XSY 2754 求和

题意 : 求 (sum_{i=1}^{n}{sum_{j=1}^{i}{sum_{k=1}^{i}{gcd(i,j,k)}}} space mod space p); 其中(n,p leq 10^9)

下式均在(mod space p)意义下进行。

(phi*1=id)得到(sum_{i=1}^{n}{sum_{j=1}^{i}{sum_{k=1}^{i}{gcd(i,j,k)}}})

(=sum_{i=1}^{n}{sum_{j=1}^{i}{sum_{k=1}^{i}{sum_{d|gcd(i,j,k)}{phi(d)}}}})

(=sum_{d=1}^{n}{phi(d)sum_{d|i}^{n}{sum_{d|j}^{i}{sum_{d|k}^{i}{1}}}})

(=sum_{d=1}^{n}{phi(d){sum_{i=xd}^n{sum_{j=yd}^i{sum_{k=zd}^i}1}}})

(=sum_{d=1}^{n}{phi(d){sum_{i=1}^{lfloor{n/d} floor} {sum_{j=1}^{i}{sum_{k=1}^{i}}1}}})

(S(n)=sum_{i=1}^n{i^2})(S(n)=frac{n(n+1)(2n+1)}{6})

那么有 (sum_{d=1}^{n}{phi(d){sum_{i=1}^{lfloor{n/d} floor} {i^2}}})

(=sum_{d=1}^n{phi(d)S(lfloor{frac{n}{d}} floor)})

满足杜教筛形式。

原文地址:https://www.cnblogs.com/cjc030205/p/11638084.html