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莫比乌斯反演

作者:互联网

以下内容中,\(a|b\)表示\(a\)是\(b\)的因数,\(a\not|b\)表示\(a\)不是\(b\)的因数

前置知识:积性函数相关

整除分块

狄利克雷卷积

定义:任意函数\(f(n),g(n)\)的狄利克雷卷积为\((f*g)(n)=\sum\limits_{d|n}f(d)g(\dfrac{n}{d})\)

其中\(\sum\limits_{d|n}\)是要枚举\(n\)的因数并累加

换一种写法就是\(\sum\limits_{xy=n}f(x)g(y)\)

还可以用代码写成:

int dirichlet(int n) {
	int ans=0;
	for(int i=1; i<=n; ++i) {
		if(n%i==0) {
			ans+=f[i]*g[n/i];//f[i]为f(i),g[n/i]为g(n/i)
		}
	}
	return ans;
}

这个卷积满足:

  1. 交换律\(f*g=g*f\)
  2. 结合律\(f*g*h=f*(g*h)\)
  3. 分配率\(f*(g+h)=f*g+f*h\)
  4. 当\(f(x),g(x)\)都为积性函数时,\((f*g)(x)\)为积性函数
  5. 当\((f*g)(x),g(x)\)都为积性函数时,\(f(x)\)为积性函数

常见的狄利克雷卷积有:

  1. \(\varphi*I=id\)
  2. \(\mu*I=\epsilon\)
  3. \(\epsilon*f=f\)
  4. \(id*I=\sigma\)
  5. \(I*I=d\)
  6. \((\lambda*I)(n)=[\sqrt{n}\in Z]\)

简单证明:(3,4,5见相关函数的定义)

  1. \(\varphi*I=id\)(下面以\(n=6\)为例)

将\([1,n]\)的整数都除以\(n\):

\(1\div6=\dfrac{1}{6},5\div6=\dfrac{5}{6}\),对应与\(6\)互质的数,个数为\(\varphi(6)\)

\(2\div6=\dfrac{1}{3},4\div6=\dfrac{2}{3}\),对应与\(3\)互质的数,个数为\(\varphi(3)\)

\(3\div6=\dfrac{1}{2}\),对应与\(2\)互质的数,个数为\(\varphi(2)\)

\(6\div6=\dfrac{1}{1}\),对应与\(1\)互质的数,个数为\(\varphi(1)\)

即\(\varphi(1)+\varphi(2)+\varphi(3)+\varphi(6)=6\)(\(\sum\limits_{d|6}\varphi(d)=6\))

  1. \(\mu*I=\epsilon\)

设\(n\)质因数分解后为\(\prod\limits_{i=1}^kp_i^{c_i}\)

当\(n>1\)时:

\(\sum\limits_{d|n}\mu(d)\\=\mu(1)+\mu(p_1)+\mu(p_1^2)+\cdots+\mu(p_i^{c_i})+\cdots+\mu(p_1^{c_1}p_2^{c_2}\cdots p_k^{c_k})\\=\mu(1)+\mu(p_1)+\mu(p_1^2)+\cdots+\mu(p_i^{c_i})+\cdots+\mu(p_1^{c_1})\mu(p_2^{c_2})\cdots \mu(p_k^{c_k})\\=\prod\limits_{i=1}^k\sum\limits_{j=0}^{c_i}\mu(p_i^j)\\=\prod\limits_{i=1}^k\sum\limits_{j=0}^{1}\mu(p_i^j)\\=\prod\limits_{i=1}^k(1-1)\\=0\)

当\(n=1\)时,\(\sum\limits_{d|n}\mu(d)=\mu(1)=1\)

  1. \((\lambda*I)(n)=[\sqrt{n}\in Z]\)

同样设\(n\)质因数分解后为\(\prod\limits_{i=1}^kp_i^{c_i}\)

\(\sum\limits_{d|n}\lambda(d)\\=\lambda(1)+\lambda(p_1)+\lambda(p_1^2)+\cdots+\lambda(p_i^{c_i})+\cdots+\lambda(p_1^{c_1}p_2^{c_2}\cdots p_k^{c_k})\\=\lambda(1)+\lambda(p_1)+\lambda(p_1^2)+\cdots+\lambda(p_i^{c_i})+\cdots+\lambda(p_1^{c_1})\lambda(p_2^{c_2})\cdots \lambda(p_k^{c_k})\\=\prod\limits_{i=1}^k\sum\limits_{j=0}^{c_i}\lambda(p_i^j)\\=\prod\limits_{i=1}^k\sum\limits_{j=0}^{c_i}(-1)^j\\=\prod\limits_{i=1}^k[c_i\equiv0\pmod2]\)

