eduardische's blog

By eduardische, 11 years ago, translation, In English

Today (20.10.2012) at 14:00 GMT the first contest of the Croatian Olympiad in Informatics takes place. You can login/register here. Duration: 3 hours. Good luck!

  • Vote: I like it
  • +38
  • Vote: I do not like it

»
11 years ago, # |
  Vote: I like it +6 Vote: I do not like it

How to solve problem F >_< ..

  • »
    »
    11 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Does anybody know why don't they display results immediately ? Can't wait for results ! :(

    • »
      »
      »
      11 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Your personal results are up right after the contest ends. Full results are posted in a week or so ( accoring to the organizers , this is an 'appeal period' , some contestants may complain about testcases {not meeting problem's constraints} , the grading process .. etc )

»
11 years ago, # |
  Vote: I like it 0 Vote: I do not like it

1 — easy, 2 — sort then easy, 3 — all tickets to girl one then easy, 4 — binary search, 5 — math, how to solve? 6 — thinking...

  • »
    »
    11 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    5 — one can observe that for any number smaller than k! the number in the second iteration will be smaller or equal to k. so all you have to do is finding out, which of the numbers have 2 as smallest "not divisor" which one has 3 and so an, up to 21 i think .

    • »
      »
      »
      11 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      I believe it's LCM of first k integers, instead of k!

      • »
        »
        »
        »
        11 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        but when you have a number n smaller than k!, than there is one number between 1,2,3,4,5,..,k that does not divide n, so one of the smallest should be between 1 to k.

        • »
          »
          »
          »
          »
          11 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          oh ok, if it is divisible by n, it is also divisible by 2*n and so on, so lcm should be right.

  • »
    »
    11 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    how do you want to solve 4 with binary search?

    • »
      »
      »
      11 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      never mind, i remembered the problem differently...

  • »
    »
    11 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    6 — trivial divide and conquer works, when merging two 2k groups into a 2k + 1 group, you need to multiply three 2k × 2k matrix, use ordinary O(n3) algorithm, then whole algorithm works in O(n3) either

»
11 years ago, # |
  Vote: I like it +9 Vote: I do not like it

Can anyone add this contest to the gym, please?

  • »
    »
    11 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    The problem is that it's not an ACM ICPC format but IOI format instead and as far as I know the gym cannot accept the second one yet.