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题意:

2019.02.12 bzoj3944: Sum(杜教筛)-LMLPHP


思路:直接上杜教筛。

知道怎么推导就很简单了,注意预处理的范围。

然后我因为预处理范围不对被zxyoi教育了(ldx你这个傻×两倍常数活该被卡TLE) 喜闻乐见

代码:

#include<bits/stdc++.h>
#define ri register int
using namespace std;
const int N=7500005,lim=7500000;
typedef long long ll;
namespace Sieve{
	int pri[N],tot=0,mu[N];
	ll phi[N];
	bool vis[N];
	map<int,int>mpa;
	map<int,ll>mpb;
	inline void init(){
		vis[1]=phi[1]=mu[1]=1;
		for(ri i=2;i<=lim;++i){
			if(!vis[i])pri[++tot]=i,phi[i]=i-1,mu[i]=-1;
			for(ri j=1;j<=tot&&i*pri[j]<=lim;++j){
				vis[i*pri[j]]=1;
				if(i==i/pri[j]*pri[j]){
					phi[i*pri[j]]=pri[j]*phi[i],mu[i*pri[j]]=0;
					break;
				}
				phi[i*pri[j]]=(pri[j]-1)*phi[i],mu[i*pri[j]]=-mu[i];
			}
		}
		for(ri i=2;i<=lim;++i)phi[i]+=phi[i-1],mu[i]+=mu[i-1];
	}
	inline int Mu(const int&x){
		if(x<=lim)return mu[x];
		if(mpa[x])return mpa[x];
		int ret=0;
		for(ri l=2,r;r<x&&l<=x;l=r+1)r=x/(x/l),ret+=Mu(x/l)*(r-l+1);
		return mpa[x]=1-ret;
	}
	inline ll Phi(const int&x){
		if(x<=lim)return phi[x];
		if(mpb[x])return mpb[x];
		ll ret=0;
		for(ri l=2,r;r<x&&l<=x;l=r+1)r=x/(x/l),ret+=Phi(x/l)*(r-l+1);
		return mpb[x]=(ll)x*((ll)x+1)/2-ret;
	}
}
int main(){
	freopen("lx.in","r",stdin);
	Sieve::init();
	int tt,n;
	scanf("%d",&tt);
	while(tt--)scanf("%d",&n),cout<<Sieve::Phi(n)<<' '<<Sieve::Mu(n)<<'\n';
	return 0;
}
04-30 01:21