# Transform the algebraic expression with brackets into RPN form (Reverse Polish Notation).

#codeifyoucansolve
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.

Example

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

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

//sollution

```#include <stdio.h>
#include<string.h>
int main(void) {
char q[400],stack[400],str[400],ch;
int n,top,i,front;
scanf("%d",&n);
while(n--)
{
front  = 0;
top = 0;
scanf("%s ",str);
for(i=0;i<strlen(str);i++)
{
if(str[i]!='+'&&str[i]!='-'&&str[i]!='*'&&str[i]!='/'&&str[i]!='^'&&str[i]!='('&&str[i]!=')')
{
q[front++] = str[i];
}
else
{
if(str[i]=='+')
{
while(top>0)
{
ch = stack[top-1];
if(ch != '(')
{
q[front++] = ch;
top--;
}
else
break;
}
}
else if(str[i]=='-')
{
while(top>0)
{
ch = stack[top-1];
if(ch != '+'&&ch != '(')
{
q[front++] = ch;
top--;
}
else
break;
}
}
else if(str[i]=='*')
{
while(top>0)
{
ch = stack[top-1];
if(ch != '+'&&ch != '-'&&ch != '(')
{
q[front++] = ch;
top--;
}
else
break;
}
}
else if(str[i]=='/')
{
while(top>0)
{
ch = stack[top-1];
if(ch != '+'&&ch != '-'&&ch != '*'&&ch != '(')
{
q[front++] = ch;
top--;
}
else
break;
}
}
else if(str[i]=='^')
{
while(top>0)
{
ch = stack[top-1];
if(ch != '+'&&ch != '-'&&ch != '*'&&ch != '/'&&ch != '(')
{
q[front++] = ch;
top--;
}
else
break;
}
}
else if(str[i]==')')
{
while(stack[top-1]!='(')
{
ch = stack[top-1];
q[front++] = ch;
top = top -1;
}
top = top - 1;
}
if(str[i] !=')')
stack[top++] = str[i];
}
}
while(top>0)
{
q[front++] =  stack[top-1] ;
top = top-1;
}
q[front] = '';
printf("%s\n",q);
}
return 0;
}
```