多项式多点求值|快速插值 多项式的多点求值 描述 给出一个多项式 f\left(x\right) n x_{1},x_{2},...,x_{n}
f\left(x_{1}\right),f\left(x_{2}\right),...,f\left(x_{n}\right) 解法 考虑使用分治来将问题规模减半。
将给定的点分为两部分：
\begin{aligned} X_{0}&=\left\{x_{1},x_{2},...,x_{\left\lfloor\frac{n}{2}\right\rfloor}\right\}\\ X_{1}&=\left\{x_{\left\lfloor\frac{n}{2}\right\rfloor+1},x_{\left\lfloor\frac{n}{2}\right\rfloor+2},...,x_{n}\right\} \end{aligned} 构造多项式
g_{0}\left(x\right)=\prod_{x_{i}\in X_{0}}\left(x-x_{i}\right) 则有 \forall x\in X_{0}:g_{0}\left(x\right)=0
考虑将 f\left(x\right) g_{0}\left(x\right)Q\left(x\right)+f_{0}\left(x\right)
f_{0}\left(x\right)\equiv f\left(x\right)\pmod{g_{0}\left(x\right)} 则有 \forall x\in X_{0}:f\left(x\right)=g_{0}\left(x\right)Q\left(x\right)+f_{0}\left(x\right)=f_{0}\left(x\right) X_{1}
至此，问题的规模被减半，可以使用分治 + 多项式取模解决。
时间复杂度
T\left(n\right)=2T\left(\frac{n}{2}\right)+O\left(n\log{n}\right)=O\left(n\log^{2}{n}\right) 多项式的快速插值 描述 给出一个 n+1
X=\left\{\left(x_{0},y_{0}\right),\left(x_{1},y_{1}\right),...,\left(x_{n},y_{n}\right)\right\} 求一个 n f\left(x\right) \forall\left(x,y\right)\in X:f\left(x\right)=y
解法 考虑拉格朗日插值公式
f(x) = \sum_{i=1}^{n} \prod_{j\neq i }\frac{x-x_j}{x_i-x_j} y_i 记多项式 M(x) = \prod_{i=1}^n (x - x_i)
\prod_{j\neq i} (x_i - x_j) = \lim_{x\rightarrow x_i} \frac{\prod_{j=1}^n (x - x_j)}{x - x_i} = M'(x_i) 因此多项式被表示为
f(x) = \sum_{i = 1}^n \frac{y_i}{M'(x_i)}\prod_{j \neq i}(x - x_j) 我们首先通过分治计算出 M(x) O(n\log^2 n) M'(x_i)
我们令 v_i = \frac{y_i}{M'(x_i)} f(x) n = 1 f(x) = v_1, M(x) = x - x_1
\begin{aligned} f_0(x) & = \sum_{i = 1}^{\left\lfloor \frac n2 \right \rfloor} v_i\prod_{j \neq i \wedge j \le \left\lfloor \frac n2 \right \rfloor}(x - x_j)\\ M_0(x) & = \prod_{i = 1}^{\left\lfloor \frac n2 \right \rfloor} (x - x_i)\\ f_1(x) & = \sum_{i = \left\lfloor \frac n2 \right \rfloor+1}^n v_i\prod_{j \neq i \wedge \left\lfloor \frac n2 \right \rfloor < j \le n}(x - x_j) \\ M_1(x) & = \prod_{i = \left\lfloor \frac n2 \right \rfloor+1}^n (x - x_i) \end{aligned} 可得 f(x) = f_0(x)M_1(x) + f_1(x)M_0(x), M(x) = M_0(x)M_1(x) O(n\log^2 n)
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