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Обсуждение задачи 1810. Антиуравненияlilipottter New tests // Задача 1810. Антиуравнения 11 сен 2020 17:34 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 Shen Yang is there O(P) algo? // Задача 1810. Антиуравнения 13 дек 2017 14:19 rt Buer I need help ! Prompt please! [1] // Задача 1810. Антиуравнения 17 июл 2011 18:33 if det(A) == 0 , what should I do ? xurshid_n Re: I need help ! Prompt please! // Задача 1810. Антиуравнения 14 авг 2012 18:29 If A - is not square matrix ( i.e. k != n ), how to calculate det(A) ? Phạm Quang Vũ - ConanKudo \(^(oo)^)/ any hint ? [4] // Задача 1810. Антиуравнения 19 мар 2011 12:28 Could anyone give me the hint to solve this problem ? me Re: any hint ? [3] // Задача 1810. Антиуравнения 27 мар 2011 22:07 Tools of linear algebra work with any field ( https://secure.wikimedia.org/wikipedia/en/wiki/Field_(mathematics) ), including finite ones like integers mod 3. The problem is effectively Gaussian Elimination. sklyack Re: any hint ? [2] // Задача 1810. Антиуравнения 29 мар 2011 00:50 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 me Re: any hint ? [1] // Задача 1810. Антиуравнения 29 мар 2011 13:45 Uhm, sorry. I was wrong. svr Re: any hint ? // Задача 1810. Антиуравнения 29 мар 2011 20:13 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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