∑ i = 1 n ∑ j = 1 n f ( g c d ( i , j ) ) k ∑ d = 1 n f ( d ) k ∑ i = 1 n d ∑ j = 1 n d [ g c d ( i , j ) = 1 ] ∑ d = 1 n f ( d ) k ∑ K = 1 n d μ ( k ) ( n K d ) 2 t = K d ∑ t = 1 n ( n t ) 2 ∑ d ∣ t f ( d ) k μ ( t d ) 我 们 记 f ( x ) k = F ( x ) 上 面 式 子 后 半 部 分 是 一 个 迪 利 克 雷 卷 积 形 式 : F ∗ μ 所 以 我 们 卷 上 一 个 I , 有 F ∗ μ ∗ I = F ∗ ϵ = F 得 到 后 半 部 分 的 前 缀 和 S ( n ) = ∑ i = 1 n F ( i ) − ∑ i = 2 n S ( n i ) \sum_{i = 1} ^{n} \sum_{j = 1} ^{n} f(gcd(i, j)) ^ k\\ \sum_{d = 1} ^{n} f(d) ^k \sum_{i = 1} ^{\frac{n}{d}} \sum_{j = 1} ^{\frac{n}{d}}[gcd(i, j) = 1]\\ \sum_{d = 1} ^{n} f(d) ^k \sum_{K = 1} ^{\frac{n}{d}} \mu(k) \left( \frac{n}{Kd} \right) ^2\\ t = Kd\\ \sum_{t = 1} ^{n} \left(\frac{n}{t} \right) ^ 2 \sum_{d \mid t} f(d) ^ k \mu(\frac{t}{d})\\ 我们记f(x) ^ k = F(x)\\ 上面式子后半部分是一个迪利克雷卷积形式:F * \mu\\ 所以我们卷上一个I,有F * \mu * I = F * \epsilon = F\\ 得到后半部分的前缀和S(n) = \sum_{i = 1} ^{n} F(i) - \sum_{i = 2} ^{n} S(\frac{n}{i})\\ i=1∑nj=1∑nf(gcd(i,j))kd=1∑nf(d)ki=1∑dnj=1∑dn[gcd(i,j)=1]d=1∑nf(d)kK=1∑dnμ(k)(Kdn)2t=Kdt=1∑n(tn)2d∣t∑f(d)kμ(dt)我们记f(x)k=F(x)上面式子后半部分是一个迪利克雷卷积形式:F∗μ所以我们卷上一个I,有F∗μ∗I=F∗ϵ=F得到后半部分的前缀和S(n)=i=1∑nF(i)−i=2∑nS(in)
化简到这里只需要跟上面一题类似用Min_25求 ∑ i = 1 n F ( i ) \sum\limits_{i = 1} ^{n} F(i) i=1∑nF(i),然后用杜教筛求 S ( n ) S(n) S(n)即可得到答案。
/*
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;
}
#define uint unsigned int
const int N = 1e6 + 10;
uint quick_pow(uint a, int n) {
uint ans = 1;
while(n) {
if(n & 1) ans = ans * a;
a = a * a;
n >>= 1;
}
return ans;
}
namespace Min_25 {
uint prime[N], g[N], sum[N], f[N], calc[N];
int a[N], id1[N], id2[N], n, m, k, cnt, T;
bool st[N];
int ID(int x) {
return x <= T ? id1[x] : id2[n / x];
}
void init() {
cnt = m = 0;
T = sqrt(n + 0.5);
for(int i = 2; i <= T; i++) {
if(!st[i]) {
prime[++cnt] = i;
f[cnt] = quick_pow(i, k);
sum[cnt] = sum[cnt - 1] + 1;
}
for(int j = 1; j <= cnt && 1ll * i * prime[j] <= T; j++) {
st[i * prime[j]] = 1;
if(i % prime[j] == 0) {
break;
}
}
}
for(ll l = 1, r; l <= n; l = r + 1) {
r = n / (n / l);
a[++m] = n / l;
if(a[m] <= T) id1[a[m]] = m;
else id2[n / a[m]] = m;
g[m] = a[m] - 1;
}
for(int j = 1; j <= cnt; j++) {
for(int i = 1; i <= m && 1ll * prime[j] * prime[j] <= a[i]; i++) {
g[i] -= g[ID(a[i] / prime[j])] - sum[j - 1];
}
}
for(int i = 1; i <= T; i++) {
st[i] = 0;
}
/*
非递归版本
*/
// for(int j = cnt; j >= 1; j--) {
// for(int i = 1; i <= m && 1ll * prime[j] * prime[j] <= a[i]; i++) {
// for(ll k = prime[j]; k * prime[j] <= a[i]; k *= prime[j]) {
// calc[i] += calc[ID(a[i] / k)] + (g[ID(a[i] / k)] - sum[j - 1]) * f[j];
// }
// }
// }
}
// uint solve(int x) {
// if(x <= 1) return 0;
// return calc[ID(x)] + g[ID(x)];
// }
/*
下面是递归版本
*/
// uint solve(int n, int m) {
// if(n < prime[m] || n <= 1) return 0;
// uint ans = 0;
// for(int j = m; j <= cnt && 1ll * prime[j] * prime[j] <= n; j++) {
// for(ll i = prime[j]; i * prime[j] <= n; i *= prime[j]) {
// ans += solve(n / i, j + 1) + (g[ID(n / i)] - sum[j - 1]) * f[j];
// }
// }
// return ans;
// }
// uint solve(int n) {
// if(n <= 1) return 0;
// return solve(n, 1) + g[ID(n)];
// }
}
unordered_map<int, uint> ans_s;
uint S(int n) {
if(ans_s.count(n)) return ans_s[n];
uint ans = Min_25::solve(n);
