int n;
cin>>n;
if (n==0)//deal with it
for (double i=0;i<=1/(double)n;i+=1e-9)//do something

It seems that, this code runs faster when n becomes larger...

When n = 10⁹, the code will be executed once. When n = 1, it runs 10⁹ times.

So I estimated it as O(n^- 1).

But, this means that this algorithm runs even faster than O(1).

How is it possible?