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    <p>You can consider a table as list of values. Then, if you have N tables, your problem is to find lists of N integers (each one taken from one of the N tables) whose product is equal to a value p. You can solve the problem recursively:</p> <ul> <li>given a non-empty list of tables <code>{t1, t2, t3, ...}</code></li> <li>given a product value <code>p</code> that you look for</li> <li>for every value <code>v</code> in <code>t1</code> you must look for solutions of the sub-problem with a product value <code>p / v</code> and and tables <code>{t2, t3, ...}</code> (this assumes that <code>p % v == 0</code>, because we're dealing with integers</li> <li>etc.</li> </ul> <p>Some java code below:</p> <pre><code>public class SO6026472 { public static void main(String[] args) { // define individual tables Integer[] t1 = new Integer[] {2,-2,4,7}; Integer[] t2 = new Integer[] {3,-3,-8,0}; Integer[] t3 = new Integer[] {-1,-4,-7,6}; Integer[] t4 = new Integer[] {1,5}; // build list of tables List&lt;List&lt;Integer&gt;&gt; tables = new ArrayList&lt;List&lt;Integer&gt;&gt;(); tables.add(Arrays.asList(t1)); tables.add(Arrays.asList(t2)); tables.add(Arrays.asList(t3)); tables.add(Arrays.asList(t4)); // find solutions SO6026472 c = new SO6026472(); List&lt;List&lt;Integer&gt;&gt; solutions = c.find(36, tables); for (List&lt;Integer&gt; solution : solutions) { System.out.println( Arrays.toString(solution.toArray(new Integer[0]))); } } /** * Computes the ways of computing p as a product of elements taken from * every table in tables. * * @param p the target product value * @param tables the list of tables * @return the list of combinations of elements (one from each table) whose * product is equal to p */ public List&lt;List&lt;Integer&gt;&gt; find(int p, List&lt;List&lt;Integer&gt;&gt; tables) { List&lt;List&lt;Integer&gt;&gt; solutions = new ArrayList&lt;List&lt;Integer&gt;&gt;(); // if we have no tables, then we are done if (tables.size() == 0) return solutions; // if we have just one table, then we just have to check if it contains p if (tables.size() == 1) { if (tables.get(0).contains(p)) { List&lt;Integer&gt; solution = new ArrayList&lt;Integer&gt;(); solution.add(p); solutions.add(solution); return solutions; } else return solutions; } // if we have several tables, then we take the first table T, and for // every value v in T we search for (p / v) in the rest of the tables; // we do this only if p % v is equal to 0, because we're dealing with // ints List&lt;Integer&gt; table = tables.remove(0); for (Integer value : table) { if (value != 0 &amp;&amp; p % value == 0) { List&lt;List&lt;Integer&gt;&gt; subSolutions = find(p / value, tables); if (! subSolutions.isEmpty()) { for (List&lt;Integer&gt; subSolution : subSolutions) { subSolution.add(0, value); } solutions.addAll(subSolutions); } } } tables.add(0, table); return solutions; } } </code></pre> <p>The code prints solutions for a slightly modified version of your example:</p> <pre><code>[2, 3, 6, 1] [-2, -3, 6, 1] </code></pre> <p>The solutions work for any number of tables. There are ways to improve the algorithm, for example by using memoization and dynamic programming. But I think that the recursive solution is more clear.</p>
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    1. COHi Macros , thanks for the help and the sol but I was looking for Multiplication operation to take place .Here the combination that we are geting is addition. I Hope you got my point.
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    2. CO@user756742: oh, sorry! I got confused. However, the idea is the same even for product. I've changed my code, so that is now consistent with your question: the recursive procedure gives you all the combination of elements from your table whose product is equal to a certain value. Have a look and let me know if you have questions.
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    3. CO@Macros Load of thanks for the sol.... :)
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