题解 | #N阶楼梯上楼问题#

#include <bits/stdc++.h>

using namespace std;

const int N = 300;
int f[N]; // 走到第j阶的上楼方式数 
int main(){
	int n;
	
	while (cin >> n){
		memset(f, 0, sizeof f);
		f[0] = 1;
		f[1] = 1;
		f[2] = 2;
		
		for (int i=2; i<=n; i++){
			f[i] =  f[i-1] + f[i-2];
		}
		
		cout << f[n] << endl;
	} 
	
	
	return 0;
}