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