数论2&莫&杜

积性函数:

积性函数定义ok

积性函数指对于所有互质的整数\(a\)和\(b\)有性质\(f(ab)=f(a)f(b)\)的数论函数

除数函数?

莫比乌斯函数\(\mu\)ok

\[\phi(i) =
\begin{cases}
1, & i==1 \\
(-1)^k, & i == \prod_{i=1}^{k}{p_i^1 {p}, p_i为所有质因子 \\
0, & otherwise
\end{cases}
\]

为什么是积性函数?

对任意\(a,b,gcd(a,b)=1\), \(ab\)的质因子集合可以不重不漏分到\(a,b\)的质因子集合中

分类讨论一下,满足定义

欧拉函数\(\phi\)ok

小于等于\(n\)的数中与\(n\)互质的数的数目

通式:

\[\phi(n)=n*\prod_{i=1}^{k}(1-\frac{1}{p_i})\\
\]

通式的意义:把他展开, 相当于做容斥, 筛除的所有\(n\)因数

为什么是积性函数?

对任意\(a,b,gcd(a,b)=1\), \(a,b\)没有不含有相同的\(p\), 直接带入通式得证

反正关于欧拉函数的东西盯着这个通式就好了

幂函数\(Id\)ok

\(id_k(n)=n^k\)

\(id(n)=id_1(n)=n\)

显然是积性函数

单位函数\(\epsilon\)ok

\(\epsilon(n)=[n=1]\)

显然是积性函数

线性筛

核心:保证每个数只被他最小的质数筛到一次

int tot;
bool vis[MAXN];
int p[MAXN], phi[MAXN], mu[MAXN];
void filter()
{
	phi[1] = mu[1] = 1;
	for (int i = 2; i <= n; ++ i)
		{
			if (!vis[i]) p[++ tot] = i, phi[i] = i - 1, mu[i] = -1;
			for (int j = 1; j <= tot && i * p[j] <= n; ++ j)
				{
					vis[i * p[j]] = true;
					if (i % p[j] == 0)
						{
							phi[i * p[j]] = phi[i] * p[j]; mu[i * p[j]] = 0;
                        //没有增加质因子, 结合phi的通式可得
							break;
						}
					else
						{
							phi[i * p[j]] = phi[i] * phi[p[j]];
							mu[i * p[j]] = -mu[i];
						}
				}
			printf("%d %d\n", mu[i], phi[i]);
		}
}

Dirichlet卷积

\[(f*g)(n)=\sum_{d|n}f(d)g(\frac{n}{d})
\]

交换律 \(f ∗ g = g ∗ f\)

结合律 \((f ∗ g) ∗ h = f ∗ (g ∗ h)\)

分配律 \(f ∗ (g + h) = f ∗ g + f ∗ h\)

单位元 \(f ∗ ϵ = f\)

若 \(f,g\) 均为积性函数，则 \(f*g\) 也是积性函数

证明：

对任意\(a,b,gcd(a,b)=1\), \(ab\)的质因子集合可以不重不漏分到\(a,b\)的质因子集合中

\((f*g)(ab) = \sum_{d|ab}f(d)g(\frac{ab}{d})\)

设\(d=d_a*d_b\), \(d_a,d_b\)分别由\(a,b\)的因子组成, 显然组成的方式唯一, 并且\(d\)唯一分解成\(d_a,d_b\)

\[\begin{aligned}
(f*g)(ab)&=\sum_{d_a|a, d_b|b}f(d_a)f(d_b)g(\frac{a}{d_a})g(\frac{b}{d_b})\\
&=\sum_{d_a|a, d_b|b}f(d_a)g(\frac{a}{d_a})*f(d_b)g(\frac{b}{d_b})\\
&=(f*g)(a)*(f*g)(b)
\end{aligned}
\]

已知\(f,g\),\(O(nlog_2^n)\)求\((f*g)\)

for (int i = 1; i * i <= n; ++ i)
{
    (f*g)[i*j] += f[i] * g[i];
    for (int j = i + 1; i * j <= n; ++ j)
        (f*g)[i*j] += f[i] * g[j] + f[j] * g[i];
}

