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)

 1 #include<stdio.h>
 2 
 3 void perfectNum(int num);
 4 
 5 int main(int argc, char** argv)
 6 {
 7     int num = 0;
 8 
 9     printf("please input a number:");
10     scanf("%d", &num);
11 
12     perfectNum(num);
13 
14     return 0;
15 }
16 
17 void perfectNum(int num)
18 {
19    int i = 0;
20    int j = 0;
21    int factor = 0;
22 
23    for (i = 1; i <= num; i++)
24    {
25        factor = 0;
26 
27        for (j = 1; j <= i/2; j++)
28        { 
29             if (0 == i%j)
30                 factor += j;
31        }
32 
33        if (factor == i)
34        {
35            printf("%d is perfect number
", i);
36        }
37    }
38 }