【刷题】洛谷 P3455 [POI2007]ZAP-Queries

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 aa , bb and dd , find the number of integer pairs \((x,y)\) satisfying the following conditions:

$1\le x\le a $,\(1\le y\le b\) ,\(gcd(x,y)=d\), where \(gcd(x,y)\) is the greatest common divisor of xx and yy ".

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 \(n\) (\(1\le n\le 50\ 000\) ),denoting the number of queries.

The following \(n\) lines contain three integers each: \(a\) , \(b\) and \(d\) (\(1\le d\le a,b\le 50\ 000\) ), separated by single spaces.

Each triplet denotes a single query.

输出格式：

Your programme should write nn 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

题解

这题其实还是上一篇的弱化版，f(x)和F(x)的定义一样，最后的式子是：

\[ans=f(d)=\sum_{T=1}^{min(a,b)}\mu(\frac{T}{n})\lfloor \frac{a}{T} \rfloor \lfloor \frac{b}{T} \rfloor \]

一样的整除分块，一样的前缀和

#include<bits/stdc++.h>
#define ll long long
const int MAXN=50000+10;
ll T,mu[MAXN],s[MAXN],prime[MAXN],cnt;
bool vis[MAXN];
template<typename T> inline void read(T &x)
{
	T data=0,w=1;
	char ch=0;
	while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
	if(ch=='-')w=-1,ch=getchar();
	while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();
	x=data*w;
}
template<typename T> inline void write(T x,char c='\0')
{
	if(x<0)putchar('-'),x=-x;
	if(x>9)write(x/10);
	putchar(x%10+'0');
	if(c!='\0')putchar(c);
}
template<typename T> inline void chkmin(T &x,T y){x=(y<x?y:x);}
template<typename T> inline void chkmax(T &x,T y){x=(y>x?y:x);}
template<typename T> inline T min(T x,T y){return x<y?x:y;}
template<typename T> inline T max(T x,T y){return x>y?x:y;}
inline void init()
{
	memset(vis,1,sizeof(vis));
	vis[0]=vis[1]=0;
	mu[1]=1;
	for(register int i=2;i<MAXN;++i)
	{
		if(vis[i])
		{
			prime[++cnt]=i;
			mu[i]=-1;
		}
		for(register int j=1;j<=cnt&&i*prime[j]<MAXN;++j)
		{
			vis[i*prime[j]]=0;
			if(i%prime[j])mu[i*prime[j]]=-mu[i];
			else break;
		}
	}
	for(register int i=1;i<MAXN;++i)s[i]=s[i-1]+mu[i];
}
inline ll solve(ll a,ll b,ll d)
{
	ll res=0;
	for(register ll i=1;;)
	{
		if(i>min(a,b))break;
		ll j=min(a/(a/i),b/(b/i));
		res+=(a/i)*(b/i)*(s[j/d]-s[(i-1)/d]);
		i=j+1;
	}
	return res;
}
int main()
{
	init();
	read(T);
	while(T--)
	{
		ll a,b,d;
		read(a);read(b);read(d);
		write(solve(a,b,d),'\n');
	}
	return 0;
}