hdu_3596 Lazy

题意：求算术表达式。

分析：表达式有中缀转化为后缀。设一个stack存后缀数据，一个rout栈存运算符。
方法：（1）从右向左依次取得数据ch。

            （2）如果ch是操作数，直接加进stack中。

（3）如果ch是运算符（含左右括号），则：
      a：如果ch = '('，放入堆栈rout中。
      b：如果ch = ')'，依次输出堆栈rout中的运算符，直到遇到'('为止。
      c：如果ch不是')'或者'('，那么就和堆栈rout顶点位置的运算符top做优先级比较。
          1：如果ch优先级比rtop高，那么将ch放入堆栈。
          2：如果ch优先级低于或者等于rtop，那么输出top到stack中（直到！top或者满足 1），然后将ch放入堆栈。

 注：运算符的优先级：^ , *、/，%，+ -，注意本题多个乘方如2^(-1)^2^2是左结合，而我比赛的时用的是右结合，还有就是查了很多文档，很多都是右结合，所以暂时还没弄清。不过本题考查的重点还是栈的使用和中缀转后缀表达式或者用表达树。

cpp代码：

#include <cstdio>
#include  <ctype.h>
#include  <cmath>
#include  <stdlib.h>
char rin[10100],rout[10100 ];
struct  sta{
       double  x;
       bool  digit;
       char  ch;
} stack[10100 ];
double tst[10100 ];
int  top,ttop,rtop;

int priority(char  ch)
{
    switch  (ch)
    {
           case '+'  : 
           case '-' : return 1;break ;
           case '*'  :
           case '/' : return 2;break ;
           case '^' : return 3;break ;
    }
}

int  main()
{
 //  freopen("Lazy.in","r",stdin);
 //   freopen("out.txt","w",stdout);
    int  i;
    bool  mark;
    while (scanf("%s",rin) !=  EOF)
    {
          mark = true ;
          top = 0; i = 0; rtop = 0 ;
          while  (rin[i])
          { 
                if  (mark)
                {
                         mark = false ;
                         while (rin[i] == '(' )
                         {
                              i ++ ;  
                              rout[++rtop] = '(' ;    
                         }
                         if (rin[i] == '-' || rin[i] == '+' )
                         {
                            stack[++top].x = 0 ;
                            stack[top].digit = true ;
                         }      
                         else  {
                               sscanf(&rin[i],"%lf",&stack[++ top].x);
                               stack[top].digit = true ;
                               while (! isdigit(rin[i]))
                                     i++ ;
                               while (isdigit(rin[i]) || rin[i] == '.' )
                                     i++ ;
                              }
                }
                else  {
                        mark = true ;
                        if (rin[i] == ')' )
                        {
                           while (rout[rtop] != '(' )
                           {
                                 stack[++top].ch = rout[rtop-- ];
                                 stack[top].digit = false ;
                           }
                           rtop-- ;
                           mark = false ;         
                        }  
                        else  {
                               if (!rtop || rout[rtop] == '(' )
                                  rout[++rtop] =  rin[i];
                               else  {
                                      while ( rtop&& priority(rin[i]) <  priority(rout[rtop]))
                                      {
                                        stack[++top].ch = rout[rtop-- ];
                                        stack[top].digit = false ;
                                      }
                                      if (rtop && priority(rin[i]) ==  priority(rout[rtop]))
                                      {
                                          stack[++top].ch =  rout[rtop];
                                          stack[top].digit = false ;
                                          rout[rtop] =  rin[i];        
                                      }
                                      else rout[++rtop] =  rin[i];
                                      
                                    }
                             
                             }
                        i ++ ;
                     }
          }
         while  (rtop)
          {
                stack[++top].ch = rout[rtop-- ];
                stack[top].digit = false ;
          }
   /*       for (i = 1;i <= top; i++)
          {
              if (stack[i].digit)
                 printf("%.0lf\n",stack[i].x);
              else printf("%c\n",stack[i].ch);
          }*/
          ttop  = 0; i = 1 ;
          bool flag = true ;
          while (i <= top &&  flag)
          {
                if  (stack[i].digit)
                   tst[++ttop] =  stack[i].x;
                else  {
                        double b = tst[ttop--], a =  tst[ttop];
                        switch  (stack[i].ch)
                        {
                               case '+': tst[ttop] = a +  b;
                                         break ;
                               case '-': tst[ttop] = a -  b;
                                         break ;
                               case '*': tst[ttop] = a *  b;
                                         break ;
                               case '/': if (fabs(b) < 0.00000001 )
                                               flag = false ;
                                         else tst[ttop] = a /  b;
                                         break ;
                               case '^': tst[ttop] =  pow(a,b);
                                         break ;
                        }
                     }
                i++ ;
          }
          if  (flag)
             printf("%.8lf\n" ,tst[ttop]);
          else printf("The teacher is so lazy!\n" ); 
    }
    return 0 ;
}