递归方程：
\[\begin{cases} f(n)=2f(n/2)+c\cdot &n>1\\\\ f(1)=0 &n=1 \end{cases} \]
换元：
\[\begin{array}[lcl] s令\quad k=2^n,f(n)=f(2^k)=h(k)\\\\\\ 则\quad h(k)=2h(k-1)+2^k \quad\quad\;\\\\\\ 故递归方程变为：\begin{cases} h(k)=2h(k-1)+2^k &k>0\\\\ h(0)=0 &k=0 \end{cases} \end{array} \]
构造生产函数求解：
\[\begin{array}[lcl] sG(x) =\sum_{k=0}^{+\infty}\left[h(k)\cdot x^k\right] \qquad\qquad\qquad (1)\\\\\\ 2x\cdot G(x) =\sum_{k=0}^{+\infty}\left[2h(k)\cdot x^{k+1}\right] \qquad\;\;\,(2)\\\\\\ (1)-(2)得：\\\\\\ (1-2x)\cdot G(x)=h(0)+\sum_{k=0}^{+\infty}\left\{\left[h\left(k+1\right)-2h\left(k\right)\right]\cdot x^{k+1}\right\}\\\\\\ \qquad\qquad\qquad\;\;\,=h(0)+C\cdot \sum_{k=0}^{+\infty}(2^{k+1}\cdot x^{k+1})\\\\\\ \qquad\qquad\qquad\;\;\,=h(0)+C\cdot \left(\frac{2x}{1-2x}\right)\\\\\\ \therefore G(x)=\frac{h(0)}{1-2x}+C\cdot \left[\frac{1}{\left(1-2x\right)^2}-\frac{1}{1-2x}\right]\\\\\\ \because h(0)=f(0)=0\\\\\\ \therefore G(x)=C\cdot \sum_{k=0}^{+\infty}\left[k\cdot(2x)^k\right]\\\\\\\qquad\quad\;\,= \sum_{k=0}^{+\infty}\left[c\cdot k\cdot (2x)^k\right]\\\\\\ \therefore h(k)=C \cdot k \cdot 2^k\\\\\\ 又\because n=2^k \therefore f(n)=c\cdot nlog_2n \end{array} \]