Humble Numbers - HDU 1058 递推

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本文探讨了一个特定序列中第n个谦逊数的查找方法，谦逊数是指仅包含质因数2、3、5或7的整数。通过递归算法生成并筛选出序列，最后按要求输出指定位置的谦逊数及其对应的序数表达。

Humble Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17272 Accepted Submission(s): 7504

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

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.

思路：就是一个简单的递推，注意输出格式就行了。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<map>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
map<ll,int> match;
vector<ll> vc;
ll num[6000],ret;
int main()
{ int n,i,j,k;
  num[1]=1;
  match[1]=1;
  vc.push_back(1);
  n=1;
  while(n<=5842)
  { ret=vc[n-1];
    num[n]=ret;
    if(ret*2<=2000000000 && match[ret*2]==0)
    { vc.push_back(ret*2);
      match[ret*2]=1;
    }
    if(ret*3<=2000000000 && match[ret*3]==0)
    { vc.push_back(ret*3);
      match[ret*3]=1;
    }
    if(ret*5<=2000000000 && match[ret*5]==0)
    { vc.push_back(ret*5);
      match[ret*5]=1;
    }
    if(ret*7<=2000000000 && match[ret*7]==0)
    { vc.push_back(ret*7);
      match[ret*7]=1;
    }
    n++;
  }
  sort(num+1,num+5843);
  while(~scanf("%d",&n) && n)
  { if(n%10==1 && n%100/10!=1)
     printf("The %dst humble number is %d.\n",n,num[n]);
    else if(n%10==2 && n%100/10!=1)
     printf("The %dnd humble number is %d.\n",n,num[n]);
    else if(n%10==3 && n%100/10!=1)
     printf("The %drd humble number is %d.\n",n,num[n]);
    else
     printf("The %dth humble number is %d.\n",n,num[n]);
  }
}