I - 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

虽然题目说求最小最大公约数，但其实是最小公倍数。需要用到辗转相除。

#include
#include 
using namespace std;
int division(int a,int b);
int lcm(int a,int b);
int main()
{
  int a[1000];
  int n,i,j,length,sum;
  while(scanf("%d",&n)!=EOF)
  {
   for(i=0;i>length;
    for(j=0;j>a[j];
    }
    sum=a[0];
    for(j=0;j