题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=1058
Humble Numbers
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29253 Accepted Submission(s): 12809
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
1
2
3
4
11
12
13
21
22
23
100
1000
5842
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：
如果一个数的因子只有2，3，5，7那么这个数叫差数，此题就是求第n个差数是多少？
分析：
求第n个a[n],则肯定是前面的第n-1个数中的某一个与2，3，5，7中的一个的乘积，取最小的哪一个，取的哪一个对应的指针就向前移动一位
#include<bits/stdc++.h>
#define max_v 5850
int i2=1,i3=1,i5=1,i7=1,i;
int a[max_v];
using namespace std;
int f()
{
int x=min ( min ( 2*a[i2], 3*a[i3] ),
min ( 5*a[i5], 7*a[i7] ) );
if(x==2*a[i2])
{
++i2;
}else if(x==3*a[i3])
{
++i3;
}else if(x==5*a[i5])
{
++i5;
}else if(x==7*a[i7])
{
++i7;
}
return x;
}
int main()
{
a[0]=0;
a[1]=1;
for(i=2;i<max_v;i++)
{
a[i]=f();
if(a[i]==a[i-1])
{
i--;
}
}
int n;
while(~scanf("%d",&n))
{
if(n==0)
break;
printf("The %d",n);
if(n%100==11||n%100==12||n%100==13)
{
printf("th ");
}else if(n%10==1)
{
printf("st ");
}else if(n%10==2)
{
printf("nd ");
}else if(n%10==3)
{
printf("rd ");
}else
{
printf("th ");
}
printf("humble number is %d.\n",a[n]);
}
return 0;
}