简单计算器

这道题初看感觉好简单，题意也很好理解，但是等到我真的去看的时候，就觉得还是有难度的，最开始想用数组做，一个个录入然后直接运算，但是这里还有乘除法，所以我们不能直接运算，但是我们可以用两个数组，一个录入，一个计算，只是多次循环不知道会不会超时，所以我最后还是借鉴了大神的栈的方法。这里面是边算边存数，使用两个栈，一个存数一个存符号，然后通过if这个选择计算，详情可以见代码

#include<stdio.h>
#include<string.h>
#include<stack>
using namespace std;
int main()
{
   int i;
   double a,b;
   char s[250],c;
   while(gets(s),strcmp(s,"0")!=0)
   {
       stack<char>s1;
       stack<double>s2;
       for(i=0;s[i];i++)
       {
           if(s[i]>='0'&&s[i]<='9')
           {
               a=0;
               while(s[i]>='0'&&s[i]<='9')
               {
                   a=a*10+s[i]-'0';
                   i++;
               }
               i--;
               s2.push(a);
           }
           else if(s[i]=='-'||s[i]=='+')
           {
               if(!s1.empty())
               {
                   c=s1.top();
                   s1.pop();
                   a=s2.top();
                   s2.pop();
                   b=s2.top();
                   s2.pop();
                   if(c=='+')
                       a+=b;
                   else
                       a=b-a;
                   s2.push(a);
                   s1.push(s[i]);
               }
               else
                   s1.push(s[i]);
           }
           else if(s[i]=='/')//这里就是先计算，再存数，首先找到后面的数的值，然后通过top函数引出前面一个数，然后计算，这个时候pop掉栈顶，push这个新算完的数字，存进去，这个时候就算是存完了
           {
               b=0;
               i+=2;
               while(s[i]>='0'&&s[i]<='9')
               {
                   b=b*10+s[i]-'0';
                   i++;
               }
               i--;
               a=s2.top();
               s2.pop();
               a=a/b;
               s2.push(a);
           }
           else if(s[i]=='*')
           {
               b=0;
               i+=2;
               while(s[i]>='0'&&s[i]<='9')
               {
                   b=b*10+s[i]-'0';
                   i++;
               }
               i--;
               a=s2.top();
               s2.pop();
               a=a*b;
               s2.push(a);
           }
       }
       while(!s1.empty())
       {
           c=s1.top();
           s1.pop();
           a=s2.top();
           s2.pop();
           b=s2.top();
           s2.pop();
           if(c=='+')
               a+=b;
           else
               a=b-a;
           s2.push(a);
       }
       printf("%.2f\n",s2.top());
   }
   return 0;
}