控制台程序求解有理式（带括号，适用于int类型）

C++中计算有理式

第一步，将有理式以字符串的形式输入

第二步，使用string.at()函数逐个读取有理式字符串中的字符，将数字拼在一起构成操作数，然后将操作数和操作符依次存放在一个链表中

第三步，进入有理式计算算法

（只是一般的计算算法，至于开头出现括号，出现浮点数，操作数字符串转int或double类型等等，不做进一步解释）

1、  定义两个栈，分别保存操作数和操作符

2、  定义一个判定操作符优先级的函数value(string c1,string c2)，这里由于程序需要，将操作符类型定义为string类型。

符号共有+ - * / ( ) =等7种，操作符之间的优先级为：

Value

c2

+

-

*

/

(

)

=

c1

 

+

1

1

-1

-1

0

1

1

-

1

1

-1

-1

0

1

1

*

1

1

1

1

0

1

1

/

1

1

1

1

0

1

1

(

0

0

0

0

0

0

——

数字1代表c1可以进行优先运算，-1代表c2优先级更高，0代表什么都不做，继续输入操作数即可。

3、  依次pop有理式链表中的操作数（符）进来，当检测到链表中仅剩一个 = 时，推出程序显示计算结果。

4、  遵循左算优先的原则，当某个操作数push进操作数栈后进行判断，如果value(c1,c2)<1，那么直接回到3；如果value(c1,c2)==1，那么进入5。

5、  判断此时c1是否为（，若是，就pop掉操作符栈；否则，pop出操作数栈的两个数，进行运算，然后pop掉操作符，然后回到3；

附程序：

//stack_eqution.cpp : 定义控制台应用程序的入口点。

//

 

#include "stdafx.h"

#include <string>

#include <sstream>

#include <iostream>

using namespace std;

 

//单节点定义

typedef struct nnode *PtonNode;

typedef struct nnode

{

    int a;

    PtonNode next;

}stack1,*s1;

 

typedef struct onode *PtooNode;

typedef struct onode

{

    string a;

    PtooNode next;

}stack2,*s2; 

//进栈函数

void push(intx,s1 y)

{

    s1 p = new stack1;

    p->a = x;

    p->next = y->next;

    y->next = p;

}

 

void push(stringx,s2 y)

{

    s2 p = new stack2;

    p->a = x;

    p->next = y->next;

    y->next = p;

}

 

//出栈函数

int pop(s1y)

{

    s1 z = new stack1;

    z = y->next;

    y->next = z->next;

    int x = z->a;

    free(z);

    return x;

}

 

string pop(s2y)

{

    s2 z = new stack2;

    z = y->next;

    y->next = z->next;

    string x = z->a;

    free(z);

    return x;

}

 

//string转数字类型（int,float,double）

template <classtype>

type stringtonum(conststring&str)

{

    istringstream s2num(str);

    type num;

    s2num >> num;

    return num;

}

 

//操作符比较

int value(stringcl,stringcr)

{

    int level;

    if (cl =="+" |cl =="-")

    {

        if (cr =="(")

            level = 0;

        else if (cr == "*" | cr == "/")

            level = -1;

        else

            level = 1;

    }

    else if (cl == "*" | cl == "/")

    {

        if (cr =="(")

            level = 0;

        else

            level = 1;

    }

    else if (cl == "(")//当cl出现（或者）时，

    {

        if (cr ==")")

            level = 1;

        else

            level = 0;

    }

    else

        level = 0;

    return level;

}

 

//进队列函数

void pushh(stringx,s2 y)

{

    s2 p = y;

    while (p->next != NULL)

    {

        p = p->next;

    }

    s2 z = new stack2;

    z->a = x;

    z->next = p->next;

    p->next = z;

}

//多项式字符串入栈

int into_stack(s2equation,strings)

{

    int i = 0;

    string str1;

    while (s.length() != 0)

    {

        while (s.at(i) !='+' &s.at(i) !='-' &s.at(i) !='*' &s.at(i) !='/' &s.at(i) !='(' &s.at(i) !=')' &s.at(i) !='=')

        {

            str1 += s.at(i);

            i++;

            if (s.at(i) =='+' |s.at(i) =='-' |s.at(i) =='*' |s.at(i) =='/' |s.at(i) =='(' |s.at(i) ==')' |s.at(i) =='=')

            {

                pushh(str1, equation);

                str1 = "";

            }

        }

 

        if (s.at(i) =='+' |s.at(i) =='-' |s.at(i) =='*' |s.at(i) =='/' |s.at(i) =='(' |s.at(i) ==')' |s.at(i) =='=')

        {

            str1 = "";

            str1 += s.at(i);

            pushh(str1, equation);

            str1 = "";

            if (s.at(i) =='=')

            {

                str1 += s.at(i);

                break;

            }

            else

            {

                i++;

            }

        }

        if (str1 == "=")

        {

            break;

        }  

    }

//  if (s.length() == 0)

//  {

//      cout << "无多项式输入" << endl;

//      return 0;

//  }

    return 1;

}

 

//内部运算函数

int compute(s2equ)

{

    s1 numhead = new stack1; numhead->next = NULL;

    s2 operhead = new stack2; operhead->next = NULL;

    if (equ->next ==NULL)

    {

        cout << endl << "无多项式输入" << endl << endl;

        return 0;

    }

    if (equ->next->a =="=")

    {

        cout << endl << "无多项式输入" << endl << endl;

        return 0;

    }

    string str1 = pop(equ);//用于过程中储存equ中的字符串

    int nl, nr;//用于过程中操作数的储存

    if (str1 != "+"&str1 != "-"&str1 != "*"&str1 != "/"&str1 != "("&str1 != ")"&str1 != "=")

        push(stringtonum<int>(str1),numhead);

    else

        push(str1, operhead);

 

    while (equ->next->a !="=")

    {

        str1 = pop(equ);

        if (str1 != "+"&str1 != "-"&str1 != "*"&str1 != "/"&str1 != "("&str1 != ")"&str1 != "=")

        {

            push(stringtonum<int>(str1),numhead);

            while(value(operhead->next->a,equ->next->a) == 1)

            {

                if(operhead->next->a =="(")

                {

                    pop(operhead);

                    pop(equ);

                    if (operhead->next==NULL)

                        break;

                    continue;

                }

                nr = pop(numhead);

                nl = pop(numhead);

                if(operhead->next->a =="+")

                {

                    push(nl+nr, numhead);

                }

                if(operhead->next->a =="-")

                {

                    push(nl - nr, numhead);

                }

                if(operhead->next->a =="*")

                {

                    push(nl * nr, numhead);

                }

                if(operhead->next->a =="/")

                {

                    if (nr == 0)

                    {

                        cout << endl<< "多项式有误，除数出现0" << endl << endl;

                        return 0;

                    }

                    push(nl / nr, numhead);

                }

                pop(operhead);

                if (operhead->next==NULL)

                    break;

            }

 

        }

        else

        {

            push(str1, operhead);

            if (str1 == ")")

                pop(operhead);

        }

    }

    cout << numhead->next->a<<endl;

    return 0;

}

 

 

int _tmain(int argc,_TCHAR*argv[])

{

    //多项式输入

    cout <<"输入多项式，以=结尾:" << endl;

    string str;

    getline(cin, str);

 

    //多项式入队列

    s2 equ = new stack2;  equ->next = NULL;

    into_stack(equ, str);

   

    //多项式运算（使用栈）

    compute(equ);

 

    return 0;

}