题解 | #跳台阶扩展问题#

f(n)=f(n−1)+f(n−2)+⋯+f(1)+f(0)

f(n - 1) = f(n−2)+⋯+f(1)+f(0)

f(n) = 2 * f(n - 1)

#include <iostream>
using namespace std;
const int N = 50;
int f[N];

int main() {
    int n;
    cin >> n;
    f[0] = 1;
    for(int i = 1; i <= n; i ++){
        f[i] = f[i - 1] * 2;
    }
    cout << f[n - 1] << endl;
    return 0;
}