洛谷P3768 简单的数学题（莫比乌斯反演+狄利克雷卷积+杜教筛）

传送门

不会……

两篇加在一起都看不懂……

https://www.cnblogs.com/cellular-automaton/p/8241128.html

https://www.luogu.org/blog/cjyyb/solution-p3768

 //minamoto
 #include<iostream>
 #include<cstdio>
 #include<map>
 #define ll long long
 using namespace std;
 const int N=4e6+;
 map<ll,ll> mp;
 ll sum[N],phi[N],pri[N],cnt,vis[N],mod6,mod;
 //mod是2的逆元，mod6是6的逆元
 void init(ll p){
     vis[]=sum[]=;
     for(int i=;i<N;++i){
         if(!vis[i]) pri[++cnt]=i,phi[i]=i-,sum[i]=1ll*i*i%p*phi[i]%p;
         for(int j=;j<=cnt,i*pri[j]<N;++j){
             vis[i*pri[j]]=;
             int now=i*pri[j];
             if(i%pri[j]==){
                 phi[now]=phi[i]*pri[j]%p;
                 sum[now]=phi[now]*now%p*now%p;
                 break;
             }
             phi[now]=phi[i]*(pri[j]-)%p;
             sum[now]=phi[now]*now%p*now%p;
         }
     }
     for(int i=;i<N;++i) (sum[i]+=sum[i-])%=p;
 }
 ll ksm(ll x,ll y,ll p){
     ll res=;
     while(y){
         if(y&) (res*=x)%=p;
         (x*=x)%=p,y>>=;
     }
     return res;
 }
 ll calc(ll n,ll p){
     n%=p;
     ll res=((+n)*n%p)*mod%p;
     (res*=res)%=p;
     return res;
 }
 ll calcs(ll n,ll p){
     n%=p;
     ll res=(n*(n+)%p)*(*n+)%p;
     (res*=mod6)%=p;
     return res;
 }
 ll calcsum(ll n,ll p){
     if(n<N) return sum[n];
     if(mp.count(n)) return mp[n];
     ll x=,res=calc(n,p);
     while(x<=n){
         ll y=n/(n/x);
         res=((res-(calcs(y,p)-calcs(x-,p)+p)%p*calcsum(n/x,p)%p)+p)%p;
         x=y+;
     }
     return mp[n]=res;
 }
 int main(){
 //    freopen("testdata.in","r",stdin);
     ll p,n;
     scanf("%lld%lld",&p,&n);
     mod=ksm(,p-,p),mod6=ksm(,p-,p);
     init(p);
     ll x=,ans=;
     while(x<=n){
         ll y=n/(n/x);
         (ans+=((calcsum(y,p)-calcsum(x-,p)+p)%p*calc(n/x,p)%p))%=p;
         x=y+;
     }
     printf("%lld\n",ans);
     return ;
 }