辗转相除法求最大公约数

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

题解：构建一个函数求两个数的最大公约数即可

代码c：

#include
#include

int lcm(int a,int b)
{
    int t,x,y,lcm;
    x=a;
    y=b;
    if(a         t=a;
        a=b;
        b=t;
    }
    while(b!=0){
        t=a;
        a=b;
        b=t%b;
    }
    lcm=x/a*y;
    return lcm;
}//构建函数求两个数的最小公倍数，即lcm

int main()
{
    int t,i,m,a,b;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&m);
            scanf("%d",&a);
        for(i=1;i             scanf("%d",&b);
            a=lcm(a,b);
        }
        printf("%d\n",a);
    }
    return 0;
}