#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<stack>
#include<iomanip>
#include"caltulator.h"
using namespace std;
typedef struct STK
{
double num=889277;
char note=0;
}stk;
char s[1000];//初始输入的数组
stk s1[1000];//字符串分割出来的部分
stk s3[1000];//最后的逆波兰表达式
bool if_minus(int & pos);
void turn_pi(int len);
double cut_num(int &pos)
{
int f_pos = pos; //一个数字的开始位置
double integer = 0.0;
double remainder = 0.0;
int c = 0;
while (s[pos] >= '0'&&s[pos] <= '9')
{
integer *= 10;
integer += s[pos] - '0';
pos++;
}
while (s[pos] == '.')
{
c = 1;
pos++;
while (s[pos] >= '0'&&s[pos] <= '9')
{
remainder += pow(0.1, c)*(s[pos]-'0');
c++;
pos++;
}
}
//int pre_pos = f_pos - 1; //一个数字开始之前的一个位
return remainder + integer;
}
char cut_note(int &pos)
{
if (s[pos] == '*')
{
pos++;
return '*';
}
else if (s[pos] == '+')
{
pos++;
return '+';
}
else if (s[pos] == '-')
{
pos++;
return '-';
}
else if (s[pos] == '/')
{
pos++;
return '/';
}
else if (s[pos] == '(')
{
pos++;
return '(';
}
else if (s[pos] == ')')
{
pos++;
return ')';
}
else if (s[pos] == 's')
{
pos++;
if (s[pos] == 'i')
{
pos++;
if (s[pos] == 'n')
{
pos++;
return 's';
}
else return 0;
}
else return 0;
}
else if (s[pos] == 'c')
{
pos++;
if (s[pos] == 'o')
{
pos++;
if (s[pos] == 's')
{
pos++;
return 'c';
}
else return 0;
}
else return 0;
}
else if (s[pos] == 'p')
{
pos++;
if (s[pos] == 'i')
{
pos++;
return 'p';
}
else
return 0;
}
else if (s[pos] == 't')
{
pos++;
if (s[pos] == 'a')
{
pos++;
if (s[pos] == 'n')
{
pos++;
return 't';
}
else return 0;
}
else return 0;
}
else return 0;
}
bool is_tf(char note)
{
if (note == 'c' || note == 't' || note == 's')
return 1;
else
return 0;
}
int cut_string()
{
int len = strlen(s);
int pos1 = 0; //pos1是目标逆波兰表达式所对应的位置编号
for (int pos = 0; pos < len;)
{
stk tmp;
if (s[pos] == '#')
exit(0);
else if (s[pos] >= '0'&&s[pos] <= '9')
{
tmp.num = cut_num(pos);
}
else
{
char note = cut_note(pos);
if (note) //在判断步骤已经完成了一步自增，所以要把完成的一步自增减掉
{
if (note=='-')
{
if(if_minus(--pos)) //是负号
{
s1[pos1].num = 0;
pos1++;
}
pos++;
tmp.note = note;
}
else
{
tmp.note = note;
}
}
else
{
cout << "note error " <<pos<< endl;
exit(0);
return 0;
}
}
//if (pos1 == 0&&tmp.num||tmp.note=='('||)
s1[pos1] = tmp;
pos1++;
}
return pos1; //因为数组是从零开始计数，所以最后不用自减
} //int 返回最后s1中有多少个结构体
int rate_class(char note)
{
if (note == 'c' || note == 's' || note == 't')
return 1;
else
return 2;
}
int rate_note(char n, int drct)
{
if (n == '+' || n == '-')
{
return 1;
}
else if (n == '#')
return -1;
else if (n == '*' || n == '/')
return 2;
else if (n == '(' || n == ')') //左右括号的级别都定为 0；
return 0;
else if ((n == 's' || n == 'c' || n == 't') && drct == 0) //单目操作符的右级别为0；即在逆波兰表达式中的左结核性最低
return 4;
else if ((n == 's' || n == 'c' || n == 't') && drct == 1) //左级别为3；即在逆波兰表达式中右结核性最高；
return 3;
else
{
cout << "error" << endl;
exit(0);
}
}
bool if_minus(int & pos)
{
if (s[pos] != '-')
return false;
else {
if (pos == 0 || (s[pos - 1] == 'n'&&s[pos - 2] == 'i'&&s[pos - 3] == 's') || (s[pos - 1] == 's'&&s[pos - 2] == 'o'&&s[pos - 3] == 'c') || (s[pos - 1] == 'n'&&s[pos - 2] == 'a'&&s[pos - 3] == 't') || s[pos - 1] == '(')
return true;
else return false;
}
}
