(《数论及应用1.3》NEFU 116 两仪剑法(最小公倍数&&最大公约数))

#include <iostream>using namespace std;long long gcd(long long a, long long b){   if(b == 0){    return a;   }     return gcd(b,a%b);}int main(int argc,char* argv[]){ long long  n,m;while(cin >> n >>m){   cout<<n*m/gcd(n,m)<<endl;}return 0;}