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Tough Algorithm?

Started by Gf11speed Oct 13, 2003 at 10:59 PM 9 replies 1k views

Original Post

Oct 13, 2003 at 10:59 PM

For some reason, I can't think of a way to do this. Say I have some numbers: (not necessarily in order) 0 1 2 3 4 and I want to find all the bit sequences that add up to a certain #. For example, if that # is 6. I would get the bit combitions: 11110 01110 10101 00101 .. this means that, for the bit combinations 11110, the numbers: 0 1 2 3 .. all add up to 6. For the bit combo 10101: 0 2 4 ... all add up to 6. Etc. How could I do this? I don't know if I explained this clearly, but thanks for any help you can give. [edited by - Gf11speed on October 13, 2003 12:01:17 AM]

doogle

Oct 13, 2003 at 11:26 PM

quote:
Original post by Anonymous Poster
go learn what the bitwise operators (&, |, and ~) do. this is not a difficult problem.

Do you mean ^ instead of ~? I have never heard of a ~ operator before.

Rayno

Oct 13, 2003 at 11:35 PM

~ is the bitwise NOT operator. It gives you the opposite of a binary number. For example, the bitwise NOT of 100101 would be 011010.

[edited by - rayno on October 13, 2003 12:36:44 AM]

Gf11speed

Author

Gf11speed

Oct 13, 2003 at 11:50 PM

I''m not all that advanced now

Rayno

Oct 14, 2003 at 12:12 AM

It isn''t as complicated as it sounds.

Observe:


...
unsigned short SomeVar = 5; //0000000000000101 in binary

SomeVar = ~SomeVar; //SomeVar now equals 1111111111111010 in binary because ~ just flips all the bits.

//That is 65530 in decimal numbers, by the way.

Nekhmet

Oct 14, 2003 at 2:01 AM

Does it have to be fast? And how big is the list of numbers?

- Josh

Bunnz

Oct 14, 2003 at 2:15 AM

do you want to get the bits of the binary system or of the "special" system you have used within you examples???

Because if you really want to use your "special" system, the binary operators won''t help!

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