Problem D: GCD LCM

tip:当初自己是想复杂了，什么遍历求解，枚举之类的，其实很简单的

　　要求他的最大GCD和LCM，如果GCD是LCM的因数，那么不可能存在这样的数，否则输出这俩个数就行了。

　　因为既要保证GCD最小，又要保证LCM最大。

题目链接：

　　https://uva.onlinejudge.org/external/113/11388.html

题意：

　　给两个数，求最小GCD和最大LCM

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 andL.
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

　　代码：

　　

#include<iostream>
#include<cstdio>

using namespace std;

int main()
{
    int a,b,t;
    cin>>t;
    while(t--)
    {
        cin>>a>>b;
        if(b%a==0)
            cout<<a<<" "<<b<<endl;
        else
            puts("-1");
    }
    return 0;
}