第二次周赛黄金组F题

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.

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.
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.
Sample Input
2
3 5 7 15
6 4 10296 936 1287 792 1
Sample Output
105
10296
被惊讶的无法分析：什么？？原来这道题这么简单？？为什么当时没什么人做得出来！好吧，现在冷静下来，想想。。可能是被那一大串英文吓死了吧？？？嗯…………看来下一次做题时。。。（嘿嘿嘿）我有一个大胆的想法。
咳咳，（表面）分析，用一个数组记录输入的要求LCM（原来那是公倍数的意思）的数，然后由小到大排序，接下来用一个变量a记录最下的那个，然后a一倍一倍的增加，知道可以被第二个数据整除为止，然后让a获得循环结束的a的值，然后用同样的方法增加a知道被第三个整除，不断循环，知道能被最大那个整除，然后记录当前的a的值，输出
ac代码

#include <iostream>
#include<string>
using namespace std;
int main()
{
	int a;
	cin >> a;
	for (int i = 1;i <= a;i++)
	{
		int c;
		cin >> c;
		int *b = new int[c];
		for (int i = 0;i < c;i++)
			cin >> b[i];
		for (int i = 0;i < c;i++)
		{
			int i1 = i;
			for (int i2 = i;i2 < c;i2++)
				if (b[i1] > b[i2])
					i1 = i2;
			int i3;
			i3 = b[i];
			b[i] = b[i1];
			b[i1] = i3;
		}
		int d = b[0], e=d;
		for (int i = 0;i < c-1;i++)
		{
			int i1 = 2;
			while (d%b[i+1]!=0)
			{
				d = e * i1;
				i1++;
			}
			e = d;
		}
		cout << e << endl;
		delete b;
	}
}