Let $f(x)$ is such function, that $f(x)=1$  for integer $x$ and $f(x)=0$ for non integer $x$. 
Build such function using only variable $x$, integer numbers, and operations $+,-,*,/,[.]$(plus, minus, multiply,divide and integer part)
1. We need to construct a function \( f(x) \) such that \( f(x) = 1 \) if \( x \) is an integer and \( f(x) = 0 \) if \( x \) is not an integer. We are restricted to using the variable \( x \), integer numbers, and the operations \( +, -, *, /, \lfloor \cdot \rfloor \) (integer part).

2. Consider the floor function \( \lfloor x \rfloor \), which gives the greatest integer less than or equal to \( x \). For any real number \( x \), \( \lfloor x \rfloor \) is an integer.

3. If \( x \) is an integer, then \( \lfloor x \rfloor = x \) and \( \lfloor 1 - x \rfloor = 1 - x \). Therefore, for integer \( x \):
   \[
   \lfloor x \rfloor + \lfloor 1 - x \rfloor = x + (1 - x) = 1
   \]

4. If \( x \) is not an integer, then \( \lfloor x \rfloor \) is the greatest integer less than \( x \), and \( \lfloor 1 - x \rfloor \) is the greatest integer less than \( 1 - x \). In this case, \( \lfloor x \rfloor \) and \( \lfloor 1 - x \rfloor \) are not complementary, and their sum will be less than 1. Specifically, for non-integer \( x \):
   \[
   \lfloor x \rfloor + \lfloor 1 - x \rfloor < 1
   \]

5. Therefore, the function \( f(x) = \lfloor x \rfloor + \lfloor 1 - x \rfloor \) satisfies the required conditions:
   - \( f(x) = 1 \) if \( x \) is an integer.
   - \( f(x) = 0 \) if \( x \) is not an integer.

The final answer is \( \boxed{ f(x) = \lfloor x \rfloor + \lfloor 1 - x \rfloor } \).