用c语言实现数学多项式的计算

来源：互联网发布：c语言打印金字塔图形编辑：程序博客网时间：2024/06/06 06:30

#include"stdio.h"
#include"stdlib.h"
#include"calc.h"
#include <math.h>
#define POW 1
#define MUL 2
#define DIV 3
#define MOD 4
#define DIV_INT 5
#define ADD 6
#define SUB 7
#define MOVEL 8
#define MOVER 9
#define BIG 10
#define SML 11
#define BIG_EQAUL 12
#define SML_EQAUL 13
#define NOT 14
#define EQAUL 15
#define BIT_AND 16
#define DIFFER_OR 17
#define BIT_OR 18
#define AND 19
#define OR 20
#define Lp 21
#define Rp 22
#define END 23
#define OPSUM 23
#define Epsilon 0
int ch_char_int(char *data);
int ch_int_char(int data,char *parm);
float eval(char tag,float left,float right);
int   getToken(float *nump,char *box,int *state);
int   instead(char *box,char *express,int x1,int x2,int x3, int x4);
float express_calc(char *box);
/*-------------------------------运算符号的优先级的设置--------------------------------*/
int osp[23] = {13,11,11,11,11,10,10,9,9,8,8,8,8,7,7,6,5,4,3,2,13,1,1};
int isp[21] = {12,11,11,11,11,10,10,9,9,8,8,8,8,7,7,6,5,4,3,2,1};//主要是为了"("来设置的。
struct
{
char op[2];
int code;
}opchTbl[24] =
{
{{'*','/0'},MUL},//11
{{'/','/0'},DIV},//11
{{'%','/0'},MOD},//11
{{'/','/'},DIV_INT},//11
{{'+','/0'},ADD},//10
{{'-','/0'},SUB},//10
{{'<','<'},MOVEL},//9
{{'>','>'},MOVER},//9
{{'>','/0'},BIG},//8
{{'<','/0'},SML},//8
{{'>','='},BIG_EQAUL},// 8
{{'<','='},SML_EQAUL},// 8
{{'!','='},NOT},// 7
{{'=','='},EQAUL},// 7
{{'&','/0'},BIT_AND},//6
{{'^','/0'},DIFFER_OR},//5
{{'|','/0'},BIT_OR},//4
{{'&','&'},AND},// 3
{{'|','|'},OR},// 2
{{'(','/0'},Lp},//13
{{')','/0'},Rp},// 1
{{'$','/0'},POW},// 1
{{'/n','/0'},END},//
{{' ','/0'},-1}};//

typedef struct node
{
float data;
struct node *link;
}stack;

/*----------------------------------------------
计算的方法:
------------------------------------------------*/
main()
{
   char *data = NULL;
   int number=0;
   char box[12];
   int state = 0;
   int i=0;
   int gt_re=0;
   float last_value;

   /*---------------------存储计算表达式-------------------*/
   instead(box,"x1+(x2*x3)*0.899",11,2,1,0);
   printf(box);
   /*-----------------------------------------------------*/
last_value = express_calc(box);
printf("%f",last_value);
}
/*----------------------------------------------------------
函数的功能:进入堆栈
相关功能的说明:
1,
-----------------------------------------------------------*/
int push_stack(float s_data,stack **stpoint)
{
   stack *p =(stack *)malloc(sizeof(stack));
   p->data = s_data;
   p->link = *stpoint;
   *stpoint = p;//这样才能进行传递。
   return 0;
}
/*-------------------------------------------------------------------
函数的功能：数据出堆
----------------------------------------------------------------------*/
int pop_stack(float *s_data,stack **spoint)
{
stack *p = *spoint ;
if( p==NULL) return -1;//没有数据栈
else
  *s_data = p->data;
     *spoint = p->link;
  free(p);
return 0;
}
/*--------------------------------------------------------------------------
函数的名称：int instead(char *box,char *express,int x1,int x2,int x3, int x4)
函数的参数：char *box用来存储结果
            char *express计算的表达式
   x1..4计算表达式的参数1
函数的返回值:0表示函数执行的是正确的。
函数的功能:用参数代替标号
           注意express表达式由x1,x2,x3,x4和运算符号
     '+''-''*''/'()组成
     "x1+x2*(x3+x4%5)"
--------------------------------------------------------------------------*/
int instead(char *box,char *express,int x1,int x2,int x3, int x4)
{
int i=0;
int length=0;
    int box_i=0;
int data=0;
do
{
  length++;
}while(express[length]!='/0');

for(i=0;express[i]!='/0';i++,box_i++)//因为这里已经加了，所以说box_i是最后一个位置。
{
  box[box_i]=express[i];
  if((express[i]=='x')&&((express[i+1]>='1')&&((express[i+1]<='4'))))
  {

