UVa 382 Perfection (过剩数、完美数和亏数)

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 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 , 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

完整代码，O(√n)复杂度：

/*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");
}