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Hello, Markdown!

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This is a sample code block:

function greet() {
  console.log("Hello, world!");
}

And here is an inline formula: E=mc2E = mc^2

And some block formula:

eπi+1=0e^{\pi i} + 1 = 0 (abd0ac00a)n=(annan1bnan2[ad+12(n1)bc]0anncan100an),nZ+\left( \begin{array}{ccc} a & b & d \\ 0 & a & c \\ 0 & 0 & a \\ \end{array} \right)^n = \left( \begin{array}{ccc} a^n & n a^{n-1} b & n a^{n-2} \left[a d+\frac{1}{2} (n-1) b c\right] \\ 0 & a^n & n c a^{n-1} \\ 0 & 0 & a^n \\ \end{array} \right) , \quad n \in \mathbb{Z}^+ 0xnexdx=n!\int_0^\infty x^n e^{-x} \mathrm{d} x = n!

需要注意行内公式开始的$$需要前面空一行,并且$$单独占一行,也许我可以在未来改一下配置不这么严格。

footnote

Here is a simple footnote1.

A footnote can also have multiple lines2.

A strange picture:

what

what?

Alert Example

Note

Highlights information that users should take into account, even when skimming.

Tip

Optional information to help a user be more successful.

Important

Crucial information necessary for users to succeed.

Warning

Critical content demanding immediate user attention due to potential risks.

Caution

Negative potential consequences of an action.

wao

Caution

数域XX上的两个不相交的闭集AA, BB有可能可以并成XX。 例如考虑Q\mathbb Q上的集合

A={xQx2<2},B={xQx2>2}.A = \{x\in \mathbb Q \mid x^2 <2 \}, \quad B = \{x\in \mathbb Q \mid x^2 > 2\}.

这两个集合在Q\mathbb Q上都是既开又闭的,但AB=QA\cup B=\mathbb Q. 但是由于R\mathbb R是连通的,所以这种集合不存在。

┌────────────────────────────────────────────────────┐
│ WAN                                                │
│┌────────────────────────┐                          │
││ School LAN             │   ┌────────┐             │
││┌──────────┐  ┌────────┐│   │ Server │   ┌────────┐│
│││ School   │  │        └┴───┴────────┴───┘┌──────┐││
│││ Internal │  │ Client                    │ User │││
│││ Computer │  │        ┌┬───┬────────┬───┐└──────┘││
││└──────────┘  └────────┘│   └────────┘   └────────┘│
│└────────────────────────┘                          │
└────────────────────────────────────────────────────┘
Column1 Column2
Item1.1 Item2.1
function getAllArragement(set = [1, 2]) {
  return set.length == 1
    ? [set]
    : set
        .map((_v, idx) => [...set.slice(0, idx), ...set.slice(idx + 1)])
        .map(getAllArragement)
        .map((subsets, index) =>
          subsets.map((subset, _i) => [set[index], ...subset]),
        )
        .flat();
}

Footnotes

  1. My reference.

  2. To add line breaks within a footnote, prefix new lines with many spaces. This is a second line.

// 这是脚注
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