hdu 1058:Humble Numbers（动态规划 DP）

Humble Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14584    Accepted Submission(s): 6313

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.

  

Source

University of Ulm Local Contest 1996

 

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　　简单动态规划问题。

　　思路是后面的Humble数总可以用前面求得的某个Humble数乘以{2,3,5,7}中的一个求得。

　　需要注意的是输出需要按序数的规则输出，个位数是1、2、3的后缀分别是st、nd、rd，但是后两位为11、12、13的数例外（例如11、12、13、111、112、113），它们的后缀是th。其余数后缀均为th。

　　借鉴于前人的思路：http://minda-ly.diandian.com/post/2011-02-10/17312430

 

 #include <iostream>
 using namespace std;
 int a[];
 int Min(int a,int b,int c,int d)
 {
     int t;
     t = a<b?a:b;
     t = t<c?t:c;
     t = t<d?t:d;
     return t;
 }
 int main()
 {
     int n;
     a[] = ;
     int a1=,a2=,a3=,a4=;
     int e1,e2,e3,e4;
     for(int i=;i<;i++){
         e1 = a[a1]*;
         e2 = a[a2]*;
         e3 = a[a3]*;
         e4 = a[a4]*;
         a[i] = Min(e1,e2,e3,e4);
         if(a[i]==e1)
             a1++;
         if(a[i]==e2)
             a2++;
         if(a[i]==e3)
             a3++;
         if(a[i]==e4)
             a4++;
     }
     while(cin>>n){
         if(n==)
             break;
         if(n%== || n%== || n%==)
             cout<<"The "<<n<<"th humble number is "<<a[n]<<'.'<<endl;
         else {
             int t = n%;
             switch(t){
                 case :cout<<"The "<<n<<"th humble number is "<<a[n]<<'.'<<endl;break;
                 case :cout<<"The "<<n<<"st humble number is "<<a[n]<<'.'<<endl;break;
                 case :cout<<"The "<<n<<"nd humble number is "<<a[n]<<'.'<<endl;break;
                 case :cout<<"The "<<n<<"rd humble number is "<<a[n]<<'.'<<endl;break;
                 default:cout<<"The "<<n<<"th humble number is "<<a[n]<<'.'<<endl;break;
             }
         }
     }
     return ;
 }

Freecode : www.cnblogs.com/yym2013