数论专项测试——约数个数和（lucas的数论）

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 using namespace std;
 7 typedef long long int64;
 8 const int mod=1000000007;
 9 #define maxn 2000005
10 int top,tot,d[maxn],prim[maxn],mu[maxn];
11 bool vis[maxn];
12 int64 n,f[maxn],ans;
13 void prepare(){
14     tot=top=0,memset(vis,1,sizeof(vis)),mu[0]=0,mu[1]=1,f[1]=1;
15     for (int i=2;i<maxn;i++){
16         if (vis[i]==1){
17             prim[++top]=i;
18             d[i]=i;
19             mu[i]=-1;
20             f[i]=2;
21         }
22         for (int j=1;j<=top;j++){
23             if (i*prim[j]>=maxn) break;
24             vis[i*prim[j]]=0;
25             if (i%prim[j]==0){
26                 d[i*prim[j]]=d[i]*prim[j];
27                 mu[i*prim[j]]=0;
28                 f[i*prim[j]]=f[i/d[i]]*(f[d[i]]+1);
29                 break;
30             }else{
31                 d[i*prim[j]]=prim[j];
32                 mu[i*prim[j]]=mu[i]*mu[prim[j]];
33                 f[i*prim[j]]=f[i]*f[prim[j]];
34             }
35         }
36     }
37     for (int i=2;i<maxn;i++) mu[i]+=mu[i-1];
38     for (int i=2;i<maxn;i++) f[i]=(f[i-1]+f[i])%mod;
39 }
40 #define maxp 100007
41 #define maxm 4000005
42 int now[maxp],prep[maxm];
43 int64 val[maxm],id[maxm];
44 void insert(int x,int64 y){
45     int pos=x%maxp;
46     prep[++tot]=now[pos],now[pos]=tot,val[tot]=y,id[tot]=x;
47 }
48 int64 find(int x){
49     int pos=x%maxp;
50     for (int i=now[pos];i;i=prep[i]){
51         if (id[i]==x) return val[i];
52     }
53     return -1;
54 }
55 int64 Mu(int x){
56     if (x<maxn) return mu[x];
57     int64 temp=find(x),t;
58     if (temp!=-1) return temp;
59     temp=1;
60     for (int j,i=2;i<=x;i=j+1){
61         j=x/(x/i);  t=Mu(x/i);
62         temp=((temp-1LL*(j-i+1)*t%mod)%mod+mod)%mod;
63     }
64     insert(x,temp); return temp;
65 }
66 int64 F(int x){
67     if (x<maxn) return f[x];
68     int64 temp=0;
69     for (int j,i=1;i<=x;i=j+1){
70         j=x/(x/i);
71         temp=(temp+1LL*(x/i)*(j-i+1)%mod)%mod;
72     }
73     return temp%mod;
74 }
75 int main(){
76     int64 temp;
77     prepare();
78     scanf("%lld",&n);
79     ans=0;
80     for (int j,i=1;i<=n;i=j+1){
81         j=n/(n/i); temp=F(n/i);
82         ans=(ans+1LL*(Mu(j)-Mu(i-1))%mod*temp%mod*temp%mod)%mod;
83     }
84     printf("%lld\n",(ans%mod+mod)%mod);
85     return 0;
86 }