题解 | #跳台阶扩展问题#

#include <bits/stdc++.h>
using namespace std;
int main() {
    int n;
    cin >> n;
    cout << (long long) (1 << n - 1) << endl;
    return 0;
}

f(n) = f(n-1) + f(n-2) + ... + f(1)

f(n-1) = f(n-2) + f(n-3) + ... + f(1)

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

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