莫比乌斯反演做题小结
Mophues
求有多少数对\((a,b)\)满足\(1 \le a \le n\)，\(1 \le b \le m\)，且\(\gcd(a,b)\)的质因子个数\(\le p\)。T组询问
\(1\le n,m\le 5 \times 10^5\),\(1 \le T \le 5000\)
\[\sum_{t=1}^{p}\sum_{a=1}^n\sum_{b=1}^{m}[\gcd(a,b)=t]
\]
\[\sum_{t=1}^{p}\sum_{a=1}^{n/t}\sum_{b=1}^{m/t}[\gcd(a,b)=1]
\]
\[\sum_{t=1}^{p}\sum_{a=1}^{n/t}\sum_{b=1}^{m/t}\sum_{d|\gcd(a,b)}\mu(d)
\]
\[\sum_{d=1}^n\mu(d)\sum_{t=1}^{p}\sum^{n/dt}_{a=1}\sum^{m/dt}_{b=1}1
\]
\[\sum_{d=1}^n\mu(d)\sum_{t=1}^{p}{\left\lfloor\frac{n}{dt}\right\rfloor}{\left\lfloor\frac{m}{dt}\right\rfloor}
\]
设\(D=dt\)
\[\sum_{d=1}^n\mu(d)\sum_{t=1}^{p}{\left\lfloor\frac{n}{D}\right\rfloor}{\left\lfloor\frac{m}{D}\right\rfloor}
\]
\[\sum_{D=1}^n{\left\lfloor\frac{n}{D}\right\rfloor}{\left\lfloor\frac{m}{D}\right\rfloor}\sum_{d|D}mu(d)
\]
因为质因子个数有限制，所以设\(f[i][j]\)表示质因子数目小于等于\(j\)时的前\(i\)项之和
数论分块即可
时间复杂度$$O(T\sqrt{n}+nlogn)$$
DZY loves Math
定义\(f(x)\)表示\(x\)的质因子的最大次数，求\(\sum_{i=1}^n\sum_{j=1}^mf(\gcd(i,j))\)的值，多组询问
\(T \le 10^4\) ，\(1\le n,m \le 10^7\)
\[\sum_{i=1}^n\sum_{j=1}^mf(\gcd(i,j))
\]
\[\sum_{d=1}^{n}\sum_{i=1}^n\sum_{j=1}^m[\gcd(i,j)=d]f(d)
\]
\[\sum_{d=1}^{n}\sum_{i=1}^{n/d}\sum_{j=1}^{m/d}f(d)[\gcd(i,j)=1]
\]
\[\sum_{d=1}^{n}\sum_{i=1}^{n/d}\sum_{j=1}^{m/d}f(d)\sum_{k|\gcd(i,j)}\mu(k)
\]
\[\sum_{k=1}^{n}\mu(k)\sum_{d=1}^{n}\sum_{i=1}^{n/dk}\sum_{j=1}^{m/dk}f(d)
\]
\[\sum_{k=1}^{n}\mu(k)\sum_{d=1}^{n}f(d)\sum_{i=1}^{n/dk}\sum_{j=1}^{m/dk}1
\]
\[\sum_{k=1}^{n}\mu(k)\sum_{d=1}^{n}f(d){\left\lfloor\frac{n}{dk}\right\rfloor}{\left\lfloor\frac{m}{dk}\right\rfloor}
\]
设\(T=dk\)
\[\sum_{k=1}^{n}\mu(k)\sum_{d=1}^{n}f(d){\left\lfloor\frac{n}{T}\right\rfloor}{\left\lfloor\frac{m}{T}\right\rfloor}
\]
\[\sum_{T=1}^{n}{\left\lfloor\frac{n}{T}\right\rfloor}{\left\lfloor\frac{m}{T}\right\rfloor}\sum_{k|T}\mu(k)f(T/k)
\]
经过分析，可以知道\(F(T)=\sum_{k|T}f(T/k)\mu(k)\)当且仅当\(F((p_1p_2...p_k)^t)=(-1)^{k+1}\)，其他情况都是0
对于 \(\sum_{k|T}^{n}\mu(k)f(T/k)\)，可以线性筛处理，前面部分数论分块计算即可
时间复杂度\(O(T \times \sqrt n)\)
JZPTAB
求\(\sum_{i=1}^{n}\sum_{j=1}^{m}lcm(i,j)\)答案对\(10^8+9\)取模，多组询问
\(1\le n,m\le 10^7\) \(1 \le T \le 5 \times 10^4\)
\[\sum_{i=1}^{n}\sum_{j=1}^mlcm(i,j)
\]
\[\sum_{i=1}^{n}\sum_{j=1}^m\frac{ij}{\gcd(i,j)}
\]
\[\sum_{d=1}^{n}\sum_{i=1}^{n}\sum_{j=1}^m\frac{ij}{d}[\gcd(i,j)=d]
\]