最后那个式子只有在\(c_i\)都为偶数的时候才为\(1\),所以\(n\)为完全平方数时才是\(1\)。

一道题

给出\(n(n\leq10^{12})\),求\(\sum\limits_{k=1}^n\sum\limits_{i|k}\sum\limits_{j|i}\lambda(i)\lambda(j)\pmod{998244353}\)

我们可以先枚举\(i\),再枚举满足\(i|k\)的\(k\):

\(\sum\limits_{i=1}^n\sum\limits_{i|k}\sum\limits_{j|i}\lambda(i)\lambda(j)\)

可以提出公因式:

\(\sum\limits_{i=1}^n\lambda(i)(\sum\limits_{i|k}1)\sum\limits_{j|i}\lambda(j)\)

中间的部分就是求小于等于\(n\)的\(i\)的倍数有多少,也就是\(\lfloor\dfrac{n}{i}\rfloor\)

\(\sum\limits_{i=1}^n\lambda(i)\lfloor\dfrac{n}{i}\rfloor\sum\limits_{j|i}\lambda(j)\)

后面的部分是\(\lambda*I\),上面说了,是\([\sqrt{n}\in Z]\)

\(\sum\limits_{i=1}^n\lambda(i)\lfloor\dfrac{n}{i}\rfloor[\sqrt{i}\in Z]\)

这样我们就只用考虑完全平方数了。

而且当\(i\)为完全平方数时\(\lambda(i)=1\)

\(\sum\limits_{i=1}^{\sqrt{n}}\lfloor\dfrac{n}{i^2}\rfloor\)

\(O(\sqrt{n})\)可以通过。

莫比乌斯反演

第一种形式:对于\(f*I=g\),有\(f=g*\mu\),反之亦然

证明:

两边都与\(\mu\)做卷积

\(f*I*\mu=g*\mu\)

结合律:

\(f*(I*\mu)=g*\mu\)

由\(\epsilon=1*\mu\)得:

\(f*\epsilon=g*\mu\)

由\(f*\epsilon=f\)得:

\(f=g*\mu\)

第二种形式:\(\sum\limits_{n|d}f(d)=g(n)\),则\(f(n)=\sum\limits_{n|d}g(d)\mu(\dfrac{d}{n})\)

证明:

\(\sum\limits_{n|d}g(d)\mu(\dfrac{d}{n})\)

将\(d\)除以\(n\):

\(=\sum\limits_{d=1}^{+\infty}g(dn)\mu(d)\)

把\(g\)写成\(f\)

\(=\sum\limits_{d=1}^{+\infty}\mu(d)\sum\limits_{nd|t}f(t)\)

枚举\(t\),\(d\)即是\(\dfrac{t}{n}\)的因数

\(=\sum\limits_{n|t}f(t)\sum\limits_{d|\frac{t}{n}}\mu(d)\)

后面的是\(\mu*I\)的形式,改写成\(\epsilon\)

\(=\sum\limits_{n|t}f(t)[\dfrac{t}{n}=1]=f(n)\)

来几道题:

三倍经验的题:
SP3871 GCDEX - GCD ExtremeUVA11426 拿行李(极限版) GCD - Extreme (II)UVA11424 GCD - Extreme (I)

我们换个思路,先写成这个形式:

\(\sum\limits_{i=1}^n\sum\limits_{j=1}^{i-1}\gcd(i,j)\)

根据\(P2303\)的结论\(\sum\limits_{i=1}^n\gcd(i,n)=\sum\limits_{d|n}d\varphi(\dfrac{n}{d})\)(上面有)把式子写成:

\(\sum\limits_{i=1}^n((\sum\limits_{j=1}^{i}\gcd(i,j))-\gcd(i,i))=\sum\limits_{i=1}^n(\sum\limits_{d|i}d\varphi(\dfrac{i}{d}))-i=(\sum\limits_{i=1}^n\sum\limits_{d|i}d\varphi(\dfrac{i}{d}))-\dfrac{n(n+1)}{2}\)

