Mobius inversion
Speaking of which, I was interested in Mobius at the time of the New Year's Eve in 2017...
But now it hurts to see the Mobius inversion QAQ
Mobius inversion , also known as tungsten filament reproduction , is actually two formulas:
Let the functions $f(n)$, $g(n)$ be number-theoretic functions and satisfy:
$f(n)=\sum\limits_{d|n}g(d)$
Möbius inversion
$g(n)=\sum\limits_{d|n}μ(d)f(\frac{n}{d})$
or if:
$f(n)=\sum\limits_{n|d}g(d)$
Then there are:
$g(n)=\sum\limits_{n|d}μ(\frac{d}{n})f(d)(d\le maxd)$
where $μ$ is the Möbius function, set $n=p_{1}^{k_{1}}\cdot p_{2}^{k_{2}}\cdot\cdots\cdot p_{m}^ {k_{m}}$ ($p$ is a prime number)
Then $μ(n)$ is defined as follows:
$μ(n)=
\begin{cases}
1& n=1 \\
(-1)^{m}& \prod\limits_{i=1}^{m}k_{i}=1 \\
0& otherwise(k_{i}>1)
\end{cases}$
$μ(n)$ has some very wonderful properties, the following examples illustrate:
Property 1: $\mu(n)$ is an integral function, that is, for positive integers $n$, $m$, when $n$ and $m$ are relatively prime, both $f(nm)=f(n )f(m)$;
Proof: omitted, you can consider the proof from the factorization direction
Property two:
$\sum\limits_{d|n}\mu(d)=
\begin{cases}
1& n=1 \\
0& n\not=1
\end{cases}$
Proof: The conclusion is obvious when $n=1$
When $n\not=1$, decompose $n$ into $p_{1}^{k_{1}}\cdot p_{2}^{k_{2}}\cdot\cdots\cdot p_{m }^{k_{m}}$
At this time, only the prime factors with degree 1 need to be considered, and it is easy to obtain that there are $C_{m}^{r}$ factors of which the number of prime factors is $r$, then the original formula can be transformed into:
$\sum\limits_{d|n}\mu(d)=C_{m}^{0}-C_{m}^{1}+C_{m}^{2}-\cdots+(-1)^{m}C_{m}^{m}=\sum\limits_{i=0}^{m}(-1)^{i}C_{m}^{i}$
According to the binomial theorem, we get
$(x+y)^{m}=\sum\limits_{i=0}^{m}C_{m}^{i}x^{i}y^{n-i}$
Substitute $x=-1, y=1$ into $\sum\limits_{i=0}^{m}(-1)^{i}C_{m}^{i}=0$
Certificate completed.
Now let's prove the Mobius inversion. Many of the proofs on the Internet are very simple, and I am confused. Here I will write a little more in detail.
Verification: When $f(n)=\sum\limits_{d|n}g(d)$, $g(n)=\sum\limits_{d|n}\mu(d)f(\frac{ n}{d})$
Proof: Defined by $f$,$g$, $\sum\limits_{d|n}\mu(d)f(\frac{n}{d})=\sum\limits_{d|n}\ mu(d)\sum\limits_{k|\frac{n}{d}}g(k)$
变式得$\sum\limits_{d|n}\mu(d)\sum\limits_{k|\frac{n}{d}}g(k)=\sum\limits_{k|n}g(k)\sum\limits_{d|\frac{n}{k}}\mu(d)$
And because when $n>1$ $\sum\limits_{d|n}\mu(d)=0$
So if and only if $k=n$, the value of the sum formula is $g(n)$, otherwise it is $0$, and the final result is also $g(n)$
Certificate completed.
