ACM 1136求大侠帮忙看看我的题怎么不对
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 Specification
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 Specification
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.


题目是这个
我的代码
说什么就是不对
救命啊
#include <iostream>
using namespace std;
bool isPrimeFactor(long int temp)
{
if(temp%2==0||temp%3==0||temp%5==0||temp%7==0)
return 1;
else
return 0;
}
int main()
{
long int sequence[5843],temp;
int a=1;
sequence[0]=1;
for(temp=2;temp <=2000000000;temp++)
if(isPrimeFactor(temp)==1){
sequence[a]=temp;
a++;
}
cin> > temp;
if(temp==0)
return 1;
cout < < "The temp ";
if(temp%10==1)
cout < < "st ";
else if(temp%10==2)
cout < < "nd ";
else if(temp%10==3)
cout < < "rd ";
else
cout < < "th ";
cout < < "humble number is " < <sequence[a-1] < < ", " < <endl;

return 0;
}



[解决办法]
另外还有一种简单的方法的代码:
#include <iostream.h>
#include <fstream.h>

long humble[5850]={0};

long min(long a, long b, long c, long d)
{
long m;
if(a> b)m=b;
else m=a;
if(c <m)m=c;
if(d <m)m=d;
return m;
}

int main()
{
humble[0]=1;
int p2,p3,p5,p7,i;
//ofstream out( "out.txt ");
p2=p3=p5=p7=i=0;
while(humble[i] <2000000000){
humble[++i]=min(humble[p2]*2, humble[p3]*3, humble[p5]*5, humble[p7]*7);
if(humble[p2]*2==humble[i])p2++;
if(humble[p3]*3==humble[i])p3++;
if(humble[p5]*5==humble[i])p5++;
if(humble[p7]*7==humble[i])p7++;
}
int no;
while(cin> > no, no> 0){
if(no%10==1 && no%100!=11)
cout < < "The " < < no < < "st " < < "humble number is ";
else if(no%10==2 && no%100!=12)
cout < < "The " < < no < < "nd " < < "humble number is ";
else if(no%10==3 && no%100!=13)
cout < < "The " < < no < < "rd " < < "humble number is ";
else
cout < < "The " < <no < < "th " < < "humble number is ";
cout < < humble[no-1] < < ". " < <endl;
}
return 0;
}