HDU 1058 Humble Numbers

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原文链接：http://www.cnblogs.com/372465774y/archive/2012/04/12/2443282.html

本文介绍了一种使用C语言实现的算法，该算法用于找出特定序列中的谦数，并提供了一个完整的程序示例。谦数是指一个数仅由2、3、5或7这几个质数因子组成的数。文章还展示了如何根据输入的序号打印对应的谦数。

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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.

//英语不好的孩子伤不起呀！

#include <stdio.h>
#include <string.h>
#define N 6666
__int64 dp[N];
int main()
{
    int i,j,k,n,r[4]={2,3,5,7};
    __int64 min,temp,tp;
    dp[1]=1;dp[2]=2;dp[3]=3;dp[4]=4;dp[5]=5;
    dp[6]=6;dp[7]=7;dp[8]=8;
    for(i=9;i<=5842;i++)
    {    tp=i-1;
        min=0xffffffffff;
        for(j=tp;j>0&&dp[j]*r[3]>dp[tp];j--)//从[1..i-1]中通过乘（2,3,5,7）并取min得第i个数
          for(k=0;k<4;k++)
           {
               temp=dp[j]*r[k];
              if(temp>dp[tp]&&temp<min)
                 min=temp;
           }
        dp[i]=min;
    }
   while(scanf("%d",&n),n)
    {
        if((n%100)/10==1)
           printf("The %dth humble number is %I64d.\n",n,dp[n]);
        else
           {
               switch(n%10)
               {
                   case 1:printf("The %dst humble number is %I64d.\n",n,dp[n]);break;
                   case 2:printf("The %dnd humble number is %I64d.\n",n,dp[n]);break;
                   case 3:printf("The %drd humble number is %I64d.\n",n,dp[n]);break;
                   default:printf("The %dth humble number is %I64d.\n",n,dp[n]);break;
               }
           }
    }
    return 0;
}

转载于:https://www.cnblogs.com/372465774y/archive/2012/04/12/2443282.html