AOJ 0005 GCD and LCM (最大公约数_裸题)

最新推荐文章于 2019-09-29 17:55:52 发布

原创最新推荐文章于 2019-09-29 17:55:52 发布 · 945 阅读

1 ·

CC 4.0 BY-SA版权

最大公约数_最小公倍数专栏收录该内容

6 篇文章

订阅专栏

该程序用于计算两个整数a和b的最大公约数(GCD)和最小公倍数(LCM)，其中0<a, b≤2,000,000,000，且LCM(a, b)≤2,000,000,000。输入包含多组数据，每组数据由a和b组成，用空格分隔，结束于EOF。输出为每组数据的GCD和LCM，之间用空格隔开。" 106911709,9548378,Phoenix SQL皮肤详解与快速上手,"['HBase', 'SQL接口', 'JDBC', '索引']

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Write a program which computes the greatest common divisor (GCD) and the least common multiple (LCM) of given a and b (0 < a, b ≤ 2,000,000,000). You can supporse that LCM(a, b) ≤ 2,000,000,000.

Input

Input consists of several data sets. Each data set contains a and b separated by a single space in a line. The input terminates with EOF.

Output

For each data set, print GCD and LCM separated by a single space in a line.

Sample Input

8 6
50000000 30000000

Output for the Sample Input

2 24
10000000 150000000

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

int gcd(int a,int b)
{
	if(b==0) return a;
	return gcd(b,a%b);
} 

int lcm(int a,int b)
{
	return a/gcd(a,b)*b;
}

int main()
{
	int a,b;
	while(cin>>a>>b) {
		cout<<gcd(a,b)<<" "<<lcm(a,b)<<endl;
	}
	return 0;
}