BZOJ 3561 DZY Loves Math VI
求(sum_{i=1}^{n}sum_{j=1}^{m} ext{lcm}(i,j)^{gcd(i,j)}),钦定(nleq m)
(sum_{i=1}^{n}sum_{j=1}^{m}(frac{ij}{{gcd(i,j)}})^{gcd(i,j)})
按套路,提出(gcd(i,j)),枚举的(i),(j)都除(g)
(sum_{g=1}^ng^gsum_{i=1}^{n/g}sum_{j=1}^{m/g}(ij)^g[gcd(i,j)=1])
([gcd(i,j)=1])改成约数mu之和
(sum_{g=1}^ng^gsum_{i=1}^{n/g}sum_{j=1}^{m/g}(ij)^gsum_{k|gcd(i,j)}mu(k))
(sum_{g=1}^ng^gsum_{k=1}^{n/g}mu(k)k^{2g}sum_{i=1}^{n/gk}sum_{j=1}^{m/gk}(ij)^g)
(sum_{g=1}^ng^gsum_{k=1}^{n/g}mu(k)k^{2g}sum_{i=1}^{n/gk}i^gsum_{j=1}^{m/gk}j^g)
然后就能暴力求了。23333