求解前缀表达式值

从后往前遍历前缀表达式，

1. 遇到数字放入栈中。
2. 遇到操作符，取出栈顶元素ta， tb，将操作完后的数放入栈
3. 最后栈顶元素就是答案

tips：这里用vector代替栈。另外，atof函数是将string转换成double Code：

#include <bits/stdc++.h>
#define ALL(a) (a).begin(), (a).end()
using namespace std;
using LL = long long;
typedef pair<int, int> PII;
template < typename T> inline void Max(T &a, T b) { if(a < b) a = b; }
template < typename T> inline void Min(T &a, T b) { if(a > b) a = b; }

int main() {
    cin.tie(nullptr) -> sync_with_stdio(false);
    string s;
    vector<string> a;
    vector<double> stk;
    while (cin >> s) a.push_back(s);

    for (int i = a.size() - 1; i >= 0; -- i) {
        if(a[i].size() == 1 && 
        (a[i][0] == '*' || a[i][0] == '/' || a[i][0] == '+' || a[i][0] == '-')) {
            double ta = stk.back(); stk.pop_back();
            double tb = stk.back(); stk.pop_back();
            if(a[i][0] == '+') stk.push_back(ta + tb);
            else if (a[i][0] == '/') stk.push_back(ta / tb);
            else if (a[i][0] == '*') stk.push_back(ta * tb);
            else {
                stk.push_back(ta - tb);
            }
        } else {
            double x = atof(a[i].c_str());
            stk.push_back(x);
        }
    }
    printf("%.2f\n", stk.back());
    return 0;
}