【hdu 4497】GCD and LCM 【算术基本定理】

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

Source
2013 ACM-ICPC吉林通化全国邀请赛——题目重现

Recommend
liuyiding
就是一点 算术基本定理 一个正整数 n 一定可以分解为
n=p1^r1*p2^r2*p3^r3……pn^rn （其中p1p2p3..pn 都是素数）
还有一个公式
IF gcd（a,b）==d to gcd（a/d，b/d）==1 。

这个的题解讲的很清楚了
代码

#include<bits/stdc++.h>#define LL long longusing namespace std;const int MAXN = 500+10;const int MAXM = 1e5;LL g,l;LL p[50],ge;LL r[50];void getn(int m){    ge=0;LL n=m;    memset(r,0,sizeof(r));    for(LL i=2;i*i<=n;i++){        if(n%i==0){            p[ge]=i;            while(n%i==0) {            r[ge]++;n/=i;            }            ge++;            //printf("i==%d\n",i)   ;        }    }    if(n>1) {  p[ge]=n;  r[ge++]++;  }}int main(){    int t;cin>>t;    while(t--){        scanf("%lld%lld",&g,&l);        LL cnt=1;        if(l%g) { puts("0");continue;}        getn(l/g);        for(int i=0;i<ge;i++)             cnt=cnt*6*r[i];        printf("%lld\n",cnt);    }      return 0;}