### floor, ceil

$\lfloor \sqrt{\lfloor x\rfloor}\rfloor = \lfloor \sqrt{ x}\rfloor$

$\lceil \sqrt{\lceil x\rceil}\rceil = \lceil \sqrt{x}\rceil$

continuous, monotonically increasing $f$ with property $f(x)\in Z \Rightarrow x \in Z$. Then below holds
$$
\begin{align}
    \lfloor f(\lfloor x\rfloor)\rfloor = \lfloor f(x)\rfloor,\ \lceil f(\lceil x\rceil)\rceil = \lceil f(x)\rceil
\end{align}
$$

$\lceil \sqrt{\lfloor x\rfloor}\rceil = \lceil \sqrt{ x}\rceil$, only when $m^2 < x < m^2+1$ not hold.

$$
\begin{align}
    \sum_{k=0}^{m-1} \lfloor \frac{n+k}{m} \rfloor =n= \sum_{k=0}^{m-1} \lceil \frac{n-k}{m} \rceil 
\end{align}
$$

$$
\begin{align}
    \sum_{k=0}^{n-1} \lfloor \sqrt{k} \rfloor = na - \frac{1}{3} a^3 - \frac{1}{2}a^2 - \frac{1}{6}a,\ a = \lfloor \sqrt{n} \rfloor
\end{align}
$$

$$
\begin{align}
    \sum_{k=0}^{m-1} \lfloor \frac{nk+x}{m} \rfloor = d  \lfloor \frac{x}{d} \rfloor + \frac{(m-1)(n-1)}{2} + \frac{d-1}{2} = \sum_{k=0}^{n-1} \lfloor \frac{mk+x}{n} \rfloor,\ d = \text{gcd}(m,n)
\end{align}
$$



```python

```

7060a88026a2c197de9ddf80c89f4d4ea62c76fd
ipynb
Jupyter Notebook
notes/floor_ceil.ipynb
sogapalag/problems
0ea7d65448e1177f8b3f81124a82d187980d659c
MIT
2020-04-04T14:56:12.000Z
2020-04-04T14:56:12.000Z
notes/floor_ceil.ipynb
sogapalag/problems
0ea7d65448e1177f8b3f81124a82d187980d659c
MIT
notes/floor_ceil.ipynb
sogapalag/problems
0ea7d65448e1177f8b3f81124a82d187980d659c
MIT
Qwen/Qwen-72B
1. YES
2. YES