f(n)=f(n-1)+f(n-2) f(0)=f(1)=1，求斐波那契数列第20项，分别用循环和递归的方式，比较时间效率。提示：可以使用c函数clock取出当前系统时间，计算前后各一次，两次相减除以每秒的时钟数，就可以得到以秒为单位的差距-f (n) = f (n-1)+ f (n-2) f (0) = f (1) = 1, Item 20 seeking Fibonacci number Fibonacci sequence that were circulating and recursi

Code for finding the gap in the binary representation of an integer. Solution to the codility s binary gap question. Please check wikipedia for descr iption.