Aizu 0005 GCD and LCM[전전 상 나눗셈]

원제 사이트 주소:
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

비교적 간단 한 전전 상 제법 누 드 문제...
/*
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
",tp,n/tp*m); } return 0; }

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