n=int(input())
def f(x):
    if x==1:
        return 0
    elif x==2 or x==3:
        return 1
    else:
        return f(x-1)+f(x-2)
print(f(n))

num = int(input())-1
s = (((1+5**0.5)/2)**num-((1-5**0.5)/2)**num)/(5**0.5)
print(int(s))

一道递归的水题 原本以为会炸内存

#include <bits/stdc++.h>
using namespace std;

int fab(int n)
{
	if(n == 1) return 0;
	else if(n == 2) return 1;
	else return fab(n-1)+fab(n-2);
}

int main()
{
	int n;
	cin >> n;
	cout << fab(n);
	return 0;
}

#include <bits/stdc++.h> using namespace std; int main() { int n,i,sum=1; cin>>n; int a[40]={0,1,1,2,3,5,8}; for (i=2;i<=n;i++)//控制行数n { a[i]=a[i-1]+a[i-2]; } cout<<a[n-1]; return 0; }