题目

传送门

思路

对于第一个询问，令 \(g=I,h=id\)，则满足 \(h=\varphi*g\)，带入得

\[\text{Ans}_1(n)=\frac{n(n+1)}{2}-\sum_{i=2}^n\text{Ans}_1(\frac{n}{i}) \]

默认分数下取整.

对于第二个询问，令 \(g=I,h=\epsilon\)，则满足 \(h=\mu*g\)，带入杜教筛，有

\[\begin{aligned} \text{Ans}_2(n)&=\frac{\sum_{i=1}^nh(i)-\sum_{i=2}^ng(i)\text{Ans}_2(\frac{n}{i})}{g(1)} \\ &=1-\sum_{i=2}^n\text{Ans}_2(\frac{n}{i}) \end{aligned} \]

默认分数下取整.

然后打代码就可以了.

代码

const int maxn=2e6;

map<int,ll>mphi,mmu;
ll mu[maxn+5],phi[maxn+5];
int prime[maxn>>2],pcnt;

inline void sieve(){
    phi[1]=mu[1]=1;
    rep(i,2,maxn){
        if(!phi[i])mu[i]=-1,phi[i]=i-1,prime[++pcnt]=i;
        for(int j=1;j<=pcnt && i*prime[j]<=maxn;++j){
            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]];
        }
    }
    rep(i,2,maxn)mu[i]+=mu[i-1],phi[i]+=phi[i-1];
}

ll getphi(const int n){
    if(n<=maxn)return phi[n];
    else if(mphi[n])return mphi[n];
    ll ret=1ll*n*(1ll+n)/2;
    for(ll l=2,r;l<=n;l=r+1){
        r=n/(n/l);
        ret-=1ll*(r-l+1)*getphi(n/l);
    }
    // printf("Now n == %d phi == %lld\n",n,ret);
    return mphi[n]=ret;
}

ll getmu(const int n){
    if(n<=maxn)return mu[n];
    else if(mmu[n])return mmu[n];
    ll ret=1;
    for(ll l=2,r;l<=n;l=r+1){
        r=n/(n/l);
        // printf("When n == %d, l == %lld, r == %lld\n",n,l,r);
        ret-=(r-l+1)*getmu(n/l);
    }
    return mmu[n]=ret;
}

int n;

signed main(){
    // freopen("shit.out","w",stdout);
    phi[1]=mu[1]=1;
    sieve();
    rep(i,1,readin(1)){
        n=readin(1);
        writc(getphi(n),' ');
        writc(getmu(n),'\n');
    }
    return 0;
}

用到の一些小 \(\tt trick\)

除了杜教筛板子以外好像没什么 \(\tt trick\) 可说了......