测试文章

测试abcd

分割线

数学表达式

求tan⁡(x)\tan (x),可以用

tan⁡(x)=sin⁡(x)cos⁡(x)\tan (x)=\frac{\sin (x)}{\cos (x)}

或者这个

tan⁡(x)=∑x=1∞(−1)n−122n(22n−1)B2nx2n−1(2n)!\tan (x)=\sum _{x=1}^{\infty } \frac{(-1)^{n-1} 2^{2 n} \left(2^{2 n}-1\right) B_{2 n} x^{2 n-1}}{(2 n)!}

其中B2nB_{2 n}为伯努利数。B2nB_{2 n}和1cos⁡(x)\frac{1}{\cos (x)}都用多项式求逆来做。

24=0.52c+2d+24=a\dfrac{2}{4} = 0.5 \qquad \dfrac{2}{c + \dfrac{2}{d + \dfrac{2}{4}}} = a

{abcd}\begin{Bmatrix} a & b \\ c & d \end{Bmatrix}

(123x000y0).(123x000y0)2=(4x+3xy+14x+3y+26x+32x2+x2x+3xy3xxy2xy3xy) \begin{pmatrix} 1 & 2 &3 \\ x & 0 &0 \\ 0 & y &0 \end{pmatrix} . \begin{pmatrix} 1 & 2 &3 \\ x & 0 &0 \\ 0 & y &0 \end{pmatrix}^2= \begin{pmatrix} 4x+3xy+1 & 4x+3y+2 &6x+3 \\ 2x^2+x & 2x+3xy &3x \\ xy & 2xy &3xy \end{pmatrix}

A→aBb↓↑cC=D\begin{CD} A @>a>> B \\ @VbVV @AAcA \\ C @= D \end{CD}

lim⁡x→0sin⁡x−xx3=−16\lim\limits_{x\rightarrow 0}\frac{\sin x -x}{x^3}=-\frac{1}{6}

∭Edx dy dz\iiint\limits_E dx\,dy\,dz

∫(x,y)∈Cx3 dx+4y2 dy\int_{(x,y)\in C} x^3\, dx + 4y^2\, dy

∮(x,y)∈Cx3 dx+4y2 dy\oint_{(x,y)\in C} x^3\, dx + 4y^2\, dy

${}$ 和$${}$$的区别

${}$
∫−12f(x)dx\int_{-1}^{2}f(x)dx

$${}$$

∫−12f(x)dx\int_{-1}^{2}f(x)dx

代码测试

print()函数

1
print 'Hellow world!'

使用 Ctrl+Alt+Del 重启电脑

代码块

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
from time import process_time
def gaussian_prime_count(n):
   """Prepare a lookup table to allow us to call P[x] to find number of gaussian primes <=x
   For any x that can be reached as n//i for integer i (returned in reverse order as V)"""
   r = int(n**0.5)
   ps = prime_sieve(r+1)
   assert r*r <= n and (r+1)**2 > n
   V = [n//i for i in range(1,r+1)]
   V += list(range(V[-1]-1,0,-1))  # V is interesting positions in descending order
   S1 = {i:(i-1)//4 for i in V} # Count of integers up to 2..i  of form 4k+1 after sieving primes up to p
   S2 = {i:max(0,(i+1)//4) for i in V} # Count of integers up to 2..i of form 4k+3 after sieving primes up to p
   for i,p in enumerate(ps[1:]): # Skip 2
         p2 = p*p
         if p%4==1:
            sp1 = S1[p-1]  # count of primes smaller than p
            sp2 = S2[p-1]
            for v in V:
               if v < p2: break
               S1[v] -= S1[v//p] - sp1
               S2[v] -= S2[v//p] - sp2
         else:
            sp1 = S1[p-1]  # count of primes smaller than p
            sp2 = S2[p-1]
            for v in V:
               if v < p2: break
               S1[v] -= S2[v//p] - sp2
               S2[v] -= S1[v//p] - sp1   
   return S1[n],S2[n]

def prime_sieve(n):
    sieve = [True] * (n//2)
    for i in range(3,int(n**0.5)+1,2):
        if sieve[i//2]:
            sieve[i*i//2::i] = [False] * ((n-i*i-1)//(2*i)+1)
    return [2] + [2*i+1 for i in range(1,n//2) if sieve[i]]
    
def ss(n):
    return gaussian_prime_count(n)
    
s=process_time()
a=ss(17179869184)
b=ss(8589934592)
print('4k+1:',a[0]-b[0])
print('4k+3:',a[1]-b[1])
print(process_time()-s,'secs')

图片测试

风景

其他

This is Tab 1.

This is Tab 2.

This is Tab 3.

列表

dasd sdasda sdad
1 2 3
4 5 5
6 7 8

臣亮言：先帝 创业未半，而中道崩殂 。今天下三分，益州疲敝 ，此诚危急存亡之秋 也！然侍衞之臣，不懈于内；忠志之士 ，忘身于外者，盖追先帝之殊遇，欲报之于陛下也。诚宜开张圣听，以光先帝遗德，恢弘志士之气；不宜妄自菲薄，引喻失义，以塞忠谏之路也。
宫中、府中，俱为一体；陟罚臧否，不宜异同。若有作奸 、犯科 ，及为忠善者，宜付有司，论其刑赏，以昭陛下平明之治；不宜偏私，使内外异法也。