題目：
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.

題意：humble number: 公因數都是2，3，5，7的數
求humble number 可以用這幾個數字想成來做。求第幾個數字，可以比較各個乘積的大小。定義陣列num儲存所有humble number。 然後按順序存入。

程式碼

//
//  main.cpp
//  hdu1058
//
//  Created by zhan_even on 2018/10/28.
//  Copyright © 2018年 zhan_even. All rights reserved.
//

#include <iostream>
#include <cstdio>
using namespace std;
int i,j,k,l;
int num[5843],n;
int min(int a,int b,int c,int d){
    int min = a;
    if(b<min) min = b;
    if(c<min) min = c;
    if(d<min) min = d;
    
    if(a==min) i++;
    if(b==min) j++;
    if(c==min) k++;
    if(d==min) l++;
    return min;
}

int main() {
    i=j=k=l=1;
    num[1] = 1;
    for (int t = 2; t<5843; t++) {
        num[t] = min(2*num[i], 3*num[j], 5*num[k], 7*num[l]);
    }
    while(cin>>n){
        if(n==0)
            break;
        else if(n%10==1&&n%100!=11)
            printf("The %dst humble number is %d.\n",n,num[n]);
        else if(n%10==2&&n%100!=12)
            printf("The %dnd humble number is %d.\n",n,num[n]);
        else if(n%10==3&&n%100!=13)
            printf("The %drd humble number is %d.\n",n,num[n]);
        else
            printf("The %dth humble number is %d.\n",n,num[n]);
    }
    
    return 0;
}