Define a sequence $\{a_n\}$  by\[a_0=\frac{1}{2},\ a_{n+1}=a_{n}+\frac{a_{n}^2}{2012}, (n=0,\ 1,\ 2,\ \cdots),\]  find  integer $k$ such that   $a_{k}<1<a_{k+1}.$

(September 29, 2012, Hohhot)
1. We start with the given sequence \(\{a_n\}\) defined by:
   \[
   a_0 = \frac{1}{2}, \quad a_{n+1} = a_n + \frac{a_n^2}{2012} \quad \text{for} \quad n = 0, 1, 2, \ldots
   \]

2. We can rewrite the recurrence relation as:
   \[
   a_{n+1} = a_n \left(1 + \frac{a_n}{2012}\right)
   \]

3. To find the integer \(k\) such that \(a_k < 1 < a_{k+1}\), we need to analyze the behavior of the sequence. We start by expressing \(a_{n+1}\) in terms of \(a_n\):
   \[
   a_{n+1} = a_n + \frac{a_n^2}{2012}
   \]

4. We can approximate the growth of \(a_n\) by considering the sequence in terms of its reciprocal. Let \(b_n = \frac{1}{a_n}\). Then:
   \[
   a_{n+1} = a_n + \frac{a_n^2}{2012} \implies \frac{1}{a_{n+1}} = \frac{1}{a_n + \frac{a_n^2}{2012}}
   \]

5. Using the approximation for small \(x\), \(\frac{1}{1+x} \approx 1 - x\), we get:
   \[
   \frac{1}{a_{n+1}} \approx \frac{1}{a_n} - \frac{a_n}{2012}
   \]

6. This implies:
   \[
   b_{n+1} \approx b_n - \frac{1}{2012}
   \]

7. Since \(b_0 = \frac{1}{a_0} = 2\), we can write:
   \[
   b_n \approx 2 - \frac{n}{2012}
   \]

8. We need \(a_k < 1 < a_{k+1}\), which translates to:
   \[
   b_k > 1 \quad \text{and} \quad b_{k+1} < 1
   \]

9. Solving for \(k\) when \(b_k = 1\):
   \[
   1 \approx 2 - \frac{k}{2012} \implies k \approx 2012
   \]

10. Therefore, we conclude that \(k = 2012\) is the integer such that \(a_{2012} < 1 < a_{2013}\).

The final answer is \( \boxed{ k = 2012 } \)