「题解」Luogu P5518 [MtOI2019]幽灵乐团 / 莫比乌斯反演基础练习题

P5518 [MtOI2019]幽灵乐团 / 莫比乌斯反演基础练习题

Description

• 多测。

• 在一开始给定模数 \(p\)。

• 每次给定 \(3\) 个正整数 \(a, b, c\)，求

  \[\prod_{i = 1}^a \prod_{j = 1}^b \prod_{k = 1}^c \left(\dfrac{\operatorname{lcm}(i, j)}{\gcd(i, k)} \right)^{f(type)} \bmod p \]

  其中 \(type\in \{0, 1, 2\}\)：

  \[f(0) = 1 \\ f(1) = ijk \\ f(2) = \gcd(i, j, k) \]

  每次分别回答 \(type = 0/ 1/ 2\) 时的答案。

• \(1\le a, b, c\le 10^5, 10^7 \le p \le 1.05\times 10^9, p\in \mathbb{P}, T = 70\)。

Solution

以下令 \(S(n) = \sum_{i = 1}^n i\)​。

---

参考了 peterwuyihong 大佬提供的思路，在 \(\prod\) 较多较复杂时，可以对原式去 \(\ln\)，将其变成 \(\sum\) 的形式，处理完之后再使用 \(\exp\) 转回 \(\prod\)​ 的形式。

---

\[ans = \prod_{i = 1}^a \prod_{j = 1}^b \prod_{k = 1}^c \left(\dfrac{ij}{\gcd(i, j)\gcd(i, k)} \right)^{f(type)} \]

发现每一项只与 \(2\) 个 \(\prod\) 有关，于是设：

\[g(a, b, c) = \prod_{i = 1}^a \prod_{j = 1}^b \prod_{k = 1}^c i^{f(type)} \\ h(a, b, c) = \prod_{i = 1}^a \prod_{j = 1}^b \prod_{k = 1}^c \gcd(i, j)^{f(type)} \]

则

\[ans = \dfrac{g(a, b, c) g(b, a, c)}{h(a, b, c) h(a, c, b)} \]

分类讨论。

type = 0

\[\begin{aligned} g(a, b, c) & = \prod_{i = 1}^a \prod_{j = 1}^b \prod_{k = 1}^c i^1 \\ & = (a!)^{bc} \end{aligned} \]

预处理阶乘做到单次 \(\Omicron(\log p)\)​​。

---

\[h(a, b, c) = \prod_{i = 1}^a \prod_{j = 1}^b \prod_{k = 1}^c \gcd(i, j)^1 \]

不妨设 \(a\le b\)。

取 \(\ln\) 得

\[\begin{aligned} \sum_{i = 1}^a \sum_{j = 1}^b \sum_{k = 1}^c \ln(\gcd(i, j)) & = c \sum_{d = 1}^a \ln(d) \sum_{i = 1}^a \sum_{j = 1}^b [\gcd(i, j) = d] \\ & = c \sum_{d = 1}^a \ln(d) \sum_{k = 1}^{\left\lfloor\frac{a}{d}\right\rfloor} \mu(k) \left\lfloor\dfrac{a}{dk}\right\rfloor \left\lfloor\dfrac{b}{dk}\right\rfloor \\ & = c \sum_{T = 1}^a \left\lfloor\dfrac{a}{T}\right\rfloor \left\lfloor\dfrac{b}{T}\right\rfloor \sum_{d\mid T} \ln(d) \mu\left(\dfrac{T}{d}\right) \end{aligned} \]

取 \(\exp\) 得

\[\left[\prod_{T = 1}^a \left(\prod_{d \mid T} d^{\mu\left(\frac{T}{d}\right)} \right)^{\left\lfloor\frac{a}{T}\right\rfloor \left\lfloor\frac{b}{T}\right\rfloor} \right]^c \]

中间那个可以 \(\Omicron(n\ln n\log p)\) 预处理，结合整除分块即可 \(\Omicron(\sqrt{n}\log p)\)。

type = 1

\[\begin{aligned} g(a, b, c) & = \prod_{i = 1}^a \prod_{j = 1}^b \prod_{k = 1}^c i^{ijk} \\ & = \left(\prod_{i = 1}^a i^i \right)^{S(b) S(c)} \end{aligned} \]

