测试页
Links
Relative links:
External links:
Footnote:
Character Ascension1 material.
"This body is the noblest2 and most eminent3 of all in this world.
Codeblock escaping
接下来的 $m$ 行中的第 $i$ 行包含两个正整数 $l_i$ 和 $r_i$ ($1\le l_i\le r_i\le n$),表示第 $i$ 次操作在区间 $[l_i,r_i]$ 上进行。
Codeblock tabs
float Q_rsqrt(float number)
{
long i;
float x2, y;
const float threehalfs = 1.5F;
x2 = number * 0.5F;
y = number;
i = * ( long * ) &y; // evil floating point bit level hacking
i = 0x5f3759df - ( i >> 1 ); // what the fuck?
y = * ( float * ) &i;
y = y * ( threehalfs - ( x2 * y * y ) ); // 1st iteration
// y = y * ( threehalfs - ( x2 * y * y ) ); // 2nd iteration, this can be removed
return y;
}
def rev_sqrt(number: float) -> float:
return number**(-1/2)
中文标题
附B:文章检索
如果手写了 h2 将不会出现在 toc 里。
content
H3
<h3>
H4
<h4>
H5
<h5>
H6
<h6>
Styles
basic
bold italic inlineCode
math
inline math:
code
inline: sudo rm --no-preserve-root -rf /
//Code block
#include <cstdio>
void func(){ }
int main() {
int a,b;
func();
cin>>a>>b;
cout<<a+b;
return 0;
}
pseudo code
details
Warning Details (Not parsed):
!!! warning
我们平常写的除法是向 0 取整,而这里的右移是向下取整(注意这里的区别),即当数大于等于 0 时两种方法等价,当数小于 0 时会有区别,如: num * 10 = (num<<1) + (num<<3)
Opened details:
interactor 随机选择一个 1
:询问与答案相同,你的程序应当停止询问。
0
:询问比答案小。
2
:询问比答案大。
Details with nested code block:
#include <cstdio>
Details with classname:*
注意区分 基数排序 与 桶排序
Long details with inline markdown:
题目大意:有一棵
最大。你需要保证对于你选择的一个树上结点,它的父亲一定被选中。求出这个最大的比值。
Empty details:
random_shuffle
已于 C++14 标准中被弃用,于 C++17 标准中被移除。Matryoshka doll:
- Raging Vortex
- Uprising Wrirlwind
- Sweeping Gust
- Cherishing Breezes
- Vortex Stellaris
- Interwined Winds
?? Names ??? Kiruya Momochi !!! Shifuna Agato !!!+ note Kokoro Natsume !!! warning Eustiana von Astraea
Hello? Is anyone there?
List in a box
- List Item
-
Child List Item
-
Paragraph
Directive
The Divine Priestess of Watatsumi Island. All of the island's affairs are at this young lady's fingertips.
The battlefield is unpredictable — one false move could mark utter defeat. Before the onset of battle, I bring every possible scenario into consideration, and turn my thoughts into written directives.
blocks
Reference:
Blockquote
Blockquote
Ordered List:
- item 1
- item 2
- item 3
- item 4
Unordered List:
- A
- B
- C
- D
Snippets
Import a file:
don't waste time on polyfilling pythonic features.
Import into codeblock:
/* clang-format whatever */
int babylon(int n) {
for (int i = 0; i < n; i++) {
d[i][0] = -1;
d[i][1] = -1;
d[i][2] = -1;
}
int r = 0;
for (int i = 0; i < n; i++) { //三种建法
r = max(r, babylon_sub(i, 0, n) + z[i]);
r = max(r, babylon_sub(i, 1, n) + x[i]);
r = max(r, babylon_sub(i, 2, n) + y[i]);
}
return r;
}
/* snip #2 */
int babylon(int n) {
for (int i = 0; i < n; i++) {
d[i][0] = -1;
d[i][1] = -1;
d[i][2] = -1;
}
int r = 0;
for (int i = 0; i < n; i++) { //三种建法
r = max(r, babylon_sub(i, 0, n) + z[i]);
r = max(r, babylon_sub(i, 1, n) + x[i]);
r = max(r, babylon_sub(i, 2, n) + y[i]);
}
return r;
}
Indents:
/* clang-format whatever */
int babylon(int n) {
for (int i = 0; i < n; i++) {
d[i][0] = -1;
d[i][1] = -1;
d[i][2] = -1;
}
int r = 0;
for (int i = 0; i < n; i++) { //三种建法
r = max(r, babylon_sub(i, 0, n) + z[i]);
r = max(r, babylon_sub(i, 1, n) + x[i]);
r = max(r, babylon_sub(i, 2, n) + y[i]);
}
return r;
}
/* snip #2 */
int babylon(int n) {
for (int i = 0; i < n; i++) {
d[i][0] = -1;
d[i][1] = -1;
d[i][2] = -1;
}
int r = 0;
for (int i = 0; i < n; i++) { //三种建法
r = max(r, babylon_sub(i, 0, n) + z[i]);
r = max(r, babylon_sub(i, 1, n) + x[i]);
r = max(r, babylon_sub(i, 2, n) + y[i]);
}
return r;
}
Footnotes
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