斐波那契数列的递归递推求法

#include <bits/stdc++.h>

using namespace std;

// 递归计算斐波那契数列 

int f1(int n)
{
	if (n == 1 || n == 2)
		return 1;
	return f1(n-1) + f1(n-2);
}

// 递推计算斐波那契数列

int f2(int n) 
{
	if (n == 1 || n == 2)
		return 1;
	int fa = 1, fb = 1;
	for (int i = 3; i <= n; ++i){
		int t = fa + fb;
		fa = fb;
		fb = t;
	}
	return fb;
}

int main(){
	int n;
	scanf("%d", &n);
	int ans1 = 0, ans2 = 0;
	ans1 = f1(n);
	ans2 = f2(n);
	printf("递归结果：%d\n递推结果：%d", ans1, ans2);
}

来源：https://blog.csdn.net/qq_40432713/article/details/102757682