zoukankan      html  css  js  c++  java
  • POJ 1528问题描述

    Description

    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/-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." 
    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
  • 相关阅读:
    eclipse恢复界面默认设置
    文件夹的拷贝
    文件的输入输出
    十进制转二进制,八进制,十六进制(查表法)
    数组元素的查找(折半查找)
    C++几个小函数
    C++ 内部排序(一)
    C++实现链表
    C++输出IP地址段内的合法地址
    我看软件工程师的职业规划
  • 原文地址:https://www.cnblogs.com/guochen/p/5452721.html
Copyright © 2011-2022 走看看