GCD LCM--UVA11388

Description

 

I I U C   O N L I N E   C O N T E S T   2 0 0 8

Problem D: GCD LCM

Input: standard input
Output: standard output

 

The GCD of two positive integers is the largest integer that divides both the integers without any remainder. The LCM of two positive integers is the smallest positive integer that is divisible by both the integers. A positive integer can be the GCD of many pairs of numbers. Similarly, it can be the LCM of many pairs of numbers. In this problem, you will be given two positive integers. You have to output a pair of numbers whose GCD is the first number and LCM is the second number.

 

Input

The first line of input will consist of a positive integer T. T denotes the number of cases. Each of the next T lines will contain two positive integer, G and L.

 

Output

For each case of input, there will be one line of output. It will contain two positive integers a and b, a ≤ b, which has a GCD of G and LCM of L. In case there is more than one pair satisfying the condition, output the pair for which a is minimized. In case there is no such pair, output -1.

 

Constraints

-           T ≤ 100

-           Both G and L will be less than 2³¹.

 

Sample Input

Output for Sample Input

2

1 2

3 4

1 2

-1

/*a*b/gcd=lcm此题lcm和gcd上限达到20多亿。此题一个裸的想法就是lcm*gcd得到a*b的积。然后i从gcd开始枚举，每次循环i+=gcd;用除法得到两个数。再来确定第二个数的最大公约数是不是gcd.lcm是不是这两个数的倍数。但是显然lcm*gcd会溢出。a*b=lcm*gcd;显然i只需要在gcd到sqrt(lcm*gcd)这区间内循环。b直接通过除法得到所以i*i<=lcm*gcd  即i/gcd<=lcm/i  （两边都肯定是整除）*/#include <iostream>#include <cstdio>using namespace std;int main(){int t;scanf("%d",&t);while(t--){int g,l;scanf("%d%d",&g,&l);if(l%g){cout<<-1<<endl;}else{bool flag=false;for(int i=g;i/g<=l/i;i+=g){if((l%i==0)&&(l%(l/i*g)==0)&&((l/i*g)%g==0)){cout<<i<<" "<<l/i*g<<endl;flag=true;break;}}if(!flag){cout<<-1<<endl;}}}return 0;}