莫比乌斯反演学习笔记

前言

本文只是作者学习莫比乌斯反演后做的相关笔记，可能并不适合入门。

基础

莫比乌斯函数

设 (x=p_1^{alpha_1} imes p_2^{alpha_2} imes dots imes p_k^{alpha_k})。

则：

(mu(x)=egin{cases} 1&{x=1} \ 0&{alpha_i>1(1le ile k)} \ (-1)^k & ext{otherwise} end{cases})

一个基本性质

设 (S(n)=sumlimits_{d|n}mu(d))。

则 (S(n)=[n=1])。

证明略。

一个简单结论

[[gcd(i,j)=1]Leftrightarrowsumlimits_{d|gcd(i,j)}mu(d) ]

第一个式子

[egin{aligned} &F(n)=sumlimits_{d|n}f(d) \ Rightarrow &f(n)=sumlimits_{d|n}mu(d)F(frac{n}{d}) end{aligned}]

证明：

[egin{aligned} f(n) &= sumlimits_{d|n}mu(d)F(frac{n}{d}) \ &= sumlimits_{d|n}mu(d)sumlimits_{i|frac{n}{d}}f(i) \ &= sumlimits_{i|n}f(i)sumlimits_{d|frac{n}{i}}mu(d) \ &= sumlimits_{i|n}f(i)S(frac{n}{i}) \ &= f(n) &&square end{aligned}]

第二个式子

[egin{aligned} &F(n)=sumlimits_{n|d}f(d) \ Rightarrow &f(n)=sumlimits_{n|d}mu(frac{d}{n})F(d) end{aligned}]

证明：

令 (d'=frac{d}{n})。

[egin{aligned} f(n) &= sumlimits_{n|d}mu(frac{d}{n})F(d) \ &= sumlimits_{n|d}mu(frac{d}{n})sumlimits_{d|i}f(i) \ &= sumlimits_{n|i}f(i)sumlimits_{d'|frac{n}{i}}mu(d') \ &= sumlimits_{n|i}f(i)S(frac{n}{i}) \ &= f(n) &&square end{aligned}]

值得注意的是，我们在做题中常用的是第二个式子。

例题

[HAOI2011]Problem b

题意：求 (sumlimits_{ale xle b}sumlimits_{cle yle d}[gcd(x, y)=k])。

设 (S(a, b)=sumlimits_{1le xle a}sumlimits_{1le yle b}[gcd(x, y)=k])。

可以通过二维前缀和的方式转化问题，答案即为 (S(b,d)-S(a-1,d)-S(b,c-1)+S(a-1,c-1))。

接下来有两种方法求解。

方法一：

设 (n=lfloorfrac{a}{k} floor)，(m=lfloorfrac{b}{k} floor)。

[egin{aligned} &sumlimits_{x=1}^asumlimits_{y=1}^b [gcd(x,y)=k] \ =&sumlimits_{x=1}^{n}sumlimits_{y=1}^{m}[gcd(x,y)=1] \ =&sumlimits_{x=1}^{n}sumlimits_{y=1}^{m}sumlimits_{d|gcd(x,y)}mu(d) \ =&sumlimits_{x=1}^nsumlimits_{y=1}^msumlimits_{d|x,d|y}mu(d) \ =&sumlimits_{d=1}^{min(n,m)}mu(d)sumlimits_{lfloorfrac{n}{d} floor}sumlimits_{lfloorfrac{m}{d} floor}1 \ =&sumlimits_{d=1}^{min(n,m)}mu(d)lfloorfrac{n}{d} floorlfloorfrac{m}{d} floor end{aligned}]

方法二：

设 (F(n)=sumlimits_{x=1}^asumlimits_{y=1}^b[n|gcd(x,y)]=lfloorfrac{a}{n} floorlfloorfrac{b}{n} floor)，(f(n)=sumlimits_{x=1}^asumlimits_{y=1}^b[gcd(x,y)=n])。

很显然有：(F(n)=sumlimits_{n|d}f(d))。

莫比乌斯反演得到：(f(n)=sumlimits_{n|d}mu(frac{d}{n})F(d))。

令 (d'=frac{d}{n})，(a'=frac{a}{n})，(b'=frac{b}{n})。

推式子：

[egin{aligned} f(n)&=sumlimits_{n|d}mu(frac{d}{n})F(d) \ &=sumlimits_{n|d}mu(frac{d}{n})lfloorfrac{a}{n} floorlfloorfrac{b}{n} floor \ &=sumlimits_{d'}mu(d')lfloorfrac{a}{d'n} floorlfloorfrac{b}{d'n} floor \ &=sumlimits_{d'}mu(d')lfloorfrac{a'}{d'} floorlfloorfrac{b'}{d'} floor \ end{aligned}]

然后整除分块即可。

代码：

#include <bits/stdc++.h>
#define DEBUG fprintf(stderr, "Passing [%s] line %d
", __FUNCTION__, __LINE__)
#define File(x) freopen(x".in","r",stdin); freopen(x".out","w",stdout)

using namespace std;

typedef long long LL;
typedef pair <int, int> PII;
typedef pair <int, PII> PIII;

template <typename T>
inline T gi()
{
    T f = 1, x = 0; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return f * x;
}

const int INF = 0x3f3f3f3f, N = 50003, M = N << 1;

int T;
int mu[N], sum[N];
bool isprime[N];
int pri[N], tot;

inline void pre()
{
	mu[1] = 1;
	for (int i = 2; i <= 50000; i+=1)
	{
		if (!isprime[i]) pri[++tot] = i, mu[i] = -1;
		for (int j = 1; j <= tot && 1ll * i * pri[j] <= 50000; j+=1)
		{
			isprime[i * pri[j]] = true;
			if (i % pri[j] == 0) break;
			mu[i * pri[j]] = -mu[i];
		}
	}
	for (int i = 1; i <= 50000; i+=1) sum[i] = sum[i - 1] + mu[i];
}

inline LL S(int a, int b, int k)
{
	a /= k, b /= k;
	int n = min(a, b);
	LL ans = 0;
	for (int l = 1, r; l <= n; l = r + 1)
	{
		r = min(n, min(a / (a / l), b / (b / l)));
		ans = ans + 1ll * (sum[r] - sum[l - 1]) * (a / l) * (b / l);
	}
	return ans;
}

int main()
{
    //File("");
    pre();
    T = gi <int> ();
    while (T--)
    {
    	int a = gi <int> (), b = gi <int> (), 
    		c = gi <int> (), d = gi <int> (), 
    		k = gi <int> ();
    	printf("%lld
", S(b, d, k) - S(a - 1, d, k) 
    					- S(b, c - 1, k) + S(a - 1, c - 1, k));
    }
    return 0;
}