要求输出 例：6 its factors are 1,2,3    我的是6 its factors are 1,2,3,  
#include<stdio.h>
int main()
{
    int n,i,s = 0;

    for(n = 2;n <= 1000;n++)               //1000之内2-1000
    {
        {
            for(i = 1;i < n;i++ )            //求出因子
                if(n % i == 0)
                    s = s +i;                    //累加因子
               
                if(n == s)                         //判断是否是完数
                    {   
                        printf("%d its factors are ",n);    //输出完数
                        for(i = 1;i < n;i++ )                //输出因子
                            if(n % i == 0)
                                                               

                                       printf("%d,",i);
                                        printf("\n");
                                                               
                           
                    }

            
        }
        s = 0;                //s归0 重新开始循环
   
   
    }

    return 0;
}

增加判斷條件，只有在循環確定不結束時才打印逗號。也就是說，不要把逗號跟數據連在一起輸出，先輸出數據，看是不是要結束循環了，若不是，就再輸出一個逗號，否則就不需要再輸出逗號了。多一個if()語句不會影響什麽性能的，試圖省一個if()語句恐怕會寫出更加晦澀難懂的代碼。實際工作不是參加亂碼大賽，清晰性第一，能讓自己和別人順利理解意圖比什麽效率都強。

做了一个修改，你看一下，你的代码缩进没有弄好，不容易读。

#include<stdio.h>
int main()
{
    int n,i,s = 0;
    int cnt=0;  //用来计算完数一共有几个因子
    int lep=0;  //用来计算完数一共输出了几个因子
    for(n = 2;n <= 1000;n++)               //1000之内2-1000
    {
        {
            for(i = 1;i < n;i++ )            //求出因子
                if(n % i == 0)
                {s = s +i;                    //累加因子
                 cnt++;         //用来计数完数的因子
                }
                if(n == s)                         //判断是否是完数
                    {   
                        printf("%d its factors are ",n);    //输出完数
                        for(i = 1;i < n;i++ )                //输出因子
                        {
                            if(n % i == 0)
                                                               
                            {
                                       printf("%d",i);
                                       lep++;       //计算完数一共输出了几个因子
                                       if(lep==cnt)continue;
                                       printf(",");
                            }

                              
                        }

                                        printf("\n");
                                                               
                           
                    }

            
        }
        s = 0;                //s归0 重新开始循环
        cnt=0;                //cnt归0重新开始计数
        lep=0;
    }

    return 0;
}

多谢！

了解