【BZOJ4176】Lucas的数论-杜教筛

求$$\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}f(ij)$$，其中$f(x)$表示$x$的约数个数，$0\leq n\leq 10^9$，答案膜$10^9+7$

题解

首先有个妙不可言（被hjw污染了）的结论：$$f(nm)=\sum\limits_{i|n}\sum\limits_{j|m}[gcd(i,j)=1]$$

证明：咕

那么大力推一波式子：

$$\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}f(ij)$$

$$=\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}\sum\limits_{a|i}\sum\limits_{b|j}[gcd(a,b)=1]$$

$$=\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}\sum\limits_{a|i}\sum\limits_{b|j}\sum\limits_{d|gcd(a,b)}\mu(d)$$

$$=\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}\sum\limits_{a|i}\sum\limits_{b|j}\sum\limits_{d|a\& d|b}\mu(d)$$

$$=\sum\limits_{d=1}^{n}\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{n}{d}\rfloor}\sum\limits_{a=1}^{\lfloor\frac{n}{id}\rfloor}\sum\limits_{b=1}^{\lfloor\frac{n}{jd}\rfloor}\mu(d)$$

$$=\sum\limits_{d=1}^{n}\mu(d)(\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}\lfloor\frac{n}{id}\rfloor)$$

杜教筛+莫比乌斯反演解决

时间复杂度：$O(n^{\frac{2}{3}}logn+n^{\frac{3}{4}})$

代码：

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<cmath>
 #include<map>
 #define mod 1000000007
 using namespace std;
 typedef long long ll;
 ll n,pri=,p[],miu[],pre[],ans=;
 bool isp[];
 map<ll,ll>HASH;
 void _(){
     miu[]=pre[]=;
     for(int i=;i<=;i++){
         if(!isp[i]){
             p[++pri]=i;
             miu[i]=-;
         }
         for(int j=;j<=pri&&i*p[j]<=;j++){
             isp[i*p[j]]=true;
             if(i%p[j]==){
                 miu[i*p[j]]=;
                 break;
             }
             miu[i*p[j]]=-miu[i];
         }
     }
     for(int i=;i<=;i++){
         pre[i]=(pre[i-]+miu[i]+mod)%mod;
     }
 }
 ll work1(ll x){
     if(x<=)return pre[x];
     if(HASH.count(x))return HASH[x];
     ll ret=;
     for(int i=,j;i<=x;i=j+){
         j=x/(x/i);
         ret=(ret-(j-i+)*work1(x/i)%mod+mod)%mod;
     }
     HASH[x]=ret;
     return ret;
 }
 ll work2(ll x){
     ll ret=;
     for(int i=,j;i<=x;i=j+){
         j=x/(x/i);
         ret=(ret+(x/i)*(j-i+))%mod;
     }
     return ret*ret%mod;
 }
 int main(){
     _();
     scanf("%lld",&n);
     for(int i=,j;i<=n;i=j+){
         j=n/(n/i);
         ans=(ans+(work1(j)-work1(i-)+mod)%mod*work2(n/i))%mod;
     }
     printf("%lld",ans);
     return ;
 }