程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> UVA 11388-GCD LCM（數學）

UVA 11388-GCD LCM（數學）

編輯：C++入門知識

UVA 11388-GCD LCM（數學）

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

1 2

3 4

1 2

-1

Problem setter: Shamim Hafiz

題意：給出兩個數G,L，問是否存在一對數a,b，使得gcd(a,b)==G,lcm(a,b)==L;

可以這麼想：當gcd(G,L)==G(a),lcm(G,L)==L(b)時，此時G==a,L==b,滿足上述條件。否則不成立。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define ll long long
using namespace std;
const int INF = 0x3f3f3f3f;
ll gcd(ll a,ll b)
{
	if(b==0) return a;
	else return gcd(b,a%b);
}
ll lcm(ll a,ll b)
{
	return a*b/gcd(a,b);
}
int main()
{
	int t;ll a,b;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lld%lld",&a,&b);
		ll G=gcd(a,b),L=lcm(a,b);
		if(G==a&&L==b)
			printf("%lld %lld\n",a,b);
		else
			puts("-1");
	}
	return 0;
}