for(uint l = 2, r; l <= n; l = r + 1) {
r = n / (n / l);
ans -= (r - l + 1) * S(n / l);
}
return ans_s[n] = ans;
}
int main() {
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
uint n = read(), k = read();
Min_25::n = n, Min_25::k = k;
Min_25::init();
uint ans = 0;
for(uint l = 1, r; l <= n; l = r + 1) {
r = n / (n / l);
ans += (n / l) * (n / l) * (S(r) - S(l - 1));
}
cout << ans << endl;
return 0;
}
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\sum_{i = 1} ^{n} \sum_{j = 1} ^{n}f(gcd(i, j)) ^k\\ \sum_{d = 1} ^{n} f(d) ^ k \sum_{i = 1} ^{\frac{n}{d}} \sum_{j = 1} ^{\frac{n}{d}}[gcd(i, j) == 1]\\ \sum_{d = 1} ^{n} f(d) ^ k(2 \sum_{i = 1} ^ {\frac{n}{d}} \phi(i) - 1)\\
i=1∑nj=1∑nf(gcd(i,j))kd=1∑nf(d)ki=1∑dnj=1∑dn[gcd(i,j)==1]d=1∑nf(d)k(2i=1∑dnϕ(i)−1)
充分利用上面递推Min_25,得到一个复杂度更优的算法。
/*
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;
}
#define uint unsigned int
const int N = 1e6 + 10;
uint quick_pow(uint a, int n) {
uint ans = 1;
while(n) {
if(n & 1) ans = ans * a;
a = a * a;
n >>= 1;
}
return ans;
}
namespace Min_25 {
uint prime[N], g[N], sum[N], f[N], calc[N];
int a[N], id1[N], id2[N], n, m, k, cnt, T;
bool st[N];
int ID(int x) {
return x <= T ? id1[x] : id2[n / x];
}
void init() {
cnt = m = 0;
T = 1000000;
for(int i = 2; i <= T; i++) {
if(!st[i]) {
prime[++cnt] = i;
f[cnt] = quick_pow(i, k);
sum[cnt] = sum[cnt - 1] + 1;
}
for(int j = 1; j <= cnt && 1ll * i * prime[j] <= T; j++) {
st[i * prime[j]] = 1;
if(i % prime[j] == 0) {
break;
}
}
}
for(ll l = 1, r; l <= n; l = r + 1) {
r = n / (n / l);
a[++m] = n / l;
if(a[m] <= T) id1[a[m]] = m;
else id2[n / a[m]] = m;
g[m] = a[m] - 1;
}
for(int j = 1; j <= cnt; j++) {
for(int i = 1; i <= m && 1ll * prime[j] * prime[j] <= a[i]; i++) {
g[i] -= g[ID(a[i] / prime[j])] - sum[j - 1];
}
}
for(int i = 1; i <= T; i++) {
st[i] = 0;
}
for(int j = cnt; j >= 1; j--) {
for(int i = 1; i <= m && 1ll * prime[j] * prime[j] <= a[i]; i++) {
for(ll k = prime[j]; k * prime[j] <= a[i]; k *= prime[j]) {
calc[i] += calc[ID(a[i] / k)] + (g[ID(a[i] / k)] - sum[j - 1]) * f[j];
}
}
}
}
uint solve(int x) {
if(x <= 1) return 0;
return calc[ID(x)] + g[ID(x)];
}
}
namespace Djs {
uint prime[N], phi[N], cnt;
bool st[N];
void init() {
phi[1] = 1;
for(int i = 2; i < N; i++) {
if(!st[i]) {
prime[++cnt] = i;
phi[i] = i - 1;
}
for(int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {
st[i * prime[j]] = 1;
if(i % prime[j] == 0) {
phi[i * prime[j]] = phi[i] * prime[j];
break;
}
phi[i * prime[j]] = phi[i] * (prime[j] - 1);
}
}
for(int i = 1; i < N; i++) {
phi[i] += phi[i - 1];
}
}
unordered_map<int, uint> ans_s;
uint S(int n) {
if(n < N) return phi[n];
if(ans_s.count(n)) return ans_s[n];
uint ans = 1ll * n * (n + 1) / 2;
for(uint l = 2, r; l <= n; l = r + 1) {
r = n / (n / l);
ans -= (r - l + 1) * S(n / l);
}
return ans_s[n] = ans;
}
}
int main() {
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
uint n = read(), k = read();
Min_25::n = n, Min_25::k = k;
Min_25::init();
Djs::init();
uint ans = 0;
for(uint l = 1, r; l <= n; l = r + 1) {
r = n / (n / l);
ans += (2 * Djs::S(n / l) - 1) * (Min_25::solve(r) - Min_25::solve(l - 1));
}
cout << ans << endl;
return 0;
}