JZOJ 5539 psy

psy

Description

f(i)=i∗[∑d|n10d∗μ(id)] f ( i ) = i ∗ [ ∑ d | n 10 d ∗ μ ( i d ) ]
求 ∑ni=1f(i) ∑ i = 1 n f ( i )

Data Constraint

n n <= 1010 10 10

Solution

首先显然有

Answer=∑i=1ni∗10i∑j=1niμ(j)∗j A n s w e r = ∑ i = 1 n i ∗ 10 i ∑ j = 1 n i μ ( j ) ∗ j

首先设 g(i) g ( i ) = ∑ij=1j∗10j ∑ j = 1 i j ∗ 10 j ，这个挺容易求，像求等比数列一样裂项求即可，求得

g(n)=n∗10n+1−10n+1−199 g ( n ) = n ∗ 10 n + 1 − 10 n + 1 − 1 9 9

对答案式子分块求，现在只需要求 w(i)=∑ij=1μ(j)∗j w ( i ) = ∑ j = 1 i μ ( j ) ∗ j
考虑到有 w∗ID=[n=1] w ∗ I D = [ n = 1 ] ，于是可以用杜教筛求解，得

w(n)=1−∑i=2ni∗w(ni) w ( n ) = 1 − ∑ i = 2 n i ∗ w ( n i )

接下来只需预处理出前 n23 n 2 3 的 μ(i)∗i μ ( i ) ∗ i 即可。

Code

#include
#include
#include
#include
#define fo(i,j,l) for(int i=j;i<=l;++i)
#define fd(i,j,l) for(int i=j;i>=l;--i)

using namespace std;
typedef long long ll;
const ll K=52e5,N=521e4,U=2e5,mo=1e9+7;

int ss[K],oo;
bool bz[N];
int mu[N];
ll qz[N],keep[U],n9,n;

inline ll ksm(ll o,ll t)
{
    ll y=1;
    for(;t;t>>=1,o=o*o%mo)
    if(t&1)y=y*o%mo;
    return y;
}

inline ll op(ll t)
{return ((t*ksm(10,t+1)-(ksm(10,t+1)-1)*n9)%mo+mo+1)*n9%mo;}

inline ll ext(ll l,ll r)
{
    ll o1=l+r,o2=r-l+1;
    if(!(o1&1))o1>>=1;else o2>>=1;
    return (o1%mo)*(o2%mo)%mo;
}

ll S(ll p)
{
    if(p<=K)return qz[p];
    else if(keep[n/p]!=-1)return keep[n/p];
    ll l=2,r,lj=0;
    while(l<=p){
        r=p/(p/l); ll o=p/l;
        lj=(lj+ext(l,r)*S(o))%mo;
        l=r+1;
    }
    keep[n/p]=(1-lj+mo)%mo; return keep[n/p];
}

int main()
{
    n9=ksm(9,mo-2); 
    fo(i,2,K){
        if(!bz[i])ss[++oo]=i,mu[i]=-1;
        fo(j,1,oo)if(ss[j]*i<=K){
            bz[ss[j]*i]=true;
            mu[ss[j]*i]=-mu[i];
            if(i%ss[j]==0){
                mu[ss[j]*i]=0;
                break;
            }
        }else break;
    }
    mu[1]=1;
    fo(i,1,U-2)keep[i]=-1;
    fo(i,1,K)qz[i]=(qz[i-1]+mu[i]*(i%mo)+mo)%mo;
    freopen("psy.in","r",stdin);
    freopen("psy.out","w",stdout);
    cin>>n;
    ll l=1,r,ans=0;
    while(l<=n){
        r=n/(n/l);
        ans=(ans+(S(r)-S(l-1)+mo)*op(n/l))%mo;
        l=r+1;
    }
    cout<