AOJ 0005 GCD and LCM 题解 《挑战程序设计竞赛》 � 码农场


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;
}
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