HDU ACM 1237 简单计算器(栈)

 

http://acm.hdu.edu.cn/showproblem.php?pid=1237

题意：输入一个只有 +,-,*,/和数字的算式,输出计算结果,数字与符号之间以空格隔开.数据直到输入一行为0结束

　　　输出为计算结果,结果保留小数点后两位.

 

一边输入,一边处理的方式.

用cin>>num输入第一个数后就开始处理

模拟中序表达式转换后续表达式

注: 有0+0的数据

View Code 
  1 #include <iostream>
  2 #include <stack>
  3 using namespace std;
  4 
  5 struct FUN
  6 {
  7     char ch;
  8     int lv;
  9 };
 10 int main()
 11 {
 12     double num=0;
 13     while(cin>>num)
 14     {
 15         char aa;
 16         aa = getchar();
 17         if(num == 0 && aa == '\n')
 18         {
 19             break;
 20         }
 21         stack <FUN> ch;
 22         stack <double> d;
 23 
 24         d.push(num);
 25 
 26         FUN c;
 27 
 28         char a = 0;
 29         while(1)
 30         {
 31             
 32             if(a == '\n')
 33             {
 34                 while(!ch.empty())
 35                 {
 36                     FUN mid = ch.top();
 37                     ch.pop();
 38                     double num1,num2,mid_num;
 39                     num2 = d.top();
 40                     d.pop();
 41                     num1 = d.top();
 42                     d.pop();
 43                     if(mid.ch == '+')
 44                     {
 45                         mid_num = num1 + num2;
 46                     }
 47                     if(mid.ch == '-')
 48                     {
 49                         mid_num = num1 - num2;
 50                     }
 51                     if(mid.ch == '*')
 52                     {
 53                         mid_num = num1 * num2;
 54                     }
 55                     if(mid.ch == '/')
 56                     {
 57                         mid_num = num1 / num2;
 58                     }
 59                     d.push(mid_num);
 60                 }
 61                 break;
 62             }
 63             cin>>c.ch;
 64             getchar();
 65             if(c.ch == '-' ||c.ch == '+')
 66             {
 67                 c.lv = 1;
 68             }
 69             else
 70             {
 71                 c.lv = 2;
 72             }
 73 
 74             if(!ch.empty())
 75             {
 76                 FUN mid = ch.top();
 77                 while(mid.lv >= c.lv)
 78                 {
 79                     ch.pop();
 80                     double num1,num2,mid_num;
 81                     num2 = d.top();
 82                     d.pop();
 83                     num1 = d.top();
 84                     d.pop();
 85                     if(mid.ch == '+')
 86                     {
 87                         mid_num = num1 + num2;
 88                     }
 89                     if(mid.ch == '-')
 90                     {
 91                         mid_num = num1 - num2;
 92                     }
 93                     if(mid.ch == '*')
 94                     {
 95                         mid_num = num1 * num2;
 96                     }
 97                     if(mid.ch == '/')
 98                     {
 99                         mid_num = num1 / num2;
100                     }
101                     d.push(mid_num);
102                     if(!ch.empty())
103                     {
104                         mid = ch.top();
105                     }
106                     else
107                     {
108                         break;
109                     }
110                 }
111                 ch.push(c);
112             }
113             else
114             {
115                 ch.push(c);
116             }
117             cin>>num;
118             d.push(num);
119             a = getchar();
120         }
121         printf("%.2lf\n",d.top());
122     }
123     return 0;
124 }

 

posted @ 2012-08-23 16:10  zx雄  阅读(1531)  评论(0)  编辑  收藏  举报