Statement
带标号仙人掌计数问题.（\(n \le 131071\)）
Solution
设\(x\)个点的仙人掌个数的生成函数为\(C(x)\)
-
对于与根相邻的块, 还是仙人掌, 生成函数为\(C(x)\)
-
包含根的环, 生成函数为\(\sum_{i \ge 2}\frac{C(x)^i}{2}\)
组合起来:
\[C(x) = x \exp{\frac{2C(x)-C(x)^2}{2-2C(x)}}
\]
设\(G(C(x)) = x\exp{\frac{2C(x)-C(x)^2}{2-2C(x)}}-C(x)\), 那么:
\[\small{
\begin{aligned}
G'(C(x)) &= x\left(\exp{\frac{2C(x)-C(x)^2}{2-2C(x)}}\right)'-1 \\
&= x \exp\left(\frac{2C(x)-C(x)^2}{2-2C(x)}\right)\left(\frac{2C(x)-C(x)^2}{2-2C(x)}\right)' - 1 \\
&= x \exp\left(\frac{2C(x)-C(x)^2}{2-2C(x)}\right)
\left(\frac{\left(2-2C(x)\right)^2-\left(2C(x) - C(x)^2\right)(-2)}{(2-2C(x))^2}\right)
- 1\\
&= x \exp\left(\frac{2C(x)-C(x)^2}{2-2C(x)}\right)
\left(1+\frac{4C(x) - 2C(x)^2}{(2-2C(x))^2}\right)
- 1
\end{aligned}
}
\]
牛顿迭代:
\[\begin{aligned}
C_1(x) &= C(x) - \frac{G(C(x))}{G'(C(x))} \\
&= C(x) - \frac{2x\exp\left(\frac{2C(x)-C(x)^2}{2-2C(x)}\right)-2C(x)}
{x \exp\left(\frac{2C(x)-C(x)^2}{2-2C(x)}\right)
\left(1+\frac{1}{(C(x)-1)^2}\right)
- 2}
\end{aligned}
\]
Code
https://loj.ac/submission/358148