落谷 P3455 ZAP-Queries（莫比乌斯反演入门题+分块）

####题目链接：https://www.luogu.org/problem/P3455
P3455 [POI2007]ZAP-Queries
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通过 2.32k
时间限制 2.00s
内存限制 125.00MB
题目提供者 洛谷
难度
省选/NOI-
历史分数 100
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标签
POI
高性能
2007
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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 aaa, bbb and ddd, find the number of integer pairs (x,y)(x,y)(x,y) satisfying the following conditions:

1≤x≤a1\le x\le a1≤x≤a,1≤y≤b1\le y\le b1≤y≤b,gcd(x,y)=dgcd(x,y)=dgcd(x,y)=d, where gcd(x,y)gcd(x,y)gcd(x,y) is the greatest common divisor of xxx and yyy".

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。
输入格式

The first line of the standard input contains one integer nnn (1≤n≤50 0001\le n\le 50\ 0001≤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 0001\le d\le a,b\le 50\ 0001≤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 #1

2
4 5 2
6 4 3

输出 #1

3
2

---

分析：

与hdu 1695类似 hdu 1695题解
最后得分块，hdu 1695不分块也会过，但这题不分块会超时。首先预处理下前n个mob[i]的的和。然后就是分块。学到了两个变量的分块，具体看代码：

代码：

	#include 

using namespace std;
#define ll long long 
#define fr first
#define sc second

const int N=5e4+5;
int prime[N+1];
int mob[N+1];
ll sum[N+1];

void Mobius()
{
	memset(prime,0,sizeof(prime));
	mob[1]=1;
	sum[1]=1;
	for(int i=2;i<=N;i++)
	{
		if(!prime[i])
		{
			prime[++prime[0]]=i;
			mob[i]=-1;
		}
		for(int j=1;j<=prime[0]&&i*prime[j]<=N;j++)
		{
			prime[i*prime[j]]=1;
			if(i%prime[j]==0)
			{
				mob[i*prime[j]]=0;
				break;
			}
			mob[i*prime[j]]=-mob[i];
		}
		sum[i]=sum[i-1]+mob[i];
	}
}


int main()
{
	Mobius();
	int t,a , b, d;
	ll ans;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d%d",&a,&b,&d);
		ans=0;
		a/=d,b/=d;
		int tmp = min(a,b);
		// for(int i=1;i<=tmp;i++) ans +=mob[i]*(a/i)*(b/i);
		//上面的超时了，得分块
		for(int l=1,r;l<=tmp;l=r+1)
		{
			r=min(a/(a/l),b/(b/l));// 对于两个变量，还可以这样分块，学到了
			ans += 1LL*(sum[r]-sum[l-1])*(a/l)*(b/l);
		}
		printf("%d\n",ans);
		
	}

	return 0;
}