LG1829 题解
题意
求
\[\sum_{i=1}^{n}\sum_{j=1}^{m}lcm(i,j)
\]
\(1≤n,m≤10^7\)。
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 \(O(1)\) 算
\[g(n,m)=\sum_{i=1}^{n}i\sum_{j=1}^{m}j=\frac{n(n+1)}{2}\frac{m(m+1)}{2}
\]
则
\[\sum_{k=1}^{n}\mu(k)k^2g(\lfloor \frac{n}{k} \rfloor,\lfloor \frac{m}{k} \rfloor)
\]
又可以整除分块。于是总时间复杂度 \(O(\sqrt{n}\sqrt{n})=O(n)\)