Fastest raise to power.This tip submitted by Ezzetabi on 2005-03-18 16:31:04. It has been viewed 48121 times.Rating of 6.1 with 180 votes If you need raising to power some non standard object, like matrices or something overloaded with a integer exponent you can use this algorith: a any type that know operator*, b int (a*a)^(b/2) if b is even a^b = a*(a*a)*(int)^(b/2) if b is oddusing a recursive function your calculation will be faster (overall with big exponents) than the usual way. A little example code:double mpow (double, int); double mpow2(double, int); double mpow(double b, int exp) { if (exp < 0) { b = 1/b; exp *= -1; } return mpow2(b,exp); } double mpow2(double b, int exp) { if (exp > 2) { if (exp%2) return b*mpow(b*b,(int)exp/2); else return mpow(b*b,exp/2); } else if (2 == exp) return b*b; else if (1 == exp) return b; return 1.0; // exp == 0 }This example is about floats, so it is nothing more than the usual pow(), yet if you need raising to power try to avoid a*a*a*a*...*a as that is really slow. More tips Help your fellow programmers! Add a tip! |