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
",s2.top());
   }
   return 0;
}

  

原文地址:https://www.cnblogs.com/yintoki/p/5676637.html