No, that algorithm is O(1).
50 is a constant. As n -> infinity, the algorithm runs in constant time.
Specifically, I can pick some number C such that C * f(n) > the number of operations, then the algorithm is O(f(n)).
So I choose f(n) == 1, and C == 100.
Then the runtime of the algorithm is 1,1,1,1,1...., 50,1,1,1,1,.... For all values of n, this is less than C * f(n) == 100, a constant, so it is O(f(n)). f(n) is 1, so it is O(1).
50 is a constant. As n -> infinity, the algorithm runs in constant time.
Specifically, I can pick some number C such that C * f(n) > the number of operations, then the algorithm is O(f(n)).
So I choose f(n) == 1, and C == 100.
Then the runtime of the algorithm is 1,1,1,1,1...., 50,1,1,1,1,.... For all values of n, this is less than C * f(n) == 100, a constant, so it is O(f(n)). f(n) is 1, so it is O(1).