51nod 2026 Gcd and Lcm 杜教筛

Description

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定义

$$f(n)=\sum_{d|n}\mu(d)\cdot d$$
求 $$\sum_{i=1}^{n}\sum_{j=1}^n f(lcm(i,j))\cdot f(gcd(i,j))$$

Solution

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我们知道$lcm(i,j)=\frac{i\cdot j}{gcd(i,j)}$且$gcd(\frac{i}{gcd(i,j},j)=1$
那么可以拆成

$$ans=\left(\sum_{i=1}^n f(i)\right)^2$$
考虑怎么求f的前缀和 $$S(n)=\sum_{i=1}^n\sum_{d|i}\mu(d)\cdot d=\sum_{d=1}^n \mu(d)\cdot d \cdot\lfloor\frac{n}{d}\rfloor$$
现在考虑怎么求 $f(n)=\mu(n)\cdot n$的前缀和，这是一个很经典的杜教筛问题，然后就A了

Code

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#include <stdio.h>
#include <string.h>
#include <map>
#define rep(i,st,ed) for (int i=st;i<=ed;++i)

typedef long long LL;
const int MOD=1000000007;
const int N=10000005;

std:: map <LL,LL> map;

bool not_prime[N];

LL prime[N/10],mu[N];

void pre_work(int n) {
    mu[1]=1;
    rep(i,2,n) {
        if (!not_prime[i]) {
            prime[++prime[0]]=i;
            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) {
                mu[i*prime[j]]=0;
                break;
            }
            mu[i*prime[j]]=-mu[i];
        }
    }
    rep(i,1,n) mu[i]=(mu[i-1]+mu[i]*i%MOD+MOD)%MOD;
}

LL S(LL n) {
    return ((n+1)*n/2)%MOD;
}

LL solve(LL n) {
    if (n<=N) return mu[n];
    if (map[n]) return map[n];
    LL ret=0;
    for (LL i=2,j;i<=n;i=j+1) {
        j=n/(n/i);
        LL tmp=(S(j)-S(i-1)+MOD)%MOD;
        ret=(ret+tmp*solve(n/i)%MOD)%MOD;
    }
    ret=(1+MOD-ret)%MOD;
    return map[n]=ret;
}

int main(void) {
    pre_work(N);
    LL n,ans=0; scanf("%lld",&n);
    for (int i=1,j;i<=n;i=j+1) {
        j=n/(n/i);
        LL tmp=solve(j)-solve(i-1);
        ans=(ans+tmp*(n/i)%MOD)%MOD;
    } ans=(ans*ans)%MOD;
    printf("%lld\n", ans);
    return 0;
}