Another type of proof is similar, only writing the conclusion:
当$f(n)=\sum\limits_{n|d}g(d)$时$g(n)=\sum\limits_{n|d}μ(\frac{d}{n})f(d)$
所以，没有了？
下面我们来看实现和应用。。。这东西看起来就俩式子，但是非常非常有用！
莫比乌斯反演实现与应用
1.线性筛求莫比乌斯函数
我们莫比乌斯函数是积性函数，那么就可以用线性筛来求。
具体实现就是在素数线性筛上加了几句，首先，素数的莫比乌斯函数是$-1$，当一个数的某个质因子指数大于一时就筛成0。注意实现时如果一个数的最小质因子指数大于$1$才会直接被筛为0，否则会用$\mu(x)=-\mu(i)$来完成。每个数被它的最小质因子筛去，所以是线性的。
代码：
1 int miu[100001],pri[100001],tot=0; 2 bool ntp[100001]; 3 void getmiu(){ 4 memset(ntp,0,sizeof(ntp)); 5 memset(miu,0,sizeof(miu)); 6 miu[1]=1; 7 for(int i=2;i<=100000;i++){ 8 if(!ntp[i]){ 9 pri[++tot]=i; 10 miu[i]=-1; 11 } 12 for(int j=1;j<=tot&&pri[j]*i<=100000;j++){ 13 ntp[pri[j]*i]=true; 14 if(i%pri[j]==0){ 15 miu[i*pri[j]]=0; 16 break; 17 } 18 miu[i*pri[j]]=-miu[i]; 19 } 20 } 21 }
2.求解一系列有关gcd（是最小公因数不是……）的问题
例题一：BZOJ2301&&HDU1695(这个要求x,y和y,x不重复的)
题意：$n$次询问，每次求有多少对$(x,y)$满足$a\le x\le b$且$c\le y\le d$且$gcd(x,y)=k$。
$n,a,b,c,d$都是$50000$级别。
容斥一下（不会怎么容斥的请Alt+F4百度一下）每次拆成四个询问，都是形如有多少对$(x,y)$满足$1\le x\le n$且$1\le y\le m$且$gcd(x,y)=k$（默认$n\le m$否则交换）。
这样的询问等价于询问有多少对$(x,y)$满足$1\le x\le \lfloor\frac{n}{k}\rfloor$且$1\le y\le \lfloor\frac{m}{k}\rfloor$且$x,y$互质
直接做？4*50000^3=???
考虑莫比乌斯反演，令$g(i)$等于满足$1\le x\le n$且$1\le y\le m$且$gcd(x,y)=i$的数对$(x,y)$个数，$f(i)$等于满足$1\le x\le n$且$1\le y\le m$且$i|gcd(x,y)$的数对$(x,y)$个数。
易得$f(i)=\lfloor\frac{n}{i}\rfloor\lfloor\frac{m}{i}\rfloor$
反演一下得$g(i)=\sum\limits_{i|d}\mu(\frac{d}{i})f(d)=\sum\limits_{i|d}\mu(\frac{d}{i})\lfloor\frac{n}{d}\rfloor\lfloor\frac{m}{d}\rfloor$
枚举每个$k$的倍数，就可以$O(n)$处理每个询问了。
然而50000^2=???
继续考虑优化
观察式子$\lfloor\frac{n}{d}\rfloor$，可（da）以（dan）证（cai）明（ce）发现它最多只有$O(\sqrt{n})$种取值。
具体证明如下：
考虑$1\le d\le \sqrt{n}$的情况，此时$\lfloor\frac{n}{d}\rfloor$最多只有$\sqrt{n}$种取值；
剩余当$\sqrt{n}\le d\le n$时，有$\lfloor\frac{n}{d}\rfloor<\sqrt{n}$，因此取值也不超过$\sqrt{n}$种；
综上，$\lfloor\frac{n}{d}\rfloor$最多只有$O(\sqrt{n})$种取值。
即$\lfloor\frac{n}{d}\rfloor$取值连续的段最多只有$\sqrt{n}+\sqrt{m}$段。
所以就可以$O(\sqrt{n}+\sqrt{m})$时间内枚举，然后就……做完了？
具体实现就是分块维护一个$\mu$的前缀和。
并且非常好写！
完整代码：
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<cmath> 5 using namespace std; 6 typedef long long ll; 7 int t,a,b,c,d,k,pre[100001],miu[100001],pri[100001],tot=0; 8 bool ntp[100001]; 9 void getmiu(){ 10 memset(ntp,0,sizeof(ntp)); 11 memset(miu,0,sizeof(miu)); 12 miu[1]=1; 13 for(int i=2;i<=100000;i++){ 14 if(!ntp[i]){ 15 pri[++tot]=i; 16 miu[i]=-1; 17 } 18 for(int j=1;j<=tot&&pri[j]*i<=100000;j++){ 19 ntp[pri[j]*i]=true; 20 if(i%pri[j]==0){ 21 miu[i*pri[j]]=0; 22 break; 23 } 24 