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
这道题就做完了。

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;
}