C 처음 배우는 중인데 숙제를 어찌 해야 하는지 막막해서 글 남겨봅니다

안녕하세요. 과제가 있는데 뭐 부터 해야하는지 전혀 감이 안옵니다.
완전 초보라 ㅠㅠ 대충이라도 뭐뭐 부터 해야하는지 설명좀 해주시면 고맙겠네요
- Aim
The aim of the assignment is to make you practice the use of pointers, command line argu-
ments, and dynamic memory allocation.
- General description
Your program, will accept an English description of a system of linear equations as command
line arguments. Practically, assuming that solve is the name of the executable file, this English description will be stored in a le test and the program will be run with the command
solve $(cat test)
The English description of the system of linear equations is to be translated into a formal representation and solved. The assignment does not require to build a parser: simple pattern matching can do the job.
- Detailed description
Each equation will be described by a sentence that will be generated by the following context-free grammar. The nonterminal symbols of the grammar are:
SENTENCE
EXPRESSION1
EXPRESSION
PRODUCT1
PRODUCTS
PRODUCT
CONSTANTS
CONSTANT
VARIABLE
The start symbol is SENTENCE.
The production rules are:
SENTENCE --> EXPRESSION1 is equal to EXPRESSION.
SENTENCE --> EXPRESSION1 equals EXPRESSION.
EXPRESSION1 --> EXPRESSION1 plus EXPRESSION
EXPRESSION1 --> The result of adding PRODUCTS and PRODUCT
EXPRESSION1 --> The sum of PRODUCTS and PRODUCT
EXPRESSION1 --> PRODUCT1

EXPRESSION --> EXPRESSION plus EXPRESSION
EXPRESSION --> the result of adding PRODUCTS and PRODUCT
EXPRESSION --> the sum of PRODUCTS and PRODUCT
EXPRESSION --> PRODUCT
PRODUCTS --> PRODUCTS, PRODUCT
PRODUCTS --> PRODUCT
PRODUCT1 --> The result of multiplying CONSTANT by CONSTANT
PRODUCT1 --> The result of multiplying VARIABLE by CONSTANT
PRODUCT1 --> The result of multiplying CONSTANT by VARIABLE
PRODUCT1 --> The product of CONSTANTS and CONSTANT
PRODUCT1 --> The product of VARIABLE, CONSTANTS and CONSTANT
PRODUCT1 --> The product of VARIABLE and CONSTANT
PRODUCT1 --> The product of CONSTANTS, VARIABLE, CONSTANTS and CONSTANT
PRODUCT1 --> The product of CONSTANTS, VARIABLE and CONSTANT
PRODUCT1 --> The product of CONSTANTS and VARIABLE
PRODUCT --> the result of multiplying CONSTANT by CONSTANT
PRODUCT --> the result of multiplying VARIABLE by CONSTANT
PRODUCT --> the result of multiplying CONSTANT by VARIABLE
PRODUCT --> the product of CONSTANTS and CONSTANT
PRODUCT --> the product of VARIABLE, CONSTANTS and CONSTANT
PRODUCT --> the product of VARIABLE and CONSTANT
PRODUCT --> the product of CONSTANTS, VARIABLE, CONSTANTS and CONSTANT
PRODUCT --> the product of CONSTANTS, VARIABLE and CONSTANT
PRODUCT --> the product of CONSTANTS and VARIABLE
PRODUCT --> CONSTANT times CONSTANT
PRODUCT --> VARIABLE times CONSTANT
PRODUCT --> CONSTANT times VARIABLE
PRODUCT --> CONSTANT
PRODUCT --> VARIABLE
CONSTANTS --> CONSTANTS, CONSTANT
CONSTANTS --> CONSTANT
plus the production rules of the form
CONSTANT --> C_constant_of_type_int_or_float
VARIABLE --> C_variable
Hence the terminal symbols of the grammar are all lowercase words in the production rules, which includes C constants and variables.

An English equation is a sentence generated by the grammar, namely, any sequence of terminals that can be obtained from the grammar, starting with SENTENCE (the start symbol), applying the production rules, until nothing but terminals are obtained.
The following are examples of three les, say test1, test2 and test3 respectively, that contain
English equations:
The result of multiplying 2 by x plus y times 3 plus 5 times z is equal to 10.
The product of x and 3 plus the product of 7 and y plus z times 4 is equal to 3.
The product of 2 and y plus the result of multiplying z by 2 plus x equals 3.
The sum of 6.5 times u, 5 and -2. plus the product of 3.5 and x is equal to
the result of adding x, 4.5 times y and -10.
The sum of x and y plus the result of adding z and 8.5 equals 2 times x plus
-3 times y plus the result of multiplying 5 by u plus -7..
The result of adding 2, 2 times u and the product of -3 and u plus 8.5 times y
is equal to the sum of y and the product of -5.5 and u.
The sum of 17, the product of 2., u, 1.2 and 3. and the result of
multiplying -2 by 3 plus 7 times y is equal to the result of
adding the product of y, 6 and 1.5 and .2 times 4 plus y plus
u times 2.5 plus -1 times x plus the product of 6 and z plus
the result of multiplying -1.5 by u.
The result of multiplying 2 by x plus y times 3 plus 5 times z is equal to 1.
The product of x and 2 plus the product of 3 and y plus z times 5 is equal to 1.
The product of 2 and y plus the result of multiplying z by 2 plus x equals 3.

No assumption should be made on the number of English equations provided as input, nor on the length of an English equation, nor on the length of a variable.
The number of variables is assumed to be equal to the number of equations.
Your program should:
output the variables in the equations in lexicographic order;
output a formal representation of the English equations;
output whether or not there is a solution, and in case there is one, output it.

- Sample output

$ solve $(cat test1)
The variables are:
x y z
The equations are:
2.00 x + 3.00 y + 5.00 z = 10.00
3.00 x + 7.00 y + 4.00 z = 3.00
1.00 x + 2.00 y + 2.00 z = 3.00
The system of equations has a solution:
x = 3.00
y = -2.00
z = 2.00
$ solve $(cat test2)
The variables are:
u x y z
The equations are:
6.50 u + 2.50 x + -4.50 y + 0.00 z = -13.00
-5.00 u + -1.00 x + 4.00 y + 1.00 z = -15.50
4.50 u + 0.00 x + 7.50 y + 0.00 z = -2.00
6.20 u + 1.00 x + -3.00 y + -6.00 z = -10.20
The system of equations has a solution:
u = 7.00
x = -31.44
y = -4.47
z = 5.93
$ solve $(cat test3)
The variables are:
x y z
The equations are:
2.00 x + 3.00 y + 5.00 z = 1.00
2.00 x + 3.00 y + 5.00 z = 1.00
1.00 x + 2.00 y + 2.00 z = 3.00
The system of equations does not have a solution.

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