Complexity of signing is O(1).
That is plain wrong. The more bits you have to sign the longer signing takes. The question is: how long in terms of the input length.
We would have O(n) if Sign(32_bytes) == Sign(64_bytes) / 2. Actually we have Sign(32_bytes) == Sign(3200_bytes) / 2, so it's more like O(1). Right?
Sorry, still wrong.
It looks more like O(log n) but don't think so, as the whole input has to be read. That still takes O(n).
Bad math alert!
It's true that reading input is always O(n), in fact
\omega\theta(n), but the
constant ratios for both upper and lower bound are very small compared to signing, or pretty much any other operation.
Complexity analysis only says something about asymptotic behaviour (i.e. as n grows towards infinity), it doesn't say anything about small inputs or exact performance, which is the case here (100 sigs is definitely small, and we care about every scrap of performance). For small inputs, you'll want to run the code and measure it.