![[SPOJ] DIVCNT2 - Counting Divisors (square)](/uploads/reation/bcimg4.png)
[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}^...
