试题系列二(求1000内的完数)

1.求1000内的完数。(所有因子之和等于本身的数位完数，数本身不算因子)

#include<stdio.h>

void perfectNum(int num);

int main(int argc, char** argv)
{
    int num = 0;

    printf("please input a number:");
    scanf("%d", &num);

    perfectNum(num);

    return 0;
}

void perfectNum(int num)
{
   int i = 0;
   int j = 0;
   int factor = 0;

   for (i = 1; i <= num; i++)
   {
       factor = 0;

       for (j = 1; j <= i/2; j++)
       { 
            if (0 == i%j)
                factor += j;
       }

       if (factor == i)
       {
           printf("%d is perfect number\n", i);
       }
   }
}