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．
标签：...,sum,扩展,约瑟夫,问题,beta,ia,alpha,displaystyle From： https://www.cnblogs.com/wf715/p/Ex-Josephus-Problem.html