不会证

---

常见数论函数

\(\sigma(n)\)表示整数\(n\)的所有正因数的和

\(\tau(n)\)表示整数\(n\)的所有正因数的个数

\(Id(i)=i\)

以及前面的一些积性函数

常见狄利克雷卷积及其证明

\(\epsilon = \mu *1\)

证明:

\(n\)含有多个相同质因子或\(=1\)是显然成立

考虑\(n=p_1*p_2*...*p_k\), \(p_i\)都是不同质因子

\[\begin{aligned}
\epsilon(n) & = \sum_{d|n}\mu(d) \\
& = \sum_{i=1}^{k}\binom{k}{i}*(-1)^k \\
\end{aligned}
\]

由于\((x+y)^k=\sum x^iy^{k-i}*\binom{k}{i}\), 带入\(x=-1, y=1\), 凑进去

\[\begin{aligned}
\epsilon(n)&=\sum_{i=1}^{k}0^k\\
&=
\begin{cases}
1, & k=0\ so\ n=1 \\
0, & otherwise
\end{cases}
\end{aligned}
\]

---

\(\sigma = Id*1\)

\(\tau=1*1\)

显然

---

\(\phi = \mu * Id\)

证明:

\[\begin{aligned}
\phi(n) & = \sum_{d|n}\mu(d)id(\frac{n}{d}) \\
& = 1 \cdot n + \sum_{d=\prod p}\mu(d)id(\frac{n}{d}) + 0\cdot (...)\\
& = n + \sum_{i=1}^{k}(-1)^k*n*\frac{1}{\prod{p}} \\
& = n * \prod_{i=1}^{k}(1-\frac{1}{p_i})
\end{aligned}
\]

---

\(\phi * 1= Id\)

上面一个等式两边同乘以\(1\)

---

莫比乌斯Van演

定理:\(\epsilon = \mu * 1\)

若函数 \(f, g\) 满足

\(f(n) = \sum_{d|n}g(d)\)

则

\(g(n) = \sum_{d|n}\mu(d)f(\frac{n}{d}) = \sum_{d|n}\mu(\frac{n}{d})f(d)\)

证明:$

\(f = g*1 \longleftrightarrow \mu * f = g\) $

左式同乘以\(\mu\), 右式同乘以\(1\)

常用:

运用这个卷积:

\(\epsilon = \mu * 1\) 即 \(\sum_{d|n}\mu(d) = 1\)

\([\gcd(i, j) == 1] = \sum_{k | \gcd(i,j)} \mu(k)\)

例题:

NOI2015寿司

BZOJ1257: CQOI2007余数之和

题意:

\(n, k(\le10^9)\), 求\(\sum_{i=1}^{n}(k \mod i)\)

Sol:

\(\sum_{i=1}^{n}(k \mod i) = \sum_{i=1}^{n}(k - k / i * i) = nk - \sum_{i=1}^{n}(k/i)*i\)

BZOJ1101: POI2007Zap

\(T\) 组询问，\(1 ≤ T, n, m , d≤ 50000\)

\[\sum_{x=1}^{n}\sum_{y=1}^{m}[\gcd(x,y)=d]
\]

相当于求

\[\sum_{x=1}^{n/d}\sum_{y=1}^{m/d}[\gcd(x,y)=1]
\]

n /= d, m /= d 根据\(\epsilon = \mu * 1\),即求

\[\begin{aligned}
& = \sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{d|gcd(i,j)}\mu(d) \\
& = \sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{d|i,d|j}\mu(d) \\
& = \sum_d\mu(d)\lfloor\frac{n}{d}\rfloor\lfloor\frac{m}{d}\rfloor
\end{aligned}
\]

除法分块\(O(\sqrt{n})\)

BZOJ2154: Crash的数字表格

\(1\le n, m\le10^7\), 求

\[\sum_{i=1}^{n}\sum_{j=1}^{m} lcm(i,j)
\]

