//防止数据溢出先除再乘//辗转相除法求最大公约数最小公倍数------一I

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

解题思路：

辗转相除法先求最大公约数gcd(a,b)

最小公倍数lcm(a,b)=a*b/gcd(a,b)

为了防止数据溢出，先除再乘lcm(a,b)=a/gcd(a,b)*b

#include<stdio.h>

int gcd(int a,int b)
{
int temp;
if(a<b)
{
temp=a;
a=b;
b=temp;
}
while(b!=0)
{
temp=a%b;
a=b;
b=temp;
}
return a;
}

int lcm(int a,int b)
{
int c;
c=a/gcd(a,b)*b;//防止数据溢出，先除再乘
return c;
}

int main()
{
int total,m,result,num;

scanf("%d",&total);
for(int i=1;i<=total;i++)
{
scanf("%d%d",&m,&result);
for(int j=2;j<=m;j++)
{
scanf("%d",&num);
result=lcm(num,result);
}
printf("%d\n",result);
}
}