题意： 
读入一个只包含 +, -, *, / 的非负整数计算表达式，计算该表达式的值。 
Input 
测试输入包含若干测试用例，每个测试用例占一行，每行不超过200个字符，整数和运算符之间用一个空格分隔。没有非法表达式。当一行中只有0时输入结束，相应的结果不要输出。 
Outpu 
t对每个测试用例输出1行，即该表达式的值，精确到小数点后2位。 
Sample Input 
1 + 2
4 + 2 * 5 - 7 / 11
0
Sample Output 
3.00
13.36

#include <bits/stdc++.h>
using namespace std;
int main(){
    char c;
    double a;
    while(cin >> a){
        c = getchar();         
        if(c =='\n'){
            if(a == 0) break;              //如果是0，就退出循环。
            else{
                printf("%.2lf\n",a);
                continue;
            }
        }
        stack<double> s;
        stack<char> t;
        double sum1 = a;
        while(1){
            cin >> c >> a;             
            if(c == '/') sum1 /= a;               //根据符号，分开判断。
            if(c == '*') sum1 *= a;          
            if(c == '+' || c == '-'){
                t.push(c);
                s.push(sum1);
                sum1 = a;
            }
            if(getchar()=='\n') break;
        }
        s.push(sum1);
        double sum = 0;
        while(!t.empty()){
            double y = s.top();
            s.pop();
            char b = t.top();
            t.pop();
            if(b == '-') sum -= y;
            if(b == '+') sum += y; 
        }
        sum += s.top();
        s.pop();
        printf("%.2lf\n",sum);
    }
    return 0;
}