[HDU6833]A Very Easy Math Problem

题意

求
$\sum_{a_1=1}^n\sum_{a_2=1}^n\cdots\sum_{a_x=1}^n\Big(\prod_{j=1}^xa_j^k\Big)f(\gcd(a_1,a_2,...,a_x))\cdot\gcd(a_1,a_2,...,a_x)\mod 10^9+7$

若存在 $k>1$，且 $k^2\mid x$，则 $f(x)=0$；否则 $f(x)=1$。

$1\le n\le2\times10^5,1\le k,x\le10^9$

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题解

首先 $f(x)=|\mu(x)|$；其次根据乘法分配律，有
$\begin{aligned} g(n)\overset{\text{令}}{=}&\sum_{a_1=1}^n\sum_{a_2=1}^n\cdots\sum_{a_x=1}^n\Big(\prod_{j=1}^xa_j^k\Big)\\ =&\sum_{a_1=1}^na_1^k\sum_{a_2=1}^na_2^k\cdots\sum_{a_x=1}^na_x^k\\ =&\Big(\sum_{i=1}^ni^k\Big)^x \end{aligned}$
枚举 $\gcd$化简原式
$\begin{aligned} &\sum_{a_1=1}^n\sum_{a_2=1}^n\cdots\sum_{a_x=1}^n\Big(\prod_{j=1}^xa_j^k\Big)f(\gcd(a_1,a_2,...,a_x))\cdot\gcd(a_1,a_2,...,a_x)\\ =&\sum_{d=1}^n|\mu(d)|d\sum_{a_1=1}^n\sum_{a_2=1}^n\cdots\sum_{a_x=1}^n\Big(\prod_{j=1}^xa_j^k\Big)[\gcd(a_1,a_2,...,a_x)=d]\\ =&\sum_{d=1}^n|\mu(d)|d^{kx+1}\sum_{b_1=1}^{\lfloor\frac nd\rfloor}\sum_{b_2=1}^{\lfloor\frac nd\rfloor}\cdots\sum_{b_x=1}^{\lfloor\frac nd\rfloor}\Big(\prod_{j=1}^xb_j^k\Big)[\gcd(b_1,b_2,...,b_x)=1]&b_i=\lfloor\frac{a_i}d\rfloor\\ =&\sum_{d=1}^n|\mu(d)|d^{kx+1}\sum_{b_1=1}^{\lfloor\frac nd\rfloor}\sum_{b_2=1}^{\lfloor\frac nd\rfloor}\cdots\sum_{b_x=1}^{\lfloor\frac nd\rfloor}\Big(\prod_{j=1}^xb_j^k\Big)\sum_{p\mid\gcd}\mu(p)\\ =&\sum_{d=1}^n|\mu(d)|d^{kx+1}\sum_{p=1}^{\lfloor\frac nd\rfloor}\mu(p)p^{kx}\sum_{c_1=1}^{\lfloor\frac n{pd}\rfloor}\sum_{c_2=1}^{\lfloor\frac n{pd}\rfloor}\cdots\sum_{c_x=1}^{\lfloor\frac n{pd}\rfloor}\Big(\prod_{j=1}^xc_j^k\Big)&c_i=\lfloor\frac{b_i}p\rfloor\\ =&\sum_{d=1}^n|\mu(d)|d^{kx+1}\sum_{p=1}^{\lfloor\frac nd\rfloor}\mu(p)p^{kx}g(\lfloor\frac n{pd}\rfloor)\\ =&\sum_{T=1}^ng(\lfloor\frac nT\rfloor)\sum_{d|T}|\mu(d)|d^{kx+1}\mu(\frac Td)(\frac Td)^{kx}&T=pd\\ =&\sum_{T=1}^ng(\lfloor\frac nT\rfloor)T^{kx}\sum_{d|T}|\mu(d)|\mu(\frac Td)d\\ \end{aligned}$

令 $\displaystyle h(T)=T^{kx}\sum_{d|T}|\mu(d)|\mu(\frac Td)d$，用枚举倍数的方法、降幂公式和快速幂就可以在 $O(n\log n+n\log P)$的时间内求出所有的 $h(T)$。

令 $\displaystyle S(n)=\sum_{T=1}^ng(\lfloor\frac nT\rfloor)h(T)$，到这里已经可以利用数论分块在 $O(n\log nP+T\sqrt n)$的时间内解决这道题了，但是还可以优化。

考虑如何快速求出所有的 $S(n)$。

注意到对于 $ij\le n<(i+1)j$的所有 $n$，都含有 $g(i)h(j)$这一项，所以可以枚举 $i,j$，转化为对 $S(n)$进行区间加法，差分后求前缀和即可，时间复杂度 $\displaystyle O(\sum_{i=1}^n\sum_{j=1}^{\lfloor\frac ni\rfloor}1)=O(n\log n)$。

总时间复杂度 $O(n\log n+n\log P+T)$

#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5, P = 1e9 + 7;
typedef long long ll;
int n = 2e5, k, x;
int is[N], pr[N], mu[N], g[N], h[N], S[N];
inline int FastExp(int a, int b) {
    int x = 1;
    for (; b; b >>= 1, a = (ll)a * a % P)
        if (b & 1)
            x = (ll)x * a % P;
    return x;
}
int main() {
    scanf("%*d%d%d", &k, &x);
    mu[1] = 1;
    for (int i = 2; i <= n; ++i) {
        if (!is[i])
            pr[++pr[0]] = i, mu[i] = -1;
        for (int j = 1, x; j <= pr[0] && (x = i * pr[j]) <= n; ++j) {
            is[x] = 1, mu[x] = -mu[i];
            if (!(i % pr[j])) {
                mu[x] = 0;
                break;
            }
        }
    }
    for (int i = 1; i <= n; ++i)
        g[i] = (g[i - 1] + FastExp(i, k)) % P;
    for (int i = 1; i <= n; ++i)
        g[i] = FastExp(g[i], x);
    for (int i = 1; i <= n; ++i)
        for (int j = i; j <= n; j += i)
            mu[i] ? h[j] = (h[j] + mu[j / i] * i + P) % P : 0;
    for (int i = 1; i <= n; ++i)
        h[i] = (ll)h[i] * FastExp(i, (ll)k * x % (P - 1)) % P;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; i * j <= n; ++j) {
            int m = i * j;
            S[m] = (S[m] + (ll)g[i] * h[j] % P) % P;
            if (m + j <= n)
                S[m + j] = (S[m + j] - (ll)g[i] * h[j] % P + P) % P;
        }
    for (int i = 1; i <= n; ++i)
        S[i] = (S[i] + S[i - 1]) % P;
    while (~scanf("%d", &n))
        printf("%d\n", S[n]);
    return 0;
}