题目：
Humble Numbers
**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24880 Accepted Submission(s): 10947
**
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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\
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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.\
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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.\
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Sample Input
1
2
3
4
11
12
13
21
22
23
100
1000
5842
0
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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.
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Source
University of Ulm Local Contest 1996
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Recommend
JGShining
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Statistic | Submit | Discuss | Note
思路：
这个题起初看了很长时间没看懂，题意是每个数质因子只含有2,3,5,7,问你第几个这样的数是多少，还要注意输出的时候根据英文规则，尾数为1,2,3分别用st,nd,rd来输出，参考了大牛的代码，这种打表方式前所未有，很简练，这道题完全是暴力。。。
代码：
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <stack>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define min4(a,b,c,d) min(min(a,b),min(c,d))
using namespace std;
int a[6000];
int main()
{
int n=1;
int p2=1,p3=1,p5=1,p7=1;
a[1]=1;
while(a[n]<2000000000)//打表
{
a[++n]=min4(2*a[p2],3*a[p3],5*a[p5],7*a[p7]);
if(a[n]==2*a[p2]) p2++;
if(a[n]==3*a[p3]) p3++;
if(a[n]==5*a[p5]) p5++;
if(a[n]==7*a[p7]) p7++;
}
while(scanf("%d",&n)&&n)
{
if(n%10 == 1&&n%100!=11)
printf("The %dst humble number is ",n);//判断尾数是否为1
else if(n%10 == 2&&n%100!=12)
printf("The %dnd humble number is ",n);//判断尾数是否为2
else if(n%10 == 3&&n%100!=13)
printf("The %drd humble number is ",n);//判断尾数是否为3
else
printf("The %dth humble number is ",n);
printf("%d.\n",a[n]);
}
return 0;
}
//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。
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