设\(g(i)=\sum\limits_{d|i}d\varphi(\dfrac{i}{d})\)

我们可以试着\(O(n)\)线性筛出\(g(i)\),同时算个前缀和就预处理完了。

下面写的\(i,prime_j\)为欧拉筛代码中的\(i,prime[j]\)

我们设\(i\)分解质因数后是\(\prod\limits_{j=1}^kp_j^{c_j}\),\(low_i=p_1^{c_1}\)。其中\(p_1\)是最小的质因子。

而且\(prime_j\)为\(i\times prime_j\)得最小质因子(欧拉筛的性质)

则当\(prime_j|i\)时,\(p_1=prime_j\)

可以得到,\(g(1)=1,g(p)=2p-1\),\(p\)是质数。

由于函数\(id(x)\)和\(\varphi(x)\)是积性函数

那么\(g(i)=id*\varphi\)也是积性函数

则当\(i\)与\(prime_j\)互质时,\(g(i\times prime_j)=g(i)\times g(prime_j)\),且\(low_{i\times prime_j}=prime_j\)

当\(i\)与\(prime_j\)不互质即\(prime_j|i\)时,有\(p_1=prime_j\)(上面讲了),根据\(low\)的定义得

\(\gcd(\dfrac{i}{low_i},p_1\times low_i)=1,low_{i\times prime_j}=low_i\times prime_j\)

这样就可以用积性函数的定义求\(g(i\times prime_j)\)

\(g(i\times prime_j)=g(\dfrac{i}{low_i})\times g(p_1\times low_i)\)

但是当\(i=low_i\)时,这个式子就会变成\(g(i\times prime_j)=g(1)\times g(i\times prime_j)=g(i\times prime_j)\),无法递推

特殊处理一下\(i=low_i\)的情况:

\(i=low_i\)即\(i=p_1^{c_1}\),则\(i\times prime_j=p_1^{c_1+1}\)

\(g(p_1^{c_1+1})=\sum\limits_{d|p_1^{c_1+1}}d\varphi(\dfrac{p_1^{c_1+1}}{d})\)

把\(d=1\)的一项单独提出:

\(g(p_1^{c_1+1})=\sum\limits_{d|p_1^{c_1+1}\&d>1}d\varphi(\dfrac{p_1^{c_1+1}}{d})+\varphi(p_1^{c_1+1})\)

\(\sum\)内的\(d\)此时只能是\(p_1,p_1^2,\cdots,p_1^{c_1+1}\),我们就可以把\(d\)除以\(p_1\)

\(g(p_1^{c_1+1})=\sum\limits_{d|p_1^{c_1}}dp_1\varphi(\dfrac{p_1^{c_1+1}}{dp_1})+\varphi(p_1^{c_1+1})\)

化简一下,并且提出\(p_1\)

\(g(p_1^{c_1+1})=p_1\sum\limits_{d|p_1^{c_1}}d\varphi(\dfrac{p_1^{c_1}}d)+\varphi(p_1^{c_1+1})\)

可以发现\(\sum\limits_{d|p_1^{c_1}}d\varphi(\dfrac{p_1^{c_1}}d)=g(p_1^{c_1})\),代入

\(g(p_1^{c_1+1})=p_1g(p_1^{c_1})+\varphi(p_1^{c_1+1})\)

也就是\(g(i\times prime_j)=prime_j\times g(i)+\varphi(i\times prime_j)\)

总结一下\(g(x)=\begin{cases}1\quad x=1\\2x-1\quad x\in prime\\g(i)\times g(prime_j)\quad x=i\times prime_j,\gcd(prime_j,i)=1\\prime_j\times g(i)+\varphi(i\times prime_j)\quad x=i\times prime_j,prime_j|i,low_i=i\\g(i\times prime_j)=g(\dfrac{i}{low_i})\times g(p_1\times low_i)\quad x=i\times prime_j,prime_j|i,low_i\not=i\end{cases}\)

\(low_x=\begin{cases}0\quad i=1\\x\quad x\in prime\\prime_j\quad x=i\times prime_j,\gcd(i,prime_j)=1\\low_i\times prime_j\quad x=i\times prime_j,prime_j|i\end{cases}\)

