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2013
11-11

POJ 2247 Humble Numbers [解题报告] Java

Humble Numbers

问题描述 :

A number whose only prime factors are 2,3,5 or 7 is called a humble number. The sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, … shows the first 20 humble numbers.

Write a program to find and print the nth element in this sequence.

输入:

The input consists of one or more test cases. Each test case consists of one integer n with 1 <= n <= 5842. Input is terminated by a value of zero (0) for n.

输出:

For each test case, print one line saying “The nth humble number is number.”. Depending on the value of n, the correct suffix “st”, “nd”, “rd”, or “th” for the ordinal number nth has to be used like it is shown in the sample output.

样例输入:

1
2
3
4
11
12
13
21
22
23
100
1000
5842
0

样例输出:

The 1st humble number is 1.
The 2nd humble number is 2.
The 3rd humble number is 3.
The 4th humble number is 4.
The 11th humble number is 12.
The 12th humble number is 14.
The 13th humble number is 15.
The 21st humble number is 28.
The 22nd humble number is 30.
The 23rd humble number is 32.
The 100th humble number is 450.
The 1000th humble number is 385875.
The 5842nd humble number is 2000000000.

解题代码:

//* @author: [email protected]
import java.util.*;
public class Main
{
 public static void main(String[] args)
 {
  TreeSet< Integer> t=new TreeSet< Integer>();
  for(int i=0;i< 50;i++)
  {
   if(Math.pow(2, i)>2000000000) break;
   for(int j=0;;j++)
   {
    int ans2=(int)(Math.pow(2, i)*Math.pow(3, j));
    if(ans2>2000000000) break;
    for(int k=0;;k++)
    {
     int ans1=(int)(Math.pow(2, i)*Math.pow(3, j)*Math.pow(5, k));
     if(ans1>2000000000) break;
     for(int w=0;;w++)
     {
      int ans=(int)(Math.pow(2, i)*Math.pow(3, j)*Math.pow(5, k)*Math.pow(7, w));
      if(ans>2000000000) break;
      t.add(ans);
      }
     }
    }
   }
   Integer[] arr=new Integer[6000];
   t.toArray(arr);
   Scanner in=new Scanner(System.in);
   while(true)
   {
   int n=in.nextInt();
   if(n==0) break;
   if (n % 10==1&&n%100!=11) System.out.println("The "+n+"st humble number is "+arr[n-1]+".");
   else if (n % 10==2&&n%100!=12) System.out.println("The "+n+"nd humble number is "+arr[n-1]+".");
   else if (n % 10==3&&n%100!=13) System.out.println("The "+n+"rd humble number is "+arr[n-1]+".");
   else System.out.println("The "+n+"th humble number is "+arr[n-1]+".");
   }
 }
}

  1. 我还有个问题想请教一下,就是感觉对于新手来说,递归理解起来有些困难,不知有没有什么好的方法或者什么好的建议?