[bzoj2693]jzptab

又没推出来……

不过通过这道题还是学到好多东西呢，比如积性函数，线筛什么的。

 $\sum \limits _{i=1}^{n} \sum \limits _{j=1}^{m} lcm(i,j)$

=$\sum \limits _{d=1}^{min(n,m)} \sum \limits _{i=1}^{\left \lfloor \frac{n}{d} \right \rfloor} \sum \limits _{j=1}^{\left \lfloor \frac{m}{d} \right \rfloor} i*j*d \left [ gcd(i,j)=1 \right ]$

=$\sum \limits _{d=1}^{min(n,m)} \sum \limits _{i=1}^{\left \lfloor \frac{n}{d} \right \rfloor} \sum \limits _{j=1}^{\left \lfloor \frac{m}{d} \right \rfloor} \left ( i*j*d *\sum \limits _{t\mid gcd(i,j)}u(t) \right )$

=$\sum \limits _{d=1}^{min(n,m)} d* \sum \limits _{i=1}^{\left \lfloor \frac{n}{d} \right \rfloor} \sum \limits _{j=1}^{\left \lfloor \frac{m}{d} \right \rfloor} \left ( i*j *\sum \limits _{t\mid gcd(i,j)}u(t) \right )$

=$\sum \limits _{d=1}^{min(n,m)} d* \sum \limits _{t=1}^{min\left ( \left \lfloor \frac{n}{d} \right \rfloor \left \lfloor \frac{m}{d} \right \rfloor\right )} u(t) * t^2 *\sum \limits _{i=1}^{\left \lfloor \frac{n}{d*t} \right \rfloor} \sum \limits _{j=1}^{\left \lfloor \frac{m}{d*t} \right \rfloor} i*j$

自己推到这就又不会了……

主要是没想到这玩意：$\sum \limits _{i} \sum \limits _{j} \left ( i*j  \right )=  \sum \limits _{i} \left ( i*\sum \limits _{j} j\right ) = \left ( \sum \limits _{i}i \right )*\left ( \sum \limits _{j}j \right )$

于是原式=$\sum \limits _{d=1}^{min(n,m)} \sum \limits _{t=1}^{min\left ( \left \lfloor \frac{n}{d} \right \rfloor ,\left \lfloor \frac{m}{d} \right \rfloor \right )} u(t) * t^2 *\left ( \sum \limits _{i=1}^{\left \lfloor \frac{n}{d*t} \right \rfloor}i \right ) * \left ( \sum \limits _{j=1}^{\left \lfloor \frac{m}{d*t} \right \rfloor}j \right )$

后面是一个等差数列，原式=$\sum \limits _{d=1}^{min(n,m)} d* \left ( \sum \limits _{t=1}^{min\left ( \left \lfloor \frac{n}{d} \right \rfloor \left \lfloor \frac{m}{d} \right \rfloor \right )} u(t)*t^2 *\frac{\left \lfloor \frac{n}{d*t} \right \rfloor *\left ( \left \lfloor \frac{n}{d*t} \right \rfloor +1 \right ) * \left \lfloor \frac{m}{d*t} \right \rfloor \left ( \left \lfloor \frac{m}{d*t} \right \rfloor +1 \right )}{4} \right )$

令T=d*t,原式=$\sum \limits _{T=1}^{min(n,m)} \frac{\left \lfloor \frac{n}{T} \right \rfloor * \left ( \left \lfloor \frac{m}{T} \right \rfloor +1 \right ) * \left \lfloor \frac{m}{T} \right \rfloor * \left ( \left \lfloor \frac{m}{T} \right \rfloor +1 \right )}{4} * T*\sum \limits _{t\mid T}u(t)*t $

到这里式子就推完了（稍恶心），之后设$f(n)=n*\sum \limits _{t\mid n} u(t)*t$,通过一些我看不懂的证明可以发现它是积性函数，于是我们可以用线筛搞出来它，前面的部分整除分块就可以了。

#include<iostream>
#include<cstring>
#include<cstdio>
#define int LL
#define LL long long
using namespace std;
const int mod=100000009;
bool noprime[12000000];
int prime[12000000],cnt;
int low[12000000],f[12000000],mu[12000000];
void shai(int n)
{
    f[1]=mu[1]=1;
    for(int i=2;i<=n;i++)
    {
        if(!noprime[i]) prime[++cnt]=i,mu[i]=-1,f[i]=-i+1+mod;
        for(int j=1;j<=cnt&&prime[j]*i<=n;j++)
        {    
            noprime[prime[j]*i]=1;
            if(i%prime[j]==0){f[i*prime[j]]=f[i];break;}
            f[i*prime[j]]=(f[i]*f[prime[j]])%mod;mu[i*prime[j]]=-mu[i];
        }
    }
    for(int i=1;i<=n;i++)f[i]=f[i]*i%mod;
    for(int i=1;i<=n;i++)f[i]=(f[i]+f[i-1])%mod;
}
LL get(LL x){return (x*(x+1)/2)%mod;}
int T,n,m;
signed main()
{    
    shai(12000000);
    cin>>T;
    while(T--)
    {    
        cin>>n>>m;if(n>m)swap(n,m);
        LL ans=0;
        for(int l=1,r;l<=n;l=r+1)
        {    
            r=min(n/(n/l),m/(m/l));
            ans=(ans+get(n/l)*get(m/l)%mod*((f[r]-f[l-1]+mod)%mod)%mod)%mod;
        }
        printf("%lld\n",(ans%mod+mod)%mod);
    }
}