Humble Numbers soj1029

         Humble Numbers

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 Specification

The input consists of one or more test cases. Each test case consists of one integer n with  . Input is terminated by a value of zero (0) for n.

Output Specification

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.

  题意：求出丑数。所谓的丑数就是因子只含2，3，5，7.若一个数n是丑数，则必定有n=2^a*3^b*5^c*7^d，因此只需求出所有的丑数，对其进行排序即可。

 在这里还有注意输出格式的问题，对于序数词：

 1)若n%10=1,2,3并且n%100!=11,12,13，则末尾为st,nd,rd;

 2)否则末尾为th;

#include < iostream >
#include < algorithm >
#include < stdlib.h >
#include < math.h >
#define  MAX 2000000000
using namespace  std;

int  hum[ 5842 ];

void  getHum()
{
int  a,b,c,d;
int  len = 0 ;
long long  num2,num3,num5,num7; 
for (a = 0 ;a <= 31 ;a ++ )
{
num2 = ( long long )pow( 2.0 ,( double )a);
for (b = 0 ;b <= 20 ;b ++ )
{
num3 = ( long long )pow( 3.0 ,( double )b);
if (num2 * num3 > MAX)
break ;
for (c = 0 ;c <= 14 ;c ++ )
{
num5 = ( long long )pow( 5.0 ,( double )c);
if (num2 * num3 * num5 > MAX)
break ;
for (d = 0 ;d <= 12 ;d ++ )
{ 
num7 = ( long long )pow( 7.0 ,( double )d);
if (num2 * num3 * num5 * num7 <= MAX)
hum[len ++ ] = num2 * num3 * num5 * num7;
else break ;
}
}
}
}
sort(hum,hum + len);
}



int  main( void )
{
int  n;
getHum();
while (scanf( " %d " , & n) == 1 && n != 0 )
{
if (n % 10 == 1 && n % 100 != 11 )
printf( " The %dst humble number is %d.\n " ,n,hum[n - 1 ]);
else if (n % 10 == 2 && n % 100 != 12 )
printf( " The %dnd humble number is %d.\n " ,n,hum[n - 1 ]);
else if (n % 10 == 3 && n % 100 != 13 )
printf( " The %drd humble number is %d.\n " ,n,hum[n - 1 ]);
else
printf( " The %dth humble number is %d.\n " ,n,hum[n - 1 ]);
}
return 0 ;
}

本文转载自海 子博客园博客，原文链接：http://www.cnblogs.com/dolphin0520/archive/2011/04/15/2016774.html如需转载自行联系原作者