bzoj3944sum

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杜教筛
可以知道$\sum_{i=1}^{n}\sum_{d|i}\varphi(d)==n*(n+1)/2$
把式子转化成$\sum_{i=1}^n\sum_{d=1}^{\lfloor n/i \rfloor}\varphi(d)$
这样的话每个$\varphi(d)$都被算了多次
容斥一下
式子变为$n*(n+1)/x-\sum_{i=2}^n\sum_{d=1}^{\lfloor n/i \rfloor}\varphi(d)$
可以看出后面其实是一个规模较小的问题
直接上杜教筛

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=2000000+10;
typedef long long ll;
ll phi[maxn];
ll mu[maxn];
ll prime[maxn];
ll ps=0;
ll q[maxn],p[maxn];
int vis[maxn];
ll m=2000000;
ll n;
int done[maxn];
inline void get_prime(){
    mu[1]=1;
    phi[1]=1;
    for(int i=2;i<=m;i++){
        if(!vis[i]){
            prime[++ps]=i;
            mu[i]=-1;
            phi[i]=i-1;
        }
        for(int j=1;j<=ps;j++){
            if(prime[j]*i>m)
                break;
            vis[i*prime[j]]=1;
            if(i%prime[j]==0){
                mu[i*prime[j]]=0;
                phi[i*prime[j]]=phi[i]*prime[j];
                break;
            }
            else {
                mu[i*prime[j]]=-mu[i];
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
            }
        }
    }
    for(int i=2;i<=m;i++)
        mu[i]+=mu[i-1],phi[i]+=phi[i-1];
}
inline ll get_phi(ll x){
    if(x<=m)
        return phi[x];
    return q[n/x];
}
inline ll get_mu(ll x){
    if(x<=m)
        return mu[x];
    return p[n/x];
}
inline void work(ll x){
    if(x<=m)
        return ;
    //printf("%lld\n",x);
    ll t=n/x;
    if(done[t])
        return ;
    done[t]=1;
    ll nw=2;
    p[t]=1,q[t]=(x*(x+1ll)/2);
    while(nw<=x){
        ll tmp=x/nw;
        ll r=x/tmp;
        work(tmp);
        p[t]-=get_mu(tmp)*(r-nw+1);q[t]-=get_phi(tmp)*(r-nw+1);
        nw=r+1;
    }
}
int main(){
    //freopen("a.in","r",stdin);
    //freopen("a.out","w",stdout);
    ll T;
    scanf("%lld",&T);
    get_prime();
    while(T--){
        scanf("%lld",&n);
        memset(done,0,sizeof(done));
        if(n<=m)
            printf("%lld %lld\n",phi[n],mu[n]);
        else {
            work(n);
            printf("%lld %lld\n",q[1],p[1]);
        }
    }
return 0;
}