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    <p>You're asking if the regular expression</p> <p>(xxx|xxxx|xxxxx)*</p> <p>matches xx...x, where x occurs 456 times.</p> <p>Here's a solution in O(n+a^2), where a is the smallest of the numbers on the left side (in this case 3).</p> <p>Suppose your numbers are 6,7,15. I'll call a number expressible in the form 6x+7y+15z "available". You are to check if a given number is available.</p> <p>If you're able to get some number n, then surely you will be able to get n+6, n+12, n+18 - in general, n+6k for all k >= 0. On the other side, if you are unable to get some number n, then n-6 is surely not available too (if you could get (n-6), then (n-6)+6=n would be available), this means n-12, n-18, n-6k are not available neither.</p> <p>Suppose you have determined that 15 is available but 9 is not. In our case, 15=6*0+7*0+15*1 but won't be able to get 9 in any way. So, by our previous reasoning, 15+6k is available for all k >= 0 and 9-6k for all k >= 0 is not. If you've got some number that divided by 6 gives 3 as remainder (3, 9, 15, 21, ...) you can quickly answer: numbers &lt;= 9 are not available, numbers >= 15 are.</p> <p>It is enough to determine for all possible remainders of division by 6 (that is 0,1,2,3,4,5) what is the smallest number that is available. (I just have shown that this number for the remainder 3 is 15).</p> <p>How to do it: Create a graph with vertices 0,1,2,3,4,5. For all numbers k that you are given (7,15 - we disregard 6) add an edge from x to (x + k) mod 6. Give it weight (x + k) div 6. Use <a href="http://en.wikipedia.org/wiki/Dijkstra%27s_algorithm" rel="noreferrer">Dijkstra's algorithm</a> using 0 as the initial node. The distances found by the algorithm will be exactly those numbers we are searching for.</p> <p>In our case (6,7,15) the number 7 gives rise to 0 -> 1 (weight 1), 1 -> 2 (weight 1), 2 -> 3 (weight 1), ..., 5 -> 0 (weight 1) and the number 15 gives 0 -> 3 (weight 2), 1 -> 4 (weight 2), ..., 5 -> 1 (weight 2). The shortest path from 0 to 3 has one edge - its weight is 2. So 6*2 + 3 = 15 is the smallest number that gives 3 as remainder. 6*1 + 3 = 9 is not available (well, we checked that previously by hand).</p> <p>And what is the connection to regular expressions? Well, every regular expression has an equivalent finite automaton, and I constructed one of them.</p> <p>This problem, with multiple queries allowed, appeared on the <a href="http://www.oi.edu.pl/php/show.php?ac=e180911&amp;module=show&amp;file=zadania/oi10/sum" rel="noreferrer">Polish Olympiad</a> and I translated the solution. Now, if you hear now a person saying computer science is not useful for real programmers, punch him in face.</p>
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    1. POAlgorithm to determine non-negative-values solution existance for linear diophantine equation
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    1. POAlgorithm to determine non-negative-values solution existance for linear diophantine equation
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    1. CODefinitely non-standard, but I love it! Reminds me of my days in Theory of Computation talking about Chomsky and the good ol' Pumping Lemma ;) +100 for the punch in the face ;)
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    2. COVery, very neat solution. You could also include a bit more info on how the graph can be used as an automaton. My impression is that the automaton is NFA and for example to process 6+7+15=28, 28 % 6 = 4, thus the accepting state would be the node 4, and you have (28 - 4) / 6 = 4 tokens to process, and each edge consumes as many tokens as its weight. Is this right? Also, what's the story of the problem? Is it in a textbook? Or someone came up with the solution in the olympiad?
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    3. COYour description is correct. The path goes 0 -> 1 -> 4, where the first edge means adding 7 (weight 1), and the second edge means adding 15 (weight 2); this consumes 3 tokens - the fourth one is taken by using number 6 once. In http://stackoverflow.com/questions/2912885/ gives a link with more information about the problem. I don't know the story how jury chose the task. Three participants (out of 42) scored maximum points (task "sum"): http://www.oi.edu.pl/php/show.php?ac=e180912.
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