A Very Easy Math Problem

推式子

$$\sum _{a_i=1}^n\sum _{a_2=1}^n\cdots \sum _{a_x=1}^n\left(\prod _{j=1}^x{a}_j^k\right)f(gcd(a_1,a_2,\dots ,a_x))\times gcd(a_1,a_2,\dots ,a_x)$$

$$=\sum _{d=1}^{n}df(d)\sum _{a_i=1}^n\sum _{a_2=1}^n\cdots \sum _{a_x=1}^n\left(\prod _{j=1}^x{a}_j^k\right)(gcd(a_1,a_2,\dots ,a_x)==d)$$

$$有\sum _{a_i=1}^n\sum _{a_2=1}^n\cdots \sum _{a_x=1}^n\left(\prod _{j=1}^x{a}_j^k\right)={\left(\sum _{i=1}^n{i}^k\right)}^x$$

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

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

$$=\sum _{d=1}^{n}f(d)d^{dx+1}\sum _{t=1}^{\frac{n}{d}}\mu (t)t^{kx}{\left(\sum _{i=1}^{\frac{n}{dt}}{i}^k\right)}^x$$
另$T=dt$
$$=\sum _{T=1}^n{\left(\sum _{i=1}^{\frac{n}{T}}{i}^k\right)}^x{T}^{kx}\sum _{d\mid T}d\mu (\frac{T}{d})f(d)$$
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 | 1using 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;
}