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
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.
题目大意：找出丑数，即因子只有2，3，5，7中数字的数，很容易想到较大的丑数都是由小的丑数而来，递推即可。值得注意的是输出时需要输出序数词后缀，规律是个位是1，2，3时分别输出st,nd,rd，十位11，12，13反常全是th。特判即可。
AC代码：
#include<bits/stdc++.h>
using namespace std;
int main()
{
int cs[6000]={0,1},v2,v3,v5,v7,next,n;
v2=v3=v5=v7=1;
cs[1]=1;
for(int i=2; i<=5842; i++)
{
next=min(cs[v2]*2,min(cs[v3]*3,min(cs[v5]*5,cs[v7]*7)));
if(next==cs[v2]*2)v2++;//这4个if不加else，可以去重
if(next==cs[v3]*3)v3++;
if(next==cs[v5]*5)v5++;
if(next==cs[v7]*7)v7++;
cs[i]=next;
}
while(scanf("%d",&n),n){
printf("The %d",n);
if((n0)!=11&&(n)==1)printf("st humble number is %d.\n",cs[n]);
else if((n0)!=12&&(n)==2)printf("nd humble number is %d.\n",cs[n]);
else if((n0)!=13&&(n)==3)printf("rd humble number is %d.\n",cs[n]);
else printf("th humble number is %d.\n",cs[n]);
}
return 0;
}
