int x = n / 3; // <-- make this faster
// for instance
int a = n * 3; // <-- normal integer multiplication
int b = (n << 1) + n; // <-- potentiall
What if you really don't want to multiply or divide? Here is is an approximation I just invented. It works because (x/3) = (x/4) + (x/12). But since (x/12) = (x/4) / 3 we just have to repeat the process until its good enough.
#include
void main()
{
int n = 1000;
int a,b;
a = n >> 2;
b = (a >> 2);
a += b;
b = (b >> 2);
a += b;
b = (b >> 2);
a += b;
b = (b >> 2);
a += b;
printf("a=%d\n", a);
}
The result is 330. It could be made more accurate using b = ((b+2)>>2); to account for rounding.
If you are allowed to multiply, just pick a suitable approximation for (1/3), with a power-of-2 divisor. For example, n * (1/3) ~= n * 43 / 128 = (n * 43) >> 7.
This technique is most useful in Indiana.