uva—gcd lcm

IIU 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 integersa and b, a ≤ b, which has a GCD ofG and LCM of L. In case there is more than one pair satisfying the condition, output the pair for whicha 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

题意：给出了gcd（最大公约数），lcm（最小公倍数），求满足gcd，lcm的两个数n,m，输出最小的一组

解析：设n=a*b*c*d,m=a*b*c*e,那么易知gcd=a*b*c,lcm=a*b*c*d*e;即lcm%gcd==0.

           再逆向思考给出gcd=a*b*c,lcm=a*b*c*d*e.那么有以下几种情况：

           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,且明显第一组是最小的。

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

代码：

#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;    }}