正题

题目链接:https://www.luogu.com.cn/problem/P4213

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题目大意

给出$n$，求$$\sum _{i=1}^n\phi (i)$$
和
$$\sum _{i=1}^n\mu (i)$$

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解题思路

考虑如何将$\phi$卷起来，我们可以乘上一个函数$I(I(x)=1)$，就有
$(\phi \ast I)(n)={\sum }_{d|n}\phi (d)I(\frac{n}{d})$
变成杜教筛的式子之后就是
$$I(1)S(n)=\sum _{i=1}^n(\phi \ast I)(n)-\sum _{i=2}^{n}I(i)\ast S(\lfloor \frac{n}{i}\rfloor )$$

整除分块计算下去就好了。

$\mu$同理

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$code$

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#define ll long long
using namespace std;
const ll N=1e7+1;
ll T,n,cnt,mu[N],phi[N],pri[N];
bool vis[N];
map<ll,ll> sp,sm; 
void prime(){mu[1]=phi[1]=1;for(ll i=2;i<N;i++){if(!vis[i])pri[++cnt]=i,mu[i]=-1,phi[i]=i-1;for(ll j=1;j<=cnt&&pri[j]*i<N;j++){vis[pri[j]*i]=1;if(i%pri[j]==0){phi[i*pri[j]]=phi[i]*pri[j];break;}phi[i*pri[j]]=phi[pri[j]]*phi[i];mu[i*pri[j]]=-mu[i];}}for(ll i=1;i<N;i++)phi[i]+=phi[i-1],mu[i]+=mu[i-1];return;
}
ll GetSphi(ll n){if(n<N)return phi[n];if(sp[n])return sp[n];ll rest=n*(n+1ll)/2ll;for(ll l=2ll,r;l<=n;l=r+1ll)r=n/(n/l),rest-=(r-l+1ll)*GetSphi(n/l);return (sp[n]=rest);
}
ll GetSmul(ll n){if(n<N)return mu[n];if(sm[n])return sm[n];ll rest=1ll;for(ll l=2ll,r;l<=n;l=r+1ll)r=n/(n/l),rest-=(r-l+1ll)*GetSmul(n/l);return (sm[n]=rest);
}
int main()
{scanf("%lld",&T);prime();while(T--){scanf("%lld",&n);printf("%lld %lld\n",GetSphi(n),GetSmul(n));}
}