If n = 0, return true. If n < 0, set n = -n. Iterate n <-- (n >> 2) + (n & 3) until n <= 3. If n = 3, return true; else return false.
Dave On Jun 5, 1:45 pm, divya <[email protected]> wrote: > Find if a number is divisible my 3, without using %,/ or *. You can > use atoi(). -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
