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# 题解 P5326 【[ZJOI2019]开关】

\begin{aligned} f(x) &= \sum_{k \ge 0} \left(k! \cdot [x^k]F(x)\right) x^k \\\\ &= \sum_{k \ge 0} \left(k! \cdot [x^k]\left(\sum_{i} c_ie^{ix}\right)\right) x^k \\\\ &= \sum_{k \ge 0} \left(k! \cdot [x^k]\left(\sum_{i} c_i\cdot\left(\sum_{j \ge 0} \frac1{j!}(ix)^j\right)\right)\right) x^k \\\\ &= \sum_{k \ge 0} \left(k! \left(\sum_{i} c_i\cdot \frac1{k!}i^k \right)\right) x^k \\\\ &= \sum_{k \ge 0} \left(\sum_{i} c_ii^k \right) x^k \\\\ &= \sum_{i} c_i\left(\sum_{k \ge 0} i^k x^k \right) \\\\ &= \sum_{i} \frac{c_i}{1-ix} \\\\ \end{aligned}

\begin{aligned}f^\prime(x) &= \left(\sum_{i} \frac{c_i(1-x)}{1-ix}\right)^\prime \\\\&= \sum_{i} \left(\frac{c_i-c_ix}{1-ix}\right)^\prime \\\\&= \sum_{i \ne 1} \frac{-c_i(1-ix)-(c_i - c_ix)(-i)}{(1-ix)^2} \\\\&= \sum_{i \ne 1} \frac{(i-1)c_i}{(1-ix)^2} \\\\f^\prime(1) &= \sum_{i \ne 1} \frac{(i-1)c_i}{(1-i)^2} \\\\&= \sum_{i \ne 1} \frac{c_i}{i-1} \\\\\end{aligned}

2020-03-28 16:08:03 in 题解