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
“ZIO 2023, more like ZMO 2023” - Me during the test
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
“ZIO 2023, more like ZMO 2023” - Me during the test
I am getting following answers (I smuggled out answers on my hall ticket )
(a) 30
(b) 136
(c) 441
(a) 53
(b) 33750
(c) 5751022 (probably wrong oops)
(a) 3
(b) 112
(c) 344960
(a) 181
(b) BLANK
(c) BLANK
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).
How did you get 70 for 3(b)?
How did u get the answers for 2c and 3c?
I remember getting 2*(7C4) at the end
For 2c, I just did bruteforce along with casework, and for 3c, basically do this algorithm:
It’s actually 2 \cdot \dbinom{8}{3}. (according to me, maybe it is wrong.)
Was the value of N 8 or 9 if you’d remember?
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
You are not subtracting the common parts…Think about it, maybe do some writing.
Howsopro sir. And why did you leave UFDS?? :wew:
what was the test case for 2(a)?
What is the algorithm for problem 4, parts (b) and (c) specifically?
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.
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?