HDU 6833 A Very Easy Math Problem

A Very Easy Math Problem

推式子

$\sum_{a_i = 1} ^{n} \sum_{a_2 = 1} ^{n} \dots \sum_{a_x = 1} ^{n} \left(\prod_{j = 1} ^{x} a_j ^ k \right)f(gcd(a_1, a_2, \dots, a_x))\times gcd(a_1, a_2, \dots, a_x)$

$= \sum_{d = 1} ^{n}df(d)\sum_{a_i = 1} ^{n} \sum_{a_2 = 1} ^{n} \dots \sum_{a_x = 1} ^{n} \left(\prod_{j = 1} ^{x} a_j ^ k \right)(gcd(a_1, a_2, \dots, a_x) == d)$

$有\sum_{a_i = 1} ^{n} \sum_{a_2 = 1} ^{n} \dots \sum_{a_x = 1} ^{n} \left(\prod_{j = 1} ^{x} a_j ^ k \right) = \left(\sum_{i = 1} ^{n} i ^ k \right) ^ x$

$= \sum_{d = 1} ^{n}d ^ {kd + 1} f(d)\left(\sum_{i = 1} ^{\frac{n}{d}} i ^ k \right) ^ x (gcd(a_1, a_2, \dots, a_x) == 1)$

$= \sum_{d = 1} ^{n}d ^ {kd + 1} f(d)\left(\sum_{i = 1} ^{\frac{n}{d}} i ^ k \right) ^ x \sum_{t \mid \frac{n}{d}}\mu(t)$

$= \sum_{d = 1} ^{n} f(d) d ^{dx + 1} \sum_{t = 1} ^{\frac{n}{d}} \mu(t) t^{kx} \left( \sum_{i = 1} ^{\frac{n}{dt}} i ^ k\right) ^ x$
另 $T = dt$
$= \sum\limits_{T = 1} ^{n}\left( \sum_{i = 1} ^{\frac{n}{T}}i ^ k \right) ^ x T ^ {kx} \sum_{d\mid T}d \mu(\frac{T}{d})f(d)$
最后我们只要先预处理出 $T ^ {kx} \sum_{d\mid T}d \mu(\frac{T}{d})f(d)$，就能就简简单单的进行除法分块了。

代码

/*
  Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>

#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;

inline ll read() {
    ll f = 1, x = 0;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-')    f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return f * x;
}

const int N = 2e5 + 10, mod = 1e9 + 7;

int mu[N];

ll sum1[N], sum2[N], sum3[N], f[N], k, x;

bool st[N];

vector<int> prime;

ll quick_pow(ll a, ll n, ll mod) {
    ll ans = 1;
    while(n) {
        if(n & 1) ans = ans * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return ans;
}

void mobius() {
    f[1] = st[0] = st[1] = mu[1] = 1;
    for(int i = 2; i < N; i++) {
        f[i] = 1;
        if(!st[i]) {
            prime.pb(i);
            mu[i] = -1;
        }
        for(int j = 0; j < prime.size() && i * prime[j] < N; j++) {
            st[i * prime[j]] = 1;
            if(i % prime[j] == 0) break;
            mu[i * prime[j]] = -mu[i];
        }
    }
    for(int i = 2; i * i < N; i++) {
        for(int j = i * i; j < N; j += i * i) {
            f[j] = 0;
        }
    }
    for(int i = 1; i < N; i++) {
        for(int j = i; j < N; j += i) {
            sum3[j] = (sum3[j] + i * mu[j / i] % mod * f[i] % mod + mod) % mod;
        }
    }
    for(int i = 1; i < N; i++) {
        ll temp = quick_pow(i, k, mod);
        sum1[i] = (sum1[i - 1] + temp) % mod;
        sum2[i] = (sum2[i - 1] + quick_pow(temp, x, mod) * sum3[i] % mod) % mod;
    }
}

int main() {
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int T = read();
    k = read(), x = read();
    mobius();
    while(T--) {
        ll n = read(), ans = 0;
        for(ll l = 1, r; l <= n; l = r + 1) {
            r = n / (n / l);
            ans = (ans + quick_pow(sum1[n / l], x, mod) * (sum2[r] - sum2[l - 1]) % mod) % mod;
        }
        cout << (ans % mod + mod) % mod << endl;
    }
    return 0;
}