POJ 1528 Perfection 模拟

来源：互联网发布：memcache 教程php 编辑：程序博客网时间：2024/06/03 07:31

Perfection

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 10880 Accepted: 5190

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  ABUNDANT60000  ABUNDANT   22  DEFICIENT  496  PERFECTEND OF OUTPUT

Source

Mid-Atlantic 1996

#include <cstdio>using namespace std;int main(){int n,sum,i;printf("PERFECTION OUTPUT\n");while (scanf("%d",&n)!=EOF && n){printf("%5d  ",n);sum=0;for (i=1;i<n;++i)if (n%i==0)sum+=i;if (sum<n)  printf("DEFICIENT\n");if (sum==n) printf("PERFECT\n");if (sum>n)  printf("ABUNDANT\n");}printf("END OF OUTPUT\n");return 0;}

kdwycz的网站： http://kdwycz.com/

kdwyz的刷题空间：http://blog.csdn.net/kdwycz

0 0