loj SDOI2017数字表格 先存下

Posted 2021-02-09 dsrdsr

(prod limits_{i=1}^{n}prodlimits_{j=1}^{m}f[gcd(i,j)])
(prodlimits_{k=1}^{n}f[k]^{sumlimits_{i=1}^{n}sumlimits_{j=1}^{m}[gcd(i,j)==k]})
幂我们很熟悉
就是
(g(x)=sumlimits_{i=1}^{n}sumlimits_{j=1}^{m}[gcd(i,j)==x]=sumlimits_{i=1}^{frac{n}{x}}sumlimits_{j=1}^{frac{m}{x}}[gcd(i,j)==1])
(f(x)=sumlimits_{x|d}^ng(d)=frac{n}{x}frac{m}{x})
(g(x)=sumlimits_{x|d}^nmu(frac{d}{x})f(d))
(g(x)=sumlimits_{x|d}^nmu(frac{d}{x})frac{n}{d}frac{m}{d})
(g(x)=sumlimits_{d=1}^{n/x}mu(d)frac{n}{xd}frac{m}{xd})
带回去
(prodlimits_{k=1}^{n}f[k]^{sumlimits_{d=1}^{n/k}mu(d)frac{n}{dk}frac{m}{dk}})
令i=d*k
(prodlimits_{k=1}^{n}f[k]^{sumlimits_{k|i}^{n}mu(i/k)frac{n}{i}frac{m}{i}})
(prodlimits_{k=1}^{n}prodlimits_{k|i}^{n}f[k]^{mu(i/k)frac{n}{i}frac{m}{i}})
(prodlimits_{i=1}^{n}prodlimits_{k|i}^{n}f[k]^{mu(i/k)frac{n}{i}frac{m}{i}})
设(Au(i)=prodlimits_{k|i}^{n}f[k]^{mu(i/k)})
(ans=prodlimits_{i=1}^{n}Au(i)^{frac{n}{i}frac{m}{i}})