题解 | #牛牛的Hermite多项式#
牛牛的Hermite多项式
http://www.nowcoder.com/practice/0c58f8e5673a406cb0e2f5ccf2c671d4
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() {
unsigned n, x;
scanf("%d %d", &n, &x);
printf("%d", output(n, x));
return 0;
}
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