[SPOJ] DIVCNT2 - Counting Divisors (square)

题解：

操作挺多的一道题

网上证明挺多就不打了

$\sigma_0(n^2) = \sum_{d\mid n} 2^{\omega(d)} = \sum_{d\mid n} \sum_{e\mid d} \mu^2(e) = ((\mu^2 * 1) * 1) (n)$

$(\mu * 1) * 1 = \mu * (1*1) = \mu * \sigma_0$

$\sum_{i=1}^n \mu^2(i) = \sum_{i=1}^{\sqrt{n}}\mu(i)\lfloor \frac{n}{i^2} \rfloor$

知道了这些按照杜教筛的套路处理$n^{\frac{2}{3}}$就可以了

[SPOJ] DIVCNT2 - Counting Divisors (square)

原文：https://www.cnblogs.com/yinwuxiao/p/10349446.html