Humble Numbers

最新推荐文章于 2018-11-01 19:35:49 发布

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本文介绍了一种特殊数列——谦逊数（Humble Numbers），这些数仅由2、3、5、7作为质因数构成。文章详细描述了一个算法过程，用于找出数列中的特定元素，并提供了完整的代码实现及样例输出。

Humble Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20341 Accepted Submission(s): 8869

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.

Source

University of Ulm Local Contest 1996

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分析：题目比较好理解，但是由于数据比较大，和原来做过得题类似，想到暴力，结果过了。

然后看后面分析，发现这题是个dp问题。

代码1：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
int max1=2000000000;
long long a[100000];
int cmp(int b,int c)
{
return b<c;
}
int main()
{
int i,j,k,n,l,m=0;
for(i=0;i<32&&pow(2,i)<=max1;i++)
for(j=0;j<21&&pow(2,i)*pow(3,j)<=max1;j++)
for(k=0;k<15&&pow(2,i)*pow(3,j)*pow(5,k)<=max1;k++)
for(l=0;l<13&&pow(2,i)*pow(3,j)*pow(5,k)*pow(7,l)<=max1;l++)
{
a[++m]=pow(2,i)*pow(3,j)*pow(5,k)*pow(7,l);
}
sort(a+1,a+m+1,cmp);
while(~scanf("%d",&n)&&n)
{
int flag=0;
printf("The %d",n);
if(n%10==1&&n%100!=11)
{
printf("st");flag=1;
}
if(n%10==2&&n%100!=12)
{
printf("nd");flag=1;
}
if(n%10==3&&n%100!=13)
{
printf("rd");flag=1;
}
if(!flag)
printf("th");
printf(" humble number is %lld.\n",a[n]);
}
return 0;
}