 16 hours ago, # ^ | I needed a way to calculate the number of attacks necessary to kill a given monster, but I hadn't noticed the subtlety that vgtcross pointed out, so instead I found a formula for the amount of damage dealt over m attacks, given Genos power level. Then I solved for m such that the damage dealt was equal to the monster's health, h, and took the ceiling of the solution as my number of attacks. Then Genos power level was reduced accordingly and I moved on through the list of monsters.

• 14 hours ago, # ^ | Thanks for pointing it out! Will correct it.

• 14 hours ago, # ^ | Thanks! Fixed now.

• 12 hours ago, # ^ | Use binomial formula: sum(C(n,k))=2^n where k=0 to n

• 7 hours ago, # ^ | You may refer to this comment. A similar approach was also described by satyam_343 during testing as well. Hope this helps :)