BZOJ_2693_jzptab_莫比乌斯反演

BZOJ_2693_jzptab_莫比乌斯反演

Description

Input

一个正整数T表示数据组数

接下来T行 每行两个正整数 表示N、M

Output

T行 每行一个整数 表示第i组数据的结果

Sample Input

1

4 5

Sample Output

122

HINT
T <= 10000

N, M<=10000000

---

$sumlimits_{i=1}^{n}sumlimits_{j=1}^{m}lcm(i,j)$

$=sumlimits_{i=1}^{n}sumlimits_{j=1}^{m}frac{i*j}{gcd(i,j)}$

$=sumlimits_{p=1}^{n}sumlimits_{i=1}^{lfloorfrac{n}{p} floor}
sumlimits_{j=1}^{lfloorfrac{m}{p} floor} i*j*p*[gcd(i,j)=1]$

$=sumlimits_{p=1}^{n}psumlimits_{i=1}^{lfloorfrac{n}{p} floor}
sumlimits_{j=1}^{lfloorfrac{m}{p} floor} i*j
sumlimits_{d|gcd(i,j)}mu(d)$


$=sumlimits_{p=1}^{n}p
sumlimits_{d=1}^{n/p}mu(d)*d^{2}
sumlimits_{i=1}^{lfloorfrac{n/p}{d} floor}
sumlimits_{j=1}^{lfloorfrac{m/p}{d} floor} i*j
$

设$s[n]=sumlimits_{i=1}^{n}i$

$=sumlimits_{p=1}^{n}p
sumlimits_{d=1}^{n/p}mu(d)*d^{2}*
s[lfloorfrac{n/p}{d} floor]*
s[lfloorfrac{m/p}{d} floor]
$

设$Q=d*p，先枚举Q$

$=sumlimits_{Q=1}^{n}
s[lfloorfrac{n}{Q} floor]*
s[lfloorfrac{m}{Q} floor]
sumlimits_{d|Q}mu(d)*d^{2}*lfloorfrac{Q}{d} floor
$

设$f[n]=sumlimits_{d|n}mu(d)*d^{2}*lfloorfrac{n}{d} floor
=nsumlimits_{d|n}mu(d)*d$

$=sumlimits_{Q=1}^{n}
s[lfloorfrac{n}{Q} floor]*
s[lfloorfrac{m}{Q} floor]*f[Q]
$

$然后发现f[n]=n*g[n],g[n]为 id卷mu 的积性函数$

$我们可以处理出f[n]的前缀和，然后O(sqrt{n})处理即可$

$mdlswl$