# [ZJOI2019] 开关 (一种扩展性较高的做法)

## 题解：

$\sum_{x_i} \prod [2|x_i-s_i] \prod q_i^{x_i} \sum_{i=1}^nx_i \sum_{m=0}^{\infty} \sum_{y_{1i};\exists i,y_{1i}>0 } \sum_{y_{2i};\exists i,y_{2i}>0 }... \sum_{y_{mi};\exists i, y_{mi}>0 } \sum_{{z_i};\exists i,z_i>0 }\\ \prod [2|y_{ij}] \prod_{j}[\sum_{i=1}^m y_{ij} +z_i = x_i] (-1)^m {(\sum_{i=1}^m z_i)! \over \prod_{i=1}^n z_i!}\prod_{i=1}^m {(\sum_{j=1}^n y_{ij})! \over \prod_{j=1}^n y_{ij}}$

$q_i$$p_i \over \sum p_i$

$F(x)=\prod (e^{q_ix} + q_ixye^{q_ix} +(-1)^{s_i} e^{-q_ix} - (-1)^{s_i} q_i xye^{-q_ix}) \\ =\sum_{i=0}^{\infty} {f_{0i} \over i!}x^i + {f_{1i} \over i!}x^iy$

$G(x)=\prod (e^{q_ix} + q_ixye^{q_ix} + e^{-q_ix} - q_i xye^{-q_ix}) \\ =\sum_{i=0}^{\infty}{g_{0i} \over i!}x^i + {g_{1i} \over i!}x^iy$

$f_0(x)=\sum_{i=0}^{\infty}f_{0i}x^i,f_1(x)=\sum_{i=0}^{\infty}f_{1i}x^i,f(x)=f_0(x)+f_1(x)y$
$g_0(x)=\sum_{i=0}^{\infty}g_{0i} x^i,g_1(x)=\sum_{i=0}^{\infty}g_{1i}x^i,g(x)=g_0(x)+g_1(x)y$
$F(x)$为例,$F(x)=\sum_{i=- \infty}^{\infty} a_i e^{ix} + b_i xy e^{ix}$

$\lim_{x \to 1 }[y^1]f(x)\sum_{m=0}^{\infty}(-g(x)+1)^m=\lim_{x \to 1}[y^1]{f(x) \over g(x)}\\ ={f_2(x)g_1(x) -g_2(x)f_1(x) \over g_1^2(x)}$

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posted @ 2019-05-06 16:52  Deadecho  阅读(673)  评论(7编辑  收藏