SOJ1029 Humble Numbers （枚举）

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.

思路：本题要求第i小的质因子只有2，3，5,7的数，一个一个与质数相关的题目显然会超时，所以遇到这种问题一般会提前先进行预处理，之后查询。由于题目中说明n<=5842，最大的值为2*10^9所以可以断定本题使用枚举法将结果存在一个整形数组里即可。之后就是需要考虑枚举范围，2*10^9的是2^31,3^20,5^14,7^12以内，所以2、3、5、7的枚举范围分别为31,20,14,12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;

AC代码：

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

int hum[];

void getHum()
{
    int a,b,c,d;
    int len=;
    long long num2,num3,num5,num7;
    for(a=;a<=;a++)
    {
        num2=(long long)pow(2.0,(double)a);
        for(b=;b<=;b++)
        {
            num3=(long long)pow(3.0,(double)b);
            if(num2*num3>MAX)
                break;
            for(c=;c<=;c++)
            {
                num5=(long long)pow(5.0,(double)c);
                if(num2*num3*num5>MAX)
                    break;
                for(d=;d<=;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)==&&n!=)
    {
        if(n%==&&n%!=)
            printf("The %dst humble number is %d.\n",n,hum[n-]);
        else if(n%==&&n%!=)
            printf("The %dnd humble number is %d.\n",n,hum[n-]);
        else if(n%==&&n%!=)
            printf("The %drd humble number is %d.\n",n,hum[n-]);
        else
            printf("The %dth humble number is %d.\n",n,hum[n-]);
    }
    return ;
}

参考博客：http://www.cnblogs.com/dolphin0520/archive/2011/04/15/2016774.html