Loading docs/math/mobius.md +33 −33 Original line number Diff line number Diff line Loading @@ -6,7 +6,7 @@ * * * ## 数论分块与整除相关 ## 数论分块与整除相 先补一下数学小 trick Loading Loading @@ -34,7 +34,7 @@ $$ ### 引理 2 $$ \forall n,d\in\mathbb{N},\left|\left\{\left\lfloor\frac{n}{d}\right\rfloor\right\}\right|\leq\left\lfloor2\sqrt{n}\right\rfloor \forall n \in N, \left|\left\{ \lfloor \frac{n}{d} \rfloor \mid d \in N \right\}\right| \leq \lfloor 2\sqrt{n} \rfloor $$ $|V|$ 表示集合 $V$ 的元素个数 Loading Loading @@ -372,12 +372,12 @@ int main() { } ``` ### [「SPOJ 5971」LCMSUM](https://www.luogu.org/problemnew/show/SP5971) ### [「SPOJ 5971」LCMSUM](https://www.spoj.com/problems/LCMSUM/) 求值（多组数据） $$ \sum_{i=1}^n \text{lcm}(i,n)\qquad (1\leqslant T\leqslant 3\times 10^5,1\leqslant n\leqslant 10^6) \sum_{i=1}^n \text{lcm}(i,n)\quad \text{s.t.}\ 1\leqslant T\leqslant 3\times 10^5,1\leqslant n\leqslant 10^6 $$ 易得原式即 Loading @@ -389,7 +389,7 @@ $$ 根据 $\gcd(a,n)=1$ 时一定有 $\gcd(n-a,n)=1$ ，可将原式化为 $$ \frac{1}{2}\cdot(\sum_{i=1}^{n-1}\frac{i\cdot n}{\gcd(i,n)}+\sum_{i=n-1}^{1}\frac{i\cdot n}{\gcd(i,n)})+n \frac{1}{2}\cdot \left(\sum_{i=1}^{n-1}\frac{i\cdot n}{\gcd(i,n)}+\sum_{i=n-1}^{1}\frac{i\cdot n}{\gcd(i,n)}\right)+n $$ 上述式子中括号内的两个 $\sum$ 对应的项相等，故又可以化为 Loading Loading @@ -683,7 +683,7 @@ signed main(){ 求 $$ \sum_{i=1}^n\sum_{j=1}^ni\cdot j\cdot gcd(i,j)\bmod p\\ \sum_{i=1}^n\sum_{j=1}^ni\cdot j\cdot \gcd(i,j)\bmod p\\ n\leq10^{10},5\times10^8\leq p\leq1.1\times10^9,\text{p 是质数} $$ Loading @@ -692,22 +692,23 @@ $$ 我们利用 $\varphi\ast1=ID$ 反演 $$ \begin{split} &\sum_{i=1}^n\sum_{j=1}^ni\cdot j \begin{eqnarray} && \sum_{i=1}^n\sum_{j=1}^ni\cdot j\cdot \gcd(i,j)\\ &=&\sum_{i=1}^n\sum_{j=1}^ni\cdot j \sum_{d|i,d|j}\varphi(d)\\ &\sum_{d=1}^n\sum_{i=1}^n &=&\sum_{d=1}^n\sum_{i=1}^n \sum_{j=1}^n[d|i,d|j]\cdot i\cdot j \cdot\varphi(d)\\ &\sum_{d=1}^n &=&\sum_{d=1}^n \sum_{i=1}^{\left\lfloor\frac{n}{d}\right\rfloor} \sum_{j=1}^{\left\lfloor\frac{n}{d}\right\rfloor} d^2\cdot i\cdot j\cdot\varphi(d)\\ &\sum_{d=1}^nd^2\cdot\varphi(d) &=&\sum_{d=1}^nd^2\cdot\varphi(d) \sum_{i=1}^{\left\lfloor\frac{n}{d}\right\rfloor}i \sum_{j=1}^{\left\lfloor\frac{n}{d}\right\rfloor}j\\ &\sum_{d=1}^nF^2\left(\left\lfloor\frac{n}{d}\right\rfloor\right)\cdot d^2\varphi(d) &=&\sum_{d=1}^nF^2\left(\left\lfloor\frac{n}{d}\right\rfloor\right)\cdot d^2\varphi(d) \left(F(n)=\frac{1}{2}n\left(n+1\right)\right)\\ \end{split} \end{eqnarray} $$ 对 $\sum_{d=1}^nF\left(\left\lfloor\frac{n}{d}\right\rfloor\right)^2$ 做数论分块，$d^2\varphi(d)$ 的前缀和用杜教筛处理： Loading Loading @@ -798,41 +799,40 @@ $$ 我们证明一下 $$ \begin{split} &g(n)=\sum_{i=1}^n\mu(i)t(i)f\left(\left\lfloor\frac{n}{i}\right\rfloor\right)\\ =&\sum_{i=1}^n\mu(i)t(i) \begin{eqnarray} &&g(n)=\sum_{i=1}^n\mu(i)t(i)f\left(\left\lfloor\frac{n}{i}\right\rfloor\right)\\ &=&\sum_{i=1}^n\mu(i)t(i) \sum_{j=1}^{\left\lfloor\frac{n}{i}\right\rfloor}t(j) g\left(\left\lfloor\frac{\left\lfloor\frac{n}{i}\right\rfloor}{j}\right\rfloor\right)\\ =&\sum_{i=1}^n\mu(i)t(i) &=&\sum_{i=1}^n\mu(i)t(i) \sum_{j=1}^{\left\lfloor\frac{n}{i}\right\rfloor}t(j) g\left(\left\lfloor\frac{n}{ij}\right\rfloor\right)\\ =&\sum_{T=1}^n &=&\sum_{T=1}^n \sum_{i=1}^n\mu(i)t(i) \sum_{j=1}^{\left\lfloor\frac{n}{i}\right\rfloor}[ij=T] t(j)g\left(\left\lfloor\frac{n}{T}\right\rfloor\right) &\text{【先枚举 ij 乘积】}\\ =&\sum_{T=1}^n &&\text{【先枚举 ij 乘积】}\\ &=&\sum_{T=1}^n \sum_{i|T}\mu(i)t(i) t\left(\frac{T}{i}\right)g\left(\left\lfloor\frac{n}{T}\right\rfloor\right) &\text{【}\sum_{j=1}^{\left\lfloor\frac{n}{i}\right\rfloor}[ij=T] \text{对答案的贡献为 1，于是省略】}\\ =&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right) &&\text{【}\sum_{j=1}^{\left\lfloor\frac{n}{i}\right\rfloor}[ij=T] \text{对答案的贡献为 1，于是省略】}\\ &=&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right) \sum_{i|T}\mu(i)t(i)t\left(\frac{T}{i}\right)\\ =&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right) &=&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right) \sum_{i|T}\mu(i)t(T) &\text{【t 是完全积性函数】}\\ =&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)t(T) &&\text{【t 是完全积性函数】}\\ &=&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)t(T) \sum_{i|T}\mu(i)\\ =&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)t(T) &=&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)t(T) \varepsilon(T) &\text{【}\mu\ast 1= \varepsilon\text{】}\\ =&g(n)t(1) &\text{【当且仅当 T=1,}\varepsilon(T)=1\text{时】}\\ =&g(n) &&& \square \end{split} &&\text{【}\mu\ast 1= \varepsilon\text{】}\\ &=&g(n)t(1) &&\text{【当且仅当 T=1,}\varepsilon(T)=1\text{时】}\\ &=&g(n) && \square \end{eqnarray} $$ ======= **解法二** 转化一下，可以将式子写成 Loading Loading
docs/math/mobius.md +33 −33 Original line number Diff line number Diff line Loading @@ -6,7 +6,7 @@ * * * ## 数论分块与整除相关 ## 数论分块与整除相 先补一下数学小 trick Loading Loading @@ -34,7 +34,7 @@ $$ ### 引理 2 $$ \forall n,d\in\mathbb{N},\left|\left\{\left\lfloor\frac{n}{d}\right\rfloor\right\}\right|\leq\left\lfloor2\sqrt{n}\right\rfloor \forall n \in N, \left|\left\{ \lfloor \frac{n}{d} \rfloor \mid d \in N \right\}\right| \leq \lfloor 2\sqrt{n} \rfloor $$ $|V|$ 表示集合 $V$ 的元素个数 Loading Loading @@ -372,12 +372,12 @@ int main() { } ``` ### [「SPOJ 5971」LCMSUM](https://www.luogu.org/problemnew/show/SP5971) ### [「SPOJ 5971」LCMSUM](https://www.spoj.com/problems/LCMSUM/) 求值（多组数据） $$ \sum_{i=1}^n \text{lcm}(i,n)\qquad (1\leqslant T\leqslant 3\times 10^5,1\leqslant n\leqslant 10^6) \sum_{i=1}^n \text{lcm}(i,n)\quad \text{s.t.}\ 1\leqslant T\leqslant 3\times 10^5,1\leqslant n\leqslant 10^6 $$ 易得原式即 Loading @@ -389,7 +389,7 @@ $$ 根据 $\gcd(a,n)=1$ 时一定有 $\gcd(n-a,n)=1$ ，可将原式化为 $$ \frac{1}{2}\cdot(\sum_{i=1}^{n-1}\frac{i\cdot n}{\gcd(i,n)}+\sum_{i=n-1}^{1}\frac{i\cdot n}{\gcd(i,n)})+n \frac{1}{2}\cdot \left(\sum_{i=1}^{n-1}\frac{i\cdot n}{\gcd(i,n)}+\sum_{i=n-1}^{1}\frac{i\cdot n}{\gcd(i,n)}\right)+n $$ 上述式子中括号内的两个 $\sum$ 对应的项相等，故又可以化为 Loading Loading @@ -683,7 +683,7 @@ signed main(){ 求 $$ \sum_{i=1}^n\sum_{j=1}^ni\cdot j\cdot gcd(i,j)\bmod p\\ \sum_{i=1}^n\sum_{j=1}^ni\cdot j\cdot \gcd(i,j)\bmod p\\ n\leq10^{10},5\times10^8\leq p\leq1.1\times10^9,\text{p 是质数} $$ Loading @@ -692,22 +692,23 @@ $$ 我们利用 $\varphi\ast1=ID$ 反演 $$ \begin{split} &\sum_{i=1}^n\sum_{j=1}^ni\cdot j \begin{eqnarray} && \sum_{i=1}^n\sum_{j=1}^ni\cdot j\cdot \gcd(i,j)\\ &=&\sum_{i=1}^n\sum_{j=1}^ni\cdot j \sum_{d|i,d|j}\varphi(d)\\ &\sum_{d=1}^n\sum_{i=1}^n &=&\sum_{d=1}^n\sum_{i=1}^n \sum_{j=1}^n[d|i,d|j]\cdot i\cdot j \cdot\varphi(d)\\ &\sum_{d=1}^n &=&\sum_{d=1}^n \sum_{i=1}^{\left\lfloor\frac{n}{d}\right\rfloor} \sum_{j=1}^{\left\lfloor\frac{n}{d}\right\rfloor} d^2\cdot i\cdot j\cdot\varphi(d)\\ &\sum_{d=1}^nd^2\cdot\varphi(d) &=&\sum_{d=1}^nd^2\cdot\varphi(d) \sum_{i=1}^{\left\lfloor\frac{n}{d}\right\rfloor}i \sum_{j=1}^{\left\lfloor\frac{n}{d}\right\rfloor}j\\ &\sum_{d=1}^nF^2\left(\left\lfloor\frac{n}{d}\right\rfloor\right)\cdot d^2\varphi(d) &=&\sum_{d=1}^nF^2\left(\left\lfloor\frac{n}{d}\right\rfloor\right)\cdot d^2\varphi(d) \left(F(n)=\frac{1}{2}n\left(n+1\right)\right)\\ \end{split} \end{eqnarray} $$ 对 $\sum_{d=1}^nF\left(\left\lfloor\frac{n}{d}\right\rfloor\right)^2$ 做数论分块，$d^2\varphi(d)$ 的前缀和用杜教筛处理： Loading Loading @@ -798,41 +799,40 @@ $$ 我们证明一下 $$ \begin{split} &g(n)=\sum_{i=1}^n\mu(i)t(i)f\left(\left\lfloor\frac{n}{i}\right\rfloor\right)\\ =&\sum_{i=1}^n\mu(i)t(i) \begin{eqnarray} &&g(n)=\sum_{i=1}^n\mu(i)t(i)f\left(\left\lfloor\frac{n}{i}\right\rfloor\right)\\ &=&\sum_{i=1}^n\mu(i)t(i) \sum_{j=1}^{\left\lfloor\frac{n}{i}\right\rfloor}t(j) g\left(\left\lfloor\frac{\left\lfloor\frac{n}{i}\right\rfloor}{j}\right\rfloor\right)\\ =&\sum_{i=1}^n\mu(i)t(i) &=&\sum_{i=1}^n\mu(i)t(i) \sum_{j=1}^{\left\lfloor\frac{n}{i}\right\rfloor}t(j) g\left(\left\lfloor\frac{n}{ij}\right\rfloor\right)\\ =&\sum_{T=1}^n &=&\sum_{T=1}^n \sum_{i=1}^n\mu(i)t(i) \sum_{j=1}^{\left\lfloor\frac{n}{i}\right\rfloor}[ij=T] t(j)g\left(\left\lfloor\frac{n}{T}\right\rfloor\right) &\text{【先枚举 ij 乘积】}\\ =&\sum_{T=1}^n &&\text{【先枚举 ij 乘积】}\\ &=&\sum_{T=1}^n \sum_{i|T}\mu(i)t(i) t\left(\frac{T}{i}\right)g\left(\left\lfloor\frac{n}{T}\right\rfloor\right) &\text{【}\sum_{j=1}^{\left\lfloor\frac{n}{i}\right\rfloor}[ij=T] \text{对答案的贡献为 1，于是省略】}\\ =&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right) &&\text{【}\sum_{j=1}^{\left\lfloor\frac{n}{i}\right\rfloor}[ij=T] \text{对答案的贡献为 1，于是省略】}\\ &=&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right) \sum_{i|T}\mu(i)t(i)t\left(\frac{T}{i}\right)\\ =&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right) &=&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right) \sum_{i|T}\mu(i)t(T) &\text{【t 是完全积性函数】}\\ =&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)t(T) &&\text{【t 是完全积性函数】}\\ &=&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)t(T) \sum_{i|T}\mu(i)\\ =&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)t(T) &=&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)t(T) \varepsilon(T) &\text{【}\mu\ast 1= \varepsilon\text{】}\\ =&g(n)t(1) &\text{【当且仅当 T=1,}\varepsilon(T)=1\text{时】}\\ =&g(n) &&& \square \end{split} &&\text{【}\mu\ast 1= \varepsilon\text{】}\\ &=&g(n)t(1) &&\text{【当且仅当 T=1,}\varepsilon(T)=1\text{时】}\\ &=&g(n) && \square \end{eqnarray} $$ ======= **解法二** 转化一下，可以将式子写成 Loading
