[bzoj4176]Lucas的数论

Orz w_yqts

Orz xudyh

Orz popoqqq

求∑nx=1∑ny=1d(ij)

n<=109

首先证明d(nm)=∑i|n∑j|m[gcd(i,j)=1]

若有n=n′∗pk1,m=m′∗pk2

则素数p对左式d(nm)的贡献为k1+k2+1

对于右式则有(pk1,1),(pk1−1,1),...,(p,1),(1,1),(1,p),...,(1,pk2−1),(1,pk2)

贡献也是k1+k2+1

所以等号成立

直接把这个式子代进去反演一下就好了……?



μ的部分用杜教筛做…

#include <bits/stdc++.h>
using namespace std;
#define p 1000000007
#define ll long long
#define N 2500000
map <int,int> f;
int pn,pr
,flag
,miu
;
int n;
inline int sqr(int x)
{
return (ll)(x)*x%p;
}
inline int calc(int n)
{
int res=0;
for (int i=1,pos;i<=n;i=pos+1)
{
pos=n/(n/i);
res+=(ll)(pos-i+1)*(n/i)%p;
res%=p;
}
return res;
}
inline int calcmiu(int n)
{
if (n<N) return miu
;
if (f
) return f
;
int res=1;
for (int i=2,pos;i<=n;i=pos+1)
{
pos=n/(n/i);
res-=calcmiu(n/i)*(pos-i+1);
}
return f
=res;
}
void init()
{
miu[1]=flag[1]=1;
for (int i=2;i<N;++i)
{
if (!flag[i]) pr[++pn]=i,miu[i]=-1;
for (int j=1;j<=pn && i*pr[j]<N;++j)
{
flag[i*pr[j]]=1;
if (i%pr[j]==0) {miu[i*pr[j]]=0;break;}
miu[i*pr[j]]=-miu[i];
}
}
for (int i=1;i<N;++i) miu[i]+=miu[i-1];
}
inline int mobius(int n)
{
int res=0;
for (int i=1,pos;i<=n;i=pos+1)
{
pos=n/(n/i);
res+=(ll)(calcmiu(pos)-calcmiu(i-1))*sqr(calc(n/i))%p;
res%=p;
}
return (res+p)%p;
}
main()
{
init();
cin>>n;
//for (int i=1;i<=n;++i) cout<<miu[i]<<' ';cout<<endl;
cout<<mobius(n)<<endl;
}