代码

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN=4e6+100;
int prime[MAXN],pcnt,phi[MAXN+1],n;
ll g[MAXN+1],low[MAXN+1],sum[MAXN+1];
bool isp[MAXN+1];
int main() {
	sum[1]=g[1]=phi[1]=1;
	for(int i=2; i<=MAXN; ++i) {//筛出phi,g,low
		if(!isp[i]) {
			prime[++pcnt]=i;
			phi[i]=i-1;
			low[i]=i;
			g[i]=(i<<1)-1;//素数
		}
		for(int j=1,x; j<=pcnt&&(x=i*prime[j])<=MAXN; ++j) {
			isp[x]=true;
			if(i%prime[j]) {
				g[x]=g[i]*g[prime[j]];
				phi[x]=phi[i]*phi[prime[j]];
				low[x]=prime[j];//三个都要算上
			} else {
				low[x]=low[i]*prime[j];
				phi[x]=phi[i]*prime[j];
				g[x]=(low[i]==i)?g[i]*prime[j]+phi[x]:g[i/low[i]]*g[low[i]*prime[j]];
			}//记得特判i=low i
		}
		sum[i]=sum[i-1]+g[i];//前缀和
	}
	while(~scanf("%d",&n)&&n) {
		printf("%lld\n",sum[n]-((n+1ll)*n>>1));//输出
	}
	return 0;
}

P2257 YY的GCD

这题是要求:\(\sum\limits_{p\in prime}\sum\limits_{i=1}^n\sum\limits_{j=1}^m[\gcd(i,j)=p]\)

把\(i,j\)除以\(p\)

\(\sum\limits_{p\in prime}\sum\limits_{i=1}^{\lfloor\frac{n}{p}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{p}\rfloor}[\gcd(i,j)=1]\)

可以对后面的部分进行莫比乌斯反演

设\(f(x)\)表示\(\gcd(i,j)\)为\(x\)的二元组的个数,\(g(x)\)表示\(\gcd(i,j)\)为\(x\)的倍数的二元组的个数,

把它俩用式子写就是:

\(f(x)=\sum\limits_{i=1}^{\lfloor\frac{n}{p}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{p}\rfloor}[\gcd(i,j)=x]\)

\(g(x)=\sum\limits_{i=1}^{\lfloor\frac{n}{p}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{p}\rfloor}[x|\gcd(i,j)]=\sum\limits_{i=1}^{\lfloor\frac{n}{p}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{p}\rfloor}\sum\limits_{x|d}[\gcd(i,j)=d]\)

后面的\(\sum\limits_{x|d}\)可以提前

\(g(x)=\sum\limits_{x|d}\sum\limits_{i=1}^{\lfloor\frac{n}{p}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{p}\rfloor}[\gcd(i,j)=d]=\sum\limits_{x|d}f(x)\)

根据莫比乌斯反演第二形式,可得:\(f(x)=\sum\limits_{x|d}g(d)\mu(\dfrac{d}{x})\)

\(f(1)=\sum\limits_{1|d}g(d)\mu(\dfrac{d}{1})=\sum\limits_{i=1}^{\lfloor\frac{n}{p}\rfloor}g(i)\mu(i)\)

再看\(g(x)=\sum\limits_{i=1}^{\lfloor\frac{n}{p}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{p}\rfloor}[x|\gcd(i,j)]\)这个式子,把\(i,j\)除以\(x\)可以得到

\(g(x)=\sum\limits_{i=1}^{\lfloor\frac{n}{px}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{px}\rfloor}[1|\gcd(i,j)]=\sum\limits_{i=1}^{\lfloor\frac{n}{px}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{px}\rfloor}1=\lfloor\frac{n}{px}\rfloor\lfloor\frac{m}{px}\rfloor\)

代回\(f(1)=\sum\limits_{i=1}^{\lfloor\frac{n}{p}\rfloor}g(i)\mu(i)=\sum\limits_{i=1}^{\lfloor\frac{n}{p}\rfloor}\lfloor\frac{n}{px}\rfloor\lfloor\frac{m}{px}\rfloor\mu(i)\)

提出\(\lfloor\frac{n}{px}\rfloor\lfloor\frac{m}{px}\rfloor\)

\(f(1)=\lfloor\frac{n}{px}\rfloor\lfloor\frac{m}{px}\rfloor\sum\limits_{i=1}^{\lfloor\frac{n}{p}\rfloor}\mu(i)\)

