题解 | #牛牛的Hermite多项式#
牛牛的Hermite多项式
http://www.nowcoder.com/practice/0c58f8e5673a406cb0e2f5ccf2c671d4
def h(n, x): if n== 0: return 1 elif n==1: return 2 else: return 2 * x * h(n-1, x) - 2 * (n-1) * h(n-2, x)
a, b = map(int, input().split()) print(h(a, b))
