扩展约瑟夫问题

看到具体数学上面写了就手推一下

定义：

\[f(x) = \alpha_x (1 \leq x < a), \]

\[f(ax + l) = bf(x) + \beta_l (0 \leq l < a, x \geq 1) \]

求证：

\[f((c_m c_{m - 1} ... c_1 c_0)_a) = (\alpha_{c_m} \beta_{c_{m - 1}} \beta_{c_{m - 2}} ... \beta_{c_2} \beta_{c_1})_b \]

证明：

列表：

\(x\)	\(f(x)\)
\(1\)	\(\alpha_1\)
\(a - 1\)	\(\alpha_{a - 1}\)
\(a\)	\(b\alpha_1 + \beta_0\)
\(a + (a - 1)\)	\(b\alpha_1 + \beta_{a - 1}\)
\(a^2\)	\(bf(a) + \beta_0 = b(b\alpha_1 + \beta_0) + \beta_0\)

从中可发现一些规律．

设 \(x = \displaystyle\sum_{i = 0}^{m} c_ia^i (0 \leq c_i < a)\)，则 \(x = (c_m c_{m - 1} ... c_1 c_0)_a\)．

\[f(x) = f(\displaystyle\sum_{i = 0}^{m} c_ia^i) \]

\[= bf(\displaystyle\sum_{i = 1}^{m} c_ia^{i - 1}) + \beta_{c_0} \]

\[= b(bf(\displaystyle\sum_{i = 2}^{m} c_ia^{i - 2}) + \beta_{c_1}) + \beta_{c_0} \]

\[= ... \]

\[= b^mf(c_m) + \displaystyle\sum_{i = 0}^{m - 1} b^i\beta_{c_i} \]

\[= b^m\alpha_{c_m} + b^{m - 1}\beta_{m - 1} + b^{m - 2}\beta_{m - 1} + ... + b^1\beta_1 + \beta_0 \]

\[= (\alpha_{c_m} \beta_{c_{m - 1}} \beta_{c_{m - 2}} ... \beta_{c_1} \beta_{c_0})_b \]

证毕．