再代回原来的式子:

\(\sum\limits_{p\in prime}\lfloor\frac{n}{px}\rfloor\lfloor\frac{m}{px}\rfloor\sum\limits_{i=1}^{\lfloor\frac{n}{p}\rfloor}\mu(i)\)

先用\(T\)枚举\(px\),再枚举\(p|T\)

\(\sum\limits_{i=1}^T\lfloor\frac{n}{T}\rfloor\lfloor\frac{m}{T}\rfloor\sum\limits_{p|T\& p\in prime}\mu(i)\)

由于\(10000128\)以内的素数只有\(664583\)个,所以我们暴力计算后面的部分都不会超时。

前面的部分就可以用整除分块优化

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN=10000129;
int prime[MAXN>>3],pcnt;
bool vis[MAXN];
int mu[MAXN];
ll sum[MAXN],g[MAXN];
int read() {
    int x=0,c=getchar();
    bool f=1;
    while(c<=47||c>=58) {
        f=f&&(c^45);
        c=getchar();
    }
    while(c>=48&&c<=57) {
        x=(x<<3)+(x<<1)+(c&15);
        c=getchar();
    }
    return f?x:-x;
}
int main() {
    vis[0]=vis[1]=1;
    mu[1]=1;
    for(int i=2; i<MAXN; ++i) {//预处理mu
        if(!vis[i]) {
            prime[++pcnt]=i;
            mu[i]=-1;
        }
        for(int j=1; i*1ll*prime[j]<MAXN&&j<=pcnt; ++j) {
            vis[i*prime[j]]=1;
            if(i%prime[j]==0)break;
            else mu[i*prime[j]]=-mu[i];
        }
    }
    for(int i=1; i<=pcnt; ++i) {
        for(int j=1; j*1ll*prime[i]<MAXN; ++j) {
            g[j*prime[i]]+=mu[j];
        }
    }
    for(int i=1; i<MAXN; ++i) {//预处理后面部分的前缀和
        sum[i]=sum[i-1]+g[i];
    }
    int t=read(),n,m,l,r;
    ll ans;
    while(t--) {
        n=read(),m=read();
        if(n>m)swap(n,m);
        ans=0;
        for(l=1; l<=n; l=r+1) {//整除分块
            r=min(n/(n/l),m/(m/l));
            ans+=(n/l)*(sum[r]-sum[l-1]+0ll)*(m/l);
        }
        printf("%lld\n",ans);
    }
    return 0;
}

P1829 [国家集训队]Crash的数字表格 / JZPTAB

这题是要求:\(\sum\limits_{i=1}^n\sum\limits_{j=1}^m\operatorname{lcm}(i,j)\)

根据\(\operatorname{lcm}\)的性质:

\(\sum\limits_{i=1}^n\sum\limits_{j=1}^m\dfrac{ij}{\gcd(i,j)}\)

加入一个\(d=\gcd(i,j)\):

\(\sum\limits_{d=1}^n\sum\limits_{i=1}^n\sum\limits_{j=1}^m[\gcd(i,j)=d]\dfrac{ij}{d}\)

把\(i,j\)都除以\(d\):

\(\sum\limits_{d=1}^n\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{d}\rfloor}[\gcd(i,j)=1]ijd\)

提出\(d,i\),把\(\epsilon=I*\mu\)代入:

\(\sum\limits_{d=1}^nd\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}i\sum\limits_{j=1}^{\lfloor\frac{m}{d}\rfloor}j\sum\limits_{k|\gcd(i,j)}\mu(k)\)

\(k|\gcd(i,j)\)其实就是\(k|i\&k|j\)(手玩一下可以知道)

\(\sum\limits_{d=1}^nd\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}i\sum\limits_{j=1}^{\lfloor\frac{m}{d}\rfloor}j\sum\limits_{k|i\&k|j}\mu(k)\)

直接把\(k\)提前:

\(\sum\limits_{d=1}^nd\sum\limits_{k=1}^{\lfloor\frac{n}{d}\rfloor}\mu(k)\sum\limits_{k|i\&i<\lfloor\frac{n}{d}\rfloor}i\sum\limits_{k|j\&j<\lfloor\frac{m}{d}\rfloor}j\)

把\(i,j\)都除以\(k\)

