最小公倍数之和

题目大意

求$\sum_{i=1}^{n-1}lcm(i,n)$(n<=$10^9$)

有t组询问(t<=50000)

欧拉函数

$\sum_{i=1}^{n-1}lcm(i,n)$

$=n*\sum_{d|n}\sum_{i=1}^{[\frac{n-1}{d}]}[gcd(i,\frac{n}{d})=1]*i$

$=n+n*\sum_{d|n(d\neq{n})}\frac{\varphi(\frac{n}{d})*\frac{n}{d}}{2}$

$=n+\frac{n}{2}*\sum_{d|n(d\neq{1})}\varphi(d)*d$

去掉欧拉函数

$n=\Pi_{i=1}^Tp[i]^{k[i]}$p[i]表示互异的素数。

$\sum_{d|N}\varphi(d)*d$=$\Pi_{i=1}^T\sum_{j=0}^{k[i]}\varphi(p[i]^j)*p[i]^j$

=$\Pi_{i=1}^T1+\sum_{j=1}^{k[i]}\varphi(p[i]^j)*p[i]^j$

=$\Pi_{i=1}^T1+\sum_{j=1}^{k[i]}p[i]^{j-1}*(p[i]-1)*p[i]^j$

=$\Pi_{i=1}^T1+(p[i]-1)*\sum_{j=1}^{k[i]}p[i]^{2*j-1}$

=$\Pi_{i=1}^T1+(p[i]-1)*\frac{p[i]^{2*k[i]+1}-p[i]}{p[i]^2-1}$

=$\Pi_{i=1}^T1+\frac{p[i]^{2*k[i]+1}-p[i]}{p[i]+1}$

然后是快速分解就好了。