[Luogu3455][POI2007]ZAP-Queries

BZOJ（权限题）

Luogu

题目描述

Byteasar the Cryptographer works on breaking the code of BSA (Byteotian Security Agency). He has alreadyfound out that whilst deciphering a message he will have to answer multiple queries of the form"for givenintegers aaa, bbb and ddd, find the number of integer pairs (x,y) satisfying the following conditions:

1≤x≤a,1≤y≤b,gcd(x,y)=d, where gcd(x,y)is the greatest common divisor of x and y".

Byteasar would like to automate his work, so he has asked for your help.

TaskWrite a programme which:

reads from the standard input a list of queries, which the Byteasar has to give answer to, calculates answers to the queries, writes the outcome to the standard output.

FGD正在破解一段密码，他需要回答很多类似的问题：对于给定的整数a,b和d，有多少正整数对x,y，满足x<=a，y<=b，并且gcd(x,y)=d。作为FGD的同学，FGD希望得到你的帮助。

输入输出格式

输入格式：

The first line of the standard input contains one integer nnn (1≤n≤50 000),denoting the number of queries.

The following nnn lines contain three integers each: aaa, bbb and ddd(1≤d≤a,b≤50 000), separated by single spaces.

Each triplet denotes a single query.

输出格式：

Your programme should write nnn lines to the standard output. The iii'th line should contain a single integer: theanswer to the iii'th query from the standard input.

输入输出样例

输入样例#1：

2

4 5 2

6 4 3

输出样例#1：

3

2

sol

参见莫比乌斯反演总结中的举个栗子、骚操作以及奇技淫巧部分

这个题我们要求$$\sum_{i=1}^{{a}\sum_{j-1}}{b}[gcd(i,j)==d]$$

首先使用骚操作把式子化成

\[\sum_{i=1}^{a/d}\sum_{j=1}^{b/d}[gcd(i,j)==1]
\]

以下令\(a/d=n\)，\(b/d=m\)，且\(n\le m\)。

接着设两个函数\(f(x)\)，\(F(x)\)，其中

\[f(x)=\sum_{i=1}^{n}\sum_{j-1}^{m}[gcd(i,j)==x]
\]

\[F(x)=\sum_{x|d}^{n}f(d)
\]

那么我们可以手动脑补出这个关系

\[F(x)=\lfloor\frac nx\rfloor\lfloor\frac mx\rfloor
\]

所以\(F(x)\)是可以\(O(1)\)算的，处理出全部的\(F(x)\)，总时间复杂度是\(O(n)\)

又根据莫比乌斯反演可得

\[f(x)=\sum_{x|d}^{n}\mu(\frac dx)F(d)
\]

因为我们需要求的是\(f(1)\)所以把\(x=1\)代入得

\[f(1)=\sum_{d=1}^{n}\mu(d)F(d)
\]

所以我们只要搞出所有的\(F(x)\)跟所有的\(\mu(x)\)就可以了

但是每组数据都这么搞一遍，\(O(Tn)\)的时间复杂度承受不了啊怎么办。

这里就运用到奇技淫巧中提到的数论分块。

可以证明，\(F(x)\)在\([1,n]\)上只有\(O(\sqrt n)\)种取值。

我们维护一个\(\mu(x)\)的前缀和，然后把\(F(x)\)值相同的放在一起算贡献。

总复杂度变为\(O(T\sqrt n)\)

具体实现参见代码。

code

#include<cstdio>
#include<algorithm>
using namespace std;
#define ll long long
const int N = 50005;
const int n = 50000;
int gi()
{
	int x=0,w=1;char ch=getchar();
	while ((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
	if (ch=='-') w=0,ch=getchar();
	while (ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
	return w?x:-x;
}
int mu[N],pri[N],tot,s[N];
bool zhi[N];
void Mobius()
{
	zhi[1]=true;mu[1]=1;
	for (int i=2;i<=n;i++)
	{
		if (!zhi[i]) pri[++tot]=i,mu[i]=-1;
		for (int j=1;j<=tot&&i*pri[j]<=n;j++)
		{
			zhi[i*pri[j]]=true;
			if (i%pri[j]) mu[i*pri[j]]=-mu[i];
			else {mu[i*pri[j]]=0;break;}
		}
	}
	for (int i=1;i<=n;i++) s[i]=s[i-1]+mu[i];
}
ll calc(int a,int b,int k)
{
	a/=k;b/=k;
	if (a>b) swap(a,b);
	int i=1,j;ll ans=0;
	while (i<=a)
	{
		j=min(a/(a/i),b/(b/i));
		ans+=1ll*(s[j]-s[i-1])*(a/i)*(b/i);
		i=j+1;
	}
	return ans;
}
int main()
{
	int T=gi();
	Mobius();
	while (T--)
	{
		int a=gi(),b=gi(),k=gi();
		printf("%lld\n",calc(a,b,k));
	}
	return 0;
}