Wednesday, 5 August 2015

Uva - 382 - Perfection UVa Problem Solution Using C++.

/*

382 - Perfection

Time limit: 3.000 seconds

 Perfection 

From the article Number Theory in the 1994 Microsoft Encarta: ``If abc are integers such that a = bc, a is called a multiple of b or ofc, and b or c is called a divisor or factor of a. If c is not tex2html_wrap_inline41 , 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."

Problem Statement

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

TAG: C++, UVa ,Easy Problem, Easy Math Problem, Algorithm ,Easy  Solution , 382 - Perfection UVa.

*/

Solution:

#include<bits/stdc++.h>
using namespace std;
/*
    *
    * Prosen Ghosh
    * American International University - Bangladesh (AIUB)
    *
*/
int main(){

    int n;
    cout << "PERFECTION OUTPUT" << '\n';
    while(cin >> n && n!=0){
        int tmp = n,sum = 0;
        for(int i = 1; i <= n/2;i++){
            if(tmp%i == 0)sum+=i;
        }
        if(n == sum)printf("%5d  PERFECT\n",n);
        if(sum > n)printf("%5d  ABUNDANT\n",n);
        if(sum < n)printf("%5d  DEFICIENT\n",n);
    }
    cout << "END OF OUTPUT" << '\n';
    return 0;
}

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