51nod 1584加权约数和

学到了好多东西啊这题。。。

https://blog.csdn.net/sdfzyhx/article/details/72968468

 #include<bits/stdc++.h>
 using namespace std;
 const int N=1e6+,mod=1e9+;
 typedef long long ll;
 ll miu[N],v[N],p[N],cnt,sum1[N],sum2[N],mi1[N],mi2[N],h[N],g[N],f[N],T,n;
 ll inc(ll a,ll b)
 {
     a=a+b;
     if(a>mod)a%=mod;
     return a;
 }
 ll dec(ll a,ll b)
 {
     a=a-b;
     if(a<)a+=mod;
     return a%mod;
 }
 ll qmod(ll x,ll y)
 {
     int ans=;
     while(y)
     {
         if(y&)ans=1ll*ans*x%mod;
         x=1ll*x*x%mod;y>>=;
     }
     return ans;
 }
 void init()
 {
     miu[]=sum1[]=sum2[]=mi1[]=mi2[]=;
     for(int i=;i<=1e6;++i)
     {
         if(!v[i])
         {
             p[++cnt]=i;
             miu[i]=-;
             sum1[i]=i+;
             sum2[i]=inc(1ll*i*i%mod+i,);
             mi1[i]=mi2[i]=i;
         }
         for(int j=;j<=cnt&&i*p[j]<=1e6;++j)
         {
             int x=i*p[j];v[x]=;
             if(i%p[j])
             {
                 mi1[x]=mi2[x]=p[j];
                 miu[x]=-miu[i];
                 sum1[x]=1ll*sum1[i]*sum1[p[j]]%mod;
                 sum2[x]=1ll*sum2[i]*sum2[p[j]]%mod;
             }
             else
             {
                 mi1[x]=p[j];
                 mi2[x]=1ll*mi2[i]*p[j]%mod;
                 sum1[x]=inc(sum1[i],1ll*p[j]*mi2[i]%mod*sum1[i/mi2[i]]%mod);//求约数和
                 sum2[x]=inc(sum2[i],1ll*inc(1ll*mi2[i]*mi2[i]%mod*mi1[i]%mod,1ll*mi2[i]*mi2[i]%mod*mi1[i]%mod*mi1[i]%mod)*sum2[i/mi2[i]]%mod);//求平方约数和
                 break;
             }
         }
     }
     for(int i=;i<=1e6;++i)
     {
         h[i]=inc(h[i-],sum1[i]);
         g[i]=1ll*i*sum1[i]%mod*h[i]%mod;
         sum2[i]=1ll*sum2[i]*i%mod;
         sum2[i]=inc(sum2[i],sum2[i-]);
     }
     for(int i=;i<=1e6;++i)
     {

         for(int j=i,k=;j<=1e6;j+=i,k++)
         f[j]=inc(f[j],1ll*miu[i]%mod*i%mod*i%mod*g[k]%mod);
         f[i]=inc(f[i],f[i-]);
     }
     return;
 }
 int main()
 {
     init();scanf("%d",&T);
     for(int i=;i<=T;++i)
     {
         scanf("%d",&n);
         printf("Case #%d: %lld\n",i,(dec(2ll*f[n]%mod,sum2[n])+mod)%mod);
     }
     return ;
 }