   switch(express[i+1])
   {
   case '1':
      data = x1;
      break;
   case '2':
        data =x2;
     break;
   case '3':
        data = x3;
     break;
   case '4':
        data =x4;
     break;
   }
   /*-------复合语句实现的是把数据存储在box[box_i-1]后面---------------*/
   {
    int k=0;//存储data的位数
    int temp=0;//存储data的中间变量
                int j=0;
    temp=data;
    while(temp!=0)
    {
     k++;
     temp=temp/10;
    }
    temp=data;
    for(j=0;j<k;j++)
    {
     box[box_i-1+k-j] = temp%10+48;
     temp=temp/10;
    }
    box_i = box_i-1+k;
   }
   /*-------------------------------------------------------------------*/
   i=i+1;
  }
}
box[box_i]='/n';//整个计算式的最后一个字母是换行符。
return 0;
}
/*--------------------------------------------------------------------------
函数的功能：计算表达式
函数实现的过程:
----------------------------------------------------------------------------*/
float express_calc(char *box)
{
int state=0;//记录box当前的位置信息
float number; //出堆栈的数据存储
float tag=0.0;
float right=0.0;
float left=0.0;
int gt_re_value;
float last_value;
    stack *num_stack = NULL;//用于存放数字
stack *tab_stack = NULL;//用于存放计算符号

push_stack(Lp,&tab_stack);//首先把左括号放到符号堆栈里面
while(1)
{
    gt_re_value=getToken(&number,box,&state);
       if(gt_re_value==0)//表示的是返回是数值
     push_stack(number,&num_stack);//存储数值
    else//返回的是标号
    {
     if(osp[gt_re_value - 1] > isp[(int)tab_stack->data - 1])//进行优先级的判断
    push_stack((float)gt_re_value,&tab_stack);//对标号进行存储
     else
     {   /*--在这里进行判断*/
      while(osp[gt_re_value - 1] <= isp[(int)tab_stack->data - 1] && tab_stack->data <= OPSUM-3)
      {
     if(pop_stack(&tag,&tab_stack))   return -1;
     if(pop_stack(&right,&num_stack)) return -1;
     if(pop_stack(&left,&num_stack)) return -1;
        last_value = eval((char)tag,left,right);
     push_stack(last_value,&num_stack);
      }
      if(gt_re_value == END) break;
      if(gt_re_value == Rp)
      {
       do
       {
        if(pop_stack(&tag,&tab_stack))
        return CalcError;
       }
       while((char)tag != Lp);
      }
      else
     push_stack((float)gt_re_value,&tab_stack);
     }
    }
}
     if(pop_stack(&last_value,&num_stack)) return -1;
while(num_stack != NULL) pop_stack(&number,&num_stack);
while(tab_stack != NULL) pop_stack(&number,&tab_stack);
return last_value;
}
/*----------------------------------------------------------------------------
函数的功能:1,
------------------------------------------------------------------------------*/
int getToken(float *nump,char *box,int *state)
{
float data = 0;
int num = 0;
int i=0;
int j=0;
int flag=0;
while((box[*state]<'0')||(box[*state]>'9'))
{
  for(i=0;i<OPSUM-1;i++)
  {
   if(opchTbl[i].op[0]==box[*state])
   {
    for(j=0;j<OPSUM-1;j++)
    {
     if((opchTbl[j].op[0]==box[*state])&&(opchTbl[j].op[1]==box[*state+1]))
     {
      flag=1;//说明是双目运算
     }
    }
      if(j==(OPSUM-1)) flag=0;//说明是单目运算符
    break;
   }
  }
  if(flag==1) *state = *state + 2;
  if(flag==0) *state = *state +1;

  if(i==OPSUM-1)
  {
   if(opchTbl[i].op[0]=='/n')//说明是最后一个
   {
    return opchTbl[i].code;
   }
   else return -1;//表示的是表达式是错误的
  }
  return opchTbl[i].code;
}

while((box[*state]>='0')&&(box[*state]<='9'))//说明是一个字
{
  data = data*10 + box[*state] - '0';
        *state = *state + 1;
}
if(box[*state]=='.')
{
  *state = *state+1;
  {
   int dex=10;
   float count_down = (float)1/dex;
   while((box[*state]>='0')&&(box[*state]<='9'))
   {
    data = ((float)(box[*state]-'0'))*(count_down) + data;
    count_down = count_down/dex;
    *state = *state+1;
   }
  }
}
//printf("%f",data);
*nump = data;
return 0;//0表示的是返回的是值。
}
/*---------------------------------------------------------------------------
函数的功能是实现:左值和右值的计算
-----------------------------------------------------------------------------*/
float eval(char tag,float left,float right)
{
int n;
float result;
switch(tag)
{
case POW :
  for(n = 1,result = left;n < right;n++)
   result *= left;
  return result;
case ADD: return left + right;
case SUB: return left - right;
case MUL: return left * right;
case MOD: return (float)((int)left % (int)right);
case DIV:
  if(fabs(right) <= Epsilon)
  {
   printf("除0出错!/n");
   exit(1);
  }
  return left/right;
case DIV_INT:
  if(fabs(right) <= Epsilon)
  {
   printf("除0出错!/n");
   exit(1);
  }
  return (float)((int)left/(int)right);
case BIG: return (float)(left > right?1:0);
case SML: return (float)(left < right?1:0);
case EQAUL: return (float)(left == right?1:0);
case NOT: return (float)(left != right?1:0);
case AND: return (float)(left && right);
case OR: return (float)(left || right);
case MOVEL: return (float)((int)left <<(int) right);
case MOVER: return (float)((int)left >> (int)right);
case BIG_EQAUL: return (float)((left > right) || (left == right));
case SML_EQAUL: return (float)((left < right) || (left == right));
case BIT_AND: return (float)((int)left & (int)right);
case DIFFER_OR: return (float)((int)left ^ (int)right);
case BIT_OR: return (float)((int)left | (int)right);

}
printf("表达式有错!/n");
return 1.0;
}