【反演复习计划】【bzoj2154】Crash的数字表格

膜拜cdc……他的推导详细到我这种蒟蒻都能看得懂！

膜拜的传送门

所以我附一下代码就好了。

#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+)/)%yql)*((y*(y+)/)%yql)%yql;
}
inline void calcmu(){
    memset(vis,,sizeof(vis));cnt=;mu[]=;
    for(int i=;i<=maxn;i++){
        if(vis[i]){prime[++cnt]=i;mu[i]=-;}
        for(int j=;j<=cnt;j++){
            int t=i*prime[j];if(t>maxn)break;
            vis[t]=;
            if(i%prime[j]==){mu[t]=;break;}
            mu[t]=-mu[i];
        }
    }
    for(ll i=;i<=maxn;i++)s[i]=(s[i-]+(i*i*mu[i])%yql)%yql;
}
inline ll F(ll x,ll y){
    ll ans=;ll n=min(x,y);
    for(ll i=,j=;i<=n;i=j+){
        j=min(x/(x/i),y/(y/i));
        ans=(ans+(s[j]-s[i-])*sum(x/i,y/i)%yql)%yql;
    }
    return ans;
}
int main(){
    cin>>n>>m;maxn=min(n,m);
    calcmu();
    for(ll i=,j=;i<=maxn;i=j+){
        j=min(n/(n/i),m/(m/i));
        ans=(ans+(i+j)*(j-i+)/%yql*F(n/i,m/i)%yql)%yql;
    }
    printf("%lld\n",(ans+yql)%yql);
}