Aizu 0005 GCD and LCM【辗转相除法】

原题网址：
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0005
GCD and LCM
Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld
& %llu
Submit Status Practice Aizu
0005

Appoint description: 
rupak_biswas  (2014-03-13)System Crawler  (2016-04-09)

Description

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

比较简单的辗转相除法裸题....

/* http://blog.csdn.net/liuke19950717 */
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int gcd(int a,int b)
{
if(!b)
{
return a;
}
return gcd(b,a%b);
}
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
int tp=gcd(n,m);
printf("%d %d\n",tp,n/tp*m);
}
return 0;
}