GCD LCM--UVA11388

最新推荐文章于 2022-02-28 08:39:54 发布

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Description

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

/*
a*b/gcd=lcm
此题lcm和gcd上限达到20多亿。此题一个裸的想法就是lcm*gcd得到a*b的积。然后i从gcd开始枚举，每次循环i+=gcd;
用除法得到两个数。再来确定第二个数的最大公约数是不是gcd.lcm是不是这两个数的倍数。
但是显然lcm*gcd会溢出。a*b=lcm*gcd;显然i只需要在gcd到sqrt(lcm*gcd)这区间内循环。b直接通过除法得到
所以i*i<=lcm*gcd  即i/gcd<=lcm/i  （两边都肯定是整除）*/
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int g,l;
		scanf("%d%d",&g,&l);
		if(l%g)
		{
			cout<<-1<<endl;
		}
		else
		{
			bool flag=false;
			for(int i=g;i/g<=l/i;i+=g)
			{
				if((l%i==0)&&(l%(l/i*g)==0)&&((l/i*g)%g==0))
				{
					cout<<i<<" "<<l/i*g<<endl;
					flag=true;
					break;
				}
			}
			if(!flag)
			{
				cout<<-1<<endl;
			}
		}
	}
	return 0;
}