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#include<cstdio>
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){
return b == 0 ? a : gcd(b ,a%b);
}
ll lcm(ll a,ll b){
return a / gcd(a,b) * b;
}
int main(){
ll a,b;
scanf("%lld%lld",&a,&b);
printf("%lld\n",lcm(a,b));
return 0;
}
[2016-05-09][51nod][1012 最小公倍数LCM]
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原文地址:http://www.cnblogs.com/qhy285571052/p/c0f13497a60fd316ee1b414fb40905ef.html