51nod1012 最小公倍数LCM

#include<iostream>
#include<cstdio>
using namespace std;
long long gcd(long long a, long long b)
{
	if(a<b)
	swap(a,b);
	return b == 0 ? a : gcd(b, a % b);
}
int main()
{
	long long m, n, t;
	while(~scanf("%lld%lld", &m, &n))
	{
		t = gcd(m,n);
		printf("%lld\n", m / t * n);
	}
}

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转载自blog.csdn.net/love20165104027/article/details/81583179