bzoj3944 Sum

Description

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给定一个正整数$n<=2^{31}-1$，求$\sum^{n}_{i=1}{\mu{(i)}}$和$\sum^{n}_{i=1}{\varphi(i)}$，多组询问

Solution

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杜教筛裸题
先考虑μ的前缀和怎么求。我们知道$\sum_{i|n}\mu(i)=[n==1]$，那么不妨设$S(n)=\sum^{n}_{i=1}\mu(i)$，有$\sum_{i=1}^{n}\sum_{d|i}{\mu(d)}=\sum_{i=1}^{n}\sum_{d|i}{\mu(\frac{i}{d})}=\sum_{d=1}^{n}\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}{\mu(i)}=\sum_{d=1}^{n}S(\lfloor \frac{n}{d} \rfloor)=1$
观察一下式子，发现当d=1时n/d=n，那么单独拉出来得到$S(n)=1-\sum_{d=2}^{n}S(\lfloor \frac{n}{d} \rfloor)$，显然这可以根号跳着做，一部分前缀和用线性筛，剩下的递推做即可

Code

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#include <stdio.h>
#include <map>
typedef long long LL;
const int N=5000000;
std:: map<LL,LL>s1,s2;
bool not_prime[N+5];
LL phi[N+5],mu[N+5];
int prime[N+5];
void init() {
    phi[1]=mu[1]=1;
    for (int i=2;i<=N;i++) {
        if (!not_prime[i]) {
            prime[++prime[0]]=i;
            phi[i]=i-1;
            mu[i]=-1;
        }
        for (int j=1;i*prime[j]<=N&&j<=prime[0];j++) {
            not_prime[i*prime[j]]=1;
            if (i%prime[j]==0) {
                phi[i*prime[j]]=phi[i]*prime[j];
                mu[i*prime[j]]=0;
                break;
            }
            phi[i*prime[j]]=phi[i]*phi[prime[j]];
            mu[i*prime[j]]=-mu[i];
        }
    }
    for (int i=1;i<=N;i++) phi[i]+=phi[i-1];
    for (int i=1;i<=N;i++) mu[i]+=mu[i-1];
}
LL get_mu(LL n) {
    if (n<=N) return mu[n];
    if (s1.count(n)) return s1[n];
    LL ret=1;
    for (LL i=2,j;i<=n;i=j+1) {
        j=n/(n/i);
        ret-=(j-i+1)*get_mu(n/i);
    }
    s1[n]=ret;
    return ret;
}
LL get_phi(LL n) {
    if (n<=N) return phi[n];
    if (s2.count(n)) return s2[n];
    LL ret=(n+1)*n/2;
    for (LL i=2,j;i<=n;i=j+1) {
        j=n/(n/i);
        ret-=(j-i+1)*get_phi(n/i);
    }
    s2[n]=ret;
    return ret;
}
int main(void) {
    init();
    int T; scanf("%d",&T);
    while (T--) {
        LL x; scanf("%lld",&x);
        printf("%lld %lld\n", get_phi(x),get_mu(x));
    }
    return 0;
}