bzoj4176. Lucas的数论 杜教筛

标签：mmx   ret   bit   std   base   inline   span   make   space   

题意:求\(\sum_{i=1}^n\sum_{j=1}^nd(ij),d是约数个数函数\)
题解:首先有一个结论\(d(ij)=\sum_{x|i}\sum_{y|j}[(i,j)==1]\)
那么\(\sum_{i=1}^n\sum_{j=1}^n\sum_{x|i}\sum_{y|j}\sum_{d|(i,j)}\mu(d)\)
枚举d,\(\sum_{i=1}^n\sum_{j=1}^n\sum_{d|i,d|j}\mu(d)d(\frac{i}{d})d(\frac{j}{d})=\sum_{d=1}^n\mu(d)\sum_{d|i}\sum_{d|j}d(\frac{i}{d})d(\frac{j}{d})=\sum_{d=1}^n\mu(d)\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}d(i)\sum_{j=1}^{\lfloor \frac{n}{d} \rfloor}d(j)\)
这里我们需要分块求\(\mu\)和\(d\)的前缀和,\(\mu\)很好求,对于d,我们考虑\(d=I*I\),考虑把杜教筛中的\(g(x)=\mu\),\(I*I*\mu=I*e=e\),那么\(S(n)=\sum_{i=1}^n1-\sum_{d=2}\mu(d)S(\lfloor \frac{n}{d} \rfloor)\)

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
template<typename T>
inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>
inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=5000000+10,maxn=3000000+10,inf=0x3f3f3f3f;

int prime[N],cnt;
ll d[N],num[N],mu[N];
bool mark[N];
map<ll,ll>dd,muu;
void init()
{
    mu[1]=d[1]=1;
    for(int i=2;i<N;i++)
    {
        if(!mark[i])prime[++cnt]=i,mu[i]=-1,d[i]=2,num[i]=1;
        for(int j=1;j<=cnt&&i*prime[j]<N;j++)
        {
            mark[i*prime[j]]=1;
            if(i%prime[j]==0)
            {
                mu[i*prime[j]]=0;
                num[i*prime[j]]=num[i]+1;
                d[i*prime[j]]=d[i]/num[i*prime[j]]*(num[i*prime[j]]+1);
                break;
            }
            mu[i*prime[j]]=-mu[i];
            d[i*prime[j]]=d[i]<<1;
            num[i*prime[j]]=1;
        }
    }
    for(ll i=1;i<N;i++)
    {
        add(d[i],d[i-1]);
        mu[i]=(mu[i]+mod)%mod;
        add(mu[i],mu[i-1]);
    }
}
ll getmu(ll n)
{
    if(n<N)return mu[n];
    if(muu.find(n)!=muu.end())return muu[n];
    ll ans=1;
    for(ll i=2,j;i<=n;i=j+1)
    {
        j=n/(n/i);
        ll te=(j-i+1)%mod;
        sub(ans,te*getmu(n/i)%mod);
    }
    return muu[n]=ans;
}
ll getd(ll n)
{
    if(n<N)return d[n];
    if(dd.find(n)!=dd.end())return dd[n];
    ll ans=n%mod;
    for(ll i=2,j;i<=n;i=j+1)
    {
        j=n/(n/i);
        ll te=(getmu(j)-getmu(i-1)+mod)%mod;
        sub(ans,te*getd(n/i)%mod);
    }
    return dd[n]=ans;
}
int main()
{
    init();
    ll n;scanf("%lld",&n);
    ll ans=0;
    for(ll i=1,j;i<=n;i=j+1)
    {
        j=n/(n/i);
        ll te=(getmu(j)-getmu(i-1)+mod)%mod;
        add(ans,te*getd(n/i)%mod*getd(n/i)%mod);
    }
    printf("%lld\n",ans);
    return 0;
}
/********************

********************/

bzoj4176. Lucas的数论 杜教筛

标签：mmx   ret   bit   std   base   inline   span   make   space   

原文地址：https://www.cnblogs.com/acjiumeng/p/9744905.html