#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stack> 
using namespace std;
stack<double>num;
int main() 
{
    double n;
    while(scanf("%lf", &n))
    {
        char c,k;
        c = getchar();
        if(c==‘\n‘ && n==0)
            break;
        num.push(n);
        double m;
        while(1)
        {
            cin>>k;
            cin>>n;
            if(k == ‘*‘)
            {
                m = num.top();
                m*=n;
                num.pop();
                num.push(m);    
            }
            else if(k == ‘/‘)
            {
                m = num.top();
                m/=n;
                num.pop();
                num.push(m);
            }
            else if(k == ‘+‘)
            {
                num.push(n);
            }
            else if(k == ‘-‘)
            {
                num.push(0-n);
            }
            if(getchar()==‘\n‘)    //本行输入完毕 
                break;
        } 
        double sum = 0;
        while(!num.empty())
        {
            sum+=num.top();
            num.pop();         //使用完后栈一定要清空
        }
        printf("%.2lf\n", sum);

    } 
    return 0;
}

#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
int main()
{
    double a[1000];
    double s, ss;
    char k, f;
    while (cin >> s)//题目要求多次输入
    {
        int flag = 1, i = 0;
        memset(a, 0, sizeof(a));
        a[0] = s;
        f = getchar();
        if (s == 0 && f == ‘\n‘)
        {
            flag = 0;
            break;
        }
        while(1)//优先处理*、/运算
        {
            cin >> k;//输入运算符
            cin >> ss;
            if (k == ‘*‘)
                a[i] = a[i] * ss;
            if (k == ‘/‘)
                a[i] = a[i] / ss;
            if (k == ‘+‘)
                a[++i] = ss;
            if (k == ‘-‘)
                a[++i] = (-1)*ss;
            if (getchar() == ‘\n‘)
                break;
        }
        if (flag == 1)
        {
            double sum = 0;
            for (int j = 0; j <= i; j++)
                sum = sum + a[j];
            printf("%.2lf\n", sum);
        }
    }
    return 0;
}