LaTeX测试

首先输出个\(\LaTeX\ \)，看上去非常高端！

然后上论文，测试以后发现不行QAQQQ

貌似只能插入一个公式来着。。。比如：$\theta(\vec{u},\ \vec{v}) = arccos(\frac {\vec{u} \cdot \vec{v}} {|\vec{u}| \ |\vec{v}|})$

不过貌似加载公式挺快的！恩恩恩

以后就有写题解更简单了！赞

$$y = \frac{g} {-2 * cos^{2}\theta * v_{0}^ {2}} * x ^ {2} + tan\theta * x$$

$$A = \frac{y \ + \frac{g} {v0 ^ {2}}* x ^ {2}} {\sqrt{x ^{2} + y ^{2}}}$$

$$\phi = arctan \frac {y} {x}$$

$$\theta = \frac{arcsin A + \phi} {2} or \frac {\pi - arcsin A + \phi} {2}$$

$$\begin{align} ans &= ans \\\end{align}$$

$$\begin{align} ans &= \sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{d|i, d|j}\mu(d)\lfloor \frac{n}{i} \rfloor \lfloor \frac{m}{j} \rfloor \\&= \sum_{d=1}^{n}\sum_{i=1}^{n/d}\sum_{j=1}^{m/d}\mu(d) \lfloor \frac{n}{id} \rfloor \lfloor \frac{m}{jd} \rfloor ~~~~~(*) \\&= \sum_{d=1}^{n} \mu(d) (\sum_{i=1}^{n/d} \lfloor \frac{n}{id} \rfloor \times \sum_{j=1}^{m/d} \lfloor \frac{m}{jd} \rfloor) \\&= \sum_{d=1}^{n} \mu(d) f(\frac{n}{d}) f(\frac{m}{d}) \\ \end{align}$$

$Solution 1 (9\ points):\\We\ try\ all\ possible\ n\ values (starting\ from\ smallest - 1).\\With\ each\ n,\ we\ check\ if\ it\ is\ the\ answer\ using\ brute-force -\ checking\ if\ all\ sequence\ numbers\ contains\ required\ digits.\\Time\ complexity:\ o(N*K).\\N -\ correct\ answer.\\Since\ k <= 1000\ and\ n <= 1000\ in\ the\ first\ subtask,\ this\ solution\ is\ enough\ for\ it.\\Solution 2 (25\ points):\\We\ are\ assuming\ that\ all\ elements\ of\ the\ given\ sequence\ are\ equal (subtask 3).\\Lets\ donate\ this\ digit\ as\ d.\\If 1 <=\ d <= 8,\ than\ our\ answer\ n =\ d*10^x =\ d0..0.\\If\ d = 9,\ than\ n = 8..89 (some\ number\ of\ digits '8'\ and\ one\ digit '9'\ at\ the\ end).\\If\ d = 0,\ than\ n = 10^x = 10..0.\\The\ length\ of\ n\ depends\ on\ k.\\For\ example,\ if\ k = 5\ and\ our\ sequence\ is '3 3 3 3 3',\ than\ d = 3\ and\ n =\ d0..0 = 30..0 = 30.\\If\ k = 11\ and\ d = 3,\ than\ n =\ d0..0 = 30..0 = 300.\\Time\ complexity:\ o(1).\\Solution 3 (100\ points):\\We'll\ solve\ our\ task\ by\ solving\ a\ more\ complex\ task\ first:\ for\ each\ i-th\ sequence\ element\ we'll\ declare\ sequence\ a_i.\\A_i\ is\ the\ sequence\ of\ digits\ which\ i-th\ sequence\ element\ has\ to\ have.\\For\ our\ initial\ task,\ a_i = {d_i},\ where\ d_i - ...\\Lets\ donate\ n = (X)y,\ where y =\ n\ mod 10\ is\ the\ last\ digit\ and\ x =\ n\ div 10.\\Now\ we\ have\ to\ try\ all\ possible y\ values (0, 1, ..., 9).\\After\ we\ have\ locked y\ value\ with\ one\ of\ possible\ values,\ our\ sequence\ looks\ like\ this: (X)y, (X)y+1, ..., (X)8, (X)9, (X+1)0, (X