题解 | #简单计算器#

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

using namespace std;

stack<char> op; //运算符
stack<double> d; //操作数

double getnum(string str,
              int& t) { //将多个字符转化为数字,t为引用类型
    double ans = 0;
    while (isdigit(str[t])) {
        ans = ans * 10 + str[t] - '0';
        t++;
    }
    return ans;
}

int priority(char c) {
    if (c == '#') return 0;
    else if (c == '$') return 1;
    else if (c == '+' || c == '-') return 2;
    else return 3;
}

double calculate(double x, double y, char op) {
    double res = 0;
    if (op == '+') res =  x + y;
    else if (op == '-')  res = x - y;
    else if (op == '*')  res = x * y;
    else if (op == '/')  res = x / y;
    return res;
}

int main() {
    string str;
    while (getline(cin, str)) {
        if (str == "0") break;
        op.push('#'); //#为最低优先级
        str += '$'; //处理结束的标志
        int t = 0;
        while (t < str.size()) {
            if (str[t] == ' ') t++; //跳过空格
            else if (isdigit(str[t])) d.push(getnum(str,
                                                        t)); //数字入栈，可能有多位
            else {
                if (priority(op.top()) < priority(
                            str[t])) { //优先级比运算符栈顶大，则入栈
                    op.push(str[t]);
                    t++;
                } else {
                    double y = d.top(); //取操作数
                    d.pop();
                    double x = d.top();
                    d.pop();
                    d.push(calculate(x, y, op.top())); //后进先出
                    op.pop(); //弹出运算符
                }
            }
        }
        printf("%.2f\n", d.top());
    }
    return 0;
}