HDU 6588 Function

 

【传送门】

求$$\sum_{i=1}^{n} \gcd(\lfloor \sqrt[3]{i} \rfloor, i)$$
题解写的很清楚，自己重新推一推。

$$\sum_{i=1}^{n} \gcd(\lfloor \sqrt[3]{i} \rfloor, i)$$

$$=\sum_{a=1}^{\lfloor\sqrt[3]{n}\rfloor}\sum_{i=1}^{n}\gcd(a, i)[\sqrt[3]{i}=a]$$

$$=\sum_{a=1}^{\lfloor\sqrt[3]{n}\rfloor}\sum_{i=a^3}^{\min\{(a+1)^3-1,n\}}\gcd(a,i)$$

$$=\sum_{i=\lfloor \sqrt[3]{n} \rfloor ^3}^{n}\gcd(\sqrt[3]{n}, i)+\sum_{a=1}^{r}\sum_{i=a^3}^{(a+1)^3-1}\gcd(a,i)$$

其中 $r = \sqrt[3]{n}-1$
设 $f(n,a)=\sum_{i=1}^{n}\gcd(i, a)$
$$f(n,a)=\sum_{i=1}^{n}\gcd(i,a)$$
$$=\sum_{d|a}d\sum_{i=1}^{n}[\gcd(i,a)=d]$$
$$=\sum_{d|a}d\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}[\gcd(i, \frac{a}{d}) =1]$$
$$=\sum_{d|a}d\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sum_{p|\gcd(i,\frac{a}{d})}\mu(p)$$
$$=\sum_{d|a}d\sum_{p|\frac{a}{d}}\mu(p)\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}[p|i]$$
$$=\sum_{d|a}d\sum_{p|\frac{a}{d}}\mu(p)\lfloor \frac{n}{pd} \rfloor$$
$$=\sum_{T|a}\lfloor \frac{n}{T}\rfloor\sum_{p|T}\mu(p)\frac{T}{p}$$
$$=\sum_{T|a}\lfloor \frac{n}{T}\rfloor \varphi(T)$$
第一部分可以 $O(\sqrt{n})$ 解决。

将 $f(n, a)$ 带入第二部分得
$$\sum_{a=1}^r \sum_{i=a}^{(a+1)^3-1}\gcd(a,i)$$
$$=\sum_{a=1}^{r}\sum_{T|a}(\lfloor\frac{(a+1)^3-1}{T}\rfloor - \lfloor\frac{a^3-1}{T}\rfloor)\varphi(T)$$
$$=\sum_{T = 1}^{r}\varphi(T)\sum_{b=1}^{\lfloor\frac{r}{T}\rfloor}(\lfloor\frac{(bT+1)^3-1}{T} \rfloor-\lfloor\frac{(bT)^3-1}{T} \rfloor)$$
$$=\sum_{T=1}^{r}\varphi(T)\sum_{b=1}^{\lfloor\frac{r}{T}\rfloor}(\lfloor b^3T^2+3b^2T+3b\rfloor-\lfloor b^3T^2-\frac{1}{T}\rfloor)$$
$$=\sum_{T=1}^{r}\varphi(T)\sum_{b=1}^{\lfloor\frac{r}{T}\rfloor}(3b^2T+3b+1)$$
$$=\sum_{T=1}^{r}\varphi(T)(3T\sum_{b=1}^{\lfloor\frac{r}{T}\rfloor}b^2+3\sum_{b=1}^{\lfloor\frac{r}{T}\rfloor}b + \lfloor\frac{r}{T}\rfloor)$$

这部分 $O(r)$ 解决。

用太多int128会T。

#include <bits/stdc++.h>

namespace IO {
    void read() {}
    template <typename T, typename... T2>
    inline void read(T &x, T2 &... oth) {
        T f = 1; x = 0;
        char ch = getchar();
        while (!isdigit(ch)) { if (ch == '-') f = -1; ch = getchar(); }
        while (isdigit(ch)) { x = x * 10 + ch - 48; ch = getchar(); }
        x *= f;
        read(oth...);
    }
}
#define read IO::read
#define print IO::print
#define ll long long
#define int128 __int128

const int MOD = 998244353;
const int inv6 = 166374059, inv2 = 499122177;
const int N = 1e7 + 7;
int prime[N], phi[N], prin;
bool vis[N];

void init() {
    phi[1] = 1;
    for (int i = 2; i < N; i++) {
        if (!vis[i]) { prime[++prin] = i; phi[i] = i - 1; }
        for (int j = 1; j <= prin && i * prime[j] < N; j++) {
            vis[i * prime[j]] = 1;
            if (i % prime[j] == 0) {
                phi[i * prime[j]] = prime[j] * phi[i];
                break;
            }
            phi[i * prime[j]] = phi[i] * phi[prime[j]];
        }
    }
}

int root3(int128 n) {
    int l = 1, r = 1e7 + 7;
    int ans = 1;
    while (l <= r) {
        int128 mid = (l + r) / 2;
        if (mid * mid * mid <= n) l = mid + 1, ans = mid;
        else r = mid - 1;
    }
    return ans;
}

template<class T>
T gcd(T a, T b) {
    while (b) {
        a %= b;
        std::swap(a, b);
    }
    return a;
}

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

int f(int128 n, int a) {
    int ans = 0;
    for (int i = 1; 1LL * i * i <= a; i++) {
        if (a % i) continue;
        int128 temp = n / i * phi[i] % MOD;
        M(ans += temp);
        if (a == i * i) continue;
        int j = a / i;
        temp = n / j * phi[j] % MOD;
        M(ans += temp);
    }
    return ans;
}

int sum_squr(int n) {
    int ans = 1LL * n * (n + 1) % MOD * (2 * n + 1) % MOD * inv6 % MOD;
    return ans;
}

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

int main() {
    init();
    int T;
    read(T);
    while (T--) {
        int128 n;
        read(n);
        if (n <= 7) {
            int ans = 0;
            for (int i = 1; i <= n; i++)
                ans += gcd(1, i);
            printf("%d\n", ans);
            continue;
        }
        int r = root3(n);
        int128 rr = (int128)r * r * r;
        int ans = f(n, r) - f(rr - 1, r);
        M(ans);
        r--;
        for (int i = 1; i <= r; i++) {
            int y = r / i;
            int temp = 3LL * i * sum_squr(y) % MOD;
            M(temp += 3LL * sum(y) % MOD);
            M(temp += y);
            temp = 1LL * temp * phi[i] % MOD;
            M(ans += temp);
        }
        printf("%d\n", ans);
    }
    return 0;
}