用栈来实现分解一个数成素因子乘积的形式。

stack.h源文件：
#include <math.h>
#include <iostream.h>
const int maxentry=20;
enum error_code{fail,success,overflow,underflow};
template<class type>
class stack
{
private:
int count;
type entry[maxentry];
public:
stack()
{count=0;}
error_code push(const type &a);
error_code pop();
bool empty()const;
error_code top(type &a)const;
int numbers();
};
template<class type>
error_code stack<type>::push(const type &a)
{
error_code outcome=success;
if(count<maxentry)
{
entry[count++]=a;
}
else outcome=fail;
return outcome;
}
template<class type>
error_code stack<type>::pop()
{
error_code outcome=success;
if(count>0)
{
count--;
}
else outcome=fail;
return outcome;
}
template<class type>
bool stack<type>::empty() const
{
return count==0;
}
template<class type>
error_code stack<type>::top(type &a) const
{
error_code outcome=success;
if(count>0)
{
a=entry[count-1];
}
else outcome=fail;
return outcome;
}
template<class type>
int stack<type>::numbers()
{
return count;
}

bool isprime(int n)
{
if (n==1)
return false;
int k;
k=sqrt(n);

for (int i=2;i<=k;i++)
{
if (n%i==0)
{return false;}
if (i>=k+1)
{return true;}
else {return false;}
}

}

void prime(stack<int> num)
{
int m=2;
//int a[255]={0,};
//int i;
int n;
num.top(n);
//num.pop();
//cout<<n<<endl;
if (n==1||n==0||n<0)
{cout<<"您所输入的数没有素因子。"<<endl;return;}
while (n>1)
{

loop1:{
if (isprime(n))
{goto loop2;}
else if (n%m==0)
{
//cout<<num.top()<<endl;
num.pop();
n=n/m;
num.push(m);
if (n!=1)
num.push(n);

goto loop1;

}
else m++;
if (!isprime(m))
m++;
else goto loop1;
}

}

/*if (num.top(i)==1)
num.pop();*/

loop2:{ int x=num.numbers();
for (int j=0;j<=x-1;j++)
{
int m;
num.top(m);
cout<<m<<" ";
num.pop();
}

cout<<endl;
}

}

prime.cpp:
#include <iostream>
using namespace std;
#include "stack.h"
#include <math.h>
void prime(stack<int>);
bool isprime(int);

void main()
{
stack<int> num;
int n;
char s;
loop1:{cout<<"Input a Number:"<<endl;
cin>>n;
cout<<"The prime of the number is:"<<endl;
//cout<<isprime(n)<<endl;
num.push(n);
prime(num);}
cout<<"是否继续？"<<endl;
cin>>s;
if (s=='y')
{num.pop();
goto loop1;}
else return;

}