51Nod 1238 最小公倍数之和 V3

题目大意

给定 $n,m$，令
$$Ans=\sum _{i=1}^n\sum _{j=1}^{m}lcm(i,j),Ans\equiv x(\mathrm{m}\mathrm{o}\mathrm{d}10^9+7)$$

求最小的非负 $x$。
数据范围　$1⩽n,m⩽10^{10}$

题解

$$\begin{array}{rl}Ans& =\sum _{i=1}^n\sum _{j=1}^m\frac{i\times j}{(i,j)}\\ & =\sum _{d=1}^n\sum _{i=1}^n\sum _{j=1}^m\frac{i\times j}{d}[(i,j)=d]\\ & =\sum _{d=1}^{\min (n,m)}\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{d}\rfloor }\frac{d\cdot i\times d\cdot j}{d}[(i,j)=1]\\ & =\sum _{d=1}^{\min (n,m)}d\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{d}\rfloor }ij[(i,j)=1]\end{array}$$

套莫比乌斯反演
$$\begin{array}{rl}Ans& =\sum _{d=1}^{\min (n,m)}d\times \sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{d}\rfloor }ij\sum _{k|i,k|j}\mu (k)\\ & =\sum _{d=1}^{\min (n,m)}d\times \sum _{k=1}^{\lfloor \frac{n}{d}\rfloor }\mu (k)\sum _{k|i}^{i⩽\lfloor \frac{n}{d}\rfloor }i\sum _{k|j}^{j⩽\lfloor \frac{m}{d}\rfloor }j\\ & =\sum _{d=1}^{\min (n,m)}d\times \sum _{k=1}^{\lfloor \frac{\min (n,m)}{d}\rfloor }\mu (k)\left(\sum _{k|i}^{i⩽\lfloor \frac{n}{d}\rfloor }i\right)\left(\sum _{k|j}^{j⩽\lfloor \frac{m}{d}\rfloor }j\right)\\ & =\sum _{d=1}^{\min (n,m)}d\times \sum _{k=1}^{\lfloor \frac{\min (n,m)}{d}\rfloor }\mu (k)\left(\sum _{i=1}^{\lfloor \frac{n}{dk}\rfloor }ik\right)\left(\sum _{j=1}^{\lfloor \frac{m}{dk}\rfloor }jk\right)\\ & =\sum _{d=1}^{\min (n,m)}d\times \sum _{k=1}^{\lfloor \frac{\min (n,m)}{d}\rfloor }{k}^2\mu (k)\left(\frac{\left(1+\lfloor \frac{n}{dk}\rfloor \right)\lfloor \frac{n}{dk}\rfloor }{2}\right)\left(\frac{\left(1+\lfloor \frac{m}{dk}\rfloor \right)\lfloor \frac{m}{dk}\rfloor }{2}\right)\end{array}$$

即使用杜教筛筛出 $k^2\mu (k)$，这样时间仍然是超过线性的，无法通过本题。

更换枚举项，枚举 $T=kd$，易得其上界为 $n$，并且 $k|T$，则有
$$\begin{array}{rl}Ans& =\sum _{T=1}^{\min (n,m)}\sum _{k|T}\frac{T}{k}\times k^2\times \mu (k)\times \left(\frac{\left(1+\lfloor \frac{n}{T}\rfloor \right)\lfloor \frac{n}{T}\rfloor }{2}\right)\left(\frac{\left(1+\lfloor \frac{m}{T}\rfloor \right)\lfloor \frac{m}{T}\rfloor }{2}\right)\\ & =\sum _{T=1}^{\min (n,m)}\left(\frac{\left(1+\lfloor \frac{n}{T}\rfloor \right)\lfloor \frac{n}{T}\rfloor }{2}\right)\left(\frac{\left(1+\lfloor \frac{m}{T}\rfloor \right)\lfloor \frac{m}{T}\rfloor }{2}\right)\sum _{k|T}{k}^2\times \mu (k)\times \frac{T}{k}\end{array}$$

