GCD and LCM

最新推荐文章于 2021-07-18 23:58:10 发布

zhengde152

最新推荐文章于 2021-07-18 23:58:10 发布

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分类专栏：数论 - 唯一定理

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GCD and LCM

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

题解：给你三个数x,y,z的最大公约数G和最小公倍数L，问你三个数字一共有几种可能。注意123和321算两种情况。

运用唯一定理。

代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define maxn 10000005
int t,g,l,prime[maxn];
int main() 
{
    scanf("%d",&t);
    while(t--) 
	{
        scanf("%d%d",&g,&l);
        if(l%g)
        {
            printf("0\n");
            continue;
        }
        int k=l/g;
        memset(prime,0,sizeof(prime));
        int m=0;
        for(int i=2; i<=k; i++)
		 {
            if(k%i==0) 
			{
                while(k%i==0) 
				{
                    prime[m]++;
                    k/=i;
                }
               m++;
            }
        }
        if(k>1)  
		  prime[m++]++;
        int sum=1;
        for(int i=0; i<m; i++) 
		{
            if(prime[i])
               sum*=(prime[i]*6);
        }
        printf("%d\n",sum);
    }
    return 0;
}