Zadana vrijednost n pronađite n-tu parnu vrijednost Fibonaccijev broj .
Primjeri:
Ulazni n = 3
Izlaz 34
Obrazloženje Prva 3 parna Fibonaccijeva broja su 0 2 8 34 144, a treći je 34.Ulazni n = 4
Izlaz 144
Obrazloženje Prva 4 parna Fibonaccijeva broja su 0 2 8 34 144, a četvrti je 144.
[Naivni pristup] Provjerite svaki Fibonacci broj jedan po jedan
Mi generirati sve Fibonaccijeve brojeve i provjerite svaki broj jedan po jedan je li ikada ili nije
[Učinkovit pristup] Korištenje izravne formule - O(n) vremena i O(1) prostora
Fibonaccijev niz parnih brojeva je 0 2 8 34 144 610 2584.... Iz ovog niza možemo dobiti ideju da svaki treći broj u nizu je paran a niz slijedi sljedeću rekurzivnu formulu.
Ponavljanje parnog Fibonaccijevog niza je:
Eefn = 4fn-1 + Efn-2
Kako radi gornja formula?
Pogledajmo originalnu Fibonaccijevu formulu i zapišimo je u obliku Fn-3 i Fn-6 jer je svaki treći Fibonaccijev broj paran.
Fn = Fn-1 + Fn-2 [Proširenje oba pojma]
= Fn-2 + Fn-3 + Fn-3 + Fn-4
= Fn-2 + 2Fn-3 + Fn-4 [Proširenje prvog člana]
= Fn-3 + Fn-4 + 2Fn-3 + Fn-4
= 3Fn-3 + 2Fn-4 [Proširenje jednog Fn-4]
= 3Fn-3 + Fn-4 + Fn-5 + Fn-6 [kombiniranje Fn-4 i Fn-5]
= 4Fn-3 + Fn-6
Budući da je svaki treći Fibonaccijev broj paran, pa ako je Fn
čak i tada je Fn-3 paran i Fn-6 je također paran. Neka je Fn
x-ti parni element i označite ga kao EFx.
string usporediti s JavomAko je Fn EFx tada je Fn-3 prethodni parni broj, tj. EFx-1
a Fn-6 je prethodni od EFx-1 tj. EFx-2
Dakle, Fn = 4Fn-3 + Fn-6
što znači
EFx = 4EFx-1 + EFx-2
Ispod je jednostavna implementacija ideje
C++#include using namespace std; // Optimized function to calculate the nth // even Fibonacci number int nthEvenFibonacci(int n) { // Base case: the first even Fibonacci number is 2 if (n == 1) return 2; // Start with the first two even Fibonacci numbers 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; } int main() { int n = 2; int result = nthEvenFibonacci(n); cout << result << endl; return 0; }
Java 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); } }
Python # 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)
C# 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 } }
JavaScript // 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);
Izlaz
8