\(\Omicron(n\log p)\) 预处理，\(\Omicron(\log p)\)​ 回答。

---

\[h(a, b, c) = \prod_{i = 1}^a \prod_{j = 1}^b \prod_{k = 1}^c \gcd(i, j)^{ijk} \]

不妨设 \(a\le b\)。

取 \(\ln\) 得

\[\begin{aligned} \sum_{i = 1}^a \sum_{j = 1}^b \sum_{k = 1}^c \ln(\gcd(i, j)) ijk & = \sum_{d = 1}^a \ln(d) \sum_{i = 1}^a i \sum_{j = 1}^b j \sum_{k = 1}^c k [\gcd(i, j) = d] \\ & = S(c) \sum_{d = 1}^a \ln(d) \sum_{i = 1}^{\left\lfloor\frac{a}{d}\right\rfloor} id \sum_{j = 1}^{\left\lfloor\frac{b}{d}\right\rfloor} jd [\gcd(i, j) = 1] \\ & = S(c) \sum_{d = 1}^a \ln(d) \sum_{k = 1}^{\left\lfloor\frac{a}{d}\right\rfloor} \mu(k) \sum_{i = 1}^{\left\lfloor\frac{a}{dk}\right\rfloor} idk \sum_{j = 1}^{\left\lfloor\frac{b}{dk}\right\rfloor} jdk \\ & = S(c) \sum_{d = 1}^a \ln(d) d^2 \sum_{k = 1}^{\left\lfloor\frac{a}{d}\right\rfloor} \mu(k) k^2 S\left(\left\lfloor\dfrac{a}{dk}\right\rfloor \right) S\left(\left\lfloor\dfrac{b}{dk}\right\rfloor \right) \\ & = S(c) \sum_{T = 1}^a S\left(\left\lfloor\dfrac{a}{T}\right\rfloor \right) S\left(\left\lfloor\dfrac{b}{T}\right\rfloor \right) \sum_{d\mid T} \ln(d) d^2 \mu\left(\dfrac{T}{d}\right) \left(\dfrac{T}{d}\right)^2 \\ & = S(c) \sum_{T = 1}^a T^2 S\left(\left\lfloor\dfrac{a}{T}\right\rfloor\right) S\left(\left\lfloor\dfrac{b}{T}\right\rfloor\right) \sum_{d\mid T} \ln(d) \mu\left(\dfrac{T}{d}\right) \end{aligned} \]

取 \(\exp\) 得

\[\left[\prod_{T = 1}^a \left(\prod_{d\mid T} d^{\mu\left(\frac{T}{d}\right) T^2} \right)^{S\left(\left\lfloor\frac{a}{T}\right\rfloor \right) S\left(\left\lfloor\frac{b}{T}\right\rfloor \right)} \right]^{S(c)} \]

照样是 \(\Omicron(n\ln n\log p)\) 的预处理和 \(\Omicron(\sqrt{n} \log p)\) 的回答。

type = 2

直接计算原式（注意原式不可调换 \(a, b\)）

\[\prod_{i = 1}^a \prod_{j = 1}^b \prod_{k = 1}^c \left(\dfrac{\operatorname{lcm}(i, j)}{\gcd(i, k)} \right)^{\gcd(i, j, k)} \]

取 \(\ln\) 得

\[\begin{aligned} \sum_{i = 1}^a \sum_{j = 1}^b \sum_{k = 1}^c \ln\left(\dfrac{\operatorname{lcm}(i, j)}{\gcd(i, k)} \right) \gcd(i, j, k) & = \sum_{d = 1}^{\min(a, b, c)} \varphi(d) \sum_{i = 1}^{\left\lfloor\frac{a}{d}\right\rfloor} \sum_{j = 1}^{\left\lfloor\frac{b}{d}\right\rfloor} \sum_{k = 1}^{\left\lfloor\frac{c}{d}\right\rfloor} \ln\left(\dfrac{\operatorname{lcm}(id, jd)}{\gcd(id, kd)} \right) \\ & = \sum_{d = 1}^{\min(a, b, c)} \varphi(d) \sum_{i = 1}^{\left\lfloor\frac{a}{d}\right\rfloor} \sum_{j = 1}^{\left\lfloor\frac{b}{d}\right\rfloor} \sum_{k = 1}^{\left\lfloor\frac{c}{d}\right\rfloor} \ln\left(\dfrac{\operatorname{lcm}(i, j)}{\gcd(i, k)} \right) \end{aligned} \]

