Change the suffix of an infix

Change the suffix of an infix

Time Limit: 1000 MS | memory limit: 65535 KB

Difficulty: 3

Description

People's Daily habit is to write an arithmetic expression as an infix, but it is more "used" for machines, general Data Structure books have relevant content for reference. I will not go into details here. Now your task is to change the infix into a suffix.

Input

Enter an integer N in the first line. There are N groups of test data (n <10 ).
Each group of test data has only one row. It is a string of no more than 1000 characters, indicating the infix of this formula. Each formula ends with "=. This expression only contains the +-*/and parentheses. Parentheses can be nested. Data ensures that no negative numbers are displayed in the input operations.
Data guarantee that the divisor is not 0

Output

Each group outputs the suffix corresponding to the infix in this group. Adjacent operand operators are required to be separated by spaces.

Sample Input

2

1.000 + 2/4 =

(1 + 2) * 5 + 1)/4 =

Sample output

1.000 2 4/+ =

1 2 + 5*1 + 4/=

Question: Add spaces to the specified position in the array postexp.

Program code:

# Include <stdio. h>

# Include <stdlib. h>

 

# Define Max size 1000

# Define maxop 7

 

Struct // set the operator priority

{

Char ch; // Operator

Int PRI; // priority

} Lpri [] = {'=', 0}, {'(', 1}, {'*', 5}, {'/', 5 }, {'+', 3}, {'-', 3}, {')', 6 }},

Rpri [] = {'=', 0}, {'(', 6}, {'*', 4}, {'/', 4 }, {'+', 2}, {'-', 2}, {')', 1 }};

Int leftpri (char OP) // evaluate the op priority of the Left Operator

{

Int I;

For (I = 0; I <maxop; I ++)

If (lpri [I]. CH = OP)

Return lpri [I]. PRI;

}

 

Int rightpri (char OP) // evaluate the op priority of the Right Operator

{

Int I;

For (I = 0; I <maxop; I ++)

If (rpri [I]. CH = OP)

Return rpri [I]. PRI;

}

 

Int inop (char ch) // determines if ch is an operator

{

If (CH = '(' | CH = ') '| CH =' + '| CH ='-'| CH =' * '| CH = '/')

Return 1;

Else

Return 0;

}

 

Int precede (char OP1, char OP2) // comparison result of OP1 and OP2 operator priority

{

If (leftpri (OP1) = rightpri (OP2 ))

Return 0;

Else if (leftpri (OP1) <rightpri (OP2 ))

Return-1;

Else

Return 1;

}

 

Void trans (char * exp, char postexp [])

// Convert the arithmetic expression exp to the suffix expression postexp

{

Struct

{

Char data [maxsize]; // storage Operator

Int top; // Stack pointer

} Op; // defines the operator Stack

Int I = 0; // I is the subscript of postexp

Op. Top =-1;

Op. Top ++; // Add '=' to the stack

Op. Data [op. Top] = ';

While (* exp! = '\ 0') // loop when the EXP expression is not completely scanned

{

If (* exp = ')

Break;

If (! Inop (* exp) // number

{

While (* exp> = '0' & * exp <= '9' | * exp = '.') // determines it as a number.

{

Postexp [I ++] = * exp;

Exp ++;

}

// Postexp [I ++] = '#'; // use # to mark the end of a numeric string

Postexp [I ++] = '';

}

Else // Operator

Switch (precede (op. Data [op. Top], * exp ))

{

Case-1: // low priority of the stack top operator: Stack entry

Op. Top ++;

Op. Data [op. Top] = * exp;

Exp ++; // Continue scanning other characters

Break;

Case 0: // only parentheses meet this situation

Op. Top --; // move (back to stack

Exp ++; // Continue scanning other characters

Break;

Case 1: // return the stack and output it to postexp

Postexp [I ++] = op. Data [op. Top];

Postexp [I ++] = '';

Op. Top --;

Break;

}

} // While (* exp! = '\ 0 ')

While (op. Data [op. Top]! = ')

// At this time, the exp scan is complete and the stack is pushed back to '= '.

{

Postexp [I ++] = op. Data [op. Top];

Postexp [I ++] = '';

Op. Top --;

}

Postexp [I ++] = '; // Add an end ID to the postexp expression

Postexp [I ++] = '\ 0 ';

}

 

Int main ()

{

Int N;

Scanf ("% d", & N );

Getchar ();

While (n --)

{

Char exp [maxsize];

Scanf ("% s", exp );

Char postexp [maxsize];

Trans (exp, postexp );

Printf ("% s \ n", postexp );

}

System ("pause ");

Return 0;

}