HDU 1058  Humble Numbers

Humble Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5941    Accepted Submission(s): 2570

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

 

Recommend

JGShining
我就纳闷了，这题真的很好么？做了四五遍了，还能碰到，用数组存下所有的值，最后输出就可以了
humble number从1为"始祖",剩下的所有数,其实都是在此基础上乘以2,3,5,7演化出来的
代码主要语句:num [q ]= max (num [i ]* 2 , max (num [j ]* 3 , max (num [k ]* 5 ,num [l ]* 7 ))); 如果大家觉得比较抽象,顺着代码遛一遍,在纸上一步步手动算到num15],肯定可以深入理解了
代码如下：
#include<stdio.h>
int max(int a,int b)
{
      if(b>a)return a;
      else return b;
}
int main()
{
      int num[6000];
      int i,j,k,l,q,n;
      i=j=k=l=1;
      num[1]=1;
      for(q=2;q<=5845;q++)
      {
            num[q]=max(num[i]*2,max(num[j]*3,max(num[k]*5,num[l]*7)));
            if(num[q]==num[i]*2)
                  i++;
            if(num[q]==num[j]*3)
                  j++;
            if(num[q]==num[k]*5)
                  k++;
            if(num[q]==num[l]*7)
                  l++;
      }
      while(scanf("%d",&n),n!=0)
      {
            if(n% 10==1&&n% 1 0 0!=11)//(这里是n对10取余等于1并且对100取余不等于11，下同，)
                  printf("The %dst humble number is %d.\n",n,num[n]);
            else if(n% 10==1&&n% 1 0 0!=12)//郁闷，不知道为什么改不过来
                  printf("The %dnd humble number is %d.\n",n,num[n]);
            else if(n% 10==1&&n% 1 0 0!=13)
                  printf("The %drd humble number is %d.\n",n,num[n]);
            else
                  printf("The %dth humble number is %d.\n",n,num[n]);
      }
      return 0;
}