Description

Given an infix Boolean expression with parentheses, which has at most 4 variables ‘A’, ’B’, ‘C’, and ‘D’, and two operators ‘&’ and ‘|’. Build a corresponding syntax tree for it.

To parse the infix expression with parentheses, use the grammar below.

EXPR = FACTOR | EXPR OP FACTOR

FACTOR = ID | (EXPR)

EXPR is the expression, ID is one of ‘A’, ‘B’, ’C’, or ‘D’, and OP is one of ‘&’ or ‘|’.

 

You will be provided with main.c and function.h. main.c contains the implementation of functions printPrefix and freeTree. function.h contains the definition of tree node and variables. You only need to implement the following three functions in function.c.

BTNode* makeNode(char c)   // create a node without any child

BTNode* EXPR()   // parse an infix expression and generate a syntax tree

BTNode* FACTOR()   // use the parse grammar FACTOR = ID | (EXPR) to deal with parentheses

 

For OJ submission:

        Step 1. Submit only your function.c into the submission block.(Please choose C compiler)

        Step 2. Check the results and debug your program if necessary.

Input

The input contains N infix expressions, which has at most 4 variables ‘A’, ’B’, ‘C’, and ‘D’, two operators ‘&’ and ‘|’, and parentheses. All parentheses are matched. 

Output

The output contains N prefix expressions without parentheses, which are preorders of syntax trees.

Sample Input  Download

Sample Output  Download

Partial Judge Code

10966.c

Partial Judge Header

10966.h

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Description

Given an prefix Boolean expression, which has at most 4 variables ‘A’, ’B’, ‘C’, and ‘D’, and two operators ‘&’ and ‘|’. Output its infix presentation with necessary parenthesis.

• Ex: input: ||A&BCD
  output: A|(B&C)|D

Hint : About necessary parenthesis, you can observe the syntax tree which has been constructed, then you will find out the rule. For example, the infix expression of |&|AB&CDA with necessary parenthesis is A|B&(C&D)|A, rather than (A|B)&(C&D)|A .

• Syntax tree of |&|AB&CDA

You will be provided with main.c and function.h, and asked to implement function.c.

For OJ submission:

        Step 1. Submit only your function.c into the submission block.(Please choose C compiler)

        Step 2. Check the results and debug your program if necessary.

Input

The first line will have a number N with the number of test cases, followed by N lines of input, each contain the prefix Boolean expression.

Output

There are N lines infix expression with necessary parenthesis.

Sample Input  Download

Sample Output  Download

Partial Judge Code

10968.c

Partial Judge Header

10968.h

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Discuss

Description

Given an infix Boolean expression with parentheses, which has at most 4 variables ‘A’, ’B’, ‘C’, and ‘D’, and two operators ‘&’ and ‘|’. Please remove unnecessary parentheses and print the infix expression. Existence of unnecessary parentheses doesn’t affect the result of expression. For example,

(A&B)|(C&D) → A&B|(C&D)

(((A|B))) → A|B

Hint: You can combine two homework. Build a syntax tree and print the infix expression with necessary parentheses.

 

For OJ submission:

       Step 1. Submit your main.c into the submission block.(Please choose C compiler)

       Step 2. Check the results and debug your program if necessary.

 

Input

The input is an infix expression, which has at most 4 variables ‘A’, ’B’, ‘C’, and ‘D’, two operators ‘&’ and ‘|’, and parentheses. The length of the infix expression is less than 256.

Output

The output is an infix expression without unnecessary parentheses.

Sample Input  Download

Sample Output  Download

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