#include<bits/stdc++.h> using namespace std; //寒冰射手原创！可抄！ int main(){ int a,b[9999]={1}; cin>>a;//首先，输入数据a，定义数据b，用于存储n项是多少 for (int i = 2;i <= a+1;i++){ //遍历a次，算出斐波那契数列第a项 b[i] = b[i-2] + b[i-1];//把斐波那契数列的第i项存储到b数组中 } cout<<b[a+1];//输出斐波那契数列第a项 return 0; }

#include<bits/stdc++.h> using namespace std; int n; int r[1000]; int dfs(int n){ if(n<=2) return r[n]; return (dfs(n-1)+dfs(n-2)); } int main(){ r[1]=1; r[2]=1; cin>>n; cout<<dfs(n); return 0; }

#include<bits/stdc++.h> using namespace std; const int N=10001; int n,a[N]; int main() { cin>>n; a[1] = a[2] = 1; for(int i=3;i<=n;i++) { a[i] = a[i-1] + a[i-2]; } cout<<a[n]; return 0; }

#include<iostream> using namespace std; int F(int n){ if(n<=2) return 1; return F(n-1)+F(n-2); } int main(){ int n; cin>>n; cout<<F(n); }

#include<iostream> using namespace std; int fid(int n){ if(n1 || n2){ return 1; } return fid(n-1)+fid(n-2); } int main(){ int n; cin>>n; cout<<fid(n); return 0; }

#include<bits/stdc++.h> using namespace std; int a[31]; 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]; cout << a[n]; return 0; }

n=int(input()) a=1 b=1 if n1 or n2: print(1) else: for i in range(3,n+1): c=a+b a=b b=c print(c)

//递归——斐波那契数列 #include<bits/stdc++.h> using namespace std; //构造递归函数 int f(int n) { if(n<=2){ return 1; } else return f(n-1)+f(n-2); } int main(){ int n; cin>>n; cout<<f(n); }

#include <bits/stdc++.h> using namespace std; int f(int n){ if(n1 or n2) return 1; else return f(n-1)+f(n-2); } int main(){ int n; cin>>n; cout<<f(n); }

def f(n): if n<=2: return 1 return f(n-1)+f(n-2) n=int(input()) print(f(n))

#include<bits/stdc++.h> using namespace std; int fib(int n){ if(n1 || n2){ return 1; } return fib(n-1)+fib(n-2); } int main(){ int n; cin>>n; cout<<fib(n); return 0; }

#include<bits/stdc++.h> using namespace std; int a(int b){ if(b1||b2) return 1; return a(b-1)+a(b-2); } int main(){ int b; cin>>b; cout<<a(b); return 0; }

#include<bits/stdc++.h> using namespace std; int a(int b){ if(b1||b2) return 1; return a(b-1)+a(b-2); } int main(){ int b; cin>>b; cout<<a(b); return 0; }

#include<iostream> using namespace std; int a[1010], m; void fei(int n){ if(m<=n){ cout<<a[n-1]; return ; } a[n] = a[n-1] + a[n-2]; fei(n+1); } int main(){ a[0] = 1; a[1] = 1; cin>>m; fei(2); return 0; }