Glede na vrednost n poiščite sodo n-to Fibonaccijevo število .
Primeri:
Vnos n = 3
Izhod 34
Razlaga Prva 3 soda Fibonaccijeva števila so 0 2 8 34 144, tretje pa 34.Vnos n = 4
Izhod 144
Razlaga Prva 4 soda Fibonaccijeva števila so 0 2 8 34 144, četrto pa 144.
[Naivni pristop] Preverite vsako Fibonaccijevo število enega za drugim
mi ustvari vsa Fibonaccijeva števila in preveri vsako številko eno za drugo, če je kdaj ali ne
[Učinkovit pristop] Uporaba neposredne formule – O(n) čas in O(1) prostor
Fibonaccijevo zaporedje sodih števil je 0 2 8 34 144 610 2584.... Iz tega zaporedja lahko razberemo, da vsako tretje število v zaporedju je sodo in zaporedje sledi naslednji rekurzivni formuli.
Ponovitev za celo Fibonaccijevo zaporedje je:
Eefn = 4fn-1 + Efn-2
Kako deluje zgornja formula?
Oglejmo si izvirno Fibonaccijevo formulo in jo zapišimo v obliki Fn-3 in Fn-6, saj je vsako tretje Fibonaccijevo število sodo.
Fn = Fn-1 + Fn-2 [Razširitev obeh izrazov]
= Fn-2 + Fn-3 + Fn-3 + Fn-4
= Fn-2 + 2Fn-3 + Fn-4 [Razširitev prvega člena]
= Fn-3 + Fn-4 + 2Fn-3 + Fn-4
= 3Fn-3 + 2Fn-4 [razširitev enega Fn-4]
= 3Fn-3 + Fn-4 + Fn-5 + Fn-6 [česanje Fn-4 in Fn-5]
= 4Fn-3 + Fn-6
Ker je vsako tretje Fibonaccijevo število sodo Torej, če je Fn
tudi takrat je Fn-3 sodo in Fn-6 je tudi sodo. Naj bo Fn
x-ti sodi element in ga označite kot EFx.
string addČe je Fn EFx, potem je Fn-3 prejšnje sodo število, tj. EFx-1
in Fn-6 je prejšnji od EFx-1, tj. EFx-2
Torej Fn = 4Fn-3 + Fn-6
kar pomeni
EFx = 4EFx-1 + EFx-2
Spodaj je preprosta izvedba ideje
C++#include
public class GfG { // Function to calculate the nth even Fibonacci // number using dynamic programming public static int nthEvenFibonacci(int n) { // Base case: the first even // Fibonacci number is 2 if (n == 1) return 2; // Start with the first two Fibonacci // numbers (even ones) int prev = 0; // F(0) int curr = 2; // F(3) // We need to find the nth even Fibonacci number for (int i = 2; i <= n; i++) { // Next even Fibonacci number is 4 // times the previous even Fibonacci // number plus the one before that int nextEvenFib = 4 * curr + prev; prev = curr; curr = nextEvenFib; } return curr; } public static void main(String[] args) { int n = 2; int result = nthEvenFibonacci(n); System.out.println(result); } }
# Function to calculate the nth even # Fibonacci number using dynamic programming def nthEvenFibonacci(n): # Base case: the first even Fibonacci number is 2 if n == 1: return 2 # Start with the first two Fibonacci numbers (even ones) prev = 0 # F(0) curr = 2 # F(3) # We need to find the nth even Fibonacci number for i in range(2 n + 1): # Next even Fibonacci number is 4 times the # previous even Fibonacci number plus the # one before that next_even_fib = 4 * curr + prev prev = curr curr = next_even_fib return curr # Driver code if __name__ == '__main__': n = 2 # Setting n to 2 result = nthEvenFibonacci(n) print(result)
using System; class GfG { // Function to calculate the nth even Fibonacci // number using dynamic programming public int NthEvenFibonacci(int n) { // Base case: the first even Fibonacci number is 2 if (n == 1) return 2; // Start with the first two Fibonacci numbers (even ones) int prev = 0; // F(0) int curr = 2; // F(3) // We need to find the nth even Fibonacci number for (int i = 2; i <= n; i++) { // Next even Fibonacci number is 4 times the // previous even Fibonacci number plus the // one before that int nextEvenFib = 4 * curr + prev; prev = curr; curr = nextEvenFib; } return curr; } static void Main() { GfG gfg = new GfG(); int n = 2; int result = gfg.NthEvenFibonacci(n); Console.WriteLine(result); // Output: The nth even Fibonacci number } }
// Function to calculate the nth even Fibonacci number using dynamic programming function nthEvenFibonacci(n) { // Base case: the first even Fibonacci number is 2 if (n === 1) return 2; // Start with the first two Fibonacci numbers (even ones) let prev = 0; // F(0) let curr = 2; // F(3) // We need to find the nth even Fibonacci number for (let i = 2; i <= n; i++) { // Next even Fibonacci number is 4 times // the previous even Fibonacci number plus // the one before that let nextEvenFib = 4 * curr + prev; prev = curr; curr = nextEvenFib; } return curr; } // Example usage: const n = 2; // Setting n to 2 const result = nthEvenFibonacci(n); console.log(result);
Izhod
8