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  • 【LOJ6202】—叶氏筛法(min_25筛)

    传送门


    和素数个数差不多
    f(p)=1f(p)=1改成f(p)=pf(p)=p就可以了

    不会min25min_{25}的看这个

    #include<bits/stdc++.h>
    using namespace std;
    const int RLEN=1<<20|1;
    inline char gc(){
    	static char ibuf[RLEN],*ib,*ob;
    	(ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    	return (ib==ob)?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 pb push_back
    #define ll long long
    #define cs const
    #define re register
    #define pii pair<int,int>
    #define fi first
    #define se second
    #define db double
    int mod;
    inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
    inline void Add(int &a,int b){(a+=b)>=mod?a-=mod:0;}
    inline int dec(int a,int b){return a>=b?a-=b:a-b+mod;}
    inline void Dec(int &a,int b){(a-=b)<0?a+=mod:0;}
    inline int mul(int a,int b){return 1ll*a*b>=mod?1ll*a*b%mod:a*b;}
    inline void Mul(int &a,int b){a=mul(a,b);}
    inline int ksm(int a,int b,int res=1){
    	for(;b;b>>=1,Mul(a,a))(b&1)&&(res=mul(res,a));return res;
    }
    cs int N=1000006;
    cs db eps=1e-10;
    inline db C(ll x){return (double)(x+1)*(double)x/2;}
    ll l,r;
    double f1[N],f2[N];
    inline db query(ll k){
    	if(!k)return 0;
    	int lim=sqrt(k);
    	for(int i=1;i<=lim;i++)f1[i]=C(i),f2[i]=C(k/i);
    	for(int p=2;p<=lim;p++){
    		if(fabs(f1[p]-f1[p-1])<=eps)continue;
    		cs double xx=p;
    		for(int i=1;i<=lim/p;i++)f2[i]-=(f2[i*p]-f1[p-1])*xx;
    		for(int i=lim/p+1;1.0*p*p*i<=k&&i<=lim;i++)f2[i]-=(f1[k/i/p]-f1[p-1])*xx;
    		for(int i=lim;i>=1.0*p*p;i--)f1[i]-=(f1[i/p]-f1[p-1])*xx;
    	}
    	return f2[1];
    }
    int main(){
    	scanf("%lld%lld",&l,&r);
    	printf("%.0lf",query(r)-query(l-1));
    }
    
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  • 原文地址:https://www.cnblogs.com/stargazer-cyk/p/12328749.html
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