[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 ,  and , find the number of integer pairs  satisfying the following conditions:

,,, where  is the greatest common divisor of  and ".

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  (),denoting the number of queries.

The following  lines contain three integers each: ,  and (), separated by single spaces.

Each triplet denotes a single query.

输出格式：

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

莫比乌斯反演
详见如下




那么就可以在O(n)复杂度内求出
也可以优化到O(sqrt(n))
详见code

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 50010;
int a,b,d;
int n;
int mu[maxn];
int prime[maxn],is_prime[maxn];
void get_mu()
{
    int num=0;
    mu[1]=1;
    for(int i=2;i<=50000;i++)
    {
        if(!is_prime[i]){
            mu[i]=-1;prime[++num]=i;
        }
        for(int j=1;j<=num&&i*prime[j]<=50000;j++)
        {
            is_prime[i*prime[j]]=1;
            if(i%prime[j]==0)
            {
                mu[i*prime[j]]=0;break;
            }
            mu[i*prime[j]]=-mu[i];
        }
    }
    for(int i=1;i<=50000;i++){
        mu[i]=mu[i]+mu[i-1];
    }
}

int main()
{
    scanf("%d",&n);
    get_mu();
    while(n--)
    {
        scanf("%d%d%d",&a,&b,&d);
        a/=d,b/=d;
        int ans=0;int last;
        for(int i=1;i<=a&&i<=b;i=last+1)
        {
            last=min(a/(a/i),b/(b/i));
            ans+=(mu[last]-mu[i-1])*(a/i)*(b/i);
        }
        printf("%d
",ans);
    }
    return 0;
}