暴力多项式全家桶
设 \(d\) 为 \(F(x)\) 的次数。
多项式求逆
\[\begin{aligned}
F(x)G(x) &= 1\\
\sum_{i=0}^d f_ig_{n-i} &= 0\\
g_n &= -\frac{\sum_{i=1}^d f_ig_{n-i}}{f_0}\\
g_0 &= \frac{1}{f_0}
\end{aligned}
\]
复杂度 \(O(nd)\)。
多项式 \(\exp\)
\[\begin{aligned}
G(x) &= e^{F(x)}\\
G'(x) &= F'(x)e^{F(x)}\\
G'(x) &= F'(x)G(x)\\
g_n &= \sum_{i=0}^{d-1} (i+1)f_{i+1}g_{n-i}\\
g_0 &= 1
\end{aligned}
\]
复杂度 \(O(nd)\)。
多项式 \(\ln\)
\[\begin{aligned}
G(x) &= \ln{F(x)}\\
G'(x) &= \frac{F'(x)}{F(x)}\\
G'(x)F(x) &= F'(x)\\
\sum_{i=0}^d f_i (n-i) g_{n-i} &= n f_n\\
g_{n} &= f_{n} - \frac{1}{n}(n-i) g_{n-i}\\
g_0 &= 0
\end{aligned}
\]
复杂度 \(O(nd)\)。
多项式快速幂
\[\begin{aligned}
G(x) &= F(x)^k\\
\ln G(x) &= k \ln F(x)\\
\frac{G'(x)}{G(x)} &= k \frac{F'(x)}{F(x)}\\
G'(x) F(x) &= k F'(x) G(x)\\
\sum_{i=0}^d f_i (n-i)g_{n - i} &= k\sum_{i=0}^{d - 1} (i+1)f_{i+1} g_{n-i-1}\\
n g_n &= k\sum_{i=0}^{d - 1} (i+1) f_{i+1} g_{n-i-1} - \sum_{i=1}^d f_i (n-i)g_{n - i}\\
g_n &= \frac{k\sum_{i=0}^{d - 1} (i+1) f_{i+1} g_{n-i-1} - \sum_{i=1}^d f_i (n-i)g_{n - i}}{n}\\
g_0 &= 1
\end{aligned}
\]
复杂度 \(O(nd)\)。
多项式开根
\[G(x) = F(x)^{\frac{1}{2}}
\]
复杂度 \(O(nd)\)。
