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

123411121321222310010005842

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.

大致题意：若一个数的所有素因子是2、3、5、7中的一个或多个，则这个数成为Humble数。求第n个Humble数是多少。
可以用递推的方法来求得前5842项的Humble数，由题意可知，如果一个数是Humble数，那么它的 2 , 3 , 5 , 7倍也是Humble数，由此可推得Humble数的序列。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;const int size = 200010;int num[size];bool use[size];int tot = 1;int s , t , f ,se;void work(int x){    if(x == 2 * num[s])        s ++;    if(x == 3 * num[t])        t ++;    if(x == 5 * num[f])        f ++;    if(x == 7 * num[se])        se ++;}int main(){    num[1] = 1;    s = t = f = se = 1;    while(tot <= 5842)    {        int d = min(min(2 * num[s] , 3 * num[t]) , min(5 * num[f] , 7 * num[se]));        num[++tot] = d;        work(d);    }    int n;    while(scanf("%d",&n) && n)    {        printf("The %d",n);        if(n % 10 == 1 && n % 100 != 11)            printf("st");        else if(n % 10 == 2 && n % 100 != 12)            printf("nd");        else if(n % 10 == 3 && n % 100 != 13)            printf("rd");        else            printf("th");        printf(" humble number is %d.\n",num[n]);    }    return 0;}