miu[i*pri[j]]=-miu[i]; 25 } 26 } 27 } 28 ll work(int n,int m){ 29 ll ans=0; 30 int last=0; 31 if(n>m)swap(n,m); 32 for(int i=1;i<=n;i=last+1){ 33 last=min(n/(n/i),m/(m/i)); 34 ans+=(ll)(pre[last]-pre[i-1])*(n/i)*(m/i); 35 } 36 return ans; 37 } 38 int main(){ 39 getmiu(); 40 pre[0]=0; 41 for(int i=1;i<=100000;i++)pre[i]=pre[i-1]+miu[i]; 42 scanf("%d",&t); 43 for(int i=1;i<=t;i++){ 44 scanf("%d%d%d%d%d",&a,&b,&c,&d,&k); 45 a--,c--; 46 if(!k)printf("0\n"); 47 else printf("%lld\n",work(b/k,d/k)-work(a/k,d/k)-work(b/k,c/k)+work(a/k,c/k)); 48 } 49 return 0; 50 }
例题二：BZOJ2820
题目大意：求有多少对$(x,y)$满足$1\le x\le n$且$1\le y\le m$且$gcd(x,y)=$质数
定义$f(i)$，$g(i)$同上，再枚举质数，有
$ans=\sum\limits_{p}^{n}(\sum\limits_{d=1}^{\frac{n}{p}}\mu(d)\lfloor\frac{n}{pd}\rfloor\lfloor\frac{m}{pd}\rfloor)$
直接做？复杂度大概是$O(\frac{n\sqrt{n}}{logn})$，不是很行
变一下式，设$t=pd$，有
$ans=\sum\limits_{t=1}^{n}\lfloor\frac{n}{t}\rfloor\lfloor\frac{m}{t}\rfloor(\sum\limits_{p|t}\mu(\frac{t}{p}))$
预处理一下$\sum\limits_{p|t}\mu(\frac{t}{p})$，剩下的分块前缀和就和之前一模一样了。
如何预处理？
考虑线性筛，$f[i*prime[j]]$
$prime[j]|i$时显然$f[i]=\mu(i)$
否则考虑$\mu(i*prime[j]/p_{1})$，当$prime[j]=p_{1}$时为$\mu(i)$，否则所有和为$-f[i]$，因此总和即为$\mu(i)-f[i]$。
代码：
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<cmath> 5 using namespace std; 6 typedef long long ll; 7 int t,a,b,pre[10000001],miu[10000001],pri[10000001],f[10000001],tot=0; 8 bool ntp[10000001]; 9 void getmiu(){ 10 memset(ntp,0,sizeof(ntp)); 11 memset(miu,0,sizeof(miu)); 12 miu[1]=1; 13 for(int i=2;i<=10000000;i++){ 14 if(!ntp[i]){ 15 pri[++tot]=i; 16 miu[i]=-1; 17 f[i]=1; 18 } 19 for(int j=1;j<=tot&&pri[j]*i<=10000000;j++){ 20 ntp[pri[j]*i]=true; 21 if(i%pri[j]==0){ 22 miu[i*pri[j]]=0; 23 f[i*pri[j]]=miu[i]; 24 break; 25 } 26 miu[i*pri[j]]=-miu[i]; 27 f[i*pri[j]]=miu[i]-f[i]; 28 } 29 } 30 } 31 ll work(int n,int m){ 32 ll ans=0; 33 int last=0; 34 if(n>m)swap(n,m); 35 for(int i=1;i<=n;i=last+1){ 36 last=min(n/(n/i),m/(m/i)); 37 ans+=(ll)(pre[last]-pre[i-1])*(n/i)*(m/i); 38 } 39 return ans; 40 } 41 int main(){ 42 getmiu(); 43 pre[0]=0; 44 for(int i=1;i<=10000000;i++)pre[i]=pre[i-1]+f[i]; 45 scanf("%d",&t); 46 for(int i=1;i<=t;i++){ 47 scanf("%d%d",&a,&b); 48 printf("%lld\n",work(a,b)); 49 } 50 return 0; 51 }
例题三：GDOI2018Day2T1谈笑风生
题目过于暴力不予显示（顾名思义）
题目大意：n个点m条边的无向图，边权由点权计算，其中可以花费p能量把每条边权减少p，求最少要花费多少能量才能使1～n的最短路小于等于t；
其中边权计算式为$w=\sum\limits_{i=1}^{num[u]}\sum\limits_{j=1}^{num[v]}(i+j)[gcd(i,j)==1]$
待填坑……
3.还有吗？那我就不会了
总结
不存在的
就是背好两个式子吧……
同时推荐一些很好的课件（同时也是我学习的）：Orz PoPoQQQ、Sengxian's Blog