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| Показать все ветки Спрятать все ветки Показать все сообщения Спрятать все сообщения | | New tests | lilipottter | 1810. Антиуравнения | 11 сен 2020 17:34 | 1 | Hm. Seems there are no tests with small number of linear independent rows.
I have two solutions with different answers to such tests, but both got OK.
Admins, please, add some.
30 30
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2
...
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2
30 30
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
x20
30 30
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 0
x10 | | is there O(P) algo? | Shen Yang | 1810. Антиуравнения | 13 дек 2017 14:19 | 1 | | | I need help ! Prompt please! | Buer | 1810. Антиуравнения | 14 авг 2012 18:29 | 2 | if det(A) == 0 , what should I do ?
If A - is not square matrix ( i.e. k != n ), how to calculate det(A) ? | | any hint ? | Phạm Quang Vũ - ConanKudo \(^(oo)^)/ | 1810. Антиуравнения | 29 мар 2011 20:13 | 5 | any hint ? Phạm Quang Vũ - ConanKudo \(^(oo)^)/ 19 мар 2011 12:28 Could anyone give me the hint to solve this problem ? But we can't do transformations with antiequations such as with equations. For example, let we have a system:
x1 != 0 (mod 3)
x2 != 2 (mod 3).
It has a solution (2; 0)
If we add the antiequations, we'll get this one:
x1+x2 != 2 (mod 3), but it's wrong for solution (2; 0) of the system!
Am I right?
Edited by author 29.03.2011 00:53 I think that your advice is absolutely right!
Let for example det(Aij)<>0.
Then Answer is 2^k,because (A*X)i in {0,1,2}\{bi}
If det(Aij)==0
we can use gauss method to make standard worm of the matrix A. |
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