莫比乌斯反演--「HAOI 2011」Problem b

简介: 对于一些函数(f(n))，如果很难直接求出他的值，而容易其倍数或约数和(g(n))，可以通过莫比乌斯反演简化问题

前置知识：

• 数论分块

其实在之前的博客里，做余数求和的时候总结过，但是也忘得差不多啦

数论分块的过程：考虑求含有(leftlfloor frac{n}{i} ight floor)的求和式子

对于任意一个(i)((ileq n)),我们需要找到一个最大的(j)((ileq j leq n)),满足(leftlfloor dfrac{n}{i} ight floor = leftlfloor dfrac{n}{j} ight floor)，此时(j=leftlfloor frac{n}{leftlfloor frac{n}{i} ight floor} ight floor)

这样我们可以把一些值相同的数分成一大块一起计算，是不是方便很多？

证明：

(k = leftlfloor frac{n}{i} ight floor)，考虑证明当(leftlfloor frac{n}{j} ight floor = k)时，(j)的最大值为(leftlfloor frac{n}{k} ight floor)

(leftlfloor frac{n}{j} ight floor = k Longleftrightarrow k leq frac{n}{j}<k+1 Longleftrightarrow frac{1}{k+1} < frac{j}{n}leq frac{1}{k} Longleftrightarrow frac{n}{k+1} < j leq frac{n}{k+1})

又因为(j)是整数，所以(j_{max} = leftlfloor frac{n}{k} ight floor)，这样我们每次以([i,j])分块求和即可

• 莫比乌斯函数

(mu)为莫比乌斯函数，定义为：(mu (n) = egin{cases}1&n=1\0&n含有平方因子\{(-1)}^k&k为n的本质不同的质因子个数end{cases})

性质：

1. (sumlimits_{dmid n} mu (d) = egin{cases} 1&n = 1\0&n eq 1end{cases})

2. ([gcd(i,j)==1] Longleftrightarrow sumlimits_{dmid gcd(i,j)} mu (d))

code:

void init(){
	mo[1] = 1;
	for (int i = 2;i <= 1e6;i++){
		if (!death[i]) primelist[++tot] = i,mo[i] = -1;
		for (int j = 1;j <= tot&&i*primelist[j] <= 1e6;j++){
			death[i*primelist[j]] = 1;
			if (i%primelist[j] == 0){
				mo[i*primelist[j]] = 0;
				break;
			}
			mo[i*primelist[j]] = -mo[i];
		}
	}
	for (int i = 1;i <= 1e6;i++) mo[i] += mo[i-1];
}

• 莫比乌斯反演

公式：

定义(f(n))，(g(n))为两个数论函数

如果有(f(n) = sumlimits_{dmid n}g(d))，那么有(g(n) = sumlimits_{dmid n} mu (d)f(frac{n}{d}))
如果有(f(n) = sumlimits_{dmid n}g(d))，那么有(g(n) = sumlimits_{dmid n} mu (frac{n}{d})f(d))

例题：

problem:

求符合(gcd(x,y) = k)的((x,y))的个数

solution:

(sumlimits_{i=1}^nsumlimits_{j=1}^m [gcd(i,j) = k] Longleftrightarrow sumlimits_{i=1}^{leftlfloorfrac{n}{k} ight floor}sumlimits_{j=1}^{leftlfloorfrac{m}{k} ight floor}[gcd(i,j) = 1] Longleftrightarrow sumlimits_{i=1}^{leftlfloorfrac{n}{k} ight floor}sumlimits_{j=1}^{leftlfloorfrac{m}{k} ight floor} sumlimits_{dmid gcd(i,j)} mu (d))

交换求和顺序，枚举(dmid gcd(i,j))即可

(sumlimits_{d=1} mu (d)sumlimits_{i=1}^{leftlfloorfrac{n}{k} ight floor} [dmid i]sumlimits_{j=1}^{leftlfloorfrac{m}{k} ight floor} [dmid j]Longleftrightarrow sumlimits_{d=1} mu (d)leftlfloor frac{n}{k} ight floorleftlfloorfrac{m}{k} ight floor)

code:

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
int read(){
	int x = 1,a = 0;char ch = getchar();
	while (ch < '0'||ch > '9'){if (ch == '-') x = -1;ch = getchar();}
	while (ch >= '0'&&ch <= '9'){a = a*10+ch-'0';ch = getchar();}
	return x*a;
}
const int maxn = 5e4+10;
int n;
int a,b,c,d,k;
int death[maxn],primelist[maxn],tot,mu[maxn];
void init(){
	mu[1] = 1;
	for (int i = 2;i <= 50000;i++){
		if (!death[i]) primelist[++tot] = i,mu[i] = -1;
		for (int j = 1;j <= tot&&i*primelist[j] <= 50000;j++){
			death[i*primelist[j]] = 1;
			if (i%primelist[j] == 0){
				mu[i*primelist[j]] = 0;
				break;
			}
			mu[i*primelist[j]] = -mu[i];
		}
	}
	for (int i = 1;i <= 50000;i++) mu[i] += mu[i-1];
}
int solve(int n,int m){
	int res = 0;
	for (int l = 1,r;l <= min(n,m);l = r+1){
		r = min(n/(n/l),m/(m/l));
		res += (mu[r]-mu[l-1])*(n/l)*(m/l);
	}
	return res;
}
int main(){
	n = read();init();
	for (int i = 1;i <= n;i++){
		a = read(),b = read(),c = read(),d = read(),k = read();
		 printf("%d
", solve(b/k,d/k)-solve(b/k,(c-1)/k)-solve((a-1)/k,d/k)+solve((a-1)/k,(c-1)/k));
	}
	return 0;
}