前缀式计算

实现方法：对于一个已知的前缀式，从右向左读取，如果遇到操作数，处理并入栈，遇到操作符，则将栈顶两个元素分别出栈进行计算。（注意减法和除法时的先后顺序）

题目练习：nyoj128

#include
#include
#include
#include
#include
#include
using namespace std;

string prefix;
double num[200000];
double cal()
{
    stack s;
    while(!s.empty())
    {
        s.pop();
    }
    memset(num,0,sizeof(num));

    double x,y;
    int i = 0;
    int k = 0;

    while(i < prefix.length())
    {
        if(prefix[i] == ' ')
        {
            i++;
            continue;
        }
        else if(prefix[i] == '+' || prefix[i] == '-' || prefix[i] == '*' || prefix[i] == '/')
        {
            if(prefix[i] == '+')
                num[k] = -999999;
            else if(prefix[i] == '-')
                num[k] = -999998;
            else if(prefix[i] == '*')
                num[k] = -999997;
            else if(prefix[i] == '/')
                num[k] = -999996;
            i++;
            k++;
        }
        else
        {
            x = 0;
            y = 0;
            while(prefix[i] >= '0' && prefix[i] <= '9')
            {
                x *= 10;
                x += prefix[i]-'0';
                i++;
            }
            if(prefix[i] == '.')
            {
                double k = 10.00;
                i++;
                while(prefix[i] >= '0' && prefix[i] <= '9')
                {
                    y += (prefix[i]-'0')/k;
                    k *= 10.00;
                    i++;
                }
            }
            x += y;
            num[k++] = x;
        }
    }

    i = k-1;
    for(i = k - 1; i >= 0; i--)
    {
        if(num[i] == -999999)
        {
            x = s.top();
            s.pop();
            x += s.top();
            s.pop();
            s.push(x);
        }
        else if(num[i] == -999998)
        {
            x = s.top();
            s.pop();
            x = x - s.top();
            s.pop();
            s.push(x);

        }
        else if(num[i] == -999997)
        {
            x = s.top();
            s.pop();
            x = s.top() * x;
            s.pop();
            s.push(x);

        }
        else if(num[i] == -999996)
        {
            x = s.top();
            s.pop();
            x = x / s.top();
            s.pop();
            s.push(x);
        }
        else
        {
            s.push(num[i]);
        }
    }
    return s.top();
}

int main()
{
    while(getline(cin,prefix))
    {
        printf("%.2f\n",cal());
    }
}