[C] Series title language do count them

After defining number: If a positive integer greater than 1 and equal to the sum of all the factors of its own, called the number is the number of complete, such as 6,28 are completed: 6 + 2 + 1 = 3; 28 = 1 + 2 + 4 + 7 + 14.

This task is to determine the number of question number End between two positive integers.
Input
input data comprising a plurality of rows, the first row is a positive integer n, the number of test cases, and n is the test case, one row for each instance, two positive integers num1 and num2 composition, (1 <num1 , num2 <10000).
Output
For each set of test data, and output between num1 num2 (including num1 and num2) number End number exists.
The Input the Sample
2
2. 5
. 5. 7
the Sample the Output
0
. 1

On the code:

#include <stdio.h>
int judge(int n)   //判断是否是完数的函数
{
    int i;
    int sum = 0;
    for(i=1; i<=n/2; i++)
    {
        if(n%i==0)
            sum+=i;
    }
    if(sum == n)
        return 1;
    else
        return 0;
}
int main()
{
    int n;
    int i;
    int num1, num2;
    while(scanf("%d", &n)!=EOF)
    {
        while(n--)
        {
            int count = 0;
            scanf("%d%d", &num1, &num2);
            int temp;
            if(num1>num2)        //比较大小，保证小数在前，大数在后
            {
                temp=num1;
                num1=num2;
                num2=temp;
            }
            for(i=num1; i<=num2; i++)
                if(judge(i))
                    count++;
            printf("%d\n", count);
        }
    }
    return 0;
}

I believe we can achieve the algorithm, but this problem has a relatively pit place is the need to compare the size of the input two numbers and transposition, to ensure that the decimal in front, large numbers in prior to his A out.