HDU 1058 Humble Numbers (DP)

Humble Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18238    Accepted Submission(s): 7934

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.

 

 

说DP有点牵强，不过也算是用前面的状态推出后面的状态，英语知识是个易错点。

思路：2,5,7,9是丑数，丑数的2,5,7,9倍也是丑数，所以从1开始，对每个数计算计算它的2,5,7,9倍，算出来的结果又再翻倍。难免会遇到这种考察其它乱七八糟的知识点的题，输出时的后缀很容易错，在英语里如果一个数的个位是1，那后面跟st，2跟nd，3跟rd,其它跟th，但是注意,后面两位是11,12,13的除外，这些数后面都跟th，比如12,10013,1111，我WA了三次，算是英语差的后果吧。

 1 #include<stdio.h>

 2 #include<string.h>

 3 #include<stdlib.h>

 4 #define    MAX    5845

 5 

 6 int    my_min(int a,int b,int c,int d);

 7 int    main(void)

 8 {

 9     int    n;

10     int    s[MAX] = {1};

11     int    i,i_2,i_3,i_5,i_7;

12     int    box,box_2,box_3,box_5,box_7;

13     char    cha[5];

14     i = i_2 = i_3 = i_5 = i_7 = 0;

15 

16     while(i < MAX - 2)

17     {

18         box_2 = s[i_2] * 2;

19         box_3 = s[i_3] * 3;

20         box_5 = s[i_5] * 5;

21         box_7 = s[i_7] * 7;

22 

23         box = my_min(box_2,box_3,box_5,box_7);    //每次把最小的那个插入到数组里，就不用排序了

24         if(box == box_2)            //要一直用if，因为结果可能相同

25             i_2 ++;

26         if(box == box_3)

27             i_3 ++;

28         if(box == box_5)

29             i_5 ++;

30         if(box == box_7)

31             i_7 ++;

32 

33         i ++;

34         s[i] = box;

35     }

36 

37     while(scanf("%d",&n) && n)

38     {

39         box = n % 100;

40         if(box == 11 || box == 12 || box == 13)

41         {

42             strcpy(cha,"th");

43             goto    label;

44         }

45 

46         box = n;

47         while(box / 10)

48             box %= 10;

49         switch(box % 10)

50         {

51             case    1:strcpy(cha,"st");break;

52             case    2:strcpy(cha,"nd");break;

53             case    3:strcpy(cha,"rd");break;

54             default     :strcpy(cha,"th");break;

55         }

56 

57         label:

58         printf("The %d%s humble number is %d.\n",n,cha,s[n - 1]);

59     }

60 

61     return    0;

62 }

63 

64 int    my_min(int a,int b,int c,int d)

65 {

66     int    min = a;

67 

68     min = min < b ? min : b;

69     min = min < c ? min : c;

70     min = min < d ? min : d;

71 

72     return    min;

73 }