BLMontgomery

摘要： "题目链接" 以下除法均指除后下取整 $$\prod_{i=1}^n \prod_{j=1}^mf(gcd(i,j))\\ =\prod_{x}f(x)^{\sum_i \sum_j [gcd(i,j)=x] } \\ =\prod_{x}f(x)^{ \sum_{x|d} \mu(\frac{d} 阅读全文