考虑到\(lcm(i,j)=\frac{ij}{gcd(i,j)}\)
\(\sum_{i=1}^n\sum_{j=1}^m\frac{ij}{gcd(i,j)}\)
\(\sum_{d=1}^{n}\sum_{i=1}^n\sum_{j=1}^m[gcd(i,j)==d]\frac{ij}{d}\)
\(\sum_{d=1}^{n}\sum_{i=1}^{n/d}\sum_{j=1}^{m/d}[gcd(i,j)==1]{ijd}\)
\(=\sum_{d=1}^{n}d\sum_{i=1}^{n/d}\sum_{j=1}^{m/d}[gcd(i,j)==1]{ij}\)
看后面
\(\sum_{i=1}^{n/d}\sum_{j=1}^{m/d}[gcd(i,j)==1]{ij}\)
\(=\sum_{i=1}^{x}\sum_{j=1}^{y}[gcd(i,j)==1]{ij}\)
考虑反演
\(f(d)=\sum_{i=1}^{x}\sum_{j=1}^{y}[gcd(i,j)==d]{ij}\)
\(G(d)=\sum_{i=1}^{x}\sum_{j=1}^{y}[d|gcd(i,j)]{ij}\)
\(G(d)=d^2\sum_{i=1}^{x/d}\sum_{j=1}^{y/d}[1|gcd(i,j)]{ij}\)
\(G(d)=d^2\sum_{i=1}^{x/d}\sum_{j=1}^{y/d}{ij}\)
于是
\(f(1)=\sum_{i=1}^x\mu(i)G(i)\)
\(ans=\sum_{d=1}^{n}d\sum_{i=1}^{n/d}\sum_{j=1}^{m/d}[gcd(i,j)==1]{ij}\)
在这里做第一次分块，而后
\(f(1)=\sum_{i=1}^x\mu(i)G(i)\)
\(=\sum_{i=1}^x\mu(i)i^2\sum_{i=1}^{x/d}\sum_{j=1}^{y/d}{ij}\)
然后内外各做一次整除分块即可
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 12000005;
const int mod = 20101009;
int mu[N+5],sum[N+5],pr[N+5],is[N+5],cnt;
int h(int n,int m) {
int l=1,r,ans=0;
while(l<=n) {
r=min(n/(n/l),m/(m/l));
ans += (sum[r]-sum[l-1]+mod)%mod
* ((n/l)*(n/l+1)/2%mod)%mod
* ((m/l)*(m/l+1)/2%mod)%mod;
ans %= mod;
l=r+1;
}
return ans;
}
signed main() {
mu[0]=mu[1]=1; is[1]=1;
for(int i=2;i<=N;i++) {
if(is[i]==0) {
pr[++cnt]=i;
mu[i]=-1;
}
for(int j=1; j<=cnt&&pr[j]*i<N; ++j) {
is[pr[j]*i]=1;
if(i%pr[j]==0) {
mu[pr[j]*i]=0;
break;
}
else {
mu[pr[j]*i]=-mu[i];
}
}
}
for(int i=1;i<=N;i++) sum[i]=(sum[i-1]+mu[i]*i*i%mod)%mod;
int n,m;
cin>>n>>m;
if(n>m) swap(n,m);
int l=1,r,ans=0;
while(l<=n) {
r=min(n/(n/l),m/(m/l));
ans+=(r-l+1)*(l+r)/2%mod*h(n/l,m/l)%mod;
ans%=mod;
l=r+1;
}
cout<<ans<<endl;
}
来源：https://www.cnblogs.com/mollnn/p/12263960.html