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GCD and LCM

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 3103 Accepted Submission(s): 1356

Problem Description Given two positive integers G and L, could you tell me how many solutions of (x, y, z) there are, satisfying that gcd(x, y, z) = G and lcm(x, y, z) = L?
Note, gcd(x, y, z) means the greatest common divisor of x, y and z, while lcm(x, y, z) means the least common multiple of x, y and z.
Note 2, (1, 2, 3) and (1, 3, 2) are two different solutions.

Input First line comes an integer T (T <= 12), telling the number of test cases.
The next T lines, each contains two positive 32-bit signed integers, G and L.
It’s guaranteed that each answer will fit in a 32-bit signed integer.

Output For each test case, print one line with the number of solutions satisfying the conditions above.

Sample Input 2 6 72 7 33

Sample Output 72 0

//Pro: hdu 4497 GCD and LCM

//先寫點東西： 
//如果有非負整數a,b, 將他們分解質因數,得到：
//a=p1^k1 * p2^k2 * p3^k3 *...* pn^kn
//b=p1^q1 * p2^q2 * p3^q3 *...* pn^qn (p>0, q>=0, k>=0, k,q不同時為0)
//顯然地， 
//gcd(a,b)=p1^(min(k1,p1)) * p2(min(k2,p2)) *...* pn^(min(kn,qn))
//lcm(a,b)=p1^(max(k1,p1)) * p2(max(k2,p2)) *...* pn^(max(kn,qn))
 
//=> a*b=p1^(k1+q1) * p2^(k2+q2) * p3^(k3+q3) *...* pn^(kn+qn)=gcd(a,b) * lcm(a,b)

//題意：
//三個未知數x,y,z，它們的gcd為G，lcm為L，G和L已知，求(x,y,z)三元組的個數 
//由上邊那個東東可以知道，將x,y,z,G,L質因數分解之後
//他們的相同的因子Pi的次數 ki_G<=ki_x,ki_y,ki_z<=ki_L,
//且min(ki_x,ki_y,ki_z)==ki_G && max(x,y,z)==ki_L
//所以x，y，z裏有兩個數的某一質因子的個數是固定的，分別等於ki_L或ki_G,
//剩下的第三個數的該質因子的個數可以在[ki_G,ki_L]中任選 
//(ki_g<ki_a<ki_l)的排列一共有6種，如果ki_a==ki_g或者ki_a==ki_l，各有三種排列，共2*3=6種
//那麽(x,y,z)的個數一共有6*(l-g-1)+2*3=6*(l-g)種 


#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std; 

const int N=1e7+5;

int T;
int l,g,a;

bool nprime[N];
int prime[N],cnt;
void init()
{
    nprime[1]=1;
    for(int i=2;i<N;++i)
    {
        if(!nprime[i])
            prime[++cnt]=i;
        for(int j=1,d;j<=cnt&&(d=i*prime[j])<N;++j)
        {
            nprime[d]=1;
            if(i%prime[j]==0)
                break;
        }
    }
}

int ans;
int main()
{
    init();
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&g,&l);
        if(l%g)
        {
            puts("0");
            continue;
        }
        l/=g;    //把相同的因子約一下
        ans=1;
        for(int i=1,k;i<=cnt;++i)
        {
            if(l%prime[i])
                continue;
            k=0;
            while(l%prime[i]==0)
            {
                ++k;
                l/=prime[i];
            }
            ans*=6*k;
        }
        printf("%d\n",ans);
    }
    return 0;
}

HDU 4497 GCD and LCM