取 \(\exp\) 得E

\[\prod_{d = 1}^{\min(a, b, c)} \left(\prod_{i = 1}^{\left\lfloor\frac{a}{d}\right\rfloor} \prod_{j = 1}^{\left\lfloor\frac{b}{d}\right\rfloor} \prod_{k = 1}^{\left\lfloor\frac{c}{d}\right\rfloor} \dfrac{\operatorname{lcm}(i, j)}{\gcd(i, k)} \right)^{\varphi(d)} \]

发现没？中间那一块就是 \(type = 0\) 时的答案！

于是整除分块套整除分块实现 \(\Omicron(\sqrt{n} \log p \cdot \sqrt{n}\log p) = \Omicron(n\log^2 p)\)。

Code

tips : 注意哪些是指数，要 \(\bmod p - 1\)。

// 18 = 9 + 9 = 18.
#include <iostream>
#include <cstdio>
#include <cstring>
#define Debug(x) cout << #x << "=" << x << endl
typedef long long ll;
using namespace std;

namespace IO
{
}
using IO::read;

int p, q;

int add1(int a, int b) {return (a + b) % p;}
int add2(int a, int b) {return (a + b) % q;}
int sub1(int a, int b) {return (a - b + p) % p;}
int sub2(int a, int b) {return (a - b + q) % q;}
int mul1(int a, int b) {return (ll)a * b % p;}
int mul2(int a, int b) {return (ll)a * b % q;}
int qpow(int a, int b) {if (b < 0) b += q; int base = a, ans = 1; while (b) {if (b & 1) ans = mul1(ans, base); base = mul1(base, base); b >>= 1;} return ans;}
int inv(int a) {return qpow(a, p - 2);}

const int MAXN = 1e5 + 5;
typedef int arr[MAXN];
const int N = 1e5;

arr prime, mu, phi, phiS, fac, dyy, dyyP, dyyPP, dyyPPI, dyyPI, indexP, xhj, xhjP, xhjPP, xhjPPI, xhjPI;
bool vis[MAXN];

void pre()
{
	mu[1] = phi[1] = 1;
	for (int i = 2; i <= N; i++)
	{
		if (!vis[i])
		{
			prime[++prime[0]] = i;
			mu[i] = -1;
			phi[i] = i - 1;
		}
		for (int j = 1; j <= prime[0] && i * prime[j] <= N; j++)
		{
			vis[i * prime[j]] = true;
			if (i % prime[j] == 0)
			{
				mu[i * prime[j]] = 0;
				phi[i * prime[j]] = phi[i] * prime[j];
				break;
			}
			mu[i * prime[j]] = mu[i] * mu[prime[j]];
			phi[i * prime[j]] = phi[i] * phi[prime[j]];
		}
	}
	
	for (int i = 1; i <= N; i++)
	{
		phiS[i] = add2(phiS[i - 1], phi[i]);
	}
	
	fac[0] = 1;
	for (int i = 1; i <= N; i++)
	{
		fac[i] = mul1(fac[i - 1], i);
	}
	
	for (int i = 1; i <= N; i++)
	{
		dyy[i] = 1;
	}
	for (int i = 1; i <= N; i++)
	{
		for (int j = 1; i * j <= N; j++)
		{
			dyy[i * j] = mul1(dyy[i * j], qpow(i, mu[j]));
		}
	}
	dyyP[0] = dyyPP[0] = 1;
	for (int i = 1; i <= N; i++)
	{
		dyyP[i] = mul1(dyyP[i - 1], dyy[i]);
		dyyPP[i] = mul1(dyyPP[i - 1], dyyP[i]);
	}
	dyyPPI[N] = inv(dyyPP[N]);
	dyyPI[0] = 1;
	for (int i = N - 1; i >= 0; i--)
	{
		dyyPPI[i] = mul1(dyyPPI[i + 1], dyyP[i + 1]);
		dyyPI[i + 1] = mul1(dyyPPI[i + 1], dyyPP[i]);
	}
	