这里最好不要化简后面的 $$\sum _{k|T}{k}^2\times \mu (k)\times \frac{T}{k}$$

最好直接把它转换为卷积
$$\begin{array}{rl}Ans& =\sum _{T=1}^{\min (n,m)}\left(\frac{\left(1+\lfloor \frac{n}{T}\rfloor \right)\lfloor \frac{n}{T}\rfloor }{2}\right)\left(\frac{\left(1+\lfloor \frac{m}{T}\rfloor \right)\lfloor \frac{m}{T}\rfloor }{2}\right)((i{d}^2\cdot \mu )\ast id)(T)\end{array}$$

现在只需要求出函数 $((i{d}^2\cdot \mu )\ast id)$ 的前缀和即可。因此考虑 $g=id$。

等等！

为什么要考虑 $g=id$？为了让后续的 乱猜 更方便？如果 $g=id$ 能做到些什么的话，为什么里面那个 $id$ 不能做到？
考虑到卷积有交换律，我们可以忽略后面那个 $\ast id$ 而直接考虑前面的 $(i{d}^2\cdot \mu )$，很容易发现，对于这个用点乘得到函数只需要取 $g=i{d}^2$ 即可。
$$\begin{array}{rl}\left(\right((i{d}^2\cdot \mu )\ast id)\ast i{d}^2)(n)& =\left(\right((i{d}^2\cdot \mu )\ast i{d}^2)\ast id)(n)\\ & =\sum _{d|n}((i{d}^2\cdot \mu )\ast i{d}^2)(d)\times \frac{n}{d}\\ & =\sum _{d|n}\sum _{k|d}{k}^2\cdot \mu (k)\times \frac{d^2}{k^2}\times \frac{n}{d}\\ & =n\sum _{d|n}d\sum _{k|d}\mu (k)\\ & =n\sum _{d|n}d[d=1]\\ & =n\\ & =id(n)\end{array}$$

的确是个让人省心的函数。
因而有 $$\begin{array}{rl}n& =\left(\right((i{d}^2\cdot \mu )\ast id)\ast i{d}^2)(n)\\ & =\sum _{d|n}((i{d}^2\cdot \mu )\ast id)(d)\times \frac{n^2}{d^2}\\ & =\sum _{d|n,d\ne n}((i{d}^2\cdot \mu )\ast id)(d)\times \frac{n^2}{d^2}+((i{d}^2\cdot \mu )\ast id)(n)\end{array}$$

移项，得
$$\begin{array}{r}((i{d}^2\cdot \mu )\ast id)(n)=n-\sum _{d|n,d\ne n}((i{d}^2\cdot \mu )\ast id)(d)\times \frac{n^2}{d^2}\end{array}$$
求和
$$\begin{array}{rl}{\text{Sum}}_{(i{d}^2\cdot \mu )\ast id}(n)& =\sum _{i=1}^n((i{d}^2\cdot \mu )\ast id)(i)\\ & =\sum _{i=1}^{n}i-\sum _{i=1}^n\sum _{d|i,d\ne i}((i{d}^2\cdot \mu )\ast id)(d)\times \frac{i^2}{d^2}\\ & =\frac{(1+n)n}{2}-\sum _{k=2}^n\sum _{d=1}^{\lfloor \frac{n}{k}\rfloor }((i{d}^2\cdot \mu )\ast id)(d)\times \frac{k^2{d}^2}{d^2}\\ & =\frac{(1+n)n}{2}-\sum _{k=2}^n{k}^2\times {\text{Sum}}_{(i{d}^2\cdot \mu )\ast id}(n)\left(\lfloor \frac{n}{k}\rfloor \right)\end{array}$$

递归除法分块处理即可，时间复杂度为 $T(n)=\Theta (n^{\frac{3}{4}})$
前面莫比乌斯反演的部分除法分块处理，总时间复杂度仍然为 $T(n)=\Theta (n^{\frac{3}{4}})$
如果我们能线性筛出前面 $n^{\frac{2}{3}}$ 部分的 $(i{d}^2\cdot \mu )\ast id$ 就大功告成了。
这需要计算其三条特点：
$$((i{d}^2\cdot \mu )\ast id)(1)=1$$

