傳送門

不會……

兩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

 1 //minamoto
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<map>
 5 #define ll long long
 6 using namespace std;
 7
 
 const int N=4e6+5;
 8 map<ll,ll> mp;
 9 ll sum[N],phi[N],pri[N],cnt,vis[N],mod6,mod;
10 //mod是2的逆元，mod6是6的逆元 
11 void init(ll p){
12     vis[1]=sum[1]=1;
13     for(int i=2;i<N;++i){
14         if(!vis[i]) pri[++cnt]=i,phi[i]=i-1,sum[i]=1ll*i*i%p*phi[i]%p;
15         for(int j=1;j<=cnt,i*pri[j]<N;++j){
 
16             vis[i*pri[j]]=1;
17             int now=i*pri[j];
18             if(i%pri[j]==0){
19                 phi[now]=phi[i]*pri[j]%p;
20                 sum[now]=phi[now]*now%p*now%p;
21                 break;
22             }
23             phi[now]=phi[i]*(pri[j]-1)%p;
24             sum[now]=phi[now]*now%p*now%p;
 
25         }
26     }
27     for(int i=2;i<N;++i) (sum[i]+=sum[i-1])%=p;
28 }
29 ll ksm(ll x,ll y,ll p){
30     ll res=1;
31     while(y){
32         if(y&1) (res*=x)%=p;
33         (x*=x)%=p,y>>=1;
34     }
35     return res;
36 }
37 ll calc(ll n,ll p){
38     n%=p;
39     ll res=((1+n)*n%p)*mod%p;
40     (res*=res)%=p;
41     return res;
42 }
43 ll calcs(ll n,ll p){
44     n%=p;
45     ll res=(n*(n+1)%p)*(2*n+1)%p;
46     (res*=mod6)%=p;
47     return res;
48 }
49 ll calcsum(ll n,ll p){
50     if(n<N) return sum[n];
51     if(mp.count(n)) return mp[n];
52     ll x=2,res=calc(n,p);
53     while(x<=n){
54         ll y=n/(n/x);
55         res=((res-(calcs(y,p)-calcs(x-1,p)+p)%p*calcsum(n/x,p)%p)+p)%p;
56         x=y+1;
57     }
58     return mp[n]=res;
59 }
60 int main(){
61 //    freopen("testdata.in","r",stdin);
62     ll p,n;
63     scanf("%lld%lld",&p,&n);
64     mod=ksm(2,p-2,p),mod6=ksm(6,p-2,p);
65     init(p);
66     ll x=1,ans=0;
67     while(x<=n){
68         ll y=n/(n/x);
69         (ans+=((calcsum(y,p)-calcsum(x-1,p)+p)%p*calc(n/x,p)%p))%=p;
70         x=y+1;
71     }
72     printf("%lld\n",ans);
73     return 0;
74 }

洛谷P3768 簡單的數學題（莫比烏斯反演+狄利克雷卷積+杜教篩）