Description
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给定整数 \(n\)，请求出
\[\left[\prod_{i = 1}^n \prod_{j = 1}^n \dfrac{\operatorname{lcm}(i, j)}{\gcd(i, j)} \right] \bmod 104857601 \] -
对于 \(100\%\) 的数据，\(1\le n\le 10^6\)。
Solution
\[\begin{aligned} \prod_{i = 1}^n \prod_{j = 1}^n \dfrac{\operatorname{lcm}(i, j)}{\gcd(i, j)} & = \prod_{i = 1}^n \prod_{j = 1}^n \dfrac{ij}{\gcd(i, j)^2} \\ & = \dfrac{\prod\limits_{i = 1}^n \prod\limits_{j = 1}^n ij}{\left[\prod\limits_{i = 1}^n \prod\limits_{j = 1}^n \gcd(i, j) \right]^2} \end{aligned} \]对于分子
\[\begin{aligned} \prod_{i = 1}^n \prod_{j = 1}^n ij & = \prod_{i = 1}^n i^n\cdot n! \\ & = (n!)^n \cdot (n!)^n \\ & = (n!)^{2n} \end{aligned} \]对于分母（忽略 \(2\) 次方）
\[\begin{aligned} \prod_{i = 1}^n \prod_{j = 1}^n \gcd(i, j) & = \prod_{d = 1}^n \prod_{i = 1}^n \prod_{j = 1}^n d[\gcd(i, j) = d] \\ & = \prod_{d = 1}^n \prod_{i = 1}^{\left\lfloor\frac{n}{d}\right\rfloor} \prod_{j = 1}^{\left\lfloor\frac{n}{d}\right\rfloor} d[\gcd(i, j) = 1] \\ & = \prod_{d = 1}^n d^{\sum_{i = 1}^{\left\lfloor\frac{n}{d}\right\rfloor} \sum_{j = 1}^{\left\lfloor\frac{n}{d}\right\rfloor} [\gcd(i, j) = 1]} \end{aligned} \]拆指数
\[\sum_{i = 1}^{\left\lfloor\frac{n}{d}\right\rfloor} \sum_{j = 1}^{\left\lfloor\frac{n}{d}\right\rfloor} [\gcd(i, j) = 1] = \sum_{k = 1}^{\left\lfloor\frac{n}{d}\right\rfloor} \mu(k) \left\lfloor\dfrac{n}{dk}\right\rfloor^2 \]代回去
\[\prod_{d = 1}^n d^{\sum_{i = 1}^{\left\lfloor\frac{n}{d}\right\rfloor} \sum_{j = 1}^{\left\lfloor\frac{n}{d}\right\rfloor} [\gcd(i, j) = 1]} = \prod_{d = 1}^n d^{\sum_{k = 1}^{\left\lfloor\frac{n}{d}\right\rfloor} \mu(k) \left\lfloor\frac{n}{dk}\right\rfloor^2} \]指数用欧拉定理 \(\bmod (p - 1)\) 转成正数并降至 \(\Omicron(p)\) 级别。
暴力枚举 + 整除分块。
时间复杂度为 \(\Omicron\left(\sum_{i = 1}^n \sqrt{\dfrac{n}{i}} \cdot \log p \right) = \Omicron(\log p \sum_{i = 1}^n \sqrt{\dfrac{n}{i}})\)。
当积分算
\[\int_1^n \sqrt{\dfrac{n}{x}}\, dx = 2n - 2\sqrt{n} \]所以整个是 \(\Omicron(n\log p)\) 的。
注意本题卡空间，\(\mu\) 数组的前缀和直接覆盖到 \(\mu\) 数组上即可。
Code
// 18 = 9 + 9 = 18.
#include <iostream>
#include <cstdio>
#define Debug(x) cout << #x << "=" << x << endl
typedef long long ll;
using namespace std;
const int MAXN = 1e6 + 5;
const int MOD = 104857601;
int p[MAXN / 10], mu[MAXN];
bool vis[MAXN];
void pre(int n)
{
mu[1] = 1;
for (int i = 2; i <= n; i++)
{
if (!vis[i])
{
p[++p[0]] = i;
mu[i] = -1;
}
for (int j = 1; j <= p[0] && i * p[j] <= n; j++)
{
vis[i * p[j]] = true;
if (i % p[j] == 0)
{
mu[i * p[j]] = 0;
break;
}
mu[i * p[j]] = mu[i] * mu[p[j]];
}
}
for (int i = 1; i <= n; i++)
{
mu[i] = (mu[i - 1] + mu[i] + MOD - 1) % (MOD - 1);
}
}
int GetSum(int l, int r)
{
return (mu[r] - mu[l - 1] + MOD - 1) % (MOD - 1);
}
int block(int n)
{
int res = 0;
for (int l = 1, r; l <= n; l = r + 1)
{
int k = n / l;
r = n / k;
res = (res + (ll)GetSum(l, r) * k % (MOD - 1) * k % (MOD - 1)) % (MOD - 1);
}
return res;
}
int qpow(int a, int b)
{
int base = a, ans = 1;
while (b)
{
if (b & 1)
{
ans = (ll)ans * base % MOD;
}
base = (ll)base * base % MOD;
b >>= 1;
}
return ans;
}
int inv(int a)
{
return qpow(a, MOD - 2);
}
int main()
{
int n;
scanf("%d", &n);
pre(n);
int fac = 1;
for (int i = 1; i <= n; i++)
{
fac = (ll)fac * i % MOD;
}
fac = qpow(fac, 2 * n);
int ans = 1;
for (int i = 1; i <= n; i++)
{
ans = (ll)ans * qpow(i, block(n / i)) % MOD;
}
printf("%d\n", (ll)fac * inv((ll)ans * ans % MOD) % MOD);
return 0;
}