C++大数据运算{开n次方(二分法逐渐逼近求解)，n次方，小数，符号，加减乘除模！}

main.cpp

#include <iostream>
#include "queue.h"
using namespace std;
int main()
{
	using namespace work;
	Queue<int> q;
	Queue<int> p;
	Queue<int> w;
	char ch;
	int select=0;
	char *m = new char [1000];
	cout<<"欢迎来到大数据的四则运算：>"<<endl;
	cin.getline(m,1000);
	GetNumber(ch,m,&q,&p);	
//	MOD_POWER(&q,&p,&w);
//	w.view();
	 switch(ch)
		{
			case '+':
				ADD_S(&q,&p,&w);
				w.view();
				break;
			case '-':
				DEC_S(&q,&p,&w);
				w.view();
				break;
			case '*':
				MUL_S(&q,&p,&w);
				w.view();
				break;
			case '/':
				{
				cout<<"1进入整数运算，2进入小数运算:"<<endl;
			  cin>>select;
				if(select==2)
				{
					DIV_R(&q,&p,&w);
					w.view();
				}
				else
				{
				DIV_S(&q,&p,&w);
				w.view();
				}
				}
				break;
			case '%':
				MOD(&q,&p,&w);
				w.view();
				break;
			case '^':
				POWER_S(&q,&p,&w);
				w.view();
				break;
			case '@':
				MOD_POWER(&q,&p,&w);
				w.view();
			default:
				break;
		}
	return 0;
}

queue.h:

#pragma once
#include <iostream>
using namespace std;
namespace work
{
template<typename T>
class Queue;
template<typename T>
void ADD(Queue<T> *q,Queue<T> *p,Queue<T> *m);

template<typename T>
void DEC(Queue<T> *q,Queue<T> *p,Queue<T> *m);

template<typename T>
void DIV(Queue<T> *list1,Queue<T> *list2,Queue<T> *list3);

enum WHAT
{
	Plus,
	Minus,
};//确定符号位。

template<typename T>
class DList
{
	friend class Queue<T>;
	private:
		int data;//数据域
		DList *prev;
		DList *next;
};

template<typename T>
class Queue
{
	private:
	DList<T> *head;
	DList<T> *tail;
	int size;//链表长度。
	int Split;//小数点位数。
	public:
	Queue()
	{
		head = tail = new DList<T>();
		tail -> next = NULL;
		head -> prev = NULL;
		size = 0;
		Split = 0;
	}
	void Set_Split(int x)
	{
		Split = x;//设置小数点位数
  }
	int COMPLATE_SPLIT(const Queue<T> *list)
	{
		if(Split > list->Split)
		{
			return 1;
		}
		else if(Split == list->Split)
		{
			return 0;
		}
		else
		{
			return -1;
		}
	}//比较小数点位数
	int Get_Split()
	{
		return Split;
	}//得到小数点位数
	void Head_val(int x)
	{
		head->data = x;
	}//设置头节点的值（所谓的符号位）
	int Tail_Return()
	{
		return tail->data;
	}//返回尾节点值
	int Head_Return()
	{
		return head->data;
	}//返回头节点的值，也就是符号位。
	Queue(const Queue &q)
	{
    head = tail =new DList<T>();
 		tail -> next = NULL;
		head ->prev = NULL;
		size = 0;
		DList<T> *p = (q.head)->next;
		while(p!=NULL)
		{
			this->push_back(p->data);
			p=p->next;
		}
	}
	Queue operator = (const Queue &q)
	{
		head = tail = new DList<T>();
		tail -> next =NULL;
		head -> prev = NULL;
		size = 0;
		DList<T> *p = (q.head)->next;
		while(p!=NULL)
		{
		this->push_back(p->data);
		p = p->next;
		}
	}
	bool IS_empy()
	{
		return ((size>0));
	}
	int SIZE()
	{
	return size;
	}
	void CLEAR()
	{
		DList<T> *p = head->next;
		tail = head;
		tail -> next = NULL;
		head ->prev = NULL;
		while(p != NULL)
		{	
			DList<T> *q = p->next;
			delete p;
			p = q;
		}
	}
	int Get_Number()
	{
		int temp = 0;
		Queue<T> mylist = *this;
		for(int i = this->SIZE();i>0;--i)
		{
			temp = temp*10 + this->PopHead();
		}
		*this = mylist;
		return temp;
	}//将链表转化为数字。
	void POW(int x)
	{
     Queue<T> list1;
		 Queue<T> list2;
		 Queue<T> list3;
		 Queue<T> list4 = *this;
		 for(int i=0;i<x;i++)	
		 {
		   ADD(this,&list1,&list2);
			 list3 = list2;
			 list1.CLEAR();
			 list2.CLEAR();
		   list1 = list3;
			 *this = list4;
		 }
		this->CLEAR();
		*this = 	list3;
	}//list*[1-9]。
	int PopTail()
	{
		if(size==0)
		{
			return 0;
		}
		else
		{
			int x = tail->data;
			DList<T> *p = tail->prev;
			delete tail;
			tail = p;
			tail->next = NULL;
			size--;
			return x;
		}
	}
	
	int PopHead()
	{
		if(size == 0)
		{
			return 0;
		} 
		else
		{
			    T x = head->next->data;
				
			   DList<T> *p = head->next;

				 if ( p->next != NULL )