hdu 4497 GCD and LCM （素数分解+组合数学）

GCD and LCM

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)

Total Submission(s): 74    Accepted Submission(s): 42

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

思路来源于：大牛博客 点此进入  （代码写的也比我简洁易懂）

思路：

将满足条件的一组x,z,y都除以G,得到x‘，y',z',满足条件gcd(x',y',x') = 1,同时lcm(x',y',x') = G/L.
特判，当G%L ！= 0 时，无解。
然后素数分解G/L，假设G/L = p1^t1 * p2^t2 *````* pn^tn。
满足上面条件的x,y,z一定为这样的形式。
x' = p1^i1 * p2^i2 *```* pn^in.
y' = p1^j1 * p2^j2 * ```*pn^jn.
z' = p1^k1 * p2^k2 * ```*pn^kn.
为了满足上面的条件，对于p1,一定有max(i1,j1,k1) = t1.min(i1,j1,k1) =0;则当选定第一个数为0，第二个数为t1时，第三个数可以为0-t1，又由于有顺序的，只有(0,t1,t1) 和(0,t1,0)这两种情形根据顺序只能产生三种结果，其他的由于三个数都不一样，一定能产生6种，所以最后产生了6*（t1-1）+3*2 = 6*t1种，根据乘法原理以及关于素数分解的唯一性，反过来，素数组合必然也是唯一的数，一共有6*t1 * 6*t2 *`````*6*tn种选法。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#define maxn 100005
using namespace std;

int n,m,cxx;
long long ans;
int cnt[maxn];

void solve()
{
    int i,j,t,m=n;
    memset(cnt,0,sizeof(cnt));
    t=sqrt(n*1.0+0.5);
    for(i=2;i<=t;i++)
    {
        if(n%i==0)
        {
            while(n%i==0)
            {
                n=n/i;
                cnt[i]++;
            }
        }
    }
    if(n!=1) ans=6;
    else ans=1;
    for(i=1;i<=t;i++)
    {
        if(cnt[i]) ans=ans*6*cnt[i];
    }
}
int main()
{
    int i,j,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        if(m%n!=0) printf("0\n");
        else
        {
            n=m/n;
            solve();
            printf("%I64d\n",ans);
        }
    }
    return 0;
}