小清新数学题、、、
化式子
\[\begin{aligned}&\displaystyle\prod_{i=1}^n\prod_{j=1}^n{\frac{\operatorname{lcm}(i,\ j)}{\gcd(i,\ j)}}\\&=(\displaystyle\prod_{i=1}^n\prod_{j=1}^nij)(\displaystyle\prod_{i=1}^n\prod_{j=1}^n\gcd(i,\ j))^{-2}\\&=(\displaystyle\prod_{i=1}^ni^nn!)(\displaystyle\prod_{d=1}^n\prod_{i=1}^n\prod_{j=1}^nd[\gcd(i,\ j)=d])^{-2}\\&=n!^{2n}(\displaystyle\prod_{d=1}^nd^{\displaystyle\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{d}\rfloor}[\gcd(i,\ j)=1]})^{-2}\\&=n!^{2n}(\displaystyle\prod_{d=1}^nd^{2\times(\displaystyle\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\varphi(i))-1})^{-2}\end{aligned}
\]
\]
但是这道题卡时间卡空间、、、
而 \(\varphi\) 前缀和会爆 \(int\) ，所以用欧拉定理，卡卡空间卡卡常就吼辣
时间复杂度 \(O(n\log n)\)
代码
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10, P = 104857601;
int n, tot, p[maxn / 10], phi[maxn];
bitset <maxn> f;
inline int mod(int x, int P) {
return x < P ? x : x - P;
}
inline int qp(int a, int k) {
int res = 1;
for (; k; k >>= 1, a = 1ll * a * a % P) {
if (k & 1) res = 1ll * res * a % P;
}
return res;
}
inline void sieve() {
phi[1] = 1;
for (int i = 2; i <= n; i++) {
if (!f[i]) p[++tot] = i, phi[i] = i - 1;
for (int j = 1; j <= tot && i * p[j] <= n; j++) {
f[i * p[j]] = 1;
if (i % p[j] == 0) {
phi[i * p[j]] = phi[i] * p[j]; break;
}
phi[i * p[j]] = phi[i] * phi[p[j]];
}
}
for (int i = 1; i <= n; i++) {
phi[i] = mod(mod(phi[i] << 1, P - 1) + phi[i - 1], P - 1);
}
}
int main() {
scanf("%d", &n);
sieve();
int s = 1;
for (int i = 1; i <= n; i++) {
s = 1ll * s * i % P;
}
int ans = qp(s, n << 1), sum = 1;
for (int i = 1; i <= n; i++) {
sum = 1ll * sum * qp(i, mod(phi[n / i] + P - 2, P - 1)) % P;
}
sum = qp(sum, P - 2);
sum = 1ll * sum * sum % P;
printf("%d", 1ll * ans * sum % P);
return 0;
}