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i do not know.
The straight forward approach would be:

1) a = x/a;
2) b = (x%a)/b
3) c = ((x%a)%c)/d
...


But that doesnt work, as you have to be able to "undo" the algorithm, when you...

in vs this would work:





only on x86 system, ofcourse.

hi,

i am currently trying to teach my self c++. Workign fin so far, but now i have hit a break with this problem:
(translated by me, sorry)

1) you are given a large number x and a subset of...