Crash的数字表格

求\(\sum_{i=1}^N\sum_{j=1}^Mlcm(i,j)\)

解

设\(N<M\),显然有

\[\sum_{i=1}^N\sum_{j=1}^M\frac{ij}{gcd(i,j)}=\sum_{d=1}^N\frac{1}{d}\sum_{i=1}^N\sum_{j=1}^Mij(gcd(i,j)==d)\]

设

\[f(k)=\sum_{i=1}^N\sum_{j=1}^Mij(gcd(i,j)==k)\]
\[F(k)=\sum_{i=1}^{N}\sum_{j=1}^Mij(k|gcd(i,j))=k^2\sum_{i=1}^{[N/k]}\sum_{j=1}^{[M/k]}ij\]

设\(dc[k]=\sum_{i=1}^ki=\frac{(1+k)\times k}{2},F(k)=k^2dc(N/k)dc(M/k)\)

由Mobius反演定理我们有

\[f(k)=\sum_{k|d}F(d)\mu(d/k)\]

代入有

\[ans=\sum_{d=1}^N\frac{1}{d}\sum_{d|x}^{}x^2dc(N/x)dc(M/x)\mu(x/d)=\]

\[\sum_{d=1}^Nd\sum_{x=1}^{[N/d]}dc(N/xd)dc(M/xd)x^2\mu(x)\]

维护出后式\(x^2\mu(x)\),两次整除分块即可,不难得知时间复杂度\(O(n)\)。

顺便提一下，如果\(N,M\)很小，我们可以变成一下形式，变为\(O(nlogn+T\sqrt{n})\)(T为询问组数)。

\[\sum_{x=1}^Ndc(N/x)dc(M/x)\sum_{d|x}\mu(x/d)\frac{x^2}{d}\]

参考代码：

#include <iostream>
#include <cstdio>
#define il inline
#define ri register
#define ll long long
#define yyb 20101009
#define swap(x,y) x^=y^=x^=y
using namespace std;
bool check[10000001];
int prime[750000],pt,mb[10000001];
void prepare(int);
il int min(int,int),dx(int,int);
int main(){int n,m,nd,md,ndx,mdx,i,ij,j,jj,ans1(0),ans2;scanf("%d%d",&n,&m);if(n>m)swap(n,m);prepare(m);for(i=1;i<=n;i=ij+1){ij=min(n/(n/i),m/(m/i));ans2&=0,nd=n/i,md=m/i;for(j=1;j<=nd;j=jj+1)jj=min(nd/(nd/j),md/(md/j)),(ans2+=(ll)(mb[jj]-mb[j-1])*dx(1,nd/j)%yyb*dx(1,md/j)%yyb)%=yyb;(ans1+=(ll)ans2*dx(i,ij)%yyb)%=yyb;}printf("%d",(ans1+yyb)%yyb);return 0;
}
il int dx(int a,int b){return (ll)(a+b)*(b-a+1)/2%yyb;
}
void prepare(int n){int i,j;check[1]|=mb[1]|=true;for(i=2;i<=n;++i){if(!check[i])prime[++pt]=i,mb[i]=-1;for(j=1;j<=pt&&prime[j]<=n/i;++j){check[i*prime[j]]|=true;if(!(i%prime[j]))break;mb[i*prime[j]]=-mb[i];}}for(i=1;i<=n;++i)mb[i]=((ll)mb[i]*i%yyb*i+mb[i-1])%yyb;
}
il int min(int a,int b){return a<b?a:b;
}