$$((i{d}^2\cdot \mu )\ast id)(p)=p\cdot (1-p)$$

$$((i{d}^2\cdot \mu )\ast id)(p^k)=p^k\cdot (1-p)$$

证明：由于 $p^0=1,p^1=p$，因而我们只需要证明第三条性质成立即可。
$$\begin{array}{rl}((i{d}^2\cdot \mu )\ast id)(p^k)& =\sum _{d|p^k}(i{d}^2\cdot \mu )(d)\times id\left(\frac{p^k}{d}\right)\\ & =p^k\times \sum _{d|p^k}d\cdot \mu (d)\\ & =p^k\times \sum _{t=0}^k{p}^t\cdot \mu (p^t)\\ & =p^k\times \left(p^0\cdot \mu (p^0)+p^1\cdot \mu (p^1)+\sum _{t=2}^k{p}^t\cdot \mu (p^t)\right)\\ & =p^k\times \left(1-p+\sum _{t=2}^k{p}^t\times 0\right)\\ & =p^k\cdot \left(1-p\right)\end{array}$$

至此，我们能够在 $T(k)=\Theta (1)$ 的时间内求出上面三种位置的函数值，因此，我们可以线性筛了。

总时间复杂度进一步降低到 $T(n)=\Theta (T\times n^{\frac{2}{3}})$。
%https://blog.youkuaiyun.com/C20181220_xiang_m_y/article/details/84894901

#include<bits/stdc++.h>
using namespace std;
#define maxn 7841588
const long long mod=1e9+7;
const long long inv_2=500000004;
const long long inv_6=166666668;
long long low[maxn],f[maxn],pri[maxn],vis[maxn],s[maxn];
void sieve(long long n){
	memset(vis,0,sizeof vis);
	long long tot=0;
    vis[1]=low[1]=f[1]=1;
    for (long long i=2;i<=n;i++)
    {
        if(!vis[i]) low[i]=pri[++tot]=i,f[i]=(i-(i*i)%mod+mod)%mod;
        for (long long j=1;j<=tot&&i*pri[j]<=n;j++)
        {
            vis[i*pri[j]]=1;
            if (i%pri[j]==0)
            {
                low[i*pri[j]]=low[i]*pri[j];
                if (low[i]==i)
                    f[i*pri[j]]=((((1-pri[j])*i+mod)%mod)*pri[j]+mod)%mod;
                else
                    f[i*pri[j]]=(f[i/low[i]]*f[low[i]*pri[j]]+mod)%mod;
                break;
            }
            low[i*pri[j]]=pri[j];
            f[i*pri[j]]=(f[i]*f[pri[j]])%mod;
        }
    }
    for(int i=1;i<=n;i++)
    	s[i]=(s[i-1]+f[i]+mod)%mod;
}
long long ft_1(long long a){
	long long t=a%mod;
	return ((((1+(t))*(t))%mod)*inv_2)%mod;
}
long long ft_2(long long a){
	long long t=a%mod;
	return (((t*(t+1)%mod)*(2*t+1)%mod)*inv_6)%mod;
}
map<long long,long long> Sf;
long long S(long long n){
	if(n<=maxn-10) return s[n];
	{	map<long long,long long>::iterator t=Sf.lower_bound(n);
		if(t!=Sf.end()) return t->second;	}
	long long ans=ft_1(n);
	for(long long l=2,r;l<=n;l=r+1){
		r=n/(n/l);
		ans-=S(n/l)*(ft_2(r)-ft_2(l-1))%mod;
		ans=(ans+mod);
	} return ans;
}
void main2(long long n,long long m){
	long long ans=0;
	long long len=min(n,m);
	for(long long l=1,r;l<=len;l=r+1){
		r=min(n/(n/l),m/(m/l));
		if(r>n) r=n;
		ans+=ft_1(n/l)*ft_1(m/l)%mod*(S(r)-S(l-1))%mod;
		ans=(ans+mod)%mod;
	} printf("%lld",ans); 
}
int main(){
	sieve(maxn-10);
	long long n,m,ans=0;
	scanf("%lld%lld",&n,&m);
	for(long long i=1;i<=n;i++)
		for(long long j=1;j<=m;j++){
			ans+=i/__gcd(i,j)*j;
			ans%=mod;
		}
	printf("%lld\n",ans);
	main2(n,m);
	return 0;
}

%https://www.cnblogs.com/zhoushuyu/p/8275530.html

%https://blog.youkuaiyun.com/sslz_fsy/article/details/87819016