最大公约数和最小公倍数：……
2.6 数学问题的解题窍门
辗转相除法求最大公约数gcd，然后lcm = a * b / gcd 。
long long gcd(long long a, long long b){ if (b == 0)return a; return gcd(b, a%b);}long long lcm(long long a, long long b){ return a * b / gcd(a, b);
}
int main(int argc, char *argv[]){ long long a, b; while (cin >> a >> b) { cout << gcd(a, b) << " " << lcm(a, b) << endl; } return 0;}