题解 | #牛牛的Hermite多项式#
https://www.nowcoder.com/practice/0c58f8e5673a406cb0e2f5ccf2c671d4
#include <stdio.h>
long Hermite(int n,int x);
int main(){
int n,x;
scanf("%d %d",&n,&x);
printf("%ld",Hermite(n,x));
return 0;
}
long Hermite(int n,int x){
if (n == 0)
return 1;
else if (n == 1)
return 2;
else if (n > 1)
return (2*x*Hermite(n-1,x))-(2*(n-1)*Hermite(n-2,x));
else
return -1;
}
