UVALive5429 UVA382 POJ1528 HDU1323 ZOJ1284 Perfection【整除＋水题】

Perfection

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13162		Accepted: 6021

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

Source
Mid-Atlantic 1996

问题链接：UVALive5429 UVA382 POJ1528 HDU1323 ZOJ1284 Perfection。

题意简述：

  输入若干个整数n，n=0结束，计算真因数之和。若真因数之和，等于n则输出PERFECT，大于n则输出ABUNDANT，小于n则输出DEFICIENT。

问题分析：

 一个数的真因数之和等于本身的数称为完美数。例如６＝１＋２＋３，６是完美数。

  用试探法计算所有因数之和。需要注意n=1的情况。
程序说明：

  程序中编写函数getsum()用于计算因数之和。

Regionals 1996 >> North
America - Mid-Atlantic USA

AC的C语言程序如下：

/* UVALive5429 UVA382 POJ1528 HDU1323 ZOJ1284 Perfection */

#include <stdio.h>
#include <math.h>

int getsum(int n)
{
int i, sum=1;

for(i=2; i<n; i++) {
if(n % i == 0)
sum += i;
}

return sum;
}

int main(void)
{
int n, sum;

printf("PERFECTION OUTPUT\n");
while(scanf("%d", &n) != EOF && n != 0) {
if(n == 1)
printf("%5d  DEFICIENT\n", n);
else {
sum = getsum(n);
if(sum<n)
printf("%5d  DEFICIENT\n", n);
else if(sum==n)
printf("%5d  PERFECT\n", n);
else
printf("%5d  ABUNDANT\n", n);
}
}
printf("END OF OUTPUT\n");

return 0;
}