杜教筛学习笔记

前置知识

积性函数

积性函数分为积性函数和完全积性函数。

• 积性函数：\(\forall{a\perp{b}},f(a*b)=f(a)*f(b)\)。常见的有\(\varphi.\mu,\sigma,d\)等
• 完全积性函数：\(\forall{a,b},f(a*b)=f(a)*f(b)\)。常见的有\(\epsilon,I,id\)等

其中\(\epsilon(n)=[n=1],I(n)=1,id(n)=n\)

狄利克雷卷积

\((f*g)(n)=\sum\limits_{d|n}f(d)g(\frac{n}{d})\)，满足交换律和结合律。

定义\(\epsilon\)为单位元（\(f*\epsilon=f\)，根据定义易得）

\(\begin{cases}\mu*I=\epsilon\\\varphi*I=id\\\mu*id=\varphi\end{cases}\)

莫比乌斯反演

见\(Link\)

杜教筛

原理

设\(\sum\limits_{i=1}^nf(i)=S(n)\)

再找一个积性函数\(g\)，考虑\((f*g)\)的前缀和。

\[\sum\limits_{i=1}^{n}(f*g)(i) \]

\[=\sum\limits_{i=1}^n\sum\limits_{d|i}f(d)g(\frac{i}{d}) \]

\[=\sum\limits_{d=1}^ng(d)\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}f(i) \]

\[=\sum\limits_{d=1}^ng(d)S(\lfloor\frac{n}{d}\rfloor) \]

又因为

\[g(1)S(n) \]

\[=\sum\limits_{d=1}^ng(d)S(\lfloor\frac{n}{d}\rfloor)-\sum\limits_{d=2}^ng(d)S(\lfloor\frac{n}{d}\rfloor) \]

\[=\sum\limits_{i=1}^{n}(f*g)(i)-\sum\limits_{d=2}^ng(d)S(\lfloor\frac{n}{d}\rfloor) \]

如果我们能找到合适的\(g\)来快速算出\(\sum\limits_{i=1}^{n}(f*g)(i)\)，剩下的就能够整除分块了，从而可以递归求解。

复杂度为\(O(n^{\frac{3}{4}})\)，预处理前\(m\)项使复杂度达到\(O(\frac{n}{\sqrt{m}})\)，\(m\)取\(n^{\frac{2}{3}}\)时达到最优\(O(n^{\frac{2}{3}})\)

实践

若\(f(n)=\mu(n)\)，取\(g(n)=I\)，则\((f*g)(n)=\epsilon(n)\)

若\(f(n)=\varphi(n)\)，取\(g(n)=I\)，则\((f*g)(n)=id(n)\)

注意若\(n=2^{31}-1\)时，整除分块中\(l=r+1\)可能会爆\(int\)，改为\(unsigned\ int\)即可。

一定要记得预处理！！同时记得是\(\sum\limits_{d=2}^ng(d)S(\lfloor\frac{n}{d}\rfloor)\)，而不是\(\sum\limits_{d=2}^n S(d)g(\lfloor\frac{n}{d}\rfloor)\)，否则复杂度不对！

下面是模板代码：

#include <bits/stdc++.h>

using namespace std;

#define ll long long

const int lim = (1 << 22), mx = 2147483647; 

int t, n, tot, mu[4200005], phi[4200005], vis[4200005], pr[1250005], prs_mu[4200005];

ll prs_phi[4200005];

map < int, int > fmu;
map < int, ll > fphi;

int read()
{
	int x = 0, fl = 1; char ch = getchar();
	while (ch < '0' || ch > '9') { if (ch == '-') fl = -1; ch = getchar();}
	while (ch >= '0' && ch <= '9') {x = x * 10ll + ch - '0'; ch = getchar();}
	return x * fl;
}

void init()
{
	phi[1] = mu[1] = 1;
	for (int i = 2; i <= lim; i ++ )
	{
		if (!vis[i])
		{
			vis[i] = i;
			pr[ ++ tot] = i;
			phi[i] = i - 1;
			mu[i] = -1;
		}
		for (int j = 1; j <= tot; j ++ )
		{
			if (i * pr[j] > lim || pr[j] > vis[i]) break;
			vis[i * pr[j]] = pr[j];
			if (i % pr[j])
			{
				mu[i * pr[j]] = -mu[i];
				phi[i * pr[j]] = phi[i] * phi[pr[j]];
			}
			else
			{
				mu[i * pr[j]] = 0;
				phi[i * pr[j]] = phi[i] * pr[j];
			}
		}
	}
	for (int i = 1; i <= lim; i ++ )
	{
		prs_mu[i] = prs_mu[i - 1] + mu[i];
		prs_phi[i] = prs_phi[i - 1] + phi[i];
	}
	return;
}

int d_mu(unsigned int x)
{
	if (x <= lim) return prs_mu[x];
	if (fmu[x]) return fmu[x];
	int cnt = 1;
	for (unsigned int l = 2, r; l <= x; l = r + 1)
	{
		r = min(x, x / (x / l));
		cnt = cnt - (r - l + 1) * d_mu(x / l);
	}
	return fmu[x] = cnt;
}

ll d_phi(unsigned int x)
{
	if (x <= lim) return prs_phi[x];
	if (fphi[x]) return fphi[x];
	ll cnt = 1ll * x * (x + 1) / 2;
	for (unsigned int l = 2, r; l <= x; l = r + 1)
	{
		r = min(x, x / (x / l));
		cnt = cnt - (r - l + 1) * d_phi(x / l);
	}
	return fphi[x] = cnt;
}

int main()
{
	init();
	t = read();
	while (t -- )
	{
		n = read();
		printf("%lld %d\n", d_phi(n), d_mu(n));
	}
	return 0;
}