\(i*j=\gcd(i,j) * lcm(i,j)\)

\[\begin{aligned}
&= \sum_{i=1}^{n}\sum_{j=1}^{m}\frac{i*j}{\gcd(i,j)} \\
&= \sum_{i=1}^{n}\sum_{j=1}^{m}\frac{i}{\gcd(i,j)}\frac{j}{\gcd(i,j)}\gcd(i,j) \\
&= \sum_{g}g\sum_{i=1}^{n/g}\sum_{j=1}^{m/g}i*j*[\gcd(i,j)=1] \\
&= \sum_{g}g\sum_{i=1}^{n/g}\sum_{j=1}^{m/g}i*j*\sum_{d|gcd(i,j)}\mu(d) \\
&= \sum_{g}g\sum_{i=1}^{n/g}\sum_{j=1}^{m/g}i*j*\sum_{d|i,d|j}\mu(d) \\
&= \sum_{g}g\sum_{d}\sum_{i=1}^{n/gd}\sum_{j=1}^{m/gd}i*d*j*d*\mu(d) \\
&= \sum_{gd}gd\sum_{i=1}^{n/gd}\sum_{j=1}^{m/gd}i*j*d*\mu(d) \\
\end{aligned}
\]

设\(p=g*d\)

\[\begin{aligned}
&= \sum_{p}p*\sum_{d|p}d*\mu(d)\sum_{i=1}^{n/p}\sum_{j=1}^{m/p}ij
\end{aligned}
\]

完了,然后怎么做?

大爷xyyxyyx说: \(F(p)=\sum_{d|p}d*\mu(d)*p\)是积性函数

为什么?

设\(a,b,gcd(a,b)=1,d=da*db,da|a,db|b\)

\[F(ab)=\sum_{da|a,db|b}da*\mu(da)*db*\mu(db)*a*b
\]

好了, 然后\(O(n)\)筛?然后暴力\(O(n)\)求答案?

观察\(F(i)\)的式子, 套用线性筛:

• \(i\)为质数: \(F(i)=(1*\mu(1)+i*\mu(i))*i\)
• 用已经筛出来的质数\(p[j]\)去筛\(i*p[j]\)
  • \(i\mod p[j]!=0\), 互质!\(F(i*p[j])=F(i)*F(p[j])\)
  • \(i \mod p[j] == 0\),考虑对比\(F(i)\),新加入的因数必须含有\(p^2\), 而这样的因数\(\mu{}\)是等于0的!那么直接\(F(i*p[j])=F(i)*p[j]\)

然后少写了一个\(mod\), WA了,然后又T了

整出分块!!注意到在一段区间里\((n/p, m/p)\)是不变的, 所以可以对\(F\)直接区间求和(前缀和实现)

//空间注意一点
#include <bits/stdc++.h>

using namespace std;
typedef long long LL;
const int MAXN = 1e7 + 10;
const LL MOD = 20101009;

int n, m;
LL ans;

int tot;
bool vis[MAXN];
int mu[MAXN], p[664590];
LL f[MAXN];
int sum[MAXN], sumf[MAXN];
void filter()
{
    mu[1] = sumf[1] = f[1] = sum[1] = 1;
	LL tmp1 = 0;
    for (int i = 2; i <= m; ++ i)
        {
            tmp1 = 1LL * i * (i + 1) / 2 % MOD;
			sum[i] = tmp1;
            if (!vis[i]) mu[i] = -1, p[++ tot] = i, f[i] = 1LL * (1 + i * mu[i] % MOD + MOD) * i % MOD;
            for (int j = 1; j <= tot && i * p[j] <= m; ++ j)
                {
                    vis[i * p[j]] = true;
                    if (i % p[j] == 0) { mu[i * p[j]] = 0; f[i * p[j]] = f[i] * p[j] % MOD; break; }
                    else mu[i * p[j]] = -mu[i], f[i * p[j]] = f[i] * f[p[j]] % MOD;
                }
            sumf[i] = (sumf[i - 1] + f[i]) % MOD;
        }
}

int main()
{
    scanf("%d%d", &n, &m);
    if (n > m) swap(n, m);
    filter();
    for (int i = 1; i <= n; )
        {
            int r = min(n, min(n / (n / i), m / (m / i)));
            (ans += sum[n / i] % MOD * sum[m / i] % MOD * (sumf[r] - sumf[i - 1] + MOD) % MOD) %= MOD;
            i = r + 1;
        }
    printf("%lld\n", ans);
    return 0;
}

PE530: GCD of Divisors

Let f(n) be the sum of the greatest common divisor of d and n/d over all positive divisors d of n, that is \(f(n)=\displaystyle\sum\limits_{d|n}\, \text{gcd}(d,\frac n d)\)

Let F be the summatory function of f, that is \(F(k)=\displaystyle\sum\limits_{n=1}^k \, f(n)\)

You are given that \(F(10)=32\) and \(F(1000)=12776.\)

Find \(F(1015)\).

