zoukankan      html  css  js  c++  java
  • 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;
    }
    

      

  • 相关阅读:
    ReSharper 10.0.0.2 Ultimate 破解
    Image 和byte[]之间的转换
    IntelliJ IDEA 2016.1.3(64) license server 与汉化
    冒泡排序
    选择排序
    希尔排序
    插入排序
    redis学习笔记——Redis过期键的删除策略
    redis学习笔记——应用场景
    perl学习笔记——字符串和排序
  • 原文地址:https://www.cnblogs.com/yintoki/p/5676637.html
Copyright © 2011-2022 走看看