LG1829 题解

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题意

\[\sum_{i=1}^{n}\sum_{j=1}^{m}lcm(i,j) \]

\(1≤n,m≤10^7\)

题解

直接化式子。

\[\sum_{i=1}^{n}\sum_{j=1}^{m}\frac{ij}{gcd(i,j)} \]

枚举 \(gcd\)

\[\sum_{d=1}^{n}\sum_{i=1}^{n}\sum_{j=1}^{m}\frac{ij}{k}[gcd(i,j)=d] \]

\[\sum_{d=1}^{n}\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sum_{j=1}^{\lfloor \frac{m}{d} \rfloor}\frac{ijd^2}{d}[gcd(i,j)=1] \]

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

将后部拎出来,设为 \(sum(\lfloor \frac{n}{d} \rfloor,\lfloor \frac{m}{d} \rfloor)\)。可以整除分块。于是现在只需要求

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

莫比乌斯函数

\[\sum_{i=1}^{n}\sum_{j=1}^{m}ij\sum_{k|gcd(i,j)}\mu(k) \]

\(k\) 提到前面

\[\sum_{k=1}^{n}\mu(k)\sum_{i=1}^{n}\sum_{j=1}^{m}ij[k|gcd(i,j)] \]

\[\sum_{k=1}^{n}\mu(k)k^2\sum_{i=1}^{\lfloor \frac{n}{k} \rfloor}i\sum_{j=1}^{\lfloor \frac{m}{k} \rfloor}j \]

后部提出来,可以 \(O(1)\)

\[g(n,m)=\sum_{i=1}^{n}i\sum_{j=1}^{m}j=\frac{n(n+1)}{2}\frac{m(m+1)}{2} \]

\[\sum_{k=1}^{n}\mu(k)k^2g(\lfloor \frac{n}{k} \rfloor,\lfloor \frac{m}{k} \rfloor) \]

又可以整除分块。于是总时间复杂度 \(O(\sqrt{n}\sqrt{n})=O(n)\)

posted @ 2022-09-29 15:46  realFish  阅读(28)  评论(0)    收藏  举报