C 104

[murilocamargos]

How the solution works

The main advantage of this solution is that we only keep track of the first 9 and last 9 digits of the Fibonacci number (the important ones). The last 9 is easy, we only need to mod with 1e9, the first 9 is trickier. In order to do that, a math manipulation was needed.

Considering a Fibonacci number can be given by round(phi^n/sqrt(5)), we can use log10 to store log10(Fn) as double and transform it again to long taking advantage of the size limits of this C type. Another optimization was checking if a number is 1-9 pandigital using bitwise operations instead of transforming the number into a string.

Performance

C

Real time: 0.532 s
User time: 0.511 s
Sys. time: 0.017 s
CPU share: 99.35 %
Exit code: 0

[FrankKair]

Good one, @fredericojordan, I totally forgot about that 😅 @murilocamargos You can use the Ruby script add_new_problem to automagically create directories and problem description readme files.

[murilocamargos]

@FrankKair and @fredericojordan, thanks! I did forget that...

@caian-gums you gave a nice ideia about not computing pow every time but I couldn't do it with bit masks. Instead I created a list to store powers of 10 from 0 to 9. This increased a lot the performance. The new version's stats:

Real time: 0.183 s
User time: 0.150 s
Sys. time: 0.022 s
CPU share: 94.42 %
Exit code: 0

[caiangums]

Nice! I wrote, posted and thought that you cannot do that with bitmasks. That's why I deleted the comment 😅.

But it's nice to see that you think of another solution that increased the processing time!