HDOJ-1019 LeastCommon Multiple

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<<endl;
  }
}