\[\sum_{d=1}^{n}\sum_{i=1}^{n/d}\sum_{j=1}^{m/d}{dij}[\gcd(i,j)=1]
\]
\[\sum_{d=1}^{n}\sum_{i=1}^{n/d}\sum_{j=1}^{m/d}{dij}\sum_{k|\gcd(i,j)}\mu(k)
\]
\[\sum_{d=1}^{n}{d}\sum_{k=1}^{n}\mu(k) \times k^2\sum_{i=1}^{n/dk}{i}\sum_{j=1}^{m/dk}{j}
\]
记\(f(x)=\sum_{i=1}^{x}i\)
\[\sum_{d=1}^{n}{d}\sum_{k=1}^{n}\mu(k) \times k^2 \times f({\left\lfloor\frac{n}{dk}\right\rfloor}) \times f({\left\lfloor\frac{m}{dk}\right\rfloor})
\]
设\(T=dk\)
\[\sum_{d=1}^{n}{d}\sum_{k=1}^{n}\mu(k) \times k^2 \times f({\left\lfloor\frac{n}{T}\right\rfloor}) \times f({\left\lfloor\frac{m}{T}\right\rfloor})
\]
\[\sum_{T=1}^{n}f({\left\lfloor\frac{n}{T}\right\rfloor}) \times f({\left\lfloor\frac{m}{T}\right\rfloor})\sum_{d|T}\mu(d) \times d \times T
\]
观察得到\(f(i)\)可以直接计算，\(g(T)=(\sum_{d|T}\mu(d) \times d \times T)\)是积性函数，可以线性筛预处理
现在，原式变成了
\[\sum_{T=1}^{n}f({\left\lfloor\frac{n}{T}\right\rfloor}) \times f({\left\lfloor\frac{m}{T}\right\rfloor}) \times g(T)
\]
显然可以数论分块，可以做到\(O(T\sqrt n+n)\)
数表
设\(F(i,j)\)表示能同时整除\(i\)和\(j\)的所有自然数之和
求\(\sum_{i=1}^{n_k}\sum_{j=1}^{m_k} F(i,j)[F(i,j)\le a_k]\) 对\(2^{31}\)取模的结果
\(1 \le Q \le 2 \times 10^4\) ，\(1 \le n,m \le 10^5\)，\(1 \le a_i \le 10^9\)
先看看\(F(i,j)\)函数的本质，容易发现\(F(i,j)\)是\(\gcd(i,j)\)的约数和
那么\(F(i,j)=\sigma (\gcd(i,j))\)
假装没有\(a\)的限制
\[\sum_{i=1}^{n}\sum_{j=1}^{m} \sigma(\gcd(i,j))
\]
\[\sum_{d=1}^n\sum_{i=1}^{n}\sum_{j=1}^{m} \sigma(d)[\gcd(i,j)=d]
\]
\[\sum_{d=1}^n\sum_{i=1}^{n/d}\sum_{j=1}^{m/d} \sigma(d)[\gcd(i,j)=1]
\]
\[\sum_{d=1}^n\sigma(d)\sum_{i=1}^{n/d}\sum_{j=1}^{m/d} \sum_{k|\gcd(i,j)}\mu(k)
\]
\[\sum_{d=1}^n\sigma(d)\sum_{k=1}^n\mu(k)\sum_{i=1}^{n/dk}\sum_{j=1}^{m/dk} 1
\]
\[\sum_{d=1}^n\sigma(d)\sum_{k=1}^n\mu(k){\left\lfloor\frac{n}{dk}\right\rfloor}{\left\lfloor\frac{m}{dk}\right\rfloor}
\]
设\(T=dk\)
\[\sum_{d=1}^n\sigma(d)\sum_{k=1}^n\mu(k){\left\lfloor\frac{n}{T}\right\rfloor}{\left\lfloor\frac{m}{T}\right\rfloor}
\]
\[\sum_{T=1}^{n}{\left\lfloor\frac{n}{T}\right\rfloor}{\left\lfloor\frac{m}{T}\right\rfloor}\sum_{d|T}\sigma(d)\mu(T/d)
\]
线性筛预处理\(\sum_{d|T}\sigma(d)\mu(T/d)\)，对于每个询问可以\(O(\sqrt n+\sqrt m)\)计算
考虑\(a_i\)的贡献，设\(g(T)=\sum_{d|T}\sigma(d)\mu(T/d)\) ，可以发现当\(\sigma(d) \le a_i\)时才会对\(g(T)\)产生贡献
离线，把\(a\)从小到大排序，枚举询问的时候，动态修改\(g(T)\)的值，总共需要更改的个数是调和级数
时间复杂度\(O(q\sqrt n \ logn+n\ log^2n)\)
简单的数学题
求 \((\sum_{i=1}^{n}\sum_{j=1}^{n}ij\gcd(i,j))\ mod\ p\)
