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?