关于使用 递归来解决斐波拉契问题

#include <iostream>
#include <algorithm>
#include <cstring>
#define int long long

using namespace std;

const int Num = 1e9 + 7;
int n;
int a = 1 , b = 1;
void fi(int u)
{

    int temp = (a + b) % Num;
    a = b , b = temp;

    if (u == n)
    {
        cout << temp << endl;
        return ;
    }

    fi(u + 1);
}
signed main()
{
    cin >> n;

    if (n == 1 || n == 2)
    {
        cout << "1" << endl;
        return 0;
    }
    else 
        fi(3);
    
    return 0;
}

#include<iostream>

using namespace std;

const int N = 1000010;
const int mod = 1e9 + 7;
int a[N];

int main()
{
    int n;

    cin >> n;
    a[1] = 1, a[2] = 1;
    for (int i = 3; i <= n; ++i)
    {
        a[i] = (a[i - 1] + a[i - 2]) % mod;
    }
    cout << a[n] << endl;

    return 0;
}