(number theory) greatest common divisor and least common multiple problems

Title Description Description

Enter two positive integer x0,y0 (2<=x0<100000,2<=y0<=1000000) to find the number of p,q that meet the following conditions

Condition: 1.p,q 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.

Enter a description Input Description

Two positive integers x0,y0

Output description Output Description

The number of all possible two positive integers that satisfy the condition

Sample input Sample Input

3 60

Sample output Sample Output

4

Data size & Hint

3 60

60 3

12 15

15 12

Idea: y/x then decomposes factorization and then the power of 2 is the answer.

1#include <stdio.h>2#include <math.h>3 intx, y;4 BOOLZhiintx)5 {6       for(intI=2; i<=sqrt (x); + +i)7          if(x%i==0)return 0;8      return 1;9 }Ten intMain () One { Ascanf"%d%d",&x,&y); -     if(y%x!=0) -     { theprintf"0/n"); -                return 0; -     } -x=y/x; y=0; +      for(intI=2; i<=x;++i) -         if(Zhi (i)) +         {  A                    if(x%i==0) ++y; at                     while(x%i==0) x/=i; -         } -printf"%d/n",(int) Pow (2, y)); -     return 0; -}

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(number theory) greatest common divisor and least common multiple problems