On 08/12/2010 09:56 PM, Martin P. Hellwig wrote: > On 08/11/10 21:14, Baba wrote: > <cut> > > How about rephrasing that question in your mind first, i.e.: > > For every number that is one higher then the previous one*: > If this number is dividable by: > 6 or 9 or 20 or any combination of 6, 9, 20 > than this number _can_ be bought in an exact number > else > print this number >
you are allowed to mix. 15 is neither divisable by 6 nor by nine, but 9 + 6 = 15 I guess, trying to find the result with divisions and remainders is overly complicated. Simple brute force trial to find a combination shall be enough. Also: if you know for example, that you can buy 101,102,103,104,105 and 106 nuggets, then you know, that you can buy any other larger amout of nuggets. 107 = 101 + one box of six 108 = 102 + one box of six . . . As soon as you found 6 sequential solutions you can stop searching. -- http://mail.python.org/mailman/listinfo/python-list
