COMPFEST 14 Preliminary — Editorial

1725A. Accumulation of Dominoes

Author: Pyqe
Developer: nandonathaniel
Editorialist: Pyqe

1725B. Basketball Together

Author: FerdiHS
Developer: muhammadhasan01
Editorialist: Pyqe

1725C. Circular Mirror

Author: Pyqe
Developer: steven.novaryo
Editorialist: steven.novaryo

1725D. Deducing Sortability

Author: Pyqe
Developer: TakeMe, Pyqe
Editorialist: Pyqe

1725E. Electrical Efficiency

Author: steven.novaryo
Developer: steven.novaryo
Editorialist: rama_pang

1725F. Field Photography

Author: Pyqe
Developer: Pyqe
Editorialist: Pyqe

1725G. Garage

Author: Nyse
Developer: Nyse
Editorialist: Pyqe

1725H. Hot Black Hot White

Author: Pyqe
Developer: steven.novaryo
Editorialist: steven.novaryo

1725I. Imitating the Key Tree

Author: Pyqe
Developer: Pyqe
Editorialist: Pyqe

1725J. Journey

Author: gansus
Developer: gansus, steven.novaryo
Editorialist: rama_pang

1725K. Kingdom of Criticism

Author: Pyqe
Developer: Pyqe
Editorialist: rama_pang

1725L. Lemper Cooking Competition

Author: Pyqe
Developer: steven.novaryo
Editorialist: rama_pang

1725M. Moving Both Hands

Author: Pyqe
Developer: Pyqe
Editorialist: rama_pang

• 32 hours ago, # ^ | is this solution o(n)?

  • 32 hours ago, # ^ | Which solution?

    • 31 hour(s) ago, # ^ | the two pointers one for B

• 2 days ago, # ^ | ← Rev. 2 →   +1 Run shortest path algorithm. Change the direction of all edges. Run shortest path algorithm.

  • 2 days ago, # ^ | more detailed? pls

    • 44 hours ago, # ^ | You wanna find a point x that has min minway(1, x) + minway(p, x). Also, we change the direction of all edges, so now, minway(1, x) + minway(p, x) = minway(1, x) + minway(x, p'), where p' is a new point for p, that shows, that p' is p in graph with reversed edges. We can see, that every point on the shortest way from 1 to p' has min sum of that 2 ways.

      • 12 hours ago, # ^ | I don't know where my code is wrong. Can you help me find out the details? thank you 171057638

• 45 hours ago, # ^ | It is actually possible to build all sparse trees simultaneously using small-to-large, but the time complexity is worse. The intended solution uses an algorithm that runs in for each set . The algorithm is as follows: Sort the vertices in based on their euler tour traversal order. All extra vertices in the sparse tree can found as the of every pair of vertices in that are adjacent in the sorted order. Once all required vertices are found, we can find the edges by iterating the vertices (including the extra ones) in euler tour traversal order and maintaining a stack. You can optimise it further to make the total time complexity . But the time limit is not that tight that even the small-to-large solution is able to pass.

  • 35 hours ago, # ^ | Ah thanks, that's really cool. Haven't really seen many problems with this "auxiliary tree" idea, so its nice to learn good techniques for it.

int Temp = a[i];
while(Temp <= D)
{
    Temp += a[i];
}

• 35 hours ago, # ^ | It probably just runs that fast. The hot path only includes an add, a compare, and a conditional jump, which is < 3 cycles with branch prediction. Computers run at a few GHZ, so 500ms sounds right. The things/second heuristic is just a heuristic for usualish groups of operations, you can do better if u have a fast loop body (especially if the compiler avx-ifies, which might be happening here). To see exactly what is happening u can try putting it in https://godbolt.org/ with the correct compiler version/flags.

  • 20 hours ago, # ^ | ← Rev. 2 →   0 Alright, I got it. Thanks :D

• 37 hours ago, # ^ | ← Rev. 2 →   +2 we can write (b^2-a^2) as(b-a)*(b+a) // proof for 1 can not be expressed in terms of b^2-a^2 so the min positive value of a we can take is 1 and for b its 2 as (b>a) as stated in the problem so, (b-a)=1; (b+a)=3; and their multiplication would give us 3 as the min value which can be expressed in terms of b^2-a^2; and one more conclusion can be drawn is that after three all odd numbers can be expressed in the form of b^2-a^2 (because we can express every odd number (lets say a)as 1*a; and a can always be represented as a sum of 2 consecutive numbers which are always odd **** lest take a=4 ,b=5; b-a=1; b+a=9; 9*1=9; that is odd) // proof for 4 cannot be expressed in terms of b^2-a^2 to make equation even we have to make (b-a) even first so in order to make that even min value of a we can take is 1 and for b is 3 so, (b-a)=2; (b+a)=4; and their multiplication will give us 8 that is the minimum even value we can achieve and from here we can draw one more conclusion that all the even values will be multiples of 4 as no matter what we take values of a and b whenever (b-a) is even (b+a) would also be even (because to make the diffrence of 2 numbers even their parity should be same and if we add same parity numbers then result is even) so that would make the result divisble by 4

  • 37 hours ago, # ^ | Thank U

• 32 hours ago, # ^ | NO Java AC solution for M. Please test the solutions for Java, Python as well before the round to know if the same solution passess the time limit. This is not fair that although both the solutions has the time complexity, it gives TLE in Java and AC in C++

• 20 hours ago, # ^ | The comment is hidden because of too negative feedback, click here to view it

• 13 hours ago, # ^ | I could come up with this result by building a sequence, I added the difference between b^2 and a^2 in the sequence as follows: 4 — 1, 9 — 4, 16 — 9, 9 — 1, 25 — 16 .. by listing the "difference between squares" in nondecreasing order, I got the sequence: 3, 5, 7, 8, 9, 11, 12, 13, 15, 16, 17, 19, 20, 21, 23, 24, 25, ,27, 28, 29, 31.. I noticed that the first number "3" doesn't follow the pattern, so I assumed it was a special case, but the remaining numbers follow a consistent pattern that I came to figure out as: "3 + 4 * (N // 3) + N % 3"