2019北航夏令营机试 T2 复杂计算器

题意：

• 用字符串代表变量，先给出一行中缀表达式，然后按变量出现的顺序，给出变量值。
• 中缀表达式转后缀，要求输出后缀表达式，且输出运算结果。

输入输出用例：

(ab+cd+de)/(ab+cd)#
5 5 5
 
1.5

代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MOD = 1e9+7;
 
string expr;
stack<double> sym;
stack<char> ope;
map<string, double> value;
stack<string> postfix;
 
int main(){
    ios::sync_with_stdio(false);
    getline(cin, expr);
    expr.erase(remove_if(expr.begin(), expr.end(),
        [](unsigned char x){return isspace(x);}), expr.end());
    int p = 0;
    while(p<expr.length()){
        if(isalpha(expr[p])){
            int q = p+1; while(q<expr.length() && isalpha(expr[q])) ++q;
            string sy = expr.substr(p, q-p); p=q;
            cout<<"new symbol: "<<sy<<endl;
            if(value.count(sy)==0){
                double temp; cin>>temp; value[sy] = temp;
            }
            postfix.push(sy);
            sym.push(value[sy]);
        }
        else if(expr[p]=='(') ope.push(expr[p]), ++p;
        else{
            while(
                ((expr[p]==')') && (ope.top()!='(')) ||
                ((expr[p]=='+'||expr[p]=='-'||expr[p]=='#') && (!ope.empty())&&ope.top()!='(') ||
                ((expr[p]=='*'||expr[p]=='/') && (!ope.empty())&&(ope.top()!='+')&&(ope.top()!='-'))
            ){
                char operat = ope.top(); ope.pop();
                cout<<"operat = "<<operat<<endl;
                double d2 = sym.top(); sym.pop(); double d1 = sym.top(); sym.pop();
                string s2 = postfix.top(); postfix.pop();
                string s1 = postfix.top(); postfix.pop();
                if(operat=='+')
                    postfix.push(s1+s2+"+"), sym.push(d1+d2);
                else if(operat=='-')
                    postfix.push(s1+s2+"-"), sym.push(d1-d2);
                else if(operat=='*')
                    postfix.push(s1+s2+"*"), sym.push(d1*d2);
                else if(operat=='/')
                    postfix.push(s1+s2+"/"), sym.push(d1/d2);
                cout<<postfix.top()<<" "<<sym.top()<<endl;
            }
            ope.push(expr[p]); ++p;
            if(ope.top()==')') ope.pop(), ope.pop();
        }
    }
    cout<<postfix.top()<<" = "<<setprecision(2)<<sym.top();
}