[bzoj2693]jzptab

Posted 2022-12-26 al-ca

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。

 $\sum \limits _i=1^n \sum \limits _j=1^m lcm(i,j)$

=$\sum \limits _d=1^min(n,m) \sum \limits _i=1^\left \lfloor \fracnd \right \rfloor \sum \limits _j=1^\left \lfloor \fracmd \right \rfloor i*j*d \left [ gcd(i,j)=1 \right ]$

=$\sum \limits _d=1^min(n,m) \sum \limits _i=1^\left \lfloor \fracnd \right \rfloor \sum \limits _j=1^\left \lfloor \fracmd \right \rfloor \left ( i*j*d *\sum \limits _t\mid gcd(i,j)u(t) \right )$

=$\sum \limits _d=1^min(n,m) d* \sum \limits _i=1^\left \lfloor \fracnd \right \rfloor \sum \limits _j=1^\left \lfloor \fracmd \right \rfloor \left ( i*j *\sum \limits _t\mid gcd(i,j)u(t) \right )$

=$\sum \limits _d=1^min(n,m) d* \sum \limits _t=1^min\left ( \left \lfloor \fracnd \right \rfloor \left \lfloor \fracmd \right \rfloor\right ) u(t) * t^2 *\sum \limits _i=1^\left \lfloor \fracnd*t \right \rfloor \sum \limits _j=1^\left \lfloor \fracmd*t \right \rfloor i*j$

自己推到这就又不会了……

主要是没想到这玩意：$\sum \limits _i \sum \limits _j \left ( i*j  \right )=  \sum \limits _i \left ( i*\sum \limits _j j\right ) = \left ( \sum \limits _ii \right )*\left ( \sum \limits _jj \right )$

于是原式=$\sum \limits _d=1^min(n,m) \sum \limits _t=1^min\left ( \left \lfloor \fracnd \right \rfloor ,\left \lfloor \fracmd \right \rfloor \right ) u(t) * t^2 *\left ( \sum \limits _i=1^\left \lfloor \fracnd*t \right \rfloori \right ) * \left ( \sum \limits _j=1^\left \lfloor \fracmd*t \right \rfloorj \right )$

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

令T=d*t,原式=$\sum \limits _T=1^min(n,m) \frac\left \lfloor \fracnT \right \rfloor * \left ( \left \lfloor \fracmT \right \rfloor +1 \right ) * \left \lfloor \fracmT \right \rfloor * \left ( \left \lfloor \fracmT \right \rfloor +1 \right )4 * T*\sum \limits _t\mid Tu(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。

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #define int LL
 5 #define LL long long
 6 using namespace std;
 7 const int mod=100000009;
 8 bool noprime[12000000];
 9 int prime[12000000],cnt;
10 int low[12000000],f[12000000],mu[12000000];
11 void shai(int n)
12 
13     f[1]=mu[1]=1;
14     for(int i=2;i<=n;i++)
15     
16         if(!noprime[i]) prime[++cnt]=i,mu[i]=-1,f[i]=-i+1+mod;
17         for(int j=1;j<=cnt&&prime[j]*i<=n;j++)
18             
19             noprime[prime[j]*i]=1;
20             if(i%prime[j]==0)f[i*prime[j]]=f[i];break;
21             f[i*prime[j]]=(f[i]*f[prime[j]])%mod;mu[i*prime[j]]=-mu[i];
22         
23     
24     for(int i=1;i<=n;i++)f[i]=f[i]*i%mod;
25     for(int i=1;i<=n;i++)f[i]=(f[i]+f[i-1])%mod;
26 
27 LL get(LL x)return (x*(x+1)/2)%mod;
28 int T,n,m;
29 signed main()
30     
31     shai(12000000);
32     cin>>T;
33     while(T--)
34         
35         cin>>n>>m;if(n>m)swap(n,m);
36         LL ans=0;
37         for(int l=1,r;l<=n;l=r+1)
38             
39             r=min(n/(n/l),m/(m/l));
40             ans=(ans+get(n/l)*get(m/l)%mod*((f[r]-f[l-1]+mod)%mod)%mod)%mod;
41         
42         printf("%lld\n",(ans%mod+mod)%mod);
43     
44 

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