uva—gcd lcm

II U C   ONLINE   C ON TEST   2 008
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
Problem setter: Shamim Hafiz

題意：給出了gcd（最大公約數），lcm（最小公倍數），求滿足gcd，lcm的兩個數n,m，輸出最小的一組

解析：設n=a*b*c*d,m=a*b*c*e,那麼易知gcd=a*b*c,lcm=a*b*c*d*e;即lcm%gcd==0.

           再逆向思考給出gcd=a*b*c,lcm=a*b*c*d*e.那麼有以下幾種情況：

           1.n=a*b*c,m=a*b*c*d*e

           2.n=a*b*c*d,m=a*b*c*e

           3.n=a*b*c*e,m=a*b*c*d

           4.n=a*b*c*d*e,m=a*b*c

          即d和e是可以自由排列的,且明顯第一組是最小的。

 if(l%g==0)
    cout<<g<<" "<<l<<endl;
 else
     cout<<"-1"<<endl;

代碼：

#include <iostream>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int g,l;
        cin>>g>>l;
        if(l%g==0)
            cout<<g<<" "<<l<<endl;
        else
            cout<<"-1"<<endl;
    }
}