Thread: divisibility by 3

I don't think he's starting with chars which is where you two lost each other, I think he is starting from an integer and not necessarily an input from the command line. So what he need is a general purpose function that takes an int and returns if it's divisible by 3 and therefore was stating it would be difficult to parse the individual digits without using / or %

Ah, that makes sense.

For those of you that follow the contest forum I submit this:

Code:

bool divby3(int num)
{
     int dividend = (double)num * .3333 + .1;
     int remainder = num-(dividend * 3);
     return remainder == 0;
}

For those that don't follow the contest forum, this is not a serious answer(though it works).

I think that operator * is like / and %, so if this is a homework question as I suspect, Brij cannot use it anyway.

If you have an efficient operation for counting the number of nonzero bits in an unsigned int (like some processors do) you can use

Code:

int divisible3(unsigned int x) {

  do {
    x = abs(countbits(x & 0xAAAAAAAA) - countbits(x & 0x55555555)); 
  } while (x > 3);

  return x == 0;
}

Originally Posted by Darryl

For those of you that follow the contest forum I submit this:

Code:

bool divby3(int num)
{
     int dividend = (double)num * .3333 + .1;
     int remainder = num-(dividend * 3);
     return remainder == 0;
}

For those that don't follow the contest forum, this is not a serious answer(though it works).

Sorry mate,

Code:

3003 * .3333 = 1000.8999
+ .1 = 1000.9999
* 3 = 3002.9997

Originally Posted by laserlight

I think that operator * is like / and %, so if this is a homework question as I suspect, Brij cannot use it anyway.

If this is a homework assignment, it probably is assigned because there's so many different ways to solve it. I don't see how solving it with multiplication is really less valid than bit counting.

[QUOTE=QuestionC;654440]Sorry mate,

Code:

3003 * .3333 = 1000.8999
+ .1 = 1000.9999
* 3 = 3002.9997

/QUOTE]

I can always add more 3s to extend its useful range.
Plus, it was only meant as a joke based on the contest thread I linked to.

Speaking of counting the number of bits in an unsigned int, I found this interesting bit of speculation. http://www.moyogo.com/blog/2005/09/secret-opcodes.html