HDU 1058 Humble Numbers

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.

 

题意：有一个数的质因子是2，3，5，7中的任意一个或多个，那么它就是Humble 数，求第N个Humble 数是多少。

分析：若一个Humble 数，则它肯定是2，3，5，7的倍数，则a[k]=min(2*b2,3*b3,5*b5,7*b7)，而b2，b3，b5，b7不断更新，这样就按顺序排列起来了

 

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
int a[5850]= {0,1};
int main()
{
    int k=1,n;
    int b2,b3,b5,b7;
    b2=b3=b5=b7=1;
    while(k<=5842)
    {
        a[++k]=min(2*a[b2],min(3*a[b3],min(5*a[b5],7*a[b7])));
        if(a[k]==2*a[b2])
            b2++;
        if(a[k]==3*a[b3])
            b3++;
        if(a[k]==5*a[b5])
            b5++;
        if(a[k]==7*a[b7])
            b7++;
    }
    while(~scanf("%d",&n),n)
    {
        if(n%10==1&&n%100!=11)
            printf("The %dst humble number is %d.
",n,a[n]);
        else if(n%10==2&&n%100!=12)
            printf("The %dnd humble number is %d.
",n,a[n]);
        else if(n%10==3&&n%100!=13)
            printf("The %drd humble number is %d.
",n,a[n]);
        else
            printf("The %dth humble number is %d.
",n,a[n]);
    }
    return 0;
}