题解 | #牛牛的Hermite多项式#

#include <iostream>

using namespace std;

int f(int n, int x);

int main()
{
    int n, x;
    cin >> n >> x;
    int a;
    a = f(n, x);
    cout << a << endl;
}

int f(int n, int x)
{
    if (n == 0)
    {
        return 1;
    }
    if (n == 1)
    {
        return 2;
    }
    if (n > 1)
    {
        return 2 * x * f(n - 1, x) - 2 * (n - 1) * f(n - 2, x);
    }
    return 0;
}