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