【刷题】BZOJ 3944 Sum

Description

Input

一共T+1行

第1行为数据组数T(T<=10)

第2~T+1行每行一个非负整数N，代表一组询问

Output

一共T行，每行两个用空格分隔的数ans1,ans2

Sample Input

6

1

2

8

13

30

2333

Sample Output

1 1

2 0

22 -2

58 -3

278 -3

1655470 2

Solution

杜教筛裸题啊

对于 \(\mu\) ，利用与它有关的卷积 \(\mu*1=e\) ，杜教筛式子为 \(S(n)=1-\sum_{i=2}^nS(\lfloor\frac{n}{i}\rfloor)\)

对于 \(\varphi\) ，利用与它有关的卷积 \(\varphi*1=id\) ，杜教筛式子为 \(S(n)=\sum_{i=1}^ni-\sum_{i=2}^nS(\lfloor\frac{n}{i}\rfloor)\)

#include<bits/stdc++.h>
#define ui unsigned int
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
const int MAXN=3000000+10;
int t,n,vis[MAXN],prime[MAXN],cnt,phi[MAXN],mu[MAXN],smu[MAXN];
ll sphi[MAXN];
std::map<int,int> M;
std::map<int,ll> P;
namespace IO
{
    const ui Buffsize=1<<15,Output=1<<23;
    static char Ch[Buffsize],*S=Ch,*T=Ch;
    inline char getc()
	{
		return((S==T)&&(T=(S=Ch)+fread(Ch,1,Buffsize,stdin),S==T)?0:*S++);
	}
    static char Out[Output],*nowps=Out;
    inline void flush(){fwrite(Out,1,nowps-Out,stdout);nowps=Out;}
    template<typename T>inline void read(T&x)
	{
		x=0;static char ch;T f=1;
		for(ch=getc();!isdigit(ch);ch=getc())if(ch=='-')f=-1;
		for(;isdigit(ch);ch=getc())x=x*10+(ch^48);
		x*=f;
	}
	template<typename T>inline void write(T x,char ch='\n')
	{
		if(!x)*nowps++='0';
		if(x<0)*nowps++='-',x=-x;
		static ui sta[111],tp;
		for(tp=0;x;x/=10)sta[++tp]=x%10;
		for(;tp;*nowps++=sta[tp--]^48);
		*nowps++=ch;
	}
}
using namespace IO;
template<typename T> inline void chkmin(T &x,T y){x=(y<x?y:x);}
template<typename T> inline void chkmax(T &x,T y){x=(y>x?y:x);}
template<typename T> inline T min(T x,T y){return x<y?x:y;}
template<typename T> inline T max(T x,T y){return x>y?x:y;}
inline void init()
{
	memset(vis,1,sizeof(vis));
	vis[0]=vis[1]=0;
	phi[1]=mu[1]=1;
	for(register int i=2;i<MAXN;++i)
	{
		if(vis[i])
		{
			prime[++cnt]=i;
			mu[i]=-1;phi[i]=i-1;
		}
		for(register int j=1;j<=cnt&&i*prime[j]<MAXN;++j)
		{
			vis[i*prime[j]]=0;
			if(i%prime[j])
			{
				mu[i*prime[j]]=mu[i]*mu[prime[j]];
				phi[i*prime[j]]=phi[i]*phi[prime[j]];
			}
			else
			{
				phi[i*prime[j]]=phi[i]*prime[j];
				break;
			}
		}
	}
	for(register int i=1;i<MAXN;++i)smu[i]=smu[i-1]+mu[i],sphi[i]=sphi[i-1]+phi[i];
}
inline int Smu(ll x)
{
	if(x<MAXN)return smu[x];
	if(M.find(x)!=M.end())return M[x];
	int res=0;
	for(register ll i=2;;)
	{
		if(i>x)break;
		ll j=x/(x/i);
		res+=(j-i+1)*Smu(x/i);
		i=j+1;
	}
	return M[x]=1-res;
}
inline ll Sphi(ll x)
{
	if(x<MAXN)return sphi[x];
	if(P.find(x)!=P.end())return P[x];
	ll res=0;
	for(register ll i=2;;)
	{
		if(i>x)break;
		ll j=x/(x/i);
		res+=1ll*(j-i+1)*Sphi(x/i);
		i=j+1;
	}
	return P[x]=1ll*(x+1)*x/2-res;
}
int main()
{
	init();read(t);
	while(t--)read(n),write(Sphi(n),' '),write(Smu(n),'\n');
	flush();
	return 0;
}