2001 Greatest common divisor and the minimum number of conventions

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 range and Tips Data Size & Hint

Exercises

Number theory + enumeration.

Equation 1: Two numbers of least common multiple * greatest common divisor = two numbers of product.

Equation 2: Greatest common divisor gcd (A, B) =gcd (b,a%b).

var X,y,j,k,ans:int64;

I:longint;

function gcd (X,y:int64): Int64;

Begin

If Y=0 then exit (x)

else exit (gcd (y,x mod y));

End

Begin

READLN (x, y);

For I:=x to Y do

Begin

J:=x*y Div i;

K:=GCD (I,J);

if (k=x) and (i*j Div k=y) then Inc (ANS);

End

Write (ANS);

End.

2001 Greatest common divisor and the minimum number of conventions