2020HDU多校第六场：1007. A Very Easy Math Problem

题目链接

hdu6833

题目大意

给定 $t,k,x(1\le t\le 10^4,1\le k,x\le 10^9)$，计算 $t$ 组 $n(1\le n\le 2\cdot 10^5)$的函数值：

$$\sum _{a_1=1}^n\cdots \sum _{a_x=1}^n\prod _{j=1}^x{a}_j^k\cdot f(\gcd (a_1,\cdots ,a_x))\gcd (a_1,\cdots ,a_x)$$

其中 $f(x)=|\mu (x)|$ 。

解题思路

枚举 $gcd(a_1,\cdots ,a_x)=d$ ：

$$\begin{array}{rl}& =\sum _{d=1}^n\sum _{a_1=1}^n\sum _{a_2=1}^n\cdots \sum _{a_x=1}^n\left(\prod _{j=1}^x{a_j}^k\right)\cdot f(d)\cdot [gcd(a_1,\cdots ,a_x)=d]\cdot d\\ & =\sum _{d=1}^{n}f(d)\cdot d^{kx+1}\sum _{a_1=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{a_2=1}^{\lfloor \frac{n}{d}\rfloor }\cdots \sum _{a_x=1}^{\lfloor \frac{n}{d}\rfloor }\left(\prod _{j=1}^x{a_j}^k\right)\cdot [gcd(a_1,\cdots ,a_x)=1]\end{array}$$

又有设：

$$\begin{array}{rl}h(n)& =\sum _{a_1=1}^n\sum _{a_2=1}^n\cdots \sum _{a_x=1}^n\left(\prod _{j=1}^x{a_j}^k\right)\cdot [gcd(a_1,\cdots ,a_x)=1]\\ & =\sum _{a_1=1}^n\sum _{a_2=1}^n\cdots \sum _{a_x=1}^n\left(\prod _{j=1}^x{a_j}^k\right)\cdot \sum _{p|gcd(a_1,\cdots ,a_x)}\mu (p)\\ & =\sum _{p=1}^n\sum _{a_1=1}^{\lfloor \frac{n}{p}\rfloor }\sum _{a_2=1}^{\lfloor \frac{n}{p}\rfloor }\cdots \sum _{a_x=1}^{\lfloor \frac{n}{p}\rfloor }\left(\prod _{j=1}^x{\left(a_j\cdot p\right)}^k\right)\cdot \mu (p)\\ & =\sum _{p=1}^n{p}^{kx}\cdot \mu (p)\cdot {\left(\sum _{i=1}^{\lfloor \frac{n}{p}\rfloor }{i}^k\right)}^x\end{array}$$

令：

$$S(n)={\left(\sum _{i=1}^n{i}^k\right)}^x$$

则：

$$h(n)=\sum _{p=1}^n{p}^{kx}\cdot \mu (p)\cdot S\left(\lfloor \frac{n}{p}\rfloor \right)$$

所以原式：

$$\begin{array}{rl}& =\sum _{d=1}^{n}f(d)\cdot d^{kx+1}\cdot h\left(\lfloor \frac{n}{d}\rfloor \right)\\ & =\sum _{d=1}^{n}f(d)\cdot d^{kx+1}\cdot \sum _{p=1}^{\lfloor \frac{n}{d}\rfloor }{p}^{kx}\cdot \mu (p)\cdot S\left(\lfloor \frac{n}{pd}\rfloor \right)\end{array}$$

枚举 $T=pd$：

$$\begin{array}{rl}& =\sum _{T=1}^{n}S\left(\lfloor \frac{n}{T}\rfloor \right)\cdot \sum _{p|T}{p}^{kx}\cdot \mu (p)\cdot f\left(\frac{T}{p}\right)\cdot {\left(\frac{T}{p}\right)}^{kx+1}\\ & =\sum _{T=1}^{n}S\left(\lfloor \frac{n}{T}\rfloor \right)\cdot \left(\sum _{p|T}\mu (p)\cdot {\left(\mu \left(\frac{T}{p}\right)\right)}^2\cdot \frac{T}{p}\right)\cdot T^{kx}\end{array}$$

令：

$$g(T)=\left(\sum _{p|T}\mu (p)\cdot {\left(\mu \left(\frac{T}{p}\right)\right)}^2\cdot \frac{T}{p}\right)\cdot T^{kx}$$

则原式：

$$=\sum _{T=1}^{n}S\left(\lfloor \frac{n}{T}\rfloor \right)\cdot g(T)$$

对于全部的 $g(T)$，我们可以从 $1$ 到 $n$ 枚举因子 $p$，对于其全部的倍数进行处理，复杂度 $O(nlogn)$；我们可以线性筛出 $\mu (i)$ 和 $S(i)$。预处理总体复杂度 $O(nlogn)$。对于每组询问，只需要整除分块即可，复杂度 $O(T\sqrt{n})$。

AC代码

#include 
using namespace std;
typedef long long ll;
const int N = 2e5 + 50;
const int P = 1e9 + 7;
int T, n, k, x;

int ksm(int a, ll b, int c) {
    int ret = 1;
    for (; b; b >>= 1, a = 1LL * a * a % c) {
        if (b & 1) {
            ret = 1LL * ret * a % c;
        }
    }
    return ret;
}

int prime[N], num[N], mu[N], tot;
void getmu() {
    mu[1] = 1;
    for (int i = 2; i < N; i++) {
        if (!num[i]) {
            prime[++tot] = i;
            mu[i] = -1;
        }
        for (int j = 1; j <= tot && i * prime[j] < N; j++) {
            num[i * prime[j]] = 1;
            if (i % prime[j]) {
                mu[i * prime[j]] = -mu[i];
            }
            else {
                break;
            }
        }
    }
}
long long g[N], sumg[N];
void getg() {
    for (int i = 1; i < N; i++) {
        for (int j = 1; i * j < N; j++) {
            g[i * j] += mu[i] * mu[j] * mu[j] * j;
        }
    }
    for (int i = 1; i < N; i++) {
        int tmp = ksm(i, 1LL * k * x, P);
        g[i] = (g[i] + P) % P * tmp % P;
    }
    for (int i = 1; i < N; i++) {
        sumg[i] = (sumg[i - 1] + g[i]) % P;
    }
}
int s[N];
void gets() {
    for (int i = 1; i < N; i++) {
        s[i] = (s[i - 1] + ksm(i, k, P)) % P;
    }
    for (int i = 1; i < N; i++) {
        s[i] = ksm(s[i], x, P);
    }
}
void init() {
    getmu();
    getg();
    gets();
}
int work() {
    int ret = 0;
    for (int i = 1, j; i <= n; i = j + 1) {
        j = n / (n / i);
        (ret += 1LL * s[n / i] * (sumg[j] - sumg[i - 1] + P) % P) %= P;
    }
    return ret;
}

int main(int argc, const char * argv[]) {
    scanf("%d%d%d", &T, &k, &x);
    init();
    while (T--) {
        scanf("%d", &n);
        printf("%d\n", work());
    }
    return 0;
}