【LOJ #572】【LibreOJ Round #11】—Misaka Network 与求和（min_25筛+杜教筛）

传送门

首先可以随便莫反一波，可以得到

$ans=sum_{d=1}^{n}f(d)^ksum_{i=1}^{frac n d}sum_{j=1}^{frac n d}gcd(i,j)=1$

由$phi$的定义可以得到
$ans=sum_{d=1}^{n}f(d)^kg(frac n d)$
其中$g(n)=sum_{i=1}^{n}2phi(i)-1$

次大质因子的$k$次幂可以类似$Sanrd$的做法用$min_{25}$筛

后面可以用杜教筛筛一下

#include<bits/stdc++.h>
using namespace std;
const int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ob==ib)?EOF:*ib++;
}
#define gc getchar
inline int read(){
    char ch=gc();
    int res=0,f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
#define ll long long
#define re register
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define cs const
#define bg begin
#define uint unsigned int
inline void chemx(int &a,int b){a<b?a=b:0;}
inline void chemn(int &a,int b){a>b?a=b:0;}
inline uint ksm(uint x,int b){
	uint res=1;
	for(;b;b>>=1,x=x*x)if(b&1)res=res*x;
	return res;
}
cs int N=50004;
int n,k;
namespace F{
	uint f1[N],f2[N],pt[N];
	int lim,pr[N],tot;
	inline void init(int n){
		lim=sqrt(n),tot=0;
		for(int i=1;i<=lim;i++)f1[i]=i-1,f2[i]=n/i-1;
		for(int p=2;p<=lim;p++){
			if(f1[p]==f1[p-1])continue;
			pr[++tot]=p;
			for(int i=1;i<=lim/p;i++)f2[i]-=(f2[i*p]-f1[p-1]);
			for(int i=lim/p+1;1ll*i*p*p<=n&&i<=lim;i++)f2[i]-=(f1[n/i/p]-f1[p-1]);
			for(int i=lim;i>=1ll*p*p;i--)f1[i]-=(f1[i/p]-f1[p-1]);
		}
		for(int i=1;i<=tot;i++)pt[i]=ksm(pr[i],k);
	}
	inline uint F(int x){
		return x>lim?f2[n/x]:f1[x];
	}
	unordered_map<int,uint> mp[N];
	inline uint sieve(int res,int pos){
		if(mp[pos].count(res))return mp[pos][res];
		uint anc=0;
		if(res<pr[pos])return 0;
		for(int i=pos;i<=tot;i++){
			if(1ll*pr[i]*pr[i]>res)break;
			for(ll now=pr[i],xs=1;now<=res;now*=pr[i],xs++)
			if(1ll*now*pr[i]<=res)anc+=sieve(res/now,i+1),anc+=pt[i]*(F(res/now)-i+1);
		}
		return mp[pos][res]=anc;
	}
	inline uint query(int x){
		return sieve(x,1)+F(x);
	}
}
cs int mod=19260817;
struct Map{
	int key[mod];uint val[mod];
	Map(){memset(key,-1,sizeof(key));}
	int posit(int x)cs{
		int h=x%mod;
		while(key[h]!=-1&&key[h]!=x)h=h==mod-1?0:h+1;
		return h;
	}
	uint &operator[](int x){
		int p=posit(x);
		if(key[p]==-1){
			key[p]=x,val[p]=0;
		}
		return val[p];
	}
	inline bool count(int x)cs{return key[posit(x)]==x;}
};
namespace G{
	cs int N=2000006;
	Map mp;
	uint phi[N],pr[N],tot,vis[N];
	inline void init(){
		cs int len=N-6;
		phi[1]=1;
		for(int i=2;i<=len;i++){
			if(!vis[i])pr[++tot]=i,phi[i]=i-1;
			for(int j=1;j<=tot&&i*pr[j]<=len;j++){
				vis[i*pr[j]]=1;
				if(i%pr[j]==0){phi[i*pr[j]]=phi[i]*pr[j];break;}
				phi[i*pr[j]]=phi[i]*phi[pr[j]];
			}
		}
		for(int i=1;i<=len;i++)phi[i]+=phi[i-1];
	}
	inline uint S(uint x){
		return (uint)(1ll*x*(x+1)/2);
	}
	inline uint query(int x){
		if(x<=N-6)return phi[x];
		if(mp.count(x))return mp[x];
		uint res=S(x);
		for(int i=2,nxt;i<=x;i=nxt+1){
			nxt=x/(x/i);
			res-=(nxt-i+1)*query(x/i);
		}
		return mp[x]=res;
	}
}
uint res=0;
int main(){
	n=read(),k=read();
	F::init(n);G::init();
	for(int i=1,nxt;i<=n;i=nxt+1){
		nxt=n/(n/i);
		uint a=G::query(n/i)*2-1,b=(F::query(nxt)-F::query(i-1));
		res+=a*b;
	}
	cout<<res;
}