\(\sum\limits_{d=1}^nd\sum\limits_{k=1}^{\lfloor\frac{n}{d}\rfloor}\mu(k)\sum\limits_{i=1}^{\lfloor\frac{n}{dk}\rfloor}ik\sum\limits_{j=1}^{\lfloor\frac{m}{dk}\rfloor}jk\)

把两个\(k\)都提出来,把\(d\)乘进去,再设\(S(x)=1+2+\cdots+x=\dfrac{x(x+1)}{2}\):

\(\sum\limits_{d=1}^n\sum\limits_{k=1}^{\lfloor\frac{n}{d}\rfloor}dk^2\mu(k)S(\lfloor\frac{n}{dk}\rfloor)S(\lfloor\frac{m}{dk}\rfloor)\)

先枚举\(T=dk\),再枚举\(k|T\)

\(\sum\limits_{T=1}^n\sum\limits_{k|T}Tk\mu(k)S(\lfloor\frac{n}{T}\rfloor)S(\lfloor\frac{m}{T}\rfloor)\)

再提出\(TS(\lfloor\frac{n}{T}\rfloor)S(\lfloor\frac{m}{T}\rfloor)\)

\(\sum\limits_{T=1}^nTS(\lfloor\frac{n}{T}\rfloor)S(\lfloor\frac{m}{T}\rfloor)\sum\limits_{k|T}k\mu(k)\)

我们设\(f(T)=\sum\limits_{k|T}k\mu(k)\),看一下它的性质:

首先\(id,\mu\)都是积性函数,那么\(id\times\mu\)也是积性函数,而且\(I\)也是积性函数,那么\(f=(id\times\mu)*I\)也是积性函数。

那么欧拉筛这个\(f\)还需要看\(prime[j]|T\)的情况:

我们设\(T\)分解质因数后为\(p_1^{k_1}p_2^{k_2}\cdots p_c^{k_c}\)(不妨设其中\(p_1=prime[j]\))则

\(f(Tp_1)-f(T)=\sum\limits_{k|Tp_1}k\mu(k)-\sum\limits_{k|T}k\mu(k)=\sum\limits_{k|Tp_1\&k\not|T}k\mu(k)\)

\(Tp_1=p_1^{k_1+1}p_2^{k_2}\cdots p_c^{k_c}\),那么满足\(k|Tp_1\&k\not|T\)的数必须是\(p_1^{k_1+1}\)的倍数,而且\(k_1\ge1\)即\(k_1+1\ge2\)那么\(\mu(k)=0\),那么后面那部分就是\(0\),我们就得到:

\(f(Tp_1)=f(T)\)

最后总结一下:\(f(Tp_1)=\begin{cases}f(T)\qquad p_1|T\\f(T)f(p_1)\quad p_1\not|T\end{cases}\)

最后要求\(\sum\limits_{T=1}^nTS(\lfloor\frac{n}{T}\rfloor)S(\lfloor\frac{m}{T}\rfloor)f(T)\),一边线性筛一边求就行了:

#include<bits/stdc++.h>
using namespace std;
const int mod=20101009;
const int MAXN=1e7+128;
typedef long long ll;
int n,m,ans;
bool isp[MAXN+1];
int prime[MAXN+1],f[MAXN+1],pcnt;
inline int S(int x) {
	return (x*1ll*(x+1)/2)%mod;
}
int main() {
	cin>>n>>m;
	if(n>m)swap(n,m);
	f[1]=1,isp[1]=true;
	ans=(ans+S(n)*1ll*S(m))%mod;//先加上T=1时的值
	for(int i=2; i<=n; ++i) {//线性筛f(T)
		if(!isp[i]) {
			prime[++pcnt]=i;
			f[i]=mod-i+1;
		}
		for(int j=1,x; j<=pcnt&&(x=i*prime[j])<=MAXN; ++j) {
			isp[x]=true;
			if(i%prime[j]) {
				f[x]=f[i]*1ll*f[prime[j]]%mod;
			} else {
				f[x]=f[i];
				break;
			}
		}
		ans=(ans+S(n/i)*1ll*S(m/i)%mod*i%mod*f[i])%mod;//同时计算答案
	}
	printf("%d\n",ans);
	return 0;
}

标签:lfloor,frac,limits,乌斯,sum,rfloor,mu,反演,莫比
来源: https://www.cnblogs.com/mod998244353/p/14517215.html