HDU4497：GCD and LCM（数论）

GCD and LCM

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2648    Accepted Submission(s): 1162

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：若3个数的lcm为L，gcd为G，一定有G%L==0，否则直接输出结果0；根据lcm和gcd的性质，假设3个数为a，b，c，则他们一定可以整除G，为了简化问题，将3个数和L都除以G，原题变成求gcd为1，lcm为L/G的3个数的组合数，将k=L/G分解质因数，假设分解结果为p1^x1 * p2^x2 * p3^x3......* pn^xn。分析其中一个质因子p1，其数目为x1个，即这三个数各自分解质因数后必定有一个数的p1因子为0个，必定有一个数的p1因子有x1个，另外一个数的p1因子可以是[0, x1]个，根据排列组合，当另外一个数p1因子数为0时，有3种组合，当为x1时也有3种组合，为(0,x1)时，有(x1-1)*6种组合，3+3+（x1-1）*6 = 6*x1种，对于其他质因子依此类推采用乘法原理算出结果。

# include <stdio.h>
# define MAXN 500000
int a[MAXN+1]={0}, b[MAXN+1];
int main()
{
    int t, g, l, num = 0;
    for(int i=2; i<MAXN; ++i)
        if(!a[i])
        {
            b[num++] = i;
            for(int j=i; j<MAXN; j+=i)
                a[j] = 1;
        }
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&g,&l);
        int icount, ans = 1;
        if(l%g)
            puts("0");
        else
        {
            int k = l/g;
            for(int i=0; i<num&&(b[i]*b[i]<=k); ++i)
            {
                if(k%b[i]==0)
                {
                    icount = 0;
                    for(; k%b[i]==0; k/=b[i],++icount);
                    ans *= icount*6;
                }
            }
            if(k != 1)
                ans *= 6;
            printf("%d\n",ans);
        }
    }
    return 0;
}