hdu4497 GCD and LCM

GCD and LCM

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 78 Accepted Submission(s): 43

Problem DescriptionGiven 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.

InputFirst 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.

OutputFor 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

Source2013 ACM-ICPC吉林通化全國邀請賽——題目重現

Recommendliuyiding很明顯，m/n!=0的話，就直接輸出0就可以了！否剛，直接分解質因數m/n,找到，每個質因子的個數，這樣，我們，就可以得出每個質因數為a1^k1,那是題目就是要把這k1個a1分到三個數中，那麼排列組合就是k1*A（3，2）,也就是，6*k1,種，直接算出來就行了！

#include <iostream>#include <stdio.h>#include <algorithm>#include <string.h>#include <math.h>using namespace std;#define MAXN 100000int num[MAXN];int main(){    int tcase,n,m,i,ans,sum,tempm;    scanf("%d",&tcase);    while(tcase--)    {        scanf("%d%d",&n,&m);        memset(num,0,sizeof(num));        if(m%n!=0)        {            printf("0\n");            continue;        }        m=m/n;tempm=sqrt(m)+1;        for(i=2,ans=0;i<=tempm;i++)        {            if(m%i==0)            {                while(m%i==0)                {                    num[ans]++;m/=i;                }                ans++;            }        }        if(m!=1)            num[ans++]=1;        for(sum=1,i=0;i<ans;i++)            sum*=6*num[i];        printf("%d\n",sum);    }    return 0;}