AOJ_0005(GCD and LCM)

题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=30390

简单的gcd与lcm题目，ac代码如下：

#include <cstdio>
#include <iostream>
#include <string>

using namespace std;
typedef long long int ll;

ll gcd(ll a, ll b)
{
	if (b == 0)
		return a;
	else
		return gcd(b, a%b);                   //递归的求解
}
int main(void)
{
	// freopen("in.txt","r",stdin);
	ll a, b;
	while (scanf("%lld%lld", &a, &b) != EOF)
	{
		if (a<b)
			swap(a, b);
		ll gcd_ = gcd(a, b);
		ll lcm_ = (a*b) / gcd_;
		printf("%lld %lld\n", gcd_, lcm_);
	}
	return 0;
}