Noip2001-Universal Group Rematch-second question-greatest common divisor and least common multiple issues

Title DescriptionDescription Enter two positive integer x0,y0 (2<=x0<100000,2<=y0<=1000000) to find the number of p,q that meet the following conditions
Conditions:
1.p,a is a positive integer
2. Require p,q to x0 for greatest common divisor, y0 as least common multiple.
Trial: The number of all possible two positive integers that satisfy the condition. input/output format input/output Input Format:
Two positive integers x0,y0
output Format:
A number that indicates the number of p,q that satisfy the condition input and Output sample sample Input/output sample Test point # # Input Sample:3 60 Sample output: 4 idea: This problem has a small skill, because half of the answer is just the front of the upside down, so you can only enumerate to the root of the square x*y, and then the answer. this question to use to return to beg greatest common divisor, Xin loss data is not big, otherwise will "boom", haha, joking ^_^. The code is as follows:

1#include <stdio.h>2#include <math.h>3 intOJLD (intIintJ//Greatest Common divisor (recursive)4 {  5     if(i==0)returnJ; 6OJLD (j%i,i); 7 }  8 intMain ()9 {  Ten     intx,y,q,num=0, K;  One     inti; Ascanf"%d%d",&x,&y);  -k=x*y;  -q=sqrt (k);  the      for(i=x;i<=q;i++)   -     {   -         if(k%i==0&AMP;&AMP;OJLD (i,k/i) ==x) num++;  -     }    +printf"%d\n", num*2);  -     return 0;  +}

Noip2001-Universal Group Rematch-second question-greatest common divisor and least common multiple issues