\(1\le n\le 10^{10}\),\(5 \times 10^8 \le p \le 1.1 \times 10^9\)且\(p\)是质数
\[\sum_{i=1}^{n}\sum_{j=1}^{n}ij\gcd(i,j)
\]
\[\sum_{i=1}^{n}\sum_{j=1}^{n}ij\sum_{k|i,k|j}\varphi(k)
\]
\[\sum_{k=1}^{n}\varphi(k)\sum_{k|i}\sum_{k|j}ij
\]
\[\sum_{k=1}^n \varphi(k) \times k^2 \times (\sum^{n/k}_{i=1}i)^2
\]
设\({\left\lfloor\dfrac{n}{k}\right\rfloor}=D\)
\[\sum_{k=1}^n \varphi(k) \times k^2 \times (\frac{D(D+1)}{2})^2
\]
\[\sum_{k=1}^n \varphi(k) \times k^2 \times \frac{D^2(D+1)^2}{4}
\]
\[\frac{1}{4}\sum_{k=1}^n \varphi(k) \times k^2 \times {D^2(D+1)^2}
\]
对\({\left\lfloor\dfrac{n}{k}\right\rfloor}\)数论分块
对于\(\varphi(k) \times k^2\)，可以杜教筛求前缀和。
Surprise me!
给出一棵树和一个\(1...n\)的排列，求 \(\frac {1}{n(n-1)}\sum_{i=1}^{n}\sum_{j=1}^{n}\varphi(a_ia_j) \times dist(i,j)\) 的值
\(1 \le n \le 2 \times 10^5\)
\[\frac {1}{n(n-1)}\sum_{i=1}^{n}\sum_{j=1}^{n}\varphi(a_ia_j) \times dist(i,j)
\]
因为\(ai\)是\(1...n\) 的排列，所以设\(p_{a_i}=i\)
\[\frac {1}{n(n-1)}\sum_{i=1}^{n}\sum_{j=1}^{n}\varphi(ij) \times dist(p_i,p_j)
\]
\[\frac {1}{n(n-1)}\sum_{i=1}^{n}\sum_{j=1}^{n}\frac{\varphi(i)\varphi(j)\gcd(i,j)}{\varphi(\gcd(i,j))} \times dist(p_i,p_j)
\]
\[\frac {1}{n(n-1)}\sum_{d=1}^{n}\sum_{i=1}^{n}\sum_{j=1}^{n}[\gcd(i,j)=d]\frac{\varphi(i)\varphi(j)d}{\varphi(d)} \times dist(p_i,p_j)
\]
\[\frac {1}{n(n-1)}\sum_{d=1}^{n}\sum_{i=1}^{n/d}\sum_{j=1}^{n/d}[\gcd(i,j)=1]\frac{\varphi(id)\varphi(jd)d}{\varphi(d)} \times dist(p_{id},p_{jd})
\]
\[\frac {1}{n(n-1)}\sum_{d=1}^{n} \frac{d}{\varphi(d)} \sum_{i=1}^{n/d}\sum_{j=1}^{n/d}\sum_{k|\gcd(i,j)}\mu(k)\varphi(id)\varphi(jd)dist(p_{id},p_{jd})
\]
\[\frac {1}{n(n-1)}\sum_{d=1}^{n} \frac{d}{\varphi(d)} \sum_{k=1}^{n/d}\mu(k)\sum_{i=1}^{n/dk}\sum_{j=1}^{n/dk}\varphi(idk)\varphi(jdk)dist(p_{idk},p_{jdk})
\]
设\(T=dk\)
\[\frac {1}{n(n-1)}\sum_{d=1}^{n} \frac{d}{\varphi(d)} \sum_{k=1}^{n/d}\mu(k)\sum_{i=1}^{n/dk}\sum_{j=1}^{n/dk}\varphi(Tk)\varphi(Tk)dist(p_{iT},p_{jT})
\]
\[\frac {1}{n(n-1)}\sum_{T=1}^{n}\sum_{d|T} \frac{d \times \mu(T/d)}{\varphi(d)} \sum_{i=1}^{n/T}\sum_{j=1}^{n/T}\varphi(Tk)\varphi(Tk)dist(p_{iT},p_{jT})
\]
线性筛出$\sum_{d|T} \frac{d \times \mu(T/d)}{\varphi(d)} $
现在部分就是给定一棵树，求出\(\sum_{i=1}^{n}\sum_{j=1}^{n}b_ib_jdist(i,j)\)
考虑建虚树，总点数是调和级数，\(O(nlogn)\)级别的
那么树形dp即可
时间复杂度$ O(nlog^2n) $
