求助，求求你们帮帮我吧
各位高手能不能帮我写一个有关有理数四则混合运算的算法，快要考试啦，可是我还是不会做马求求你们行行好，帮帮我吧！！！
//Queue.h class Queue;
class Node{
private: Node *next; int data; int ndata;
public: Node(int d1,int n1,Node *l=NULL):data(d1),ndata(n1),next(l){}; friend class Queue;
}; class Queue{
public: Queue(); ~Queue(); void EnQue(int d,int n); void DeQue(int &d,int &n); bool IsEmpty(){return rear==front;}
void MakeEmpty(); int getlong(){return count;}
private:
Node *rear,*front; int count; };
Queue::Queue() { rear=front=NULL; count=0; }
Queue::~Queue() { MakeEmpty(); }
void Queue::EnQue(int d,int n) { Node *current=new Node(d,n,NULL);
if(rear==NULL)
rear=front=current;
else{ rear->next=current;
rear=current; } }
void Queue::DeQue(int &d,int &n) { Node *current;
assert(!IsEmpty());
current=front;
d=current->data;
n=current->ndata;
front=front->next;
delete current; }
void Queue::MakeEmpty() { Node *current;
while(front) { current=front;
front=front->next;
delete current; } } //stack.h template<typename T>class LinStack;
template<typename T>class StackNode{ private:
StackNode<T> *next;
T data; public:
StackNode(const T &item,StackNode<T> *perNext=NULL);
friend class LinStack<T>; };
template<typename T>StackNode<T>::StackNode(const T &item,StackNode<T> *perNext=NULL){ data=item; next=perNext; }
template<typename T>class LinStack{ private: StackNode<T> *top; public: LinStack(); ~LinStack(); void MakeEmpty(); void Push(const T &item); bool IsEmpty(){return top==NULL;} T pop() ; T Gettop(){ return top->data;}
void ClerStack();
}; template<typename T>LinStack<T>::LinStack(){ top=NULL; } template<typename T>LinStack<T>::~LinStack(){ MakeEmpty(); } template<typename T>void LinStack<T>::MakeEmpty(){ StackNode<T> *temp; while(top!=NULL){ temp=top; top=top->next; delete temp; } } template<typename T>void LinStack<T>::Push(const T &item){ StackNode<T> *newNode; newNode=new StackNode<T>(item,top); top=newNode; } template<typename T>T LinStack<T>::pop() {
assert(!IsEmpty());
StackNode<T> *temp; temp=top; T data=temp->data; top=top->next; delete temp; return data; } #include<iostream> #include<assert.h> #include"Queue.h" #include"stack.h" using namespace std;
typedef struct i{ int data;
int ndata; }Item;
int perence(char op) { switch(op) { case '+': case '-': return 1; case '*': case '/': return 2; default: return 0; } } void expersion(char *exp,LinStack<char> s,char *s1) { char *c=exp;
s.Push('@'); while(*c) { if(*c==' ')
c++;
else if(*c=='(') { s.Push(*c);
c++; }
else if(*c==')') { while(s.Gettop()!='(')
*s1++=s.pop();
s.pop();
c++; } else if(*c=='+'||*c=='-'||*c=='*'||*c=='/') { while(perence(*c)<perence(s.Gettop()))
*s1++=s.pop();
s.Push(*c);
c++; } else{ while(*c>='0'&&*c<='9')
*s1++=*c++; } *s1++=' '; } *s1++=' ';
char ch=s.pop();
while(ch!='@') { if(ch=='(')
exit(1); else { *s1++=ch;
*s1++=' ';
ch=s.pop(); } }
*s1='\0'; }
void calculation(char *s1,LinStack<double> sd) { double num1,num2;
char *c=s1;
double item=0.0;
while(*c) { switch(*c) { case '+': num1=sd.pop();
num2=sd.pop();
sd.Push(num2+num1);
c++;
break; case '-': num1=sd.pop();
num2=sd.pop();
sd.Push(num2-num1);
c++; break; case '*': num1=sd.pop();
num2=sd.pop();
sd.Push(num2*num1);
c++;
break; case '/': num1=sd.pop();
num2=sd.pop();
if(num2==0)
exit(1); else sd.Push(num2/num1);
c++; break; case ' ': c++; break; default: while(*c>='0'&&*c<='9') { item=item*10+*c-'0';
c++; } sd.Push(item); item=0.0;
break; } } if(!sd.IsEmpty())
cout<<"答案是:"<<sd.pop()<<endl; } void main() { Queue q;
LinStack<char> s;
LinStack<double> sd;
Item *item;
int d,n,size,i,j; char exp[255],s1[255];
long int sum,x,x1=0;
cout<<"确定多项式数组大小:"<<endl;
cin>>size;
item=new Item[size]; for(i=0;i<size;i++) { cout<<"输入系数:"<<endl; fflush(stdin); cin>>item[i].data;
cout<<"输入x的次方:"<<endl;
fflush(stdin);
cin>>item[i].ndata; }
for(i=0;i<=size;i++)
q.EnQue(item[i].data,item[i].ndata);
delete item;
cout<<"确认未知数x的值:"<<endl;
cin>>x;
x1=x;
cout<<"依照以下的数字组成中缀表达式:"<<endl; for(i=0;i<size;i++) { q.DeQue(d,n);
if(n==0) { sum=d*x;
cout<<sum<<'\t'; }
else{ for(j=0;j<n-1;j++) x*=x;
sum=d*x;
x=x1;
cout<<sum<<'\t'; }
} cout<<endl;
cout<<"输入中缀表达式:"<<endl; for(i=0;i<255;i++) { cin>>exp[i];
if(exp[i]=='=')
break; } exp[i]='\0'; expersion(exp,s,s1);
cout<<"输出后缀表达式:"<<endl;
cout<<s1<<endl;
calculation(s1,sd); } 输入格式：
例如我要计算 4x^3+2x^3+2x^2 当x＝3时的值 （^3表示3次方，^2表示平方，依此类推)
确定多项式数组大小: 3 输入系数: 4 输入x的次方: 3 输入系数: 2 输入x的次方: 3 输入系数: 2 输入x的次方: 2 确认未知数x的值: 3 依照以下的数字组成中缀表达式: 324 162 18 输入中缀表达式: 324 - 162 + 18 = 你输入x的次方为0时就可以实现你要的功能拉
