HDU 6833 A Very Easy Math Problem

A Very Easy Math Problem

推式子

∑ai=1n∑a2=1n⋯∑ax=1n(∏j=1xajk)f(gcd(a1,a2,…,ax))×gcd(a1,a2,…,ax)\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) ai=1∑na2=1∑n⋯ax=1∑n(j=1∏xajk)f(gcd(a1,a2,…,ax))×gcd(a1,a2,…,ax)

=∑d=1ndf(d)∑ai=1n∑a2=1n⋯∑ax=1n(∏j=1xajk)(gcd(a1,a2,…,ax)==d)= \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) =d=1∑ndf(d)ai=1∑na2=1∑n⋯ax=1∑n(j=1∏xajk)(gcd(a1,a2,…,ax)==d)

有∑ai=1n∑a2=1n⋯∑ax=1n(∏j=1xajk)=(∑i=1nik)x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 有ai=1∑na2=1∑n⋯ax=1∑n(j=1∏xajk)=(i=1∑nik)x

=∑d=1ndkd+1f(d)(∑i=1ndik)x(gcd(a1,a2,…,ax)==1)= \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) =d=1∑ndkd+1f(d)⎝⎛i=1∑dnik⎠⎞x(gcd(a1,a2,…,ax)==1)

=∑d=1ndkd+1f(d)(∑i=1ndik)x∑t∣ndμ(t)= \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) =d=1∑ndkd+1f(d)⎝⎛i=1∑dnik⎠⎞xt∣dn∑μ(t)

=∑d=1nf(d)ddx+1∑t=1ndμ(t)tkx(∑i=1ndtik)x= \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 =d=1∑nf(d)ddx+1t=1∑dnμ(t)tkx⎝⎛i=1∑dtnik⎠⎞x
另T=dtT = dtT=dt
=∑T=1n(∑i=1nTik)xTkx∑d∣Tdμ(Td)f(d)= \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) =T=1∑n⎝⎛i=1∑Tnik⎠⎞xTkxd∣T∑dμ(dT)f(d)
最后我们只要先预处理出Tkx∑d∣Tdμ(Td)f(d)T ^ {kx} \sum_{d\mid T}d \mu(\frac{T}{d})f(d)Tkx∑d∣Tdμ(dT)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;
}

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HDU 6833 A Very Easy Math Problem

A Very Easy Math Problem

推式子

代码

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