Created
April 27, 2020 12:33
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| #include <stdio.h> | |
| #include <time.h> | |
| unsigned long long fibo(unsigned long long f) | |
| { | |
| unsigned long long nMinusTwo = 0; | |
| unsigned long long nMinusOne = 1; | |
| unsigned long long n = 0; | |
| /* | |
| Wenn f < 2, sprich 0 oder 1 ist, return f, da diese beiden Fibonaaci-Zahlen | |
| sich etwas anders verhalten als der Rest. | |
| */ | |
| if (f < 2) | |
| { | |
| return f; | |
| } | |
| /* | |
| Die Fibonacci-Zahl n ergibt sich aus: n F_n = F_n-2 F_n-1. Somit müssen wir also | |
| jede Fibonacci-Zahl vor unserer Fibonacci-Zahl an Stelle `f` berechnen. | |
| */ | |
| for (int i = 0; i < f; i++) | |
| { | |
| /* F_n-2 */ | |
| nMinusTwo = nMinusOne; | |
| /* F_n-1 */ | |
| nMinusOne = n; | |
| /* Fibonacci Zahl an Stelle n */ | |
| n = nMinusOne + nMinusTwo; | |
| } | |
| return n; | |
| } | |
| int main() | |
| { | |
| clock_t tm1, tm2; | |
| tm1 = clock(); | |
| printf("Num: %ld\n", fibo(2999999999)); | |
| tm2 = clock(); | |
| printf("Dauer: %.2f Sekunden\n", (double)(tm2 - tm1) / CLOCKS_PER_SEC); | |
| } |
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