IIUC ONLINE CONTEST 2008 / UVa 11388 GCD LCM (数论)

11388 - GCD LCM

Time limit: 1.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=467&page=show_problem&problem=2383

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

学英语：

 You have to output a pair of numbers whose GCD is the first number and LCM is the second number.

（联系下文知这话的意思是）（对于输入的G和L，）你必须输出一对数(a,b)，a是G和L的最大公约数且等于G，b是G和L的最小公倍数且等于L。

那么只要G|L即可。

完整代码：

/*0.009s*/

#include<cstdio>

int main(void)
{
	int t, a, b;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d%d", &a, &b);
		if (b % a)
			puts("-1");
		else
			printf("%d %d\n", a, b);
	}
	return 0;
}