Find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ satisfying the following conditions :
1) $f(x+y)-f(x)-f(y) \in \{0,1\} $ for all $x,y \in \mathbb{R}$
2) $\lfloor f(x) \rfloor = \lfloor x \rfloor $ for all real $x$.
1. **Initial Assumptions and Definitions:**
   Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be a function satisfying the given conditions:
   \[
   f(x+y) - f(x) - f(y) \in \{0, 1\} \quad \forall x, y \in \mathbb{R}
   \]
   \[
   \lfloor f(x) \rfloor = \lfloor x \rfloor \quad \forall x \in \mathbb{R}
   \]
   Define \( P(x, y) \) as the assertion \( f(x+y) - f(x) - f(y) \in \{0, 1\} \).

2. **Evaluating \( P(0, 0) \):**
   \[
   P(0, 0) \implies f(0+0) - f(0) - f(0) \in \{0, 1\}
   \]
   \[
   \implies f(0) - 2f(0) \in \{0, 1\}
   \]
   \[
   \implies -f(0) \in \{0, 1\}
   \]
   Since \( \lfloor f(0) \rfloor = \lfloor 0 \rfloor = 0 \), we get \( f(0) = 0 \).

3. **Evaluating \( P(x, -x) \):**
   \[
   P(x, -x) \implies f(x + (-x)) - f(x) - f(-x) \in \{0, 1\}
   \]
   \[
   \implies f(0) - f(x) - f(-x) \in \{0, 1\}
   \]
   \[
   \implies -f(x) - f(-x) \in \{0, 1\}
   \]
   Since \( f(0) = 0 \), we have:
   \[
   f(x) + f(-x) \in \{0, -1\}
   \]

4. **Evaluating \( f(1) \) and \( f(-1) \):**
   \[
   f(1) + f(-1) \in \{0, -1\}
   \]
   If \( f(1) + f(-1) = -1 \), then \( f(1) = f(-1) + 1 \). Given \( \lfloor f(-1) \rfloor = -1 \), we have \( f(-1) \in [-1, 0) \). Thus, \( f(1) \in [0, 1) \), which contradicts \( \lfloor f(1) \rfloor = 1 \). Therefore, \( f(1) + f(-1) = 0 \).

5. **Determining \( f(1) \) and \( f(-1) \):**
   Let \( f(1) = 1 + c \) with \( c \ge 0 \) (since \( \lfloor f(1) \rfloor = 1 \)). Then:
   \[
   f(-1) = -1 - c
   \]
   \[
   f(1) + f(-1) = (1 + c) + (-1 - c) = 0
   \]
   This implies \( c = 0 \), so \( f(1) = 1 \) and \( f(-1) = -1 \).

6. **Induction for \( f(x) \) on integers:**
   Using induction, we can show that \( f(n) = n \) for all \( n \in \mathbb{Z} \).

7. **Behavior on \([0, 1)\):**
   Since \( \lfloor f(x) \rfloor = 0 \) for all \( x \in [0, 1) \), we have \( f(x) \in [0, 1) \). Given \( f(x+y) = f(x) + f(y) \) for \( x, y \in [0, 1) \), by the result of Cauchy's functional equation, \( f(x) = cx \) for some \( c \in [0, 1] \).

8. **Extending to all real numbers:**
   Using induction and the properties of \( f \), we can extend \( f(x) = x \) to all \( x \in \mathbb{R} \).

The final answer is \( \boxed{ f(x) = x } \).