hdu-题目：1058 Humble Numbes-优快云博客

本文介绍了一种使用动态规划的方法来解决求第n个谦数的问题，并提供了完整的C语言实现代码。谦数是指仅由2、3、5、7这四个质数因子构成的整数。文章首先定义了问题背景及输出格式，随后给出了具体的算法实现，包括状态转移方程的推导及程序的运行示例。

http://acm.hdu.edu.cn/showproblem.php?pid=1058

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

100

1000

5842

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.

状态转移方程：F(n)=min(f(i)*2, f(j)*3, f(k)*5, f(m)*7); n<(i,j,k,m)

注意：11th , 12th, 13th.

 1 #include <stdio.h>
 2 
 3 int min(int a, int b)
 4 {
 5     if(a>b)
 6         return b;
 7     else
 8         return a;
 9 }
10 
11 int min4(int a, int b, int c, int d)
12 {
13     return min(min(a,b), min(c,d));
14 }
15 
16 const int N=6000;
17 int F[N];
18 
19 int main()
20 {
21 
22     int n=1;
23     F[1]=1;
24 
25     int h2, h3, h5, h7;
26     h2=h3=h5=h7=1;
27 
28     while(F[n]<2000000000)
29     {
30         F[++n]=min4(2*F[h2], 3*F[h3], 5*F[h5], 7*F[h7]);
31         if(F[n]==2*F[h2]) h2++;
32         if(F[n]==3*F[h3]) h3++;
33         if(F[n]==5*F[h5]) h5++;
34         if(F[n]==7*F[h7]) h7++;
35     }
36 
37     while(scanf("%d", &n)!=EOF)
38     {
39         if(n==0) break;
40         if(n%10==1 && n%100!=11)
41             printf("The %dst humble number is %d.\n", n, F[n]);
42         else if(n%10==2 && n%100!=12)
43             printf("The %dnd humble number is %d.\n", n, F[n]);
44         else if(n%10==3 && n%100!=13)
45             printf("The %drd humble number is %d.\n", n, F[n]);
46         else
47             printf("The %dth humble number is %d.\n", n, F[n]);
48     }
49 
50     return 0;
51 }