这道题有很多种做法，但是思路大都是一样的，代码有点类似于poj2591这道题。

题意：问因子只含有2,3,5,7的第k个数是什么？

#include<stdio.h>

int f[5843],n;

int i,j,k,l;

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;

    f[1] = 1;

    for(int t=2;t<=5842;t++)

        f[t] = Min(2*f[i],3*f[j],5*f[k],7*f[l]);

    while(scanf("%d",&n),n)

    {

        if(n%100==11||n%100==12||n%100==13)

            printf("The %dth humble number is %d.\n",n,f[n]);

        else if(n%10==1)

            printf("The %dst humble number is %d.\n",n,f[n]);

        else if(n%10==2)

             printf("The %dnd humble number is %d.\n",n,f[n]);

        else if(n%10==3)

            printf("The %drd humble number is %d.\n",n,f[n]);

        else

            printf("The %dth humble number is %d.\n",n,f[n]);

    }

    return 0;

}