double operate_2(double a1, char op, double a2)
{
switch (op)
{
case'+':return a1 + a2;
case '-':return a1 - a2;
case '*':return a1*a2;
case '/':return a1 / a2;
default:cout << "error"; return 0;
}
}
double operate_1(double a, char op)
{
switch (op)
{
case's':return sin(a);
case'c':return cos(a);
case't':
{
if (abs(cos(a))<0.00000001)
{
cout << "tan illegal please reinput" <<endl;
exit(0);
}
else
return tan(a);
}
default: cout << "error"; exit(0);
}
}
int mtor()
{
int len = cut_string();
turn_pi(len);
stack<stk> s2;
stk tmp;
tmp.note = '#';
s2.push(tmp);
int pos3= 0;
for (int pos1 = 0;pos1 < len; pos1++)
{
if (!s1[pos1].note) //这个位置是数字的情况，note为空
{
s3[pos3] = s1[pos1];
pos3++;
}
else
{
if (s1[pos1].note == '(')
{
stk tmp;
tmp.note = '(';
s2.push(tmp);
}
else if (s1[pos1].note == ')')
{
int flag = 1; //是否到了（，sin，cos的需要终结的部分
while ((s2.top().note!='#')&&(flag||rate_note(s2.top().note, 1)==0|| rate_note(s2.top().note, 1) == 3))//flag为1或者是top的符号为0的时候可以继续往下做
{
if (flag&&s2.top().note == '(')
{
flag = 0;
s2.pop();
}
else
{
stk tmp;
tmp = s2.top();
s3[pos3] = tmp;
pos3++;
s2.pop();
}
}
}
else if ((s2.top().note != '#')&&rate_note(s1[pos1].note, 0) <= rate_note(s2.top().note, 1)) //需要进栈的note的左优先级要大于在栈中的右优先级时，可以进栈
{
while (rate_note(s1[pos1].note, 0) <= rate_note(s2.top().note, 1))
{
stk tmp = s2.top();
s2.pop();
s3[pos3] = tmp;
pos3++;
}
s2.push(s1[pos1]);
}
else
{
s2.push(s1[pos1]);
}
}
}
while ((s2.top().note != '#'))
{
stk tmp;
tmp = s2.top();
s2.pop();
s3[pos3] = tmp;
pos3++;
}
return pos3;
}
void turn_pi(int len)
{
for (int pos = 0; pos < len; pos++)
{
if (s1[pos].note == 'p')
{
s1[pos].note = 0;
s1[pos].num = 3.1415926;
}
}
}
double compute()
{
int len = mtor();
stack<double>s4;
for (int pos3 = 0; pos3 < len; pos3++)
{
if (s3[pos3].note) //当是操作符的时候
{
if (rate_class(s3[pos3].note) == 1) //单目操作符
{
double tmp=s4.top();
s4.pop();
tmp=operate_1(tmp, s3[pos3].note);
s4.push(tmp);
}
else
{
double tmp1 = s4.top();
s4.pop();
double tmp2 = s4.top();
s4.pop();
double tmp = operate_2(tmp2, s3[pos3].note, tmp1);
s4.push(tmp);
}
}
else
{
s4.push(s3[pos3].num);
}
}
return s4.top();
}
/*bool is_legal(int len)
{
for (int pos1 = 0; pos < len; pos++)
{
if (pos1 == 0)
if ((s1[pos1] == 's' || s1[pos1] == 'c' || s1[pos1] == 't' || if_mins(pos1))
;
else
{
cout << "note error" << pos1 << endl;
exit(0);
}
else if (pos1 < len - 1)
{
if (s1[pos1] == '*' || s1[pos1] == '-' || s1[pos1] == '/' || s1[pos1] == '+')
{
pos_r = pos + 1;
if (s1[pos_r] == ')' || s1[pos_r] == '+' || s1[pos_r] == '*' || s1[pos_r] == '-' || s1[pos_r] == '/')
{
cout << "note error" << pos1 << endl;
exit(0);
}
}
else if(s1[pos1]==)
}
}
}
*/
int main()
{
cout << "请输入一个合法的表达式（请给所有的负数都加括号）" << endl;
cout << "如果要退出请输入 ‘#’ " << endl;
cin >> s;
while (s[0] != '#')
{
//if (is_legal)
if(1)
{
double result;
result = compute();
cout << "这个表达式的计算结果为： "<< fixed << setprecision(8)<< result<<endl;
cout << "请输入一个合法的表达式（请给所有的负数都加括号）" << endl;
cout << "如果要退出请输入 ‘#’ " << endl;
}
else
{
cout << "请输入一个合法的表达式" << endl;
cout << "如果要退出请输入 ‘#’ " << endl;
}
cin >> s;
}
}//愁死我了，不会用这个编辑器，以上代码还没有写完，但是把字符串正确性检测注释掉，是可以跑的，假期填坑。这个代码太菜了，菜到自己没眼看。