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!
# | User | Rating |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3823 |
3 | Benq | 3738 |
4 | Radewoosh | 3633 |
5 | jqdai0815 | 3620 |
6 | orzdevinwang | 3529 |
7 | ecnerwala | 3446 |
8 | Um_nik | 3396 |
9 | ksun48 | 3390 |
10 | gamegame | 3386 |
# | User | Contrib. |
---|---|---|
1 | cry | 167 |
2 | Um_nik | 163 |
3 | maomao90 | 162 |
4 | atcoder_official | 161 |
5 | adamant | 159 |
6 | -is-this-fft- | 158 |
7 | awoo | 157 |
8 | TheScrasse | 154 |
9 | Dominater069 | 153 |
9 | nor | 153 |
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!
Name |
---|
How to solve problem F >_< ..
Does anybody know why don't they display results immediately ? Can't wait for results ! :(
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 )
1 — easy, 2 — sort then easy, 3 — all tickets to girl one then easy, 4 — binary search, 5 — math, how to solve? 6 — thinking...
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 .
I believe it's LCM of first k integers, instead of k!
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.
oh ok, if it is divisible by n, it is also divisible by 2*n and so on, so lcm should be right.
how do you want to solve 4 with binary search?
never mind, i remembered the problem differently...
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
Can anyone add this contest to the gym, please?
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.