【反演复习计划】【bzoj2154】Crash的数字表格

#include<bits/stdc++.h>
#define N 10000005
#define yql 20101009
using namespace std;
typedef long long ll;
int mu[N],prime[N],cnt,s[N],vis[N];
ll n,m,ans,maxn;
inline ll sum(ll x,ll y){
    return ((x*(x+1)/2)%yql)*((y*(y+1)/2)%yql)%yql;
}
inline void calcmu(){
    memset(vis,1,sizeof(vis));cnt=0;mu[1]=1;
    for(int i=2;i<=maxn;i++){
        if(vis[i]){prime[++cnt]=i;mu[i]=-1;}
        for(int j=1;j<=cnt;j++){
            int t=i*prime[j];if(t>maxn)break;
            vis[t]=0;
            if(i%prime[j]==0){mu[t]=0;break;}
            mu[t]=-mu[i];
        }
    }
    for(ll i=1;i<=maxn;i++)s[i]=(s[i-1]+(i*i*mu[i])%yql)%yql;
}
inline ll F(ll x,ll y){
    ll ans=0;ll n=min(x,y);
    for(ll i=1,j=1;i<=n;i=j+1){
        j=min(x/(x/i),y/(y/i));
        ans=(ans+(s[j]-s[i-1])*sum(x/i,y/i)%yql)%yql;
    }
    return ans;
}
int main(){
    cin>>n>>m;maxn=min(n,m);
    calcmu();
    for(ll i=1,j=1;i<=maxn;i=j+1){
        j=min(n/(n/i),m/(m/i));
        ans=(ans+(i+j)*(j-i+1)/2%yql*F(n/i,m/i)%yql)%yql;
    }
    printf("%lld\n",(ans+yql)%yql);
}