（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：

 1 #include <stdio.h>

 2 #include <math.h>

 3 #include <string.h>

 4 #include <ctype.h>

 5 

 6 char a[410],stack[400];

 7 

 8 void deal(char *s)

 9 {

10     int i,j,L=strlen(s);

11     stack[0]='#';

12     j=1;

13     for(i=0; i<L; i++)

14     {

15         if(isalpha(s[i]))

16         {

17           printf("%c",s[i]);

18         }

19         else if(s[i]=='(')

20         {

21           stack[j++]='(';

22         }

23         else

24         {

25           switch(s[i])

26             {

27             case '+':

28             case '-':

29              while(stack[j-1]!='(' && j>0)

30              {

31                printf("%c",stack[j-1]);

32                stack[j-1]=' ';

33                j--;      

34              }

35              stack[j++]=s[i];

36              break;

37              case '*':

38              case '/':

39              while(stack[j-1]!='(' && (stack[j-1]=='*' || stack[j-1]=='/' || stack[j-1]=='^') && j>0)

40              {

41                  printf("%c",stack[j-1]);

42                  stack[j-1]=' ';

43                  j--;

44              }

45              stack[j++]=s[i];

46              break;

47              case '^': stack[j++]=s[i]; break;

48              case ')':

49              while(stack[j-1]!='(' && j>0)

50              {

51                  printf("%c",stack[j-1]);

52                  stack[j-1]=' ';

53                  j--;

54              }

55              j--;

56              break;

57              default:break;

58           }

59         }

60     }

61     printf("\n");

62 }

63 void solve()

64 {

65     int t,n;

66     scanf("%d",&t);

67     getchar();

68     while(t--)

69     {

70         gets(a);

71         deal(a);

72     }

73 }

74 

75 int main()

76 {

77   solve();

78   return 0;

79 }