C++中的斐波納契系數(shù)實現(xiàn)：在斐波納契系列的情況下，下一個數(shù)字是前兩個數(shù)字的總和，例如：0,1,1,2,3,5,8,13,21等。斐波那契數(shù)列的前兩個數(shù)字是：0和1。
有兩種方法來寫斐波那契數(shù)列程序：
下面來看看看不使用遞歸在C++中實現(xiàn)斐波那契數(shù)列(fibonacci)程序。
#include <iostream>
using namespace std;
int main() {
int n1=0,n2=1,n3,i,number;
cout<<"Enter the number of elements: ";
cin>>number;
cout<<n1<<" "<<n2<<" "; //printing 0 and 1
//loop starts from 2 because 0 and 1 are already printed
for(i=2;i<number;++i)
{
n3=n1+n2;
cout<<n3<<" ";
n1=n2;
n2=n3;
}
return 0;
}
執(zhí)行上面代碼，得到以下結果 -
Enter the number of elements: 10
0 1 1 2 3 5 8 13 21 34
下面來看看看使用遞歸在C++中的斐波那契(fibonacci)數(shù)列程序。
#include<iostream>
using namespace std;
void printFibonacci(int n){
static int n1=0, n2=1, n3;
if(n>0){
n3 = n1 + n2;
n1 = n2;
n2 = n3;
cout<<n3<<" ";
printFibonacci(n-1);
}
}
int main(){
int n;
cout<<"Enter the number of elements: ";
cin>>n;
cout<<"Fibonacci Series: ";
cout<<"0 "<<"1 ";
printFibonacci(n-2); //n-2 because 2 numbers are already printed
return 0;
}
執(zhí)行上面代碼得到以下結果 -
Enter the number of elements: 15
Fibonacci Series: 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377
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