P8763 [Lanqiao Cup 2021 National ABC] XOR Transformation

Xiao Lan has a binary string $s=s_{1} s_{2} s_{3} \cdots s_{n}$. After that, at each moment, Xiao Lan will apply a transformation to this binary string. The rule of each transformation is the same. For the binary string $s=s_{1} s_{2} s_{3} \cdots s_{n}$, the transformed binary string $s^{\prime}=s_{1}^{\prime} s_{2}^{\prime} s_{3}^{\prime} \cdots s_{n}^{\prime}$ is: $$ \begin{aligned} &s_{1}^{\prime}=s_{1} \\ &s_{i}^{\prime}=s_{i-1} \oplus s_{i} \end{aligned} $$ Here $a \oplus b$ denotes the bitwise XOR of two bits. When $a$ and $b$ are the same, the result is $0$; when $a$ and $b$ are different, the result is $1$. What is the binary string after $t$ transformations?

The first line contains two integers $n, t$, representing the length of the binary string and the number of transformations. The second line contains a binary string of length $n$.

Output one line containing a binary string, which is the transformed string.

**Sample Explanation** Initially it is `10110`. After 1 transformation it becomes `11101`, after 2 transformations it becomes `10011`, and after 3 transformations it becomes `11010`. **Constraints** For $40\%$ of the testdata, $1 \leq n \leq 100$ and $1 \leq t \leq 1000$. For $80\%$ of the testdata, $1 \leq n \leq 1000$ and $1 \leq t \leq 10^{9}$. For all testdata, $1 \leq n \leq 10000$ and $1 \leq t \leq 10^{18}$. Lanqiao Cup 2021 National Contest, Group A Problem F (Group B Problem G, Group C Problem G). Translated by ChatGPT 5