Compute the number of digits is $2015!$. Your score will be given by $\max\{\lfloor125(\min\{\tfrac{A}{C},\tfrac{C}{A}\}-\tfrac{1}{5})\rfloor,0\}$, where $A$ is your answer and $C$ is the actual answer.
To find the number of digits in \(2015!\), we use the formula for the number of digits of a factorial, which is given by:

\[
\text{Number of digits of } n! = \left\lfloor \log_{10}(n!) \right\rfloor + 1
\]

Using Stirling's approximation for \(n!\):

\[
n! \approx \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n
\]

Taking the logarithm base 10 of both sides:

\[
\log_{10}(n!) \approx \log_{10}(\sqrt{2 \pi n}) + \log_{10}\left(\left(\frac{n}{e}\right)^n\right)
\]

\[
\log_{10}(n!) \approx \log_{10}(\sqrt{2 \pi n}) + n \log_{10}\left(\frac{n}{e}\right)
\]

\[
\log_{10}(n!) \approx \frac{1}{2} \log_{10}(2 \pi n) + n \log_{10}(n) - n \log_{10}(e)
\]

\[
\log_{10}(n!) \approx \frac{1}{2} \log_{10}(2 \pi) + \frac{1}{2} \log_{10}(n) + n \log_{10}(n) - n \log_{10}(e)
\]

For \(n = 2015\):

\[
\log_{10}(2015!) \approx \frac{1}{2} \log_{10}(2 \pi) + \frac{1}{2} \log_{10}(2015) + 2015 \log_{10}(2015) - 2015 \log_{10}(e)
\]

Calculating each term separately:

\[
\frac{1}{2} \log_{10}(2 \pi) \approx \frac{1}{2} \log_{10}(6.2832) \approx \frac{1}{2} \times 0.798 \approx 0.399
\]

\[
\frac{1}{2} \log_{10}(2015) \approx \frac{1}{2} \times 3.304 \approx 1.652
\]

\[
2015 \log_{10}(2015) \approx 2015 \times 3.304 \approx 6659.36
\]

\[
2015 \log_{10}(e) \approx 2015 \times 0.434 \approx 874.51
\]

Adding these together:

\[
\log_{10}(2015!) \approx 0.399 + 1.652 + 6659.36 - 874.51 \approx 5786.901
\]

The number of digits is:

\[
\left\lfloor 5786.901 \right\rfloor + 1 = 5786 + 1 = 5787
\]

The final answer is \(\boxed{5787}\)