UVa 382 Perfection (Excess number, perfect number and loss number)

382-perfection

Time limit:3.000 seconds

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=5&page=show_problem &problem=318

From the article number theory in the 1994 Microsoft Encarta: ' If a,b, c are integers-such that's = BC, A is called a mul Tiple ofb or of C, and B or C is called a divisor or factor ofa. If C is not

, 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 the integer is ' is ' isn't even, for example,-5, 1, 3, 9. A perfect number is a positive integer, which equal to the sum of "all" its positive, proper divisors; For example, 6, which equals 1 + 2 + 3, and, which equals 1 + 2 + + 4 + 7 +, are perfect numbers. A positive number that are not perfect are imperfect and are deficient or abundant according to whether the sum of it Positi ve, proper divisors is smaller or larger than the number itself. Thus, 9, with proper divisors 1, 3, are deficient; 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 would mark the end of the list.

Output

The line of output should read perfection output. The next N lines of output should list for each input integer whether it are perfect, deficient, or abundant, as shown in t He example below. Format counts:the echoed integers should is right justified within the ' the ' 5 spaces of the output line, followed by two Blank spaces, followed by the description of the integer. The final line of output should readend of output.

Sample Input

15 28 6 56 60000 22 496 0

Sample Output

Perfection OUTPUT  deficient  PERFECT
    6  PERFECT  abundant
60000  abundant  deficient
  496  PERFECT
end of OUTPUT

Full code, O (√n) Complexity:

/*0.019s*/
  
#include <cstdio>
#include <cmath>
  
int main ()
{
    int n,m,i, sum;
    Puts ("Perfection OUTPUT");
    while (scanf ("%d", &n), N)
    {
        m = (int) sqrt (n), sum = 1;
        for (i = 2; I <= m; ++i)
            if (n% i = = 0) Sum + = i + n/i;
        if (m * m = = N) sum = m;
        printf ("%5d  ", n);
        if (sum < n) puts ("deficient");
        else if (sum > N) puts ("abundant");
        Else puts ("PERFECT");
    }
    Puts ("End of OUTPUT");
}

Author: csdn Blog Synapse7

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