hdu4497GCD and LCM+数论

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

lcm(x,y,z)=L=pn11pn11pn22∗∗∗pnnn
gcd(x,y,z)=G=pm11pm11pm22∗∗∗pmnn
对于L,G都存在的质因子p1而言，x,y,z中肯定有一个的指数n1,有一个的指数m1，而另外一个可以是m1-n1中的数.而3个数可以相互交换
1.当n1,n1,m1 ans=3;
2.当n1,m1,m1 ans=3;
3.当n1,z1,m1 ans=(n1-m1-1)*6;
所以综上ans=(n1-m1)*6
但是注意当n1==m1 ans=1;
最后要注意质因数分解是有可能出现两个都是质数的情况。和mi=0的情况
//数据太水。。质因数分解枚举到L也过？(分解质因数wa了几发）fu**

#include<bits/stdc++.h>using namespace std;#define LL long longint factA[25],factB[25];int tot;void getfact(LL L,LL G){    tot=1;    LL N=(LL)(sqrt(L)+0.5);    for(LL i=2;i<=N;i++){        if(L%i==0){            while(L%i==0){                factA[tot]++;                L/=i;            }            while(G%i==0){                factB[tot]++;                G/=i;            }            tot++;        }    }    if(L>1){        factA[tot]++;        if(G>1)factB[tot]++;        tot++;    }}int main(){    int t;    scanf("%d",&t);    while(t--){        LL G,L;        scanf("%I64d %I64d",&G,&L);        if(L%G!=0){            printf("0\n");            continue;        }        else{            memset(factA,0,sizeof(factA));            memset(factB,0,sizeof(factB));            getfact(L,G);            LL ans=1;            for(int i=1;i<tot;i++){                if(factA[i]>factB[i]) ans*=(factA[i]-factB[i])*6;            }            printf("%I64d\n",ans);        }    }    return 0;}