UVA 11388 - GCD LCM

 

 

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
Problem setter: Shamim Hafiz

 

 

题目大意：已知gcd(a,b)以及lcm(a,b) 求a和b，多个的话输出a最小的那个

解题思路看代码：

 

#include <iostream>
#include <cstdio>
using namespace std;

int main(){
	int t,g,l;
	scanf("%d",&t);
	while(t-- >0){
		scanf("%d%d",&g,&l);
		if(l%g==0) printf("%d %d\n",g,l);
		else printf("-1\n");
	}
	return 0;
}