递推和递归方法求斐波那契数列

#include<iostream>
using namespace std;

int fab(int n)
{
    if (n == 1 || n == 2)
    {
        return 1;
    } else {
        return fab(n - 2) + fab(n - 1);
    }
    
}

int fab2 (int n)
{
    int a[20];
    a[1] = 1;
    a[2] = 1;
    for (int i = 3; i <= n; i++)
    {
        a[i] = a[i-2] + a[i-1];
    }
    return a[n];
}

int main()
{    
    for (int i = 1; i < 20; i++)
    {
        cout << fab (i) << endl;
    }
    cout << endl;
    for (int i = 1; i < 20; i++)
    {
        cout << fab2 (i) << endl;
    }
    return 0;
}