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2013
11-09

POJ 1528 Perfection [解题报告] Java

Perfection

问题描述 :

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.

输入:

A list of N positive integers (none greater than 60,000), with 1 <= N < 100. A 0 will mark the end of the list.

输出:

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.

样例输入:

15 28 6 56 60000 22 496 0

样例输出:

PERFECTION OUTPUT
   15  DEFICIENT
   28  PERFECT
    6  PERFECT
   56  ABUNDANT
60000  ABUNDANT
   22  DEFICIENT
  496  PERFECT
END OF OUTPUT

解题代码:

//* @author: [email protected]
import java.util.Scanner;
public class Main
{
 public static void main(String[] args)
  {
	Scanner in=new Scanner(System.in);
	System.out.println("PERFECTION OUTPUT");
	while(true)
	{
         int a=in.nextInt();
	 if(a==0) break;
	  double w=Math.sqrt(a);
	  int total=1;
	  for(int i=2;i< w;i++)
	  {
		if(a%i==0) total+=(a/i+i);
	  }
	  if(a%w==0) total+=w;
	  System.out.printf("%5d  ",a);
	  if(a==1) System.out.println("DEFICIENT");
	  else if(total>a) System.out.println("ABUNDANT");
	  else if(total< a) System.out.println("DEFICIENT");
	  else System.out.println("PERFECT");
	}
	System.out.println("END OF OUTPUT");
  }
}

  1. 嗯 分析得很到位,确实用模板编程能让面试官对你的印象更好。在设置辅助栈的时候可以这样:push时,比较要push的elem和辅助栈的栈顶,elem<=min.top(),则min.push(elem).否则只要push(elem)就好。在pop的时候,比较stack.top()与min.top(),if(stack.top()<=min.top()),则{stack.pop();min.pop();},否则{stack.pop();}.