Create fastDoublingNthFibonacci.py #160
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dnery
approved these changes
Oct 19, 2019
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| Fibonacci series: 0 1 1 2 3 ... | ||
| NthFibonacci(n, mod) -> int | ||
| mod: We find modulo of number (usually with 1e9+7), as it is too big. |
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Very interesting!
I didn't know about that programming pattern...
Thanks for sharing! 😄
| c = (a%mod * ((b*2)%mod - a%mod))%mod | ||
| d = ((a%mod * a%mod)%mod + (b%mod * b%mod)%mod)%mod | ||
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| if not n&1: |
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I am adding a function (NthFibonacci), with which we can find the nth Fibonacci number in O(logn) time. (#126)