51nod 1220 约数之和

 

$\sigma_{0}(ij)=\sum\limits_{x|i}\sum\limits_{y|j}[x\bot y]$
$\sigma_{1}(ij)=\sum\limits_{x|i}\sum\limits_{y|j}\dfrac{xj}{y}[x\bot y]$
所以 $$\begin{aligned}&\sum_{i=1}^n\sum_{j=1}^n \sigma_{1}(ij)\\=& \sum_{i=1}^n\sum_{j=1}^n \sum_{x|i}\sum_{y|j}\dfrac{xj}{y}[x\bot y]\\=& \sum_{d=1}^n\mu(d)\sum_{i=1}^n\sum_{j=1}^n\sum_{x|i}\sum_{y|j}\dfrac{xj}{y}[d|(x,y)]\\=&\sum_{d=1}^n\mu(d)\sum_{d|x}\sum_{d|y} \frac{x}{y}\sum_{x|i}\sum_{y|j}j \\=& \sum_{d=1}^n \mu(d)\sum_{d|x}\sum_{d|y}\frac{x}{y}s_0(\lfloor\frac{n}{x} \rfloor)ys_1(\lfloor\frac{n}{y} \rfloor)\\=& \sum_{i=1}^n \mu(d)d\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}is_0(\lfloor \frac{n}{id} \rfloor) \sum_{j=1}^{\lfloor \frac{n}{d} \rfloor}s_1(\lfloor \frac{n}{jd} \rfloor) \end{aligned}$$
$\sum \limits_{i=1}^n i\lfloor \dfrac{n}{i} \rfloor=\sum\limits_{i=1}^n \sigma_1(i)$
$\sum \limits_{i=1}^n s_1(\lfloor \dfrac{n}{i} \rfloor)=\sum\limits_{i=1}^n \sum\limits_{j=1}^{\lfloor \frac{n}{i} \rfloor}j=\sum_{i=1}^n i\lfloor \dfrac{n}{i} \rfloor$
二者等价
式子变成 $$\sum_{i=1}^n \mu(d)d \left(\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sigma_1(i)\right)^2$$
对于 $f(n)=n\mu(n)$，有 $f * I = \epsilon$，杜教筛之后的式子就是 $S(n)=1-\sum\limits_{i=2}^n iS(\lfloor \dfrac{n}{i} \rfloor)$
对于 $\sigma_1$ 的前缀和，小范围预处理，大的用 $\sum\limits_{i=1}^n \sigma_1(i)=\sum \limits_{i=1}^n i\lfloor \dfrac{n}{i} \rfloor$ 整除分块解决。

#include 

const int N = 1e6 + 7;
const int MOD = 1000000007;
int mu[N], prime[N], prin, d[N], t[N];
bool vis[N];

inline void M(int &ans) {
    if (ans >= MOD) ans -= MOD;
    if (ans < 0) ans += MOD;
}

int qp(int a, int b = MOD - 2) {
    int ans = 1;
    while (b) {
        if (b & 1) ans = 1LL * ans * a % MOD;
        b >>= 1;
        a = 1LL * a * a % MOD;
    }
    return ans;
}

const int inv6 = qp(6);

void init(int n) {
    mu[1] = 1;
    for (int i = 2; i <= n; i++) {
        if (!vis[i]) {
            prime[++prin] = i;
            mu[i] = -1;
        }
        for (int j = 1; j <= prin && i * prime[j] < N; j++) {
            vis[i * prime[j]] = 1;
            if (i % prime[j] == 0) {
                mu[i * prime[j]] = 0;
                break;
            }
            mu[i * prime[j]] = -mu[i];
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = i; j <= n; j += i)
            M(d[j] += i);
    }
    for (int i = 1; i <= n; i++)
        M(mu[i] = (1LL * mu[i] * i + mu[i - 1]) % MOD), M(d[i] += d[i - 1]);
}

std::unordered_map<int, int> muu, dd;

inline int sum1(int n) {
    return 1LL * n * (n + 1) / 2 % MOD;
}

inline int sum1(int i, int j) {
    int ans;
    M(ans = sum1(j) - sum1(i - 1));
    return ans;
}

inline int sum2(int n) {
    return 1LL * n * (n + 1) % MOD * (2 * n + 1) % MOD * inv6 % MOD;
}

inline int sum2(int i, int j) {
    int ans = sum2(j) - sum2(i - 1);
    M(ans);
    return ans;
}

inline int Mu(int n) {
    if (n < N) return mu[n];
    if (muu.count(n)) return muu[n];
    int ans = 1;
    for (int i = 2, j; i <= n; i = j + 1) {
        j = n / (n / i);
        M(ans -= 1LL * sum1(i, j) * Mu(n / i) % MOD);
    }
    return muu[n] = ans;
}

inline int D(int n) {
    if (n < N) return d[n];
    if (dd.count(n)) return dd[n];
    int ans = 0;
    for (int i = 1, j; i <= n; i = j + 1) {
        j = n / (n / i);
        M(ans += 1LL * sum1(i, j) * (n / i) % MOD);
    }
    return dd[n] = ans;
}

int solve(int n) {
    int ans = 0;
    for (int i = 1, j; i <= n; i = j + 1) {
        j = n / (n / i);
        M(ans += 1LL * (Mu(j) - Mu(i - 1) + MOD) % MOD * D(n / i) % MOD * D(n / i) % MOD);
    }
    return ans;
}

int main() {
    init(N - 1);
    int n;
    scanf("%d", &n);
    printf("%d\n", solve(n));
    return 0;
}

View Code