题解 | #牛牛的Hermite多项式#
牛牛的Hermite多项式
https://www.nowcoder.com/practice/0c58f8e5673a406cb0e2f5ccf2c671d4
#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));
}