Your method computes fib(n) in O(n) arithmetic operations. There's a way to do it in O(log n) arithmetic operations on integers of the same size. First note that you can obtain the two-element vector (fib(n),fib(n+1)) by applying the matrix A=((0,1),(1,1)) to the vector (fib(n-1),fib(n)). So to get from (fib(0),fib(1)) to (fib(n),fib(n+1)), you need to raise the matrix A to the nth power. That can be done in O(log n) arithmetic operations by using repeated squaring: http://en.wikipedia.org/wiki/Exponentiation_by_squaring.
Comments
Your method computes fib(n) in O(n) arithmetic operations. There's a way to do it in O(log n) arithmetic operations on integers of the same size. First note that you can obtain the two-element vector (fib(n),fib(n+1)) by applying the matrix A=((0,1),(1,1)) to the vector (fib(n-1),fib(n)). So to get from (fib(0),fib(1)) to (fib(n),fib(n+1)), you need to raise the matrix A to the nth power. That can be done in O(log n) arithmetic operations by using repeated squaring: http://en.wikipedia.org/wiki/Exponentiation_by_squaring.