本文共 1458 字，大约阅读时间需要 4 分钟。

Problem Description

The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set. For example, the LCM of 5, 7 and 15 is 105.

（一组正整数的最小公倍数（LCM）是最小正整数，其可被集合中的所有数字整除。例如，5,7和15的LCM是105。）

Input

Input will consist of multiple problem instances. The first line of the input will contain a single integer indicating the number of problem instances. Each instance will consist of a single line of the form m n1 n2 n3 … nm where m is the number of integers in the set and n1 … nm are the integers. All integers will be positive and lie within the range of a 32-bit integer. （输入将包含多个问题实例。输入的第一行将包含一个整数，表示问题实例的数量。每个实例将由m n1 n2 n3 … nm形式的单个行组成，其中m是集合中的整数数，n1 … nm是整数。所有整数都是正数，并且位于32位整数的范围内。）

Output

For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer. （对于每个问题实例，输出包含相应LCM的单行。所有结果都将位于32位整数的范围内。）

Sample Input

2 3 5 7 15 6 4 10296 936 1287 792 1

Sample Output

105 10296

用gcd逐个求出相邻两数字的最小公倍数lcm

LCM（a，b） = a * b / GCD（a，b） 注意：求解最小公倍数时，先乘后除，乘法明显可能溢出（wa了一发），所以应该先除后乘 LCM（a，b） = a / GCD（a，b） b*

#include
   
    using namespace std;typedef long long ll;ll gcd(ll a, ll b) {     return b == 0 ? a : gcd(b, a%b);}int main() {     ll r;  cin >> r;  while (r--) {   	    ll n, m1, m2, s;    cin >> n>> m1;    s = m1;    for (int i = 0; i < n-1; i++) {         cin >> m1;      s = m1/gcd(m1, s)*s;    }    cout << s<