[NowCoder5666D]Quadratic Form

题意

$$\begin{array}{rl}\underset{x}{\max }& b^{T}x\\ \mathrm{s}.\mathrm{t}.& x^{T}Ax-1\le 0\end{array}$$
求$(b^{T}x{)}^2\mathrm{m}\mathrm{o}\mathrm{d}998244353$.

$A$为$n\times n$的对称矩阵，$b,x\in {\mathbb{R}}^n$
$n\le 200,0\le |A_{i,j}|,|b_i|\le 10^9$
$\forall x\in {\mathbb{R}}^n/\{0\},x^{T}Ax>0$
$\det (A)\not\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}998244353)$

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题解

设兰格朗日函数$\mathcal{L}(x,\lambda )=b^{T}x+\lambda (x^{T}Ax-1)$，根据向量求导法则，有
$$\frac{\partial \mathcal{L}}{\partial x}=b+\lambda (A+A^T)x=b+2\lambda Ax$$
令$\frac{\partial \mathcal{L}}{\partial x}=0$，可得$x=-\frac{A^{-1}b}{2\lambda }$。由KKT条件：$\lambda (x^{T}Ax-1)=0,\lambda \le 0$。将$x$带入得
$$\begin{array}{rl}& \lambda [(-\frac{A^{-1}b}{2\lambda }{)}^{T}A(-\frac{A^{-1}b}{2\lambda })-1]=0\\ \Rightarrow & b^T(A^{-1}{)}^{T}A{A}^{-1}b=4{\lambda }^2,(A^{-1}{)}^T=A^{-1}\\ \Rightarrow & \lambda =-\frac{1}{2}\sqrt{b^T{A}^{-1}b}\end{array}$$
故
$$b^{T}x=b^T(-\frac{A^{-1}b}{2\lambda })=-\frac{b^T{A}^{-1}b}{\sqrt{b^T{A}^{-1}b}}=\sqrt{b^T{A}^{-1}b}$$
最终的答案为$(b^{T}x{)}^2=b^T{A}^{-1}b$。用高斯消元求出$A$的逆再相乘即可。

单组数据时间复杂度$O(n^3)$