Codeup2136 HDU1019 Least Common Multiple

#include<iostream>
using namespace std;

int gcd(int a, int b)
{
	if (b == 0)
		return a;
	else
		return gcd(b, a%b);
}

int lcm(int a, int b)
{
	return a / gcd(a, b)*b;//先除再乘，减少溢出
}

int main()
{
	int n;
	int m;
	int number1, number2;
	cin >> n;
	while (n--)
	{
		cin >> m;
		cin >> number1;//写的好处，省去用数组
		for (int i = 1; i < m; i++)//第一数，先输入了所以i从1开始
		{
			cin >> number2;
			number1 = lcm(number1, number2);//隐藏的递归，妙！
		}
		cout << number1 << endl;
	}

	return 0;
}