INVITATION TO CODUO-2021

general
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42

I’ve been trying to compare my solution to yours and cant really find the error. Could you take a look at it? Sadly, overflow wasn’t the cause of this problem. :frowning_face:

SUBMISSION

43

Amazing Approach.

44

Ok thanks

45

It’s showing your code’s runtime is 1.06 sec but the time limit is only 1 sec

46

That’s because CodeChef uses adaptive limits for different languages - it’s not like CodeForces where your algorithm may fail due to use of a slower language.

47

Are the test cases for Messi and his Permutations problem weak ?
https://www.codechef.com/viewsolution/45022231 this solution passed tests but gives wrong answer for the input 5 3 and all the inputs where n>y

48

Can someone please explain me the approach for Messi and his Permutations problem ?

49

Have you decided the cutoff rank?

1 Like
50

We have decided to send invites to the top 40 teams from the ranklist. And another top 10 teams from our college. The invites will have the finalised date and time and other details mentioned in it.

51

There was an error in a testcase of EZYE because of Integer Overflow,I am terribly sorry for not realizing this while testing the problem,I have fixed that testcase and all solutions have been rejudged.Thanks to sg1729 and physics0523 for asking me to review the test cases.I really hope that you all can forgive me and I will make sure that this does not happen in the Finals.

1 Like
52

Will you consider 16 seconds difference between teams with rank 40 and 41 and make the team with rank 41 unqualified​:pleading_face::pleading_face:?
image

53

For crazy string. A couple observations:

  1. The number of inversions is the same as the number of needed adjacent swaps to sort a string.
  2. A string is sorted when there are zero inversions remaining.
  3. If there are inversions remaining then there are also helpful swaps so Axios can always reduce the number of inversions by one but only one (since it is an adjacent swap).
  4. With Equinox there is a 50% chance that the number of remaining inversions increases or decreases by one.
    This gives us a recursive formula on the expected value. Denote EA[x] as the expected number of swaps until the string is sorted given the string has x inversions starting with Axios. EE[x] is the expected number of swaps until the string is sorted starting with Equinox.
    From (3) we have EA[x] = 1+EE[x-1]
    From (4) we have EE[x] = 1+1/2 * EA[x-1]+1/2 * EA[x+1]
    Solving for EA[x] gives EA[x] = 4 + EA[x-2]
    The I.C are EA[0] = 0 and EA[1] = 1
    So EA[x] = 4*floor(x/2) + (x%2) where x in the number of inversions.
54

Should we create new team with same participants on HackerEarth website for competing in finals?

55

have you sent out all the invites?

57

Yes

58

yes the invites have been sent

59

@pigmeister
Have u sent it via mail??
My team hasn’t recieved the invite yet…

60

I am unable to start the contest

61

After editing the link to add information I am able to do it, but the START NOW button redirects me to same page

62

Yes, start now button is just redirecting to the same page.