humble numbers问题

最新推荐文章于 2020-01-17 17:28:29 发布

wshe2

最新推荐文章于 2020-01-17 17:28:29 发布

阅读量360

点赞数

分类专栏： acm c++

本文链接：https://blog.youkuaiyun.com/wshe2/article/details/45032205

版权

acm 同时被 2 个专栏收录

6 篇文章

订阅专栏

c++

3 篇文章

订阅专栏

本文介绍了一种特殊的数——谦逊数，这些数仅由2、3、5、7这几个质因数组成。通过一个高效的算法实现，我们能够找出序列中的任意一项，并提供了完整的C语言实现代码，展示了如何生成和打印特定位置的谦逊数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

算法思想：

就是用2，3，5，7循环来求第i个humble[i],第i个humble[i]必定等于前i-1个数中其中一个数与｛2，3，5，7｝中其中最小的乘积.

注意：

1.在输出是st代表frist,nd代表second,rd代表third,th代表其他的数。

2.由输出的格式可以得到以11,12,13结尾的数是th表示。

3.在定义数组时要定义成long long humble[5842],如果定义int humble[5842]会越界。

代码如下：

#include <stdio.h>
#include <stdlib.h>
int prime[4]={2,3,5,7};
long long humble[5845]={0};//
int main()
{
int n,i,j,k;
humble[1]=1;
for(i=2;i<=5842;i++)
{
humble[i]=2000000001;
for(j=0;j<4;j++)
{
for(k=i-1;k>=1;k--)
{
if(humble[k]*prime[j]<=humble[i-1])
break;
if(humble[k]*prime[j]<humble[i])
humble[i]=humble[k]*prime[j];
}
}
}
while(scanf("%d",&n)&&n)
{
if(n%10==1&&n%100!=11)
{
printf("The %dst humble number is %d.\n",n,humble[n]);
}
else if(n%10==2&&n%100!=12)
{
printf("The %dnd humble number is %d.\n",n,humble[n]);
}
else if(n%10==3&&n%100!=13)
{
printf("The %drd humble number is %d.\n",n,humble[n]);
}
else
{
printf("The %dth humble number is %d.\n",n,humble[n]);
}

}
return 0;
}