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Discussion of Problem 1349. FarmIn the measurements it is written that when 1 response 1 2 3. But in the condition said "If there are several solutions, display, where a minimum. If such multiple, display, where the minimum of b and so on". So the correct answer should be 1 1 2. (В промерах написано, что при 1 ответ 1 2 3. Но в условии сказано "Если решений несколько, вывести то, где a минимально. Если и таких несколько, вывести то, где минимально b и т.д". Поэтому правильный ответ должен быть 1 1 2.) Edited by author 09.11.2014 17:33 Have you read the statement? "Output Three ***different*** integers (a, b and c) such that..." Hahaha, Timus should have something like a "Like" button for comments like yours. ;) je suis ne pas tres attentif Edited by author 01.09.2024 03:42 WHYYYYY if u have same problem solution is 3 4 5 I remember once going to see him when he was ill at Putney. I had ridden in taxi cab number 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavourable omen. "No," he replied, "it is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways." G.H.Hardy 1729 = 1^3 + 12^3 = 9^3 + 10^3 See on hint in author of the problem. One more hint: Use wiki if you know nothing about Mr. Pierre Edited by author 22.03.2016 02:06 search wikipedia for FERMAT'S LAST THEOREM and you will come to know everything about this problem.(1349 Farm) Edited by author 01.11.2011 15:24 I have discovered a truly marvellous proof of this, which this margin is too narrow to contain Thanks you are a saviour. I was applying all my mathematics to get those triplets for n > 2. in sample for n=1 -> 1 2 3, solve for n=1 -> 1 1 2 !!! who right? 'a' have to be diferent from 'b' 0 < a < b < c < 101 0 < a < b < c < 101 There is no such condition in a problem description. "Three DIFFERENT integers (a, b and c) such that..." why my clever solution is wa5 and my stupid sol is acc? Look at the author's name...))) --------------------------------- Sorry, for my bad English... Yeah! Oh yeah! I'm so ashamed that I solved it using long ariphmetics at first :( ... Three different *postitve* integers (a, b and c) ... [Code deleted] I have AC, but I can't prove it. Edited by moderator 02.10.2021 16:15 NOBODY can prove it! Elementary proof doesn't known nowadays, only in 1995 some mathematician found complicated evidence (130 pages) of this theorem. Edited by author 07.11.2009 20:07 I think that your are right! My opinion is to see on "no elementary" proves as on politic plays of big boys that can't give anything to algorithms and programming. Well, I got AC either. And my solution is very similar, the only difference is that I'm using CASE statement (: But I designed my solution being aware of Fermat's last theorem. I would be grateful if you could explain, how you managed to figure out the solution without knowing that theorem. Just by considering some sample situations? Or were you aware of that theorem either? rafailka Wouldn't high-precision be fine? This is "small Fermat theorem" Fermat told that you have no answer if n>=3 No, it's a "Great Theorem of Ferma", and he proved it only for n=4. The all theorem was proved by Andrew Wiles (England) in 1994. |
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