2020杭电多校第六场 A Very Easy Math Problem 莫比乌斯反演 (HDU 6833)

A Very Easy Math Problem

题解

$$\begin{array}{rl}& \sum _{a_1=1}^n\sum _{a_2=1}^n\cdot \cdot \cdot \sum _{a_x=1}^n\left(\prod _{j=1}^x{a}_j^k\right)f\left(gcd(a_1,a_2,...,a_n)\right)gcd(a_1,a_2,...,a_n)\\ & =\sum _{d=1}^{n}df(d)\sum _{a_1=1}^n\sum _{a_2=1}^n\cdot \cdot \cdot \sum _{a_x=1}^n\left(\prod _{j=1}^x{a}_j^k\right)\left[gcd(a_1,a_2,...,a_n)=d\right]\\ & =\sum _{d=1}^n{d}^{kx+1}f(d)\sum _{a_1=1}^{\lfloor n/d\rfloor }\sum _{a_2=1}^{\lfloor n/d\rfloor }\cdot \cdot \cdot \sum _{a_x=1}^{\lfloor n/d\rfloor }\left(\prod _{j=1}^x{a}_j^k\right)\sum _{t|a_1,t|a_2,...,t|a_n}\mu (t)\\ & =\sum _{d=1}^n{d}^{kx+1}f(d)\sum _{a_1=1}^{\lfloor n/d\rfloor }{a}_1^k\sum _{a_2=1}^{\lfloor n/d\rfloor }{a}_2^k\cdot \cdot \cdot \sum _{a_x=1}^{\lfloor n/d\rfloor }{a}_x^k\sum _{t|a_1,t|a_2,...,t|a_n}\mu (t)\\ & =\sum _{d=1}^n{d}^{kx+1}f(d){\left(\sum _{i=1}^{\lfloor n/d\rfloor }{i}^k\right)}^x\sum _{t|a_1,t|a_2,...,t|a_n}\mu (t)\\ & =\sum _{d=1}^n{d}^{kx+1}f(d)\sum _{t=1}^{\lfloor n/d\rfloor }\mu (t){\left(\sum _{i=1}^{\lfloor n/dt\rfloor }(it{)}^k\right)}^x\\ & =\sum _{d=1}^n{d}^{kx+1}f(d)\sum _{t=1}^{\lfloor n/d\rfloor }\mu (t)t^{kx}{\left(\sum _{i=1}^{\lfloor n/dt\rfloor }{i}^k\right)}^x\\ T=dt\to & =\sum _{T=1}^n{\left(\sum _{i=1}^{\lfloor n/T\rfloor }{i}^k\right)}^x{T}^{kx}\sum _{d|T}f(d)\mu (\frac{T}{d})d\\ 令g(T)=\sum _{d|T}f(d)\mu (\frac{T}{d})d\to & =\sum _{T=1}^n{\left(\sum _{i=1}^{\lfloor n/T\rfloor }{i}^k\right)}^x{T}^{kx}g(T)\end{array}$$
这里$f$函数可以$O(nlogn)$筛，筛完后$O(nlogn)$筛$g$函数，再对$g$函数做一个前缀和
$\sum _{i=1}^{\lfloor n/T\rfloor }{i}^k$可以预处理前缀和
最后计算的时候分块计算，总体复杂度$O(nlogn+T\sqrt{n})$，T为询问次数

代码

#include
using namespace std;
typedef long long ll;
const int N = 2e5 + 10;
const ll mod = 1e9 + 7;

ll qpow(ll a, ll b) {
    ll res = 1;
    while (b) {
        if (b & 1) res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}


ll T, k, x, n;
int vis[N], prime[N], num, mu[N];
ll g[N], f[N], sumG[N], sum[N];
void init(int siz) {
    f[1] = mu[1] = 1, num = 0;
    for (int i = 2; i <= siz; i++) {
        f[i] = 1;
        if (!vis[i]) prime[++num] = i, mu[i] = -1;
        for (int j = 1; j <= num && i * prime[j] <= siz; j++) {
            vis[i * prime[j]] = 1;
            if (i % prime[j] == 0) {
                mu[i * prime[j]] = 0;
                break;
            }
            else mu[i * prime[j]] = mu[i] * mu[prime[j]];
        }
    }
    for (int d = 2; d * d <= siz; d++) for (int i = d * d; i <= siz; i += d * d) f[i] = 0;//f函数
    for (int d = 1; d <= siz; d++)
        for (int i = d; i <= siz; i += d)
            g[i] = (g[i] + 1ll * d * f[d] % mod * mu[i / d] % mod + mod) % mod;//g函数
    for (int i = 1; i <= siz; i++) {
        ll t = qpow(i, k);
        sum[i] = (sum[i - 1] + t) % mod;//i^k前缀和
        sumG[i] = (sumG[i - 1] + qpow(t, x) * g[i] % mod) % mod;//g函数前缀和
    }
}


int main() {

    scanf("%lld%lld%lld", &T, &k, &x);
    init(2e5);
    while (T--) {
        scanf("%lld", &n);
        ll ans = 0;
        for (int l = 1, r; l <= n; l = r + 1) {
            r = n / (n / l);
            ans = (ans + 1ll * (sumG[r] - sumG[l - 1] + mod) % mod * qpow(sum[n / l], x) % mod) % mod;
        }
        printf("%lld\n", (ans + mod) % mod);
    }



    return 0;
}