SPOJ : DIVCNT2 - Counting Divisors (square)
设\[f(n)=\sum_{d|n}\mu^2(d)\]则\[\begin{eqnarray*}\sigma_0(n^2)&=&\sum_{d|n}f(d)\\ans&=&\sum_{i=1}^n\sigma_0(i^2)\\&=&\sum_{i=1}^n\sum_{d|i}\sum_{k|d}\mu^2(k)\\&=&a
