P3911 最小公倍数之和 题解

博客园同步

原题链接

前置知识：

本题的弱化版

不难发现，原来的：

$$\sum _{i=1}^n\sum _{j=1}^n\mathrm{lcm}(i,j)$$

变成了：

$$\sum _{i=1}^n\sum _{j=1}^n\mathrm{lcm}(a_i,a_j)$$

一言不合就开始推式子。

考虑用 $c_i$ 表示 $i$ 出现的次数，然后：

∑ i = 1 n ∑ j = 1 n lcm ⁡ ( a i , a j ) = ∑ i = 1 n ∑ j = 1 n lcm ⁡ ( i , j ) × c i × c j = ∑ i = 1 n ∑ j = 1 n i × j × c i × c j gcd ⁡ ( i , j ) = ∑ d = 1 n ∑ i = 1 ⌊ n d ⌋ ∑ j = 1 ⌊ n d ⌋ [ gcd ⁡ ( i , j ) = 1 ] d × i × j × c i d × c j d = ∑ d = 1 n ∑ i = 1 ⌊ n d ⌋ ∑ j = 1 ⌊ n d ⌋ ∑ k ∣ gcd ⁡ ( i , j ) μ ( k ) × d × i × j × c i d × c j d = ∑ d = 1 n ∑ k = 1 ⌊ n d ⌋ ∑ i = 1 ⌊ n k d ⌋ ∑ j = 1 ⌊ n k d ⌋ μ ( k ) × d × i × j × k 2 × c i d k × c j d k = ∑ T = 1 n T × ( ∑ i = 1 ⌊ n T ⌋ i × c i T ) 2 ∑ k ∣ T μ ( k ) × k \begin{aligned} & \sum_{i=1}^n \sum_{j=1}^n \operatorname{lcm}(a_i , a_j) \\ =& \sum_{i=1}^n\sum_{j=1}^n \operatorname{lcm}(i,j)\times c_i \times c_j\\ =& \sum_{i=1}^n\sum_{j=1}^n \frac{i \times j\times c_i \times c_j}{\gcd(i,j)} \\ =& \sum_{d=1}^n\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sum_{j=1}^{\lfloor \frac{n}{d}\rfloor}[\gcd(i,j)=1]d\times i \times j \times c_{id} \times c_{jd} \\ =& \sum_{d=1}^n\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sum_{j=1}^{\lfloor \frac{n}{d}\rfloor}\sum_{k|\gcd(i,j)}\mu(k) \times d\times i \times j \times c_{id} \times c_{jd} \\ =& \sum_{d=1}^n\sum_{k=1}^{\lfloor \frac{n}{d} \rfloor}\sum_{i=1}^{\lfloor \frac{n}{kd} \rfloor}\sum_{j=1}^{\lfloor \frac{n}{kd}\rfloor}\mu(k)\times d \times i \times j \times k^2 \times c_{idk} \times c_{jdk} \\ =& \sum_{T=1}^{n}T\times(\sum_{i=1}^{\lfloor \frac{n}{T} \rfloor}i\times c_{iT})^2\sum_{k|T}\mu(k)\times k \end{aligned} ======​i=1∑n​j=1∑n​lcm(ai​,aj​)i=1∑n​