bzoj3944 Sum Du Jiao Si

Template questions, pay attention to the techniques of Dujiao's screening formula:

1. Construct the sum function

2. Use the properties of multiplicative function divisors

3. Calculate the individual items by doing the difference

4. Contribution to conversion by multiples

5. Ask for quotient by block

code:

#include<iostream>
#include<cstdio>
#include<map>
#include<cmath>
using namespace std;
#define M 3000000
#define ll long long 
int tot,i,j,T,su[M+5];
ll ans,phe[M+5],n[999],N,mhe[M+5],mu[M+5],phi[M+5];
bool he[M+5];
map<int,ll>pma;
map<int,int>mma;
void eular()
{
	mu[1]=phi[1]=1;
	for(i=2;i<=N;i++)
	{
		if(!he[i])
		{
			su[++tot]=i;			
		    phi[i]=i-1;
		    mu[i]=-1;
		}
		for(j=1;j<=tot&&su[j]*i<=N;j++)
		{
		int k=su[j]*i;
		he[k]=1;
		if(i%su[j]==0)
		{phi[k]=su[j]*phi[i];
		mu[k]=0;
		break;
		}
		else
		{
			phi[k]=phi[i]*phi[su[j]];
			mu[k]=-mu[i];
		}	
		}
	}
	for(i=1;i<=N;i++)phe[i]=phe[i-1]+phi[i],mhe[i]=mhe[i-1]+mu[i];	
}
ll dfs1(ll o)
{
	if(o<=N)return phe[o]; 
	if(pma[o])return pma[o];
ll lin=o*(1+o)/2;
ll l;
for(l=2;l<=o;l++)
{
ll oo=o/l;
lin-=((o/oo)-l+1)*dfs1(oo) ;
l=o/oo;	
}	
pma[o]=lin;
return lin;
}
int dfs2(int o)
{	if(o<=N)return mhe[o]; 
	if(mma[o])return mma[o];
int lin=1;
unsigned int l;
for(l=2;l<=o;l++)
{
int oo=o/l;
lin-=((o/oo)-l+1)*dfs2(oo);
l=o/oo;	
}
mma[o]=lin;
return lin;
}
int main()
{scanf("%d",&T);
for(i=1;i<=T;i++)
{
scanf("%lld",&n[i]);
N=max(n[i],N);	
}
N=pow(N,2.000/3) ;
eular();
for(i=1;i<=T;i++)
{
printf("%lld ",dfs1(n[i]));
printf("%d\n",dfs2(n[i]));	
}
}