题意
分析
直接用线性筛来求肯定不行,我们考虑别的方法。
先来考虑ans2吧
设S(n)=∑ni=1μ(i)
根据μ的性质∑d|iμ(d)=0(i>1)或1(i=1)可得
∑ni=1∑d|iμ(d)=1
∑ni=1∑d|iμ(d)=∑ni=1∑d|iμ(d/i)=∑nd=1∑⌊nd⌋i=1μ(i)=∑nd=1S(⌊nd⌋)
从而可以得出S(n)=1−∑nd=2S(⌊nd⌋)
那么可以通过预处理[1,5000000]的ϕ和μ,大于5000000的用递推即可
ans1的话就根据∑d|iϕ(d)=i同理即可得到答案。
代码
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<map>
#define LL long long
#define M 5000000
using namespace std;
LL mu[M+5],phi[M+5];
bool not_prime[M+5];
int prime[M],tot;
map <int,LL> w1,w2;
void prework(int n)
{
mu[1]=phi[1]=1;
for (int i=2;i<=n;i++)
{
if (!not_prime[i])
{
prime[++tot]=i;
mu[i]=-1;
phi[i]=i-1;
}
for (int j=1;j<=tot&&i*prime[j]<=n;j++)
{
not_prime[i*prime[j]]=1;
if (i%prime[j]==0)
{
mu[i*prime[j]]=0;
phi[i*prime[j]]=phi[i]*prime[j];
break;
}
mu[i*prime[j]]=-mu[i];
phi[i*prime[j]]=phi[i]*phi[prime[j]];
}
}
for (int i=1;i<=n;i++)
{
phi[i]+=phi[i-1];
mu[i]+=mu[i-1];
}
}
LL get_phi(LL n)
{
if(n<=M) return phi[n];
if(w1.count(n)) return w1[n];
LL ans=(LL)n*(n+1)/2;
for (LL i=2,last;i<=n;i=last+1)
{
last=n/(n/i);
ans-=(last-i+1)*get_phi(n/i);
}
w1[n]=ans;
return ans;
}
LL get_mu(LL n)
{
if(n<=M) return mu[n];
if (w2.count(n)) return w2[n];
LL ans=1;
for (LL i=2,last;i<=n;i=last+1)
{
last=n/(n/i);
ans-=(last-i+1)*get_mu(n/i);
}
w2[n]=ans;
return ans;
}
int main()
{
prework(M);
int t;
scanf("%d",&t);
while (t--)
{
int n;
scanf("%d",&n);
printf("%lld ",get_phi(n));
printf("%lld\n",get_mu(n));
}
return 0;
}