提交答案题

\[F(n) = \sum_{i=1}^{n}\sum_{d|i}{\gcd(d, \frac{i}{d})}
\]

相当于枚举\(i\), 并将\(i\)分解成两个数的乘积, 求所有这样的\(\gcd\)的和

等价于枚举两个数, 他们乘积\(\le n\), 求他们\(\gcd\)的乘积

所以:

\[\begin{aligned}
F(n)&=\sum_{i = 1}^{n}\sum_{j=1}^{n/i}{\gcd(i,j)} \\
&=\sum_gg\sum_i\sum_j[\gcd(i,j)==1][i*g*j*g\le n]
\end{aligned}
\]

注意不能像以前那样把两个\(\sum{}\)的上界除以\(g\)

从而认为\(i*j\le n/g\)

因为两个\(\sum{}\)是有关联的,只有当第一个确定时,才能枚举第二个, 如果这么做的话相当于同时枚举两个

例如\(i*j\le n*n/i\)显然不对

那么继续

\[\begin{aligned}
F(n) &= \sum_gg\sum_i\sum_j\sum_{d|i,d|j}\mu(d)*[i*j*g^2\le n] \\
&= \sum_gg\sum_d\mu(d)\sum_i\sum_j[i*j\le\frac{n}{g^2*d^2}] \\
\end{aligned}
\]

考虑把\(g*d\)并在一起, 然后康康这个和什么卷积比较像

\[p = gd\\
\phi(x)=\sum_{d|x}Id(x)*\mu(\frac{n}{x}) \\
\]

于是就化简成了

\[F(n)=\sum_p\phi(p)*\sum_i\sum_j[i*j\le\frac{n}{p^2}]
\]

什么,要求\(10^{15}\)?不会炸空间吗?

注意后面\(\lfloor \frac{n}{p^2}\rfloor\), 可以缩小到\(\sqrt{n}\)

那么枚举\(p\), 然后每次单独分块求就好了

\[F(n)=\phi(p)*\sum_i\lfloor\frac{n}{p^2i}\rfloor
\]

复杂度呢?前半个\(O(\sqrt{n})\), 后半个\(O(ln_n)\)?不会证...

#include <bits/stdc++.h>

using namespace std;
typedef long long LL;
const int MAXN = 32e6 + 10;
const LL MOD = 20101009;

LL n;
LL ans;

int tot;
bool vis[MAXN];
LL p[MAXN], phi[MAXN];
LL f[MAXN];
void filter()
{
    phi[1] = 1;
    for (int i = 2; i < MAXN; ++ i)
        {
            if (!vis[i]) p[++ tot] = i, phi[i] = i - 1;
            for (int j = 1; j <= tot && i * p[j] < MAXN; ++ j)
                {
                    vis[i * p[j]] = true;
                    if (i % p[j] == 0) { phi[i * p[j]] = phi[i] * p[j]; break; }
                    else phi[i * p[j]] = phi[i] * phi[p[j]];;
                }
        }
}

LL calc(LL x)
{
	LL ret = 0;
	for (LL i = 1; i <= x; )
		{
			LL r = min(x / (x / i), x);
			ret += (r - i + 1) * (x / i);
			i = r + 1;
		}
	return ret;
}

int main()
{
    scanf("%lld", &n);
	n = 1e15;
    filter();
    for (LL i = 1; i * i <= n; ++ i)
        {
			ans += phi[i] * calc(n / (i * i));
        }
    printf("%lld\n", ans);
    return 0;
}

杜教筛

基本

目的:求\(f(i)\)的前缀和

套路:构造\(h=g*f\)

\[\begin{aligned}
\sum_{i=1}^{n}h(i)&=\sum_{i=1}^{n}\sum_{d|i}g(d)*f(\frac{d}{i}) \\
\end{aligned}
\]

