小 Q 与函数求和 1
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ϕ
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gcd
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gcd
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gcd
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gcd
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gcd
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ϕ
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gcd
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设
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设
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即
为
答
案
\sum_{i = 1} ^{n} \sum_{j = 1} ^{n} \phi(ij \gcd(i, j) ^ K)\\ \sum_{i = 1} ^{n} \sum_{j = 1} ^{n} \gcd(i, j) ^ K \phi(ij)\\ \sum_{i = 1} ^{n} \sum_{j = 1} ^{n} \gcd(i, j) ^ K \frac{\phi(i) \phi(j) \gcd(i, j)}{\phi(\gcd(i, j))}\\ \sum_{i = 1} ^{n} \sum_{j = 1} ^{n} \gcd(i, j) ^ {K + 1} \frac{\phi(i) \phi(j)}{ \phi(\gcd(i, j))}\\ \sum_{d = 1} ^{n} d ^{K + 1} inv(\phi(d)) \sum_{i = 1} ^{\frac{n}{d}} \sum_{j = 1} ^{\frac{n}{d}} \phi(id) \phi(jd)[\gcd(i, j) = 1]\\ \sum_{d = 1} ^{n} d ^{K + 1} inv(phi(d)) \sum_{k = 1} ^{\frac{n}{d}} \phi(k) \sum_{i = 1} ^{\frac{n}{kd}} \phi(ikd) \sum_{j = 1} ^{\frac{n}{kd}} \phi(jkd)\\ T = kd， 设f(T) = \sum_{i = 1} ^{\frac{n}{T}} \phi(it)\\ \sum_{T = 1} ^{n} f(T) ^ 2 \sum_{d \mid T} d ^{K + 1} inv(\phi(d)) \mu(\frac{T}{d})\\ 设g(n) = \sum_{i \mid n} i ^{K + 1}inv (\phi(i)) \mu(\frac{n}{i})\\ \sum_{i = 1} ^{n} f(i) ^ 2 g(i)即为答案
i=1∑nj=1∑nϕ(ijgcd(i,j)K)i=1∑nj=1∑ngcd(i,j)Kϕ(ij)i=1∑nj=1∑ngcd(i,j)Kϕ(gcd(i,j))ϕ(i)ϕ(j)gcd(i,j)i=1∑nj=1∑ngcd(i,j)K+1ϕ(gcd(i,j))ϕ(i)ϕ(j)d=1∑ndK+1inv(ϕ(d))i=1∑dnj=1∑dnϕ(id)ϕ(jd)[gcd(i,j)=1]d=1∑ndK+1inv(phi(d))k=1∑dnϕ(k)i=1∑kdnϕ(ikd)j=1∑kdnϕ(jkd)T=kd，设f(T)=i=1∑Tnϕ(it)T=1∑nf(T)2d∣T∑dK+1inv(ϕ(d))μ(dT)设g(n)=i∣n∑iK+1inv(ϕ(i))μ(in)i=1∑nf(i)2g(i)即为答案
所以预先处理
i
K
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1
i ^{K + 1}
iK+1次幂，及
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n
v
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ϕ
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inv(\phi(i))
inv(ϕ(i))，即可
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n
log
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(n \log n)
(nlogn)同时算得
f
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f(n)
f(n)，以及
g
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g(n)
g(n)，整体复杂度
O
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O(n \log n)
O(nlogn)，稍卡常，得写 add sub 函数才能过。
#include <bits/stdc++.h>
using namespace std;
const int N = 5e6 + 10, mod = 998244353;
int prime[N], inv[N], phi[N], mu[N], f[N], g[N], a[N], n, k, cnt;
bool st[N];
inline int add(int x, int y) {
return x + y < mod ? x + y : x + y - mod;
}
inline void Add(int &x, int y) {
x + y < mod ? x += y : x += y - mod;
}
inline int sub(int x, int y) {
return x >= y ? x - y : x - y + mod;
}
inline void Sub(int &x, int y) {
x >= y ? x -= y : x += mod - y;
}
int quick_pow(int a, int n) {
int ans = 1;
while (n) {
if (n & 1) {
ans = 1ll * ans * a % mod;
}
a = 1ll * a * a % mod;
n >>= 1;
}
return ans;
}
void init() {
phi[1] = mu[1] = a[1] = inv[1] = 1;
for (int i = 2; i < N; i++) {
inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
if (!st[i]) {
prime[++cnt] = i;
phi[i] = i - 1;
mu[i] = -1;
a[i] = quick_pow(i, k);
}
for (int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {
st[i * prime[j]] = 1;
a[i * prime[j]] = 1ll * a[i] * a[prime[j]] % mod;
if (i % prime[j] == 0) {
phi[i * prime[j]] = phi[i] * prime[j];
break;
}
phi[i * prime[j]] = phi[i] * (prime[j] - 1);
mu[i * prime[j]] = -mu[i];
}
}
for (int i = 1; i <= n; i++) {
for (int j = i; j <= n; j += i) {
Add(f[i], phi[j]);
if (mu[j / i] == 1) {
Add(g[j], 1ll * a[i] * inv[phi[i]] % mod);
}
else if (mu[j / i] == -1) {
Sub(g[j], 1ll * a[i] * inv[phi[i]] % mod);
}
}
}
}
int main() {
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
scanf("%d %d", &n, &k);
k++;
init();
int ans = 0;
for (int i = 1; i <= n; i++) {
Add(ans, 1ll * f[i] * f[i] % mod * g[i] % mod);
}
printf("%d\n", ans);
return 0;
}