给出一个算式字符串，计算出结果

核心思路：

1. 构造两个栈，一个栈存放数字，一个栈存放运算符
2. 运算符栈压栈时，压栈元素应优先级应大于top，小于或等于则弹出栈顶元素，进行计算，反复操作

#include<iostream>
#include<stack>
#include<string>

using namespace std;

class Solution{
    public:
    int getNumber(string str){
        int num=0;
        for(char s:str){
            num=num*10+(s-'0');
        }
        return num;
    }
    int getProtity(char c){
        if(c=='*'||c=='/'){
            return 0;
        }else if(c=='+'||c=='-'){
            return 1;
        }else if(c=='$'){
            return 2;
        }
    }
    double calculate(double num1,double num2, char operate){
        switch(operate){
            case '+': return num1+num2;
            case '-': return num2-num1;
            case '*': return num1*num2;
            case '/': return num2*1.0/num1;
        }
    }
    
};

int main(){
    Solution solution;
    string expression;
    while(getline(cin,expression)){
        if(expression.size()==1&&expression[0]=='0'){
            return 0;
        }
        stack<double> number;
        stack<char> eval;
        eval.push('$');
        string numStr = "";
        int pro;
        for(int i=0;i<expression.size();i++){
            while(expression[i]<='9'&&expression[i]>='0'){
                numStr+=expression[i];
                i++;
            }
            if(numStr!=""){
                if(i==30){
                    
                    
                    int ccc=0;
                }
                number.push(solution.getNumber(numStr));
                numStr = "";
                i--;
            }
            if(expression[i]=='+'||expression[i]=='-'||expression[i]=='*'||expression[i]=='/'){
                while(solution.getProtity(expression[i])>=solution.getProtity(eval.top())){
                    double num1 = number.top();
                    number.pop();
                    double num2 = number.top();
                    number.pop();
                    number.push(solution.calculate(num1,num2, eval.top()));
                    eval.pop();
                }
                eval.push(expression[i]);
            }
        }
        while(eval.top()!='$'){
            double num1 = number.top();number.pop();
            double num2 = number.top();number.pop();
            number.push(solution.calculate(num1, num2, eval.top()));
            eval.pop();
        }
        
        printf("%.2f\n",number.top());
    }
    return 0;
}