E24588: 后序表达式求值

Stack, http://cs101.openjudge.cn/practice/24588/

思路：按照课上讲的用栈解决即可，忘记处理单走一个数字的情形了，花了点时间找bug，用时约20min，由于代码全部手敲，因此运行时间很短，仅1ms

#include <iostream>
#include <vector>
#include <cmath>
#include <iomanip>
using namespace std;

vector<double> st;

bool isNumber(char s) 
{
    return (s >= '0' && s <= '9');
}

void calc(char c) 
{
    double a = st.back(); st.pop_back();
    double b = st.back(); st.pop_back();
    switch (c) 
    {
    case '+': st.push_back(b + a); break;
    case '-': st.push_back(b - a); break;
    case '*': st.push_back(b * a); break;
    case '/': st.push_back(b / a); break;
    }
}

double calculate() 
{
    st.clear();
    char c;
    double tmp = 0;
    bool isNum = false;
    double frac = 0;
    while ((c = getchar())) 
    {
        if (c == '\n' || c == EOF) 
        {
            if (isNum) st.push_back(tmp);
            break;
        }
        if (isNumber(c)) 
        {
            isNum = true;
            if (frac == 0) tmp = tmp * 10 + (c - '0');
            else 
            {
                tmp += (c - '0') * frac;
                frac *= 0.1;
            }
        }
        else if (c == '.') 
        {
            frac = 0.1;
        }
        else if (c == ' ') 
        {
            if (isNum) 
            {
                st.push_back(tmp);
                tmp = 0; frac = 0; isNum = false;
            }
        }
        else if (c == '+' || c == '-' || c == '*' || c == '/') 
        {
            if (isNum) 
            {
                st.push_back(tmp);
                tmp = 0; frac = 0; isNum = false;
            }
            calc(c);
        }
    }
    return st.back();
}

int main() 
{
    int N;
    cin >> N;
    cin.ignore();
    while (N--) 
    {
        double ans = calculate();
        cout << fixed << setprecision(2) << ans << endl;
    }
}