How was ZIO?

general
,
1

How did ZIO go for you guys? I felt like it was easier than last year, I attempted 65 marks, but I don’t know about silly mistakes :weary: :pensive:

“ZIO 2023, more like ZMO 2023” - Me during the test

2

I am getting following answers (I smuggled out answers on my hall ticket :smirk:)

  1. (a) 30
    (b) 136
    (c) 441

  2. (a) 53
    (b) 33750
    (c) 5751022 (probably wrong oops)

  3. (a) 3
    (b) 112
    (c) 344960

  4. (a) 181
    (b) BLANK
    (c) BLANK

3

I’m getting same answers as the user above except I didn’t attempt 2(c) and I’m getting 70 as the answer for 3(b).

4

How did you get 70 for 3(b)?

5

How did u get the answers for 2c and 3c?

6

I remember getting 2*(7C4) at the end

7

For 2c, I just did bruteforce along with casework, and for 3c, basically do this algorithm:

  • For a range [l \cdots r], find the minimum index. Say it is m.
  • Then, split at m into [l \cdots m - 1] and [m + 1 \cdots r].
  • Then, since we can get no new information by including m in any of our queries, we will now calculate the number of ways of assigning all the numbers in [l \cdots r] (except m of course, since that is fixed) to the right half. That is \dbinom{r - l}{r - m}.
  • Then multiply that by the number of solutions you get for [l \cdots m - 1] and again, multiply that by the number of solutions you get for [m + 1 \cdots r].
  • Base case is: For r - l = 1, there will always be exactly one solution. (Of course, you can break early even if r - l = 2, and there is some special case, but I’m talking about a formal definition of the recursive function)
  • Thus, we are done by calling the number of solutions of [1 \cdots n].
8

It’s actually 2 \cdot \dbinom{8}{3}. (according to me, maybe it is wrong.)

9

Was the value of N 8 or 9 if you’d remember?

10

Can someone explain the logic for the 1st question? I’m getting the B part correct but my answer for A is 36 and my answer for B is 496

11

You are not subtracting the common parts…Think about it, maybe do some writing.

12

Howsopro sir. And why did you leave UFDS?? :wew:

13

what was the test case for 2(a)?

14

What is the algorithm for problem 4, parts (b) and (c) specifically?

15

There is an algorithm based on maintaining components in small-to-large fashion, along with offset values which can solve those subtasks in reasonable time.

16

Hi bro I didn’t get what you explained in your 3rd point .
since minimum element is on m then we should calculate number of ways in [m+1 … r] as this part won’t affect previous indexes cause minimum is at m.
What do you think about it?