题目链接
我错了,杜教筛不是只背板子就够的qwq
跟着神仙yyb推了一遍式子,感觉这题还是很神仙
题意是求
∑ i = 1 n ∑ j = 1 n i j g c d ( i , j ) \sum_{i=1}^{n}\sum_{j=1}^{n}ijgcd(i,j) i=1∑nj=1∑nijgcd(i,j)
显然先把gcd枚举一下,假设gcd(i,j)==k
原式就为
∑ k = 1 n k ∑ i = 1 n ∑ j = 1 n i j [ g c d ( i , j ) = = k ] \sum_{k=1}^{n}k\sum_{i=1}^{n}\sum_{j=1}^{n}ij[gcd(i,j)==k] k=1∑nki=1∑nj=1∑nij[gcd(i,j)==k]
后面这个式子根据套路可以往外提一下
∑ k = 1 n k 3 ∑ i = 1 n / k ∑ j = 1 n / k i j [ g c d ( i , j ) = = 1 ] \sum_{k=1}^{n}k^3\sum_{i=1}^{n/k}\sum_{j=1}^{n/k}ij[gcd(i,j)==1] k=1∑nk3i=1∑n/kj=1∑n/kij[gcd(i,j)==1]
这样子看着一脸可莫比乌斯反演的样子
令 f ( x ) = ∑ i = 1 x ∑ j = 1 x i j [ g c d ( i , j ) = = 1 ] f(x)=\sum_{i=1}^{x}\sum_{j=1}^{x}ij[gcd(i,j)==1] f(x)=i=1∑xj=1∑xij[gcd(i,j)==1]
根据某引理
∑ d ∣ x μ ( d ) = [ x = 1 ] \sum_{d|x}\mu(d)=[x=1] d∣x∑μ(d)=[x=1]
f ( x ) = ∑ i = 1 x ∑ j = 1 x ∑ d ∣ g c d ( i , j ) μ ( d ) f(x)=\sum_{i=1}^{x}\sum_{j=1}^{x}\sum_{d|gcd(i,j)}\mu(d) f(x)=i=1∑xj=1∑x