题解 | #简单计算器#

简单的题还是写了那么多代码。。。

该题注意2个地方：

1. 字符串输入，sstream
2. 结果的输出，中途用stod将string转为double，保证精度

#include <iostream>
#include <stack>
#include <string>
#include <vector>
#include <map>
#include <sstream>
#include <algorithm>
using namespace std;

const int N = 10010;
stack<string> expss;
stack<double> nums;
map<string, int> exps;

//后将要压的exp1和exp2比较
bool pri(string exp1, string exp2)
{
    if (exps[exp2] - exps[exp1] > 1)
        return true;
    else
        return false;
}

//计算num1（expt）num2
double cal(string expt, double num1, double num2)
{
    double ans = 0;
    switch (exps[expt])
    {
    case 0:
        /* code */
        ans = num1 * num2;
        break;
    case 1:
        /* code */
        ans = num1 / num2;
        break;
    case 3:
        /* code */
        ans = num1 + num2;
        break;
    case 4:
        /* code */
        ans = num1 - num2;
        break;

    default:
        break;
    }
    return ans;
}
//循环
double loop(string ss)
{
    ss += " ";
    ss += "$";
    expss.push("#");

    istringstream str(ss);
    string out;
    while (str >> out)
    {
        if (out == "+" || out == "-" || out == "*" || out == "/" || out == "$")
        {
            string exp2 = expss.top();
            while (!pri(out, exp2))
            {
                double a = nums.top();
                nums.pop();
                double b = nums.top();
                nums.pop();
                double ans = cal(exp2, b, a);
                nums.push(ans);
                expss.pop();
                exp2 = expss.top();
            }
            if (pri(out, exp2))
            {
                expss.push(out);
            }
            if (expss.top() == "$")
                break;
        }
        else
        {
            nums.push(stod(out));
        }
    }
    if (!nums.empty())
        return nums.top();
    else
        return -37;
}

int main()
{
    string ss;
    exps["*"] = 0;
    exps["/"] = 1;
    exps["+"] = 3;
    exps["-"] = 4;
    exps["$"] = 6;
    exps["#"] = 8;

    while (getline(cin, ss))
    {
        if (ss == "0")
            break;
        double ans = loop(ss);
        printf("%.2lf\n", ans);
        while (!nums.empty())
            nums.pop();
        while (!expss.empty())
            expss.pop();
    }
    return 0;
}