A - Humble Numbers

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1058

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.

 

 

 1 #include<cstdio>

 2 #include<string.h>

 3 using namespace std;

 4 const int MAXN=5843;

 5 int num2=0,num3=0,num5=0,num7=0;

 6 long long s[MAXN];

 7 void Sort()//sort函数是用来求humble number数组的

 8 {

 9     s[0]=1;

10     s[num2]=1,s[num3]=1,s[num5]=1;

11     long long min=s[num2]*2;

12     int num2=0,num3=0,num5=0,num7=0;

13     for(int i=1;i<5843;i++)

14     {

15 

16        min=s[num2]*2;

17         if(min>s[num3]*3)

18             min=s[num3]*3;

19         if(min>s[num5]*5)

20             min=s[num5]*5;

21         if(min>s[num7]*7)

22             min=s[num7]*7;

23         s[i]=min;

24        // printf("%d ",s[i]);

25         //printf("%d ",min);

26         if(min==s[num2]*2)

27             num2++;

28         if(min==s[num3]*3)

29             num3++;

30         if(min==s[num5]*5)

31             num5++;

32         if(min==s[num7]*7)

33             num7++;

34     }

35 }

36 int main()

37 {

38     int n;

39     Sort();

40     while(scanf("%d",&n)!=EOF&&n)

41     {

42         if(n%10==1&&n%100!=11&&n!=11)//当时卡在n%100!=11和12,13这里了，wa了几次！

43             printf("The %dst humble number is ",n);

44         else if(n%10==2&&n%100!=12&&n!=12)

45             printf("The %dnd humble number is ",n);

46         else if(n%10==3&&n%100!=13&&n!=13)

47             printf("The %drd humble number is ",n);

48         else printf("The %dth humble number is ",n);

49         printf("%lld.\n",s[n-1]);

50     }

51     return 0;

52 }