Solution of 382 - Perfection

Problem Description
source: https://uva.onlinejudge.org/external/3/382.html

From the article Number Theory in the 1994 Microsoft Encarta: “If a, b, c are integers such that a = bc, a is called a multiple of b or of c, and b or c is called a divisor or factor of a. If c is not ±1, b is called a proper divisor of a. Even integers, which include 0, are multiples of 2, for example, -4, 0, 2, 10; an odd integer is an integer that is not even, for example, -5, 1, 3, 9. A perfect number is a positive integer that is equal to the sum of all its positive, proper divisors; for example, 6, which equals 1 + 2 + 3, and 28, which equals 1 + 2 + 4 + 7 + 14, are perfect numbers. A positive number that is not perfect is imperfect and is deficient or abundant according to whether the sum of its positive, proper divisors is smaller or larger than the number itself. Thus, 9, with proper divisors 1, 3, is deficient; 12, with proper divisors 1, 2, 3, 4, 6, is abundant.” 

     Your task is: Given a number, determine if it is perfect, abundant, or deficient. 

Input 

A list of N positive integers (none greater than 60,000), with 1 < N < 100. A ‘0’ will mark the end of the list. 

Output 

The first line of output should read ‘PERFECTION OUTPUT’. The next N lines of output should list for each input integer whether it is perfect, deficient, or abundant, as shown in the example below. Format counts: the echoed integers should be right justified within the first 5 spaces of the output line, followed by two blank spaces, followed by the description of the integer. The final line of output should read ‘END OF OUTPUT’.

Sample Input 

15 28 6 56 60000 22 496 0

Sample Output 

PERFECTION OUTPUT 
       15     DEFICIENT 
       28     PERFECT 
         6     PERFECT 
       56     ABUNDANT 
 60000     ABUNDANT 
       22     DEFICIENT 
     496     PERFECT 
END OF OUTPUT


Solution:

#include<stdio.h>

int main()
{
    long int n, i, sum, set=0;
    while(scanf("%ld",&n)==1)
    {
        if(set==0)
            printf("PERFECTION OUTPUT\n");
        if(n==0)
        {
            printf("END OF OUTPUT\n");
            break;
        }


        sum=0;
        for(i=1;i<=n/2;i++)
        {
            if(n%i==0)
                sum+=i;
        }
        if(sum<n)
            printf("%5ld  DEFICIENT\n",n);
        else if(sum == n)
            printf("%5ld  PERFECT\n",n);
        else
            printf("%5ld  ABUNDANT\n",n);

        set=1;

    }
    return 0;
}