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| Показать все ветки Спрятать все ветки Показать все сообщения Спрятать все сообщения | | Can anyone explain sample test? | Sq1 | 1805. Чапаев и шифровальная решётка | 10 авг 2022 11:59 | 2 | Can anyone explain sample test? I have smth like this:
Case n = 4, k = 1:
0000
0000
0000
1111
Case n = 4, k = 2:
0000
0000
0001
0111
Case n = 4, k = 3:
0000
0000
0001
1011
Case n = 4, k = 4:
0000
0000
0001
1101
Case n = 4, k = 15:
0000
0000
0011
1100 Matrix should not overlap itself after 90-degree rotations (no two '1' at same position) and should cover all NxN elements after 4 such rotations. | | any algo | Ibragim Atadjanov (Tashkent U of IT) | 1805. Чапаев и шифровальная решётка | 10 авг 2022 11:56 | 8 | any algo Ibragim Atadjanov (Tashkent U of IT) 12 ноя 2010 10:57 Who know how to solve it. I think all the varians will be
4^(9 + 7 + 5 + 3 + 1) = 4^25 when n = 10. So i cannot count from 1st to kth variant Re: any algo Vedernikoff 'Goryinyich' Sergey (HSE: АОП) 12 ноя 2010 13:56 O(N^4) simple algo exists. Re: any algo Ibragim Atadjanov (Tashkent U of IT) 13 ноя 2010 18:52 Can you tell anything else? I mean is the algo search(Binary Search or smth else). Please give me a clue Re: any algo Vedernikoff 'Goryinyich' Sergey (HSE: АОП) 13 ноя 2010 22:49 Suppose you have already built a part of a grille and is thinking what symbol is to put in the following cell. Can you count the number of grilles which can be built with such a beginning and symbol? I suppose that O(n^2) algorithm is even more simple. Both in inventing and coding. That's right. I track amount of remaining variations for every cycle and their total product. This information is enough at each of N^2 steps to decide if it should be '0' or '1' Given advices too weak to be helpfull.
For me key idea was forming n*n/4 classes with 4 cell in each and working
with level, where 1<=level<=n*n;
We counting all combinations under level in each classes exept whose are used already. Re: any algo Strekalovsky Oleg [Vologda SPU #1] 8 ноя 2011 00:01 Given advices too weak to be helpfull.
For me key idea was forming n*n/4 classes with 4 cell in each and working
with level, where 1<=level<=n*n;
We counting all combinations under level in each classes exept whose are used already. My algo was same as Vedernikoff 'Goryinyich' Sergey (HSE: АОП) said. Algo is very simple and wasn't too hard to implement. Edited by author 08.11.2011 00:01 |
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