题解 | #牛牛的Hermite多项式#

#include<stdio.h>
int func(int n, int x);
int main() {

    int a, b;
    scanf("%d%d", &a, &b);
    printf("%d", func(a, b));
}
int func(int n, int x) {
    if (n == 0)
        return 1;
    else if (n == 1)
        return 2;
    else 
        return  (2 * x * func(n - 1, x) - 2 * (n - 1) * func(n - 2, x));
     
}