式子好麻烦orz……大佬好腻害orz->

1 //minamoto 2 #include
     
       3 #include
      
        4 #define ll long long 5 using namespace std; 6 const int N=1e7+5,mod=20101009; 7 int n,m,vis[N],p[N],cnt,mu[N];ll sum[N]; 8 ll ans,inv2,summ; 9 void init(int lim){10     mu[1]=1;11     for(int i=2;i<=lim;++i){12         if(!vis[i]) p[++cnt]=i,mu[i]=-1;13         for(int j=1;j<=cnt&&p[j]*i<=lim;++j){14             vis[i*p[j]]=1;15             if(i%p[j]==0) break;16             mu[i*p[j]]=-mu[i];17         }18     }19     for(int i=1;i<=lim;++i)20     (sum[i]=sum[i-1]+1ll*mu[i]*i*i)%=mod;21 }22 ll calc(int mx,int l){23     return (1ll+mx/l)*(mx/l)%mod*inv2%mod;24 }25 int main(){26 //    freopen("testdata.in","r",stdin);27     scanf("%d%d",&n,&m);28     int lim;init(lim=min(n,m));29     ans=0,inv2=(mod+1)/2,summ=0;30     for(int d=1;d<=lim;++d){31         int mxx=n/d,myy=m/d,mn=min(mxx,myy);32         summ=0;33         for(int l=1,r;l<=mn;l=r+1){34             r=min(mxx/(mxx/l),myy/(myy/l));35             (summ+=(sum[r]-sum[l-1])%mod*calc(mxx,l)%mod*calc(myy,l)%mod)%=mod;36         }37         (ans+=summ*d)%=mod;38     }39     printf("%lld\n",(ans%mod+mod)%mod);40     return 0;41 }