"codevs1012" greatest common divisor and least common multiple

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

Outputs description Output Description

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

Sample input to sample

3 60

Sample output Sample Outputs

4

Analysis:

P and Q's Greatest common divisor (GCD) is x, least common multiple (LCM) is Y

So P*q=x*y

Set P=x*i,q=x*j,i and J coprime

Then p*q= (x*i) * (x*j) =x*y, then there is i*j=y/x

We can enumerate I, starting with I=1, until i*i>y/x

If I is a factor of y/x

Then j= (y/x)/I

Again to determine whether I and J coprime

Because each of the two numbers in the smaller is I, the larger the number is J,i is the small root (y/x), J is greater than the radical (y/x) so does not repeat the calculation, that count to one time, the answer will accumulate 2.

Code:

#include <iostream>using namespace std;int gcd (int x,int y) {    return (X%Y==0?Y:GCD (y,x%y));} int main () {    int x,y,ans=0;    cin>>x>>y;    if (y%x) {        cout<<0;        return 0;    }    y=y/x;    for (int i=1; i*i<=y; i++)    {        if (Y%I==0&&GCD (i,y/i) ==1)            ans+=2;    }    Cout<<ans<<endl;}

　　

"codevs1012" greatest common divisor and least common multiple