	indexP[0] = 1;
	for (int i = 1; i <= N; i++)
	{
		indexP[i] = mul1(indexP[i - 1], qpow(i, i));
	}
	
	for (int i = 1; i <= N; i++)
	{
		xhj[i] = 1;
	}
	for (int i = 1; i <= N; i++)
	{
		for (int j = 1; i * j <= N; j++)
		{
			xhj[i * j] = mul1(xhj[i * j], qpow(i, mul2(mu[j], mul2(i * j, i * j))));
		}
	}
	xhjP[0] = xhjPP[0] = 1;
	for (int i = 1; i <= N; i++)
	{
		xhjP[i] = mul1(xhjP[i - 1], xhj[i]);
		xhjPP[i] = mul1(xhjPP[i - 1], xhjP[i]);
	}
	xhjPPI[N] = inv(xhjPP[N]);
	xhjPI[0] = 1;
	for (int i = N - 1; i >= 0; i--)
	{
		xhjPPI[i] = mul1(xhjPPI[i + 1], xhjP[i + 1]);
		xhjPI[i + 1] = mul1(xhjPPI[i + 1], xhjPP[i]);
	}
}

int S(int n)
{
	return (ll)n * (n + 1) / 2 % q;
}

int GetSum_phi(int l, int r)
{
	return sub2(phiS[r], phiS[l - 1]);
}

int GetPro_dyy(int l, int r)
{
	return mul1(dyyP[r], dyyPI[l - 1]);
}

int GetPro_xhj(int l, int r)
{
	return mul1(xhjP[r], xhjPI[l - 1]);
}

struct Type0
{
	int g(int a, int b, int c)
	{
		return qpow(fac[a], mul2(b, c));
	}
	
	int h(int a, int b, int c)
	{
		if (a > b)
		{
			swap(a, b);
		}
		int res = 1;
		for (int l = 1, r; l <= a; l = r + 1)
		{
			int k1 = a / l, k2 = b / l;
			r = min(a / k1, b / k2);
			res = mul1(res, qpow(GetPro_dyy(l, r), mul2(k1, k2)));
		}
		return qpow(res, c);
	}
	
	int GetAns(int a, int b, int c)
	{
		int x = mul1(g(a, b, c), g(b, a, c)), y = mul1(h(a, b, c), h(a, c, b));
		return mul1(x, inv(y));
	}
}type0;

struct Type1
{
	int g(int a, int b, int c)
	{
		return qpow(indexP[a], mul2(S(b), S(c)));
	}
	
	int h(int a, int b, int c)
	{
		if (a > b)
		{
			swap(a, b);
		}
		int res = 1;
		for (int l = 1, r; l <= a; l = r + 1)
		{
			int k1 = a / l, k2 = b / l;
			r = min(a / k1, b / k2);
			res = mul1(res, qpow(GetPro_xhj(l, r), mul2(S(k1), S(k2))));
		}
		return qpow(res, S(c));
	}
	
	int GetAns(int a, int b, int c)
	{
		int x = mul1(g(a, b, c), g(b, a, c)), y = mul1(h(a, b, c), h(a, c, b));
		return mul1(x, inv(y));
	}
}type1;

struct Type2
{
	int GetAns(int a, int b, int c)
	{
		int res = 1, upbound = min(a, min(b, c));
		for (int l = 1, r; l <= upbound; l = r + 1)
		{
			int k1 = a / l, k2 = b / l, k3 = c / l;
			r = min(a / k1, min(b / k2, c / k3));
			res = mul1(res, qpow(type0.GetAns(k1, k2, k3), GetSum_phi(l, r)));
		}
		return res;
	}
}type2;

signed main()
{
	int t = read();
	p = read(), q = p - 1;
	pre();
	while (t--)
	{
		int a = read(), b = read(), c = read();
		printf("%d %d %d\n", type0.GetAns(a, b, c), type1.GetAns(a, b, c), type2.GetAns(a, b, c));
	}
	return 0;
}

标签：P5518,lfloor,right,int,题解,sum,MtOI2019,prod,left
来源： https://www.cnblogs.com/mangoworld/p/Solution-Luogu-P5518.html