该博客介绍了如何利用欧拉筛、莫比乌斯反演和狄利克雷卷积解决洛谷P3768题目。首先讲解了相关数学基础知识，然后通过莫反将问题转化为求解F函数，F函数可通过欧拉筛求得前缀和。由于n的范围较大，引入杜教筛优化，以O(n^1.5)的时间复杂度求解问题。最后给出了代码实现。
1.前置知识：
欧拉筛，莫比乌斯反演，狄利克雷卷积，杜教筛
2.莫反+狄利克雷卷积+欧拉筛
看到这道题的瞬间，按照DNA来一个莫反
∑ i = 1 n ∑ j = 1 n i j gcd ( i , j ) = ∑ k = 1 n k ∑ i = 1 n ∑ j = 1 n i j [ gcd ( i , j ) = k ] = ∑ k = 1 n k 3 ∑ i = 1 ⌊ n k ⌋ ∑ j = 1 ⌊ n k ⌋ i j [ gcd ( i , j ) = 1 ] = ∑ k = 1 n k 3 ∑ i = 1 ⌊ n k ⌋ ∑ j = 1 ⌊ n k ⌋ i j ∑ d ∣ gcd ( i , j ) μ ( d ) = ∑ k = 1 n k 3 ∑ d = 1 ⌊ n k ⌋ μ ( d ) ∑ i = 1 ⌊ n k ⌋ ∑ j = 1 ⌊ n k ⌋ i j [ d ∣ gcd ( i , j ) ] = ∑ k = 1 n k 3 ∑ d = 1 ⌊ n k ⌋ μ ( d ) d 2 ∑ i = 1 ⌊ n k d ⌋ i ∑ j = 1 ⌊ n k d ⌋ j \begin{aligned} &\sum_{i=1}^{n}\sum_{j=1}^{n}{ij\gcd(i,j)}\\ =&\sum_{k=1}^{n}k\sum_{i=1}^{n}\sum_{j=1}^{n}{ij[\gcd(i,j)=k]}\\ =&\sum_{k=1}^{n}{k^3}\sum_{i=1}^{\lfloor\frac{n}{k}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{k}\rfloor}{ij[\gcd(i,j)=1]}\\ =&\sum_{k=1}^{n}{k^3}\sum_{i=1}^{\lfloor\frac{n}{k}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{k}\rfloor}{ij\sum_{d|\gcd(i,j)}{\mu(d)}}\\ =&\sum_{k=1}^{n}{k^3}\sum_{d=1}^{\lfloor\frac{n}{k}\rfloor}{\mu(d)}\sum_{i=1}^{\lfloor\frac{n}{k}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{k}\rfloor}{ij[d|\gcd(i,j)]}\\ =&\sum_{k=1}^{n}{k^3}\sum_{d=1}^{\lfloor\frac{n}{k}\rfloor}{\mu(d)d^2}\sum_{i=1}^{\lfloor\frac{n}{kd}\rfloor}i\sum_{j=1}^{\lfloor\frac{n}{kd}\rfloor}{j}\\ \end{aligned} =====i=1∑nj=1∑nijgcd(i,j)k=1∑nki=1∑nj=1∑nij[gcd(i,j)=k]k=1∑nk3i=1∑⌊kn⌋j=1∑⌊kn⌋ij[gcd(i,j)=1]k=1∑nk3i=1</
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