题解 | #牛牛的Hermite多项式#

#include <stdio.h>
int Hermite(int n, int x) {
    int ret = 0;
    if (n == 0) {
        ret = 1;
    } else if (n == 1) {
        ret = 2 * n;
    } else if (n > 1) {
        ret = 2 * x * Hermite(n - 1,  x) - 2 * (n - 1) * Hermite(n - 2,  x);
    }
    return ret;
}
int main() {
    int n, x = 0;
    scanf("%d %d", &n, &x);
    printf("%d", Hermite(n, x));
    return 0;
}