HDOJ1058Humble Numbers（丑数）

Humble Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27739    Accepted Submission(s): 12234

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

1234111213212223100100058420

 

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.

 这里对丑数的定义和百科上面的有所区别，百科上面讲的是2,3,5.而这里讲的多了一个7.

百科对丑数的定义是只包含因子2,3,5的正整数被称作丑数，比如4,10,12都是丑数，而7,23,111则不是丑数，另外1也不是丑数。（这里多了7）

判断是只包含因子2,3,5,7可以使用：首先除2，直到不能整除为止，然后除5到不能整除为止，然后除3直到不能整除为止，然后除7到不能整除为止。。。最终判断剩余的数字是否为1，如果是1则为丑数，否则不是丑数。

题目是要输出指定位置的丑数。

但是这种方法我测试了下太耗时了。

代码如下：

arr = new int[5843];arr[1] = 1;int j = 2;int i = 2;while (j < 5843) {int num = i;// 判断是否有质因子2// 如果N中含有质因子2，把所有的因子2都去掉while (num % 2 == 0) {num /= 2;}while (num % 3 == 0) {num /= 3;}while (num % 5 == 0) {num /= 5;}while (num % 7 == 0) {num /= 7;}if (num == 1) {// 如果最后N除的只剩1，则N就是丑数arr[j] = i;//System.out.println(arr[j]);j++;}i++;}

我们可以使用另一种方法来算出丑数，就是从以1为"始祖",剩下的所有其他的数,其实都是在此基础上乘以2,3,5,7演化出来的。

AC代码如下：

import java.util.Scanner;public class Main{private static Scanner scanner;private static int arr[];private static int i = 1, j = 1, k = 1, l = 1;public static void main(String[] args) {scanner = new Scanner(System.in);dabiao();while (scanner.hasNext()) {int n = scanner.nextInt();if (n == 0) {break;}// 1,2,3分别是st,nd,rd,其他是thif (n % 10 == 1 && n % 100 != 11) {System.out.println("The " + n + "st humble number is " + arr[n] + ".");} else if (n % 10 == 2 && n % 100 != 12) {System.out.println("The " + n + "nd humble number is " + arr[n] + ".");} else if (n % 10 == 3 && n % 100 != 13) {System.out.println("The " + n + "rd humble number is " + arr[n] + ".");} else {System.out.println("The " + n + "th humble number is " + arr[n] + ".");}}}private static void dabiao() {arr = new int[5843];arr[1] = 1;for (int t = 2; t < arr.length; t++) {//核心代码！！arr[t] = min(2 * arr[i], 3 * arr[j], 5 * arr[k], 7 * arr[l]);}}private static int min(int a, int b, int c, int d) {int m = a;if (m > b)m = b;if (m > c)m = c;if (m > d)m = d;if (a == m)i++;if (b == m)j++;if (c == m)k++;if (d == m)l++;return m;}}