BZOJ3560 DZY Loves Math V

https://hydro.ac/p/bzoj-P3560

设 \(b_{p, i}\) 为质数 \(p\) 在 \(a_i\) 中的指数，\(e_{p, i}\) 为选出的指数，则

\[\begin{aligned} \sum_{d_1 \mid a_1} \cdots \sum_{d_n \mid a_n} \varphi(d_1 \cdots d_n) &= \sum_{\forall p,\ \substack{0 \le e_{p, 1} \le b_{p, 1} \\ \vdots \\ 0 \le e_{p, n} \le b_{p, n}}} \prod_p \varphi(p^{e_{p, 1} + \cdots + e_{p, n}}) \\ &= \prod_p \sum_{\substack{0 \le e_{p, 1} \le b_{p, 1} \\ \vdots \\ 0 \le e_{p, n} \le b_{p, n}}} \varphi(p^{e_{p, 1} + \cdots + e_{p, n}}) \\ &= \prod_p \left(\frac{1}{p} + \frac{p - 1}{p}\sum_{\substack{0 \le e_{p, 1} \le b_{p, 1} \\ \vdots \\ 0 \le e_{p, n} \le b_{p, n}}} p^{e_{p, 1} + \cdots + e_{p, n}}\right) \\ &= \prod_p \left(\frac{1}{p} + \frac{p - 1}{p}\prod_{i = 1}^n \sum_{e = 0}^{b_{p, i}} p^e\right) \\ &= \prod_p \frac{1 + (p - 1)\displaystyle\prod_{i = 1}^n \frac{p^{b_{p, i} + 1} - 1}{p - 1}}{p} \end{aligned} \]

预处理每个 \(p\) 的

\[\prod_{i = 1}^n \frac{p^{b_{p, i} + 1} - 1}{p - 1} \]

的值即可。