ЕГЭ информатика проблема во втором задании

питон?

Code Breakdown:

Python
print("x y z w") # Print header for output
for x in 0, 1: # Loop through x (0 and 1)
for y in 0, 1: # Loop through y (0 and 1)
for z in 0, 1: # Loop through z (0 and 1)
for w in 0, 1: # Loop through w (0 and 1)
if ((w or (not(x))) and (w == (not(y)) and (w <= z)) == 1:
print(x, y, z, w) # Print if condition is met
Use code with caution.

Error Explanation:

The error you're seeing (SyntaxError: invalid syntax) occurs on line 6 because of the extra colon (:). In Python, the if statement doesn't require a colon after the condition. It's only used for defining a code block within the if statement.

Corrected Code:

Python
print("x y z w") # Print header for output
for x in 0, 1: # Loop through x (0 and 1)
for y in 0, 1: # Loop through y (0 and 1)
for z in 0, 1: # Loop through z (0 and 1)
for w in 0, 1: # Loop through w (0 and 1)
if ((w or (not(x))) and (w == (not(y))) and (w <= z)): # Corrected if statement
print(x, y, z, w) # Print if condition is met
Use code with caution.

Explanation of the Condition:

The condition ((w or (not(x))) and (w == (not(y))) and (w <= z)) checks for specific combinations of x, y, z, and w:

(w or (not(x))): This ensures w is either True or x is False.
(w == (not(y))): This ensures w is the opposite of y (i.e., if y is True, then w must be False, and vice versa).
(w <= z): This guarantees w is less than or equal to z.
Output:

This corrected code will iterate through all possible combinations of x, y, z, and w (16 combinations in total) and print only the ones that satisfy the condition. However, with the given condition, no combinations will meet all three requirements, so the code won't print anything.

 print("x y z w")  
for x in 0, 1:  
    for y in 0, 1:  
        for z in 0, 1:  
            for w in 0, 1:  
                if (w or (not x)) and (w == (not y) and (w <= z)):  
                    print(x, y, z, w) 

В вашем коде ошибка синтаксиса возникает из-за отсутствия закрывающей скобки в условии if.

Обратите внимание на добавление закрывающей скобки перед двоеточием в строке с if.