I was asked this problem years ago at interview for Java developer. It was about using BigInt really, but that's not important. I later slightly ex...
For further actions, you may consider blocking this person and/or reporting abuse
That was a fun task and I could refresh some aspects of python doing it. Thanks!
Until now (programm is still executing) I just have the following solution:
X -> 92740
Y -> ???
Z -> ???
I am pretty sure that my solution is far from perfect, since it has a huge runtime and memory consumption. Therefore I'm really looking forward to a better approach.
Jannik, Hi!
Thanks for attempt!
Yep, it is just a matter of making it work without speed degradation due to number length increase :)
There a trick you can use:
last 2 and 3 digits of fibonacci repeats each 1500 terms
last 4 digits repeats each 15,000
last 5 digits repeats each 150,000
last 6 digits repeats each 1,500,000
last 7 digits repeats each 15,000,000
x = 92.740
y = 4.562.603
z = 280.555.264
Nice idea about finding "the loop length" after which "endings" repeat :) Perhaps some other puzzle could be invented from it.
However my approach was simpler - just taking modulo by
10eXwhereXis amount of digits in the number we seek. Then numbers won't grow longer than this.I suspect your and mine approaches are really two versions of the same fact, however
Hi,
so far I have
x - 92.740
y - 4.562.603
Have a nice weekend
Michael
Michael, Hi!
Thanks for try! Judging by "so far" it seems you also use straightforward implementation which becomes a bit slower as the values grew :)
Good luck!
Hi Rodion,
yes, I started with a default-Fibonacci, and that take a lot of time ;O)
With a little optimization (using only the necessary decimal places), the last value can be found within minutes.
b: 54212345 idx 155
b: 77255555 idx 92740
b: 67770777 idx 4562603
b: 20200123 idx 145555264
So I found z - 145.555.264
Is that correct?
Hi Michael!
At least my results are exactly the same! So I believe you are right and this is correct :)