bzoj2693 jzptab (莫比乌斯反演)

传送门：https://www.lydsy.com/JudgeOnline/problem.php?id=2693
题目大意：
求${\sum }_{i=1}^n{\sum }_{j=1}^{m}lcm(i,j)$

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$solution$
这种题无脑推式子就行了
假定$n\ge m$
$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^{m}lcm(i,j) \\ =\sum _{i=1}^n\sum _{j=1}^m\frac{ij}{gcd(i,j)} \\ =\sum _{d=1}^m\frac{1}{d}\sum _{d|i}\sum _{d|j}ij\ast [gcd(i,j)==d] \\ =\sum _{d=1}^{m}d\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{d}\rfloor }ij\ast [gcd(i,j)==1] \\ =\sum _{d=1}^{m}d\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{d}\rfloor }ij\sum _{t|gcd(i,j)}\mu (t) \\ =\sum _{d=1}^{m}d\sum _{t=1}^{\lfloor \frac{m}{d}\rfloor }\mu (t)t^2\sum _{i=1}^{\lfloor \frac{n}{td}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{td}\rfloor }ij \\ =\sum _{T=1}^m\sum _{i=1}^{\lfloor \frac{n}{T}\rfloor }i\sum _{j=1}^{\lfloor \frac{m}{T}\rfloor }j\sum _{t|T}\mu (t)\ast t^2\ast \frac{T}{t} \\ =\sum _{T=1}^m\left(\frac{(1+\lfloor \frac{n}{T}\rfloor )\lfloor \frac{n}{T}\rfloor }{2}\right)\ast \left(\frac{(1+\lfloor \frac{m}{T}\rfloor )\lfloor \frac{m}{T}\rfloor }{2}\right)\sum _{t|T}T\mu (t)t \\ =\sum _{T=1}^{m}f(\lfloor \frac{n}{T}\rfloor ,\lfloor \frac{m}{T}\rfloor )\ast g(T) \end{gathered}$$
$g(x)$显然是个积性函数，直接线性筛
其他就是莫比乌斯反演的套路了
$O(T\sqrt{n}+n)$

#include<bits/stdc++.h>
using namespace std;
#define rep(i,j,k) for(int i = j;i <= k;++i)
#define repp(i,j,k) for(int i = j;i >= k;--i)
#define rept(i,x) for(int i = linkk[x];i;i = e[i].n)
#define P pair<int,int>
#define Pil pair<int,ll>
#define Pli pair<ll,int>
#define Pll pair<ll,ll>
#define pb push_back 
#define pc putchar
#define mp make_pair 
#define file(k) memset(k,0,sizeof(k))
#define ll long long
#define N 10000000
int rd()
{
	int num = 0;char c = getchar();bool flag = true;
	while(c < '0'||c > '9') {if(c == '-') flag = false;c = getchar();}
	while(c >= '0' && c <= '9') num = num*10+c-48,c = getchar();
	if(flag) return num;else return -num;
}
const int p = 1e8+9;
int n,m,inv;
int f[N+10000],prime[N+10000],miu[N+10000];
inline int mo(int a){return a>=p?a-p:a;}
inline int calc(int a,int b){return (a+=b)>=p?a-=p:a;}
inline int mul(int a,int b){return 1ll*a*b%p;}
inline int del(int a,int b){return (a-=b)<0?a+=p:a;}
void pre()
{
	miu[1] = 1;f[1] = miu[1]*1;
	rep(i,2,N)
	{
		if(!f[i]) miu[i] = p-1,f[i] = calc(f[1],mul(miu[i],i)),prime[++prime[0]] = i;
		rep(j,1,prime[0])
		{
			if(i*prime[j]>N) break;
			miu[i*prime[j]] = i%prime[j]?mo(p-miu[i]):0;
			f[i*prime[j]] = i%prime[j]?mul(f[i],f[prime[j]]):f[i];
			if(i%prime[j] == 0) break;
		}
	}
	rep(i,1,N) f[i] = mul(f[i],i);
	rep(i,1,N) f[i] = calc(f[i],f[i-1]);
}
void work()
{
	n = rd();m = rd();
	if(n < m) swap(n,m);
	int i = 1;
	int ans = 0;
	while(i<=m)
	{
		int j = min(n/(n/i),m/(m/i)),tmpn = n/i,tmpm = m/i;
		int v = mul(inv,mul(1+tmpn,tmpn));v = mul(v,mul(inv,mul(1+tmpm,tmpm)));
		ans = calc(ans,mul(v,del(f[j],f[i-1])));
		i = j+1;
	}
	printf("%d\n",ans);
}
int main()
{
	inv = (p+1)/2;
	pre();
	int T = rd();
	while(T--) work();
	return 0;
}