HDOJ Humble Numbers（java）

Humble Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16520    Accepted Submission(s): 7168

Problem Description

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

 

Input

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.

 

Output

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.

 

Sample Input

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

 

Sample Output

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.

分析：本题是求解整数范围内前5842个丑数（只包含素数因子2,3,5,7），主要是迭代循环依次求解这5842个丑数，求第i个丑数f（i），它等于前i-1个丑数中的一个乘以素数数组中的某一个，其实就是将前i-1个数分别与素数因子相乘，然后得出比第i-1个丑数大但是最小的值作为第i个丑数!

AC代码（java）：

import java.util.Scanner;

/**
 * 寻找丑数问题
 * @author Administrator
 *
 */
public class Main {


	public static void main(String[] args) {
		
		Scanner in = new Scanner(System.in);
		long[] array = new long[5845];
		int[] prime = {2,3,5,7};
		array[1] = 1;
		for(int i=2; i<=5842; i++){
			
			array[i] = Integer.MAX_VALUE;
			for(int j=0; j<4; j++){
				
				for(int k=i-1; k>=1; k--){
					
					if(array[k]*prime[j] <= array[i-1])
						break;
					if(array[k]*prime[j] < array[i]){
						array[i] = array[k]*prime[j];
					}
						
						
				}
				
			}
			
		}
		while(in.hasNext()){
			
			int n = in.nextInt();
			if(n == 0)
				break;
			
			if(n%10 == 1 && n%100 != 11){
				System.out.println("The "+n+"st humble number is "+array[n]+".");
			}else if(n%10 == 2 && n%100 != 12){
				System.out.println("The "+n+"nd humble number is "+array[n]+".");
			}else if(n%10 == 3 && n%100 != 13){
				System.out.println("The "+n+"rd humble number is "+array[n]+".");
			}else{
				System.out.println("The "+n+"th humble number is "+array[n]+".");
			}
		}
		
		
	}

}