UVA - 11388 GCD LCM

最新推荐文章于 2020-02-22 15:53:53 发布

原创最新推荐文章于 2020-02-22 15:53:53 发布 · 581 阅读

0 ·

CC 4.0 BY-SA版权

3.GCD+LCM 专栏收录该内容

14 篇文章

订阅专栏

本文介绍了一种快速解决给定两个整数G和L的情况下寻找一对整数a和b的方法，使得a和b的最大公约数为G，最小公倍数为L。通过判断L是否能被G整除来确定是否存在这样的a和b，并给出了一段实现该算法的C++代码。

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
2
1 2
3 4
Sample Output
1 2
-1

已知G,L，找出（a,b）,a最小；

AC代码：

#include<cstdio>

typedef long long LL;

int main()
{
	int T; scanf("%d",&T);
	while(T--) {
		int G,L; scanf("%d%d",&G,&L);
		if(L%G) printf("-1\n");
		else printf("%d %d\n",G,L);
	}
	return 0;
}