即等同于枚举两个因数，他们乘积\(\le n\),计算他们\(g\)和\(f\)的乘积的和

\[\begin{aligned}
&=\sum_{d=1}^{n}g(d)\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}f({i}) \\
\end{aligned}
\]

设\(S(n)=\sum f(i)\)

\[\begin{aligned}
\sum_{i=1}^{n}h(i)&=\sum_{d=1}^{n}g(d)S(\lfloor\frac{n}{d}\rfloor) \\
\sum_{i=1}^{n}h(i)&=g(1)*S(n)+\sum_{d=2}^{n}g(d)S(\lfloor\frac{n}{d}\rfloor) \\
g(1)*S(n)&=\sum_{i=1}^{n}h(i)-\sum_{d=2}^{n}g(d)S(\lfloor\frac{n}{d}\rfloor)\\
\end{aligned}
\]

所以只要凑出方便的\(h, g\)就可以套用最后一行式子了

\[g(1)*S(n)=\sum_{i=1}^{n}h(i)-\sum_{d=2}^{n}g(d)S(\lfloor\frac{n}{d}\rfloor)
\]

求\(\sum_{i=1}^{n}\mu(i)\)

\(\epsilon = \mu * 1, h=\epsilon,g=1\)

\[S(n)=1-\sum_{i=2}^{n}S(\lfloor\frac{n}{i}\rfloor)
\]

求\(\sum_{i=1}^{n}\phi(i)\)

\(Id=\phi*1, h=Id, g=1\)

\[S(n)=\frac{n*(n+1)}{2}-\sum_{i=2}^{n}S(\lfloor\frac{n}{i}\rfloor)
\]

先看一道例题

例题BZOJ3944: Sum

\(T(\le 10)\)组数据, \(N(\le 2^{31}-1)\), 求\(\sum_{i=1}^{n}\mu(i), \sum_{i=1}^{n}\phi(i)\)

两个长得很像可以一起做, \(\lfloor\frac{n}{i}\rfloor\)可以整除分块

递归实现, 但是要先预先筛一部分, 不然会\(T\), 记忆化搜索用\(map\)记录

#include <cstdio>
#include <map>

using namespace std;
typedef long long LL;
typedef pair<int, LL> PIL;
#define mkpr make_pair
const int MAXN = 5e6 + 2;

inline LL in()
{
    LL x = 0, flag = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') flag = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
    return x * flag;
}

int n;

int tot;
bool vis[MAXN];
int p[348516], mu[MAXN]; //348513
LL phi[MAXN];
void filter()
{
	phi[1] = mu[1] = 1;
	for (int i = 2; i < MAXN; ++ i)
		{
			if (!vis[i]) { mu[i] = -1, phi[i] = i - 1, p[++ tot] = i; }
			for (int j = 1; j <= tot && i * p[j] < MAXN; ++ j)
				{
					vis[i * p[j]] = true;
					if (i % p[j] == 0)
						{ mu[i * p[j]] = 0, phi[i * p[j]] = phi[i] * p[j]; break; }
					else
						mu[i * p[j]] = -mu[i], phi[i * p[j]] = phi[i] * phi[p[j]];
				}
		}
	for (int i = 2; i < MAXN; ++ i) mu[i] += mu[i - 1], phi[i] += phi[i - 1];
}

map <LL, PIL> f; //92680
PIL solve(LL x) // Smu[x] = 1 - [2, n]Smu[x / i],
{
	if (x < MAXN) return mkpr(mu[x], phi[x]);
	if (f.count(x)) return f[x];
	int mu = 1; LL phi = x * (x + 1) / 2;
	for (LL i = 2, nex = i; i <= x; i = nex + 1)
		{
			nex = x / (x / i);
			PIL val = solve(x / i);
			mu -= (nex - i + 1) * val.first;
			phi -= (nex - i + 1) * val.second;
		}
	return f[x] = mkpr(mu, phi);
}

int main()
{
	filter();
	//	printf("%d\n", tot);
	int Test = in();
	while (Test --)
		{
			n = in();
			PIL ans = solve(n);
			printf("%lld %d\n", ans.second, ans.first);
		}
	return 0;
}

复杂度

不会证, 据说预处理\(n^{\frac{2}{3}}\), 复杂度为\(O(n^{\frac{2}{3}})\)