#include"stdio.h"
int main()
{
    int n,m,i,j,k,t,temp,ji,start;
    while(EOF!=scanf("%d",&n)&&n>0)
    {
    for(i=0;i<n;i++)
    {
    start=1;
    scanf("%d",&m);
    for(j=0;j<m;j++)
    {
      scanf("%d",&k);
      ji=k*start;
    if(start>k)
    {
        t=start;
        start=k;
        k=t;
    }
    while(start!=0)
    {
    temp=k%start;
    k=start;
    start=temp;
    }
    start=ji/k;
    }
    printf("%d\n",start);
    }
}
    return 0;
}

简单说说：
EOF!=scanf("%d",&n)&&n>0  通过第一个条件推出，明显不能成立

简单说说：
EOF!=scanf("%d",&n)&&n>0  通过第一个条件推出，明显不能成立

能具体说说么

题目是求一组数据的最小公倍数
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.