hdu 1058 Humble Numbers (DP)

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最新推荐文章于 2024-08-25 08:45:00 发布

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文章标签： java 操作系统 c/c++

原文链接：http://www.cnblogs.com/GO-NO-1/p/3684865.html

本文介绍了一个算法，用于找到给定序列中指定位置的谦逊数。谦逊数是仅由质因数2、3、5或7构成的数。通过使用动态规划方法，可以高效地计算出序列中的任一谦逊数。

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Humble Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15978 Accepted Submission(s): 6915

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.

Source

University of Ulm Local Contest 1996

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简单DP，有点巧妙：

 1 //93 MS    228 KB    GNU C++
 2 #include<stdio.h>
 3 int ans[6000];
 4 inline int Min(int a,int b)
 5 {
 6     return a<b?a:b;
 7 }
 8 void init()
 9 {
10     int m1=0,m2=0,m3=0,m4=0;
11     int t1,t2,t3,t4;
12     ans[0]=1;
13     int i=1;
14     while(i<5842){
15         t1=ans[m1]*2;
16         t2=ans[m2]*3;
17         t3=ans[m3]*5; 
18         t4=ans[m4]*7;
19         int tmp=Min(Min(t1,t2),Min(t3,t4));
20         ans[i++]=tmp;
21         if(tmp==t1) m1++;
22         if(tmp==t2) m2++;
23         if(tmp==t3) m3++;
24         if(tmp==t4) m4++; 
25     }
26 }
27 int main(void)
28 {
29     int n;
30     init();
31     while(scanf("%d",&n)!=EOF && n)
32     {
33         if(n%10==1 && n%100!=11) printf("The %dst humble number is ",n);
34         else if(n%10==2 && n%100!=12) printf("The %dnd humble number is ",n);
35         else if(n%10==3 && n%100!=13) printf("The %drd humble number is ",n);
36         else printf("The %dth humble number is ",n);
37         printf("%d.\n",ans[n-1]);
38     }
39     return 0;
40 }