牛客练习赛76 F phi and phi（莫比乌斯反演）

牛客练习赛76 F phi and phi（莫比乌斯反演）

题解：

$前置知识：$$$\sum _{i=1}^n\sum _{j=1}^n\varphi (ij)=\sum _{i=1}^n\sum _{j=1}^n\frac{\varphi (i)\varphi (j)gcd(i,j)}{\varphi (gcd(i,j))}$$
$证明：$
$$n=\prod _{i=1}^m{p}_i^{{\alpha }_i}$$$$\varphi (n)=n\prod _{i=1}^m(1-\frac{1}{p_i})=n\prod _{p|n}(1-\frac{1}{p})$$$$\varphi (i)=i\prod _{p_1|i}(1-\frac{1}{p_1})，\varphi (j)=j\prod _{p_2|j}(1-\frac{1}{p_2})$$$$\varphi (ij)=ij\prod _{p|ij}(1-\frac{1}{p})=\frac{i{\prod }_{p_1|i}(1-\frac{1}{p_1})\times j{\prod }_{p_2|j}(1-\frac{1}{p_2})}{{\prod }_{p_3|i,p_3|j}(1-\frac{1}{p_3})}$$$$=\frac{i{\prod }_{p_1|i}(1-\frac{1}{p_1})\times j{\prod }_{p_2|j}(1-\frac{1}{p_2})}{gcd(i,j){\prod }_{p_3|gcd(i,j)}(1-\frac{1}{p_3})}gcd(i,j)$$$$=\frac{\varphi (i)\varphi (j)gcd(i,j)}{\varphi (gcd(i,j))}$$
$证完后就可以用上面的公式来推式子了：$
$$\sum _{i=1}^n\sum _{j=1}^n\varphi (ij)\varphi (gcd(i,j))=\sum _{i=1}^n\sum _{j=1}^n\varphi (i)\varphi (j)gcd(i,j)$$
$枚举gcd(i,j)：$
$$\sum _{d=1}^n\sum _{i=1}^n\sum _{j=1}^n\varphi (i)\varphi (j)d[gcd(i,j)==d]$$$$=\sum _{d=1}^n\sum _{i=1}^{n/d}\sum _{j=1}^{n/d}\varphi (id)\varphi (jd)d[gcd(i,j)==1]$$$$=\sum _{d=1}^{n}d\sum _{i=1}^{n/d}\sum _{j=1}^{n/d}\varphi (id)\varphi (jd)\sum _{x|d}\mu (x)$$$$=\sum _{d=1}^{n}d\sum _{x=1}^{n/d}\mu (x)\sum _{i=1}^{n/dx}\sum _{j=1}^{n/dx}\varphi (idx)\varphi (jdx)$$
$令T=dx：$
$$\sum _{T=1}^n\sum _{d|T}d\mu (\frac{T}{d})(\sum _{i=1}^{n/T}\varphi (iT){)}^2$$
$前面一部分用狄利克雷卷积可以求出来：\mu \ast id=\varphi ，于是就可以进一步化简了：$
$$\sum _{T=1}^n\varphi (T)(\sum _{i=1}^{n/T}\varphi (iT){)}^2$$
$令f(d,n)={\sum }_{i=1}^{n/d}\varphi (id)，我们发现对于n\in [i\times d,(i+1)\times d)，\varphi (d)\times f(d,n{)}^2的贡献是相同的，即这一区间都要加上这一贡献，所以就可以用差分+前缀和来求了，时间复杂度大概为O(nlogn).$

代码：

#include
using namespace std;
using ll = long long;
const int N = 1e6 + 5, mod = 1e9 + 7;
int n, phi[N], p[N], cnt, ans[N];
bool vis[N];
void get() {
    phi[1] = 1;
    for(int i = 2; i <= n; i++) {
        if(!vis[i]) {
            p[cnt++] = i;
            phi[i] = i - 1;
        }
        for(int j = 0; p[j] <= n / i; j++) {
            vis[i * p[j]] = 1;
            if(i % p[j] == 0) {
                phi[i * p[j]] = phi[i] * p[j];
                break;
            }
            phi[i * p[j]] = phi[i] * (p[j] - 1);
        }
    }
}
int main() {
    scanf("%d", &n); get();
    for(int i = 1; i <= n; i++) {
        ll sum = 0;
        for(int j = 1; j * i <= n; j++) {
            int x = i * j, y = i * (j + 1);
            sum = (sum + phi[x]) % mod;
            ll tmp = sum * sum % mod * phi[i] % mod;
            ans[x] = (ans[x] + tmp) % mod;
            if(y <= n) ans[y] = (ans[y] - tmp + mod) % mod;
        }
    }
    for(int i = 1; i <= n; i++) {
        ans[i] = (ans[i] + ans[i - 1]) % mod;
        printf("%d\n", ans[i]);
    }
	return 0;
}