牛客练习赛76 phi and phi 莫比乌斯反演+差分

原题链接：https://ac.nowcoder.com/acm/contest/10845/F

题意

$$ans(n)=\sum _{i=1}^n\sum _{j=1}^n\varphi (ij)\varphi (gcd(i,j))$$

化简

先引入一个定理
$$\varphi (ij)=\frac{\varphi (i)\varphi (j)gcd(i,j)}{\varphi (gcd(i,j))}$$

然后就变成非常一般的形式

$$ans(n)=\sum _{i=1}^n\sum _{j=1}^n\varphi (i)\varphi (j)gcd(i,j)$$

$$=\sum _{d=1}^{n}d\sum _{i=1}^n\sum _{j=1}^n\varphi (i)\varphi (j)[gcd(i,j)=d]$$

$$=\sum _{d=1}^{n}d\sum _{i=1}^{n/d}\sum _{j=1}^{n/d}\varphi (id)\varphi (jd)[gcd(i,j)=1]$$

$$=\sum _{d=1}^{n}d\sum _{i=1}^{n/d}\sum _{j=1}^{n/d}\varphi (id)\varphi (jd)\sum _{t|i,t|j}\mu (t)$$

$$=\sum _{d=1}^{n}d\sum _{t=1}^{n/d}\mu (t)\sum _{i=1}^{n/td}\sum _{j=1}^{n/td}\varphi (idt)\varphi (jdt)$$

$$设T=td$$

$$=\sum _{T=1}^n\sum _{i=1}^{n/T}\varphi (iT)\sum _{j=1}^{n/T}\varphi (jT)\sum _{d|T}d\mu (\frac{T}{d})$$

$$设F(n,T)=\sum _{i=1}^{n/T}\varphi (iT)$$

$$=\sum _{T=1}^{n}F(n,d{)}^2\varphi (T)$$

接着就可以直接写了，同样使用差分，考虑F函数对不同n的贡献，把两端点算出来差分一下就可以了

Code

#include 
using namespace std;
#define fi first
#define se second
#define re register
typedef long long ll;
typedef pair<int, int> PII;
const int N = 1e6 + 100, M = 5e5 + 5, INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
ll prime[N], mu[N], k, ans[N];
ll phi[N];
bool is_prime[N];
inline void init(int n) {
    memset(is_prime, true, sizeof is_prime);
    mu[1] = 1; phi[1] = 1;
    for (int i = 2; i < n; ++i) {
        if (is_prime[i]) prime[++k] = i, mu[i] = -1, phi[i] = i-1;
        for (int j = 1; j <= k && i * prime[j] < n; ++j) {
            is_prime[i * prime[j]] = false;
            if (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 T = 1; T < n; T++) {
        ll tmp = 0;
        for (int i = T; i < n; i += T) {
            int l = i;
            int r = min(n, l + T - 1);
            tmp = (tmp + phi[i]) % MOD;
            ans[l] = (ans[l] + tmp * tmp % MOD * phi[T] % MOD) % MOD;
            ans[r + 1] = (ans[r + 1] - tmp * tmp % MOD * phi[T] % MOD + MOD) % MOD;
        }
    }
    for (int i = 1; i < n; i++) ans[i] = ((ans[i] + ans[i-1]) % MOD + MOD) % MOD;
}

void solve() {
    init(N);
    int n; cin >> n;
    for (int i = 1; i <= n; i++) printf("%lld\n", ans[i]);
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
#ifdef ACM_LOCAL
    freopen("input", "r", stdin);
    freopen("output", "w", stdout);
#endif
    solve();
}