【题解】Luogu-P5221 Product

P5221 Product

Description

• 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;
}