（SPOJ4）Transform the Expression

Transform the algebraic expression with brackets into RPN form (Reverse Polish Notation). Two-argument operators: +, -, *, /, ^ (priority from the lowest to the highest), brackets ( ). Operands: only letters: a,b,...,z. Assume that there is only one RPN form (no expressions like a*b*c).

Input

t [the number of expressions <= 100]
expression [length <= 400]
[other expressions]

Text grouped in [ ] does not appear in the input file.

Output

The expressions in RPN form, one per line.

ExampleInput:

3
(a+(b*c))
((a+b)*(z+x))
((a+t)*((b+(a+c))^(c+d)))

Output:
abc*+
ab+zx+*
at+bac++cd+^*

算法描述：
栈底放‘#’，从左至右逐字读取中缀式：
1.当前字符为数字时，直接输出；
2.当前字符为"("时，将其压栈；
3.当前字符为")"时，则弹出堆栈中最上的"("之前的所有运算符并输出，然后删除堆栈中的"(" ；
4.当前字符为运算符时，则依次弹出堆栈中优先级大于等于当前运算符的(到"("之前为止)，输出，再将当前运算符压栈；

AC code：

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <ctype.h>

char a[410],stack[400];

void deal(char *s)
{
    int i,j,L=strlen(s);
    stack[0]='#';
    j=1;
    for(i=0; i<L; i++)
    {
        if(isalpha(s[i]))
        {
          printf("%c",s[i]);
        }
        else if(s[i]=='(')
        {
          stack[j++]='(';
        }
        else
        {
          switch(s[i])
            {
            case '+':
            case '-':
             while(stack[j-1]!='(' && j>0)
             {
               printf("%c",stack[j-1]);
               stack[j-1]=' ';
               j--;      
             }
             stack[j++]=s[i];
             break;
             case '*':
             case '/':
             while(stack[j-1]!='(' && (stack[j-1]=='*' || stack[j-1]=='/' || stack[j-1]=='^') && j>0)
             {
                 printf("%c",stack[j-1]);
                 stack[j-1]=' ';
                 j--;
             }
             stack[j++]=s[i];
             break;
             case '^': stack[j++]=s[i]; break;
             case ')':
             while(stack[j-1]!='(' && j>0)
             {
                 printf("%c",stack[j-1]);
                 stack[j-1]=' ';
                 j--;
             }
             j--;
             break;
             default:break;
          }
        }
    }
    printf("\n");
}
void solve()
{
    int t,n;
    scanf("%d",&t);
    getchar();
    while(t--)
    {
        gets(a);
        deal(a);
    }
}

int main()
{
  solve();
  return 0;
}