洛谷P3455 [POI2007]ZAP-Queries

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本文介绍了一种用于破解密码的算法，该算法通过计算特定条件下的整数对数量来解决相关问题。利用莫比乌斯反演简化计算过程，并采用整除分块技术优化效率。

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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 $a$ , $b$ and $d$ , find the number of integer pairs $(x, y)$ satisfying the following conditions:

$1 \leq x \leq a$ , $1 \leq y \leq b$ , $g c d (x, y) = d$ , where $g c d (x, y)$ 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 $n$ ( $1 \leq n \leq 50000$ ),denoting the number of queries.

The following $n$ lines contain three integers each: $a$ , $b$ and $d$ ( $1 \leq d \leq a, b \leq 50000$ ), separated by single spaces.

Each triplet denotes a single query.

输出格式：

Your programme should write $n$ lines to the standard output. The $i$ 'th line should contain a single integer: theanswer to the

$i$ '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

我们设：

$f (k) = i = 1 \sum a j = 1 \sum b [g c d (i, j) = k]$

$F (n) = n ∣ k \sum f (k) = ⌊ \frac{a}{n} ⌋ ⌊ \frac{b}{n} ⌋$

则可以由莫比乌斯反演可以推出：f(n)=n∣k∑μ(⌊nk⌋)F(k)

设完这两个函数之后，我们便惊喜的发现， $A n s = f (d)$

于是就直接开始推答案：

$A n s = d ∣ k \sum μ (⌊ \frac{k}{d} ⌋) F (k)$

枚举 $⌊ \frac{k}{d} ⌋$ ，设为 $t ，则$

$A n s = t = 1 \sum m i n (a, b) μ (t) ⌊ \frac{a}{t d} ⌋ ⌊ \frac{b}{t d} ⌋$

这时候，这个式子已经可以做到 $O (n)$ 的时间复杂度了，但是因为有多组数据，所以我们再用一下整除分块，这题就可以做到 $O (sqrt(n))$ 了。

附代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#define MAXN (1<<16)
using namespace std;
long long k=0,prime[MAXN],mu[MAXN],sum[MAXN];
bool vis[MAXN];
inline int read(){
	int date=0,w=1;char c=0;
	while(c<'0'||c>'9'){if(c=='-')w=-1;c=getchar();}
	while(c>='0'&&c<='9'){date=date*10+c-'0';c=getchar();}
	return date*w;
}
void make(){
	int m=MAXN-10;
	mu[1]=1;
	for(int i=2;i<=m;i++){
		if(!vis[i]){
			mu[i]=-1;
			prime[++k]=i;
		}
		for(int j=1;j<=k&&prime[j]*i<=m;j++){
			vis[prime[j]*i]=true;
			if(i%prime[j]==0)break;
			else mu[prime[j]*i]=-mu[i];
		}
	}
	for(int i=1;i<=m;i++)sum[i]=sum[i-1]+mu[i];
}
void work(){
	long long a,b,d,maxn,ans=0;
	a=read();b=read();d=read();
	maxn=min(a,b);
	for(long long l=1,r;l<=maxn;l=r+1){
		r=min(a/(a/l),b/(b/l));
		ans+=(a/(l*d))*(b/(l*d))*(sum[r]-sum[l-1]);
	}
	printf("%lld\n",ans);
}
int main(){
	int t=read();
	make();
	while(t--)work();
	return 0;
}