The 5842nd humble number is 2000000000.
#include<iostream>using
namespace
std;long
int
x[5900];int
num2=0,num3=0,num5=0,num7=0;void
t(){
x[0]=1; for(int
i=1;i<5900;i++) { int
min=x[num2]*2,b=x[num3]*3,c=x[num5]*5,d=x[num7]*7; if(min>b) { min=b; } if(min>c) { min=c; } if(min>d) { min=d; } x[i]=min; if(x[i]==x[num2]*2) num2++; if(x[i]==x[num3]*3) num3++; if(x[i]==x[num5]*5) num5++; if(x[i]==x[num7]*7) num7++; }}int
main(){ int
n; t(); while(cin>>n&&n!=0) { if(n%10==1&&n!=11) cout<<"The"<<"
"<<n<<"st"<<"
"<<"humble
number is "<<x[n-1]<<"."<<endl; else
if(n%10==2&&n!=12) cout<<"The"<<"
"<<n<<"nd"<<"
"<<"humble
number is "<<x[n-1]<<"."<<endl; else
if(n%10==3&&n!=13) cout<<"The"<<"
"<<n<<"rd"<<"
"<<"humble
number is "<<x[n-1]<<"."<<endl; else
cout<<"The"<<"
"<<n<<"th"<<"
"<<"humble
number is "<<x[n-1]<<"."<<endl; } return
0;}一直因子只能是2,3,5,7，所以分别用四个量来表示，通过将最小值放入数组中构造成功
另外需要考虑输出形式，在11，12,13处分别有小陷阱需注意
原文：http://blog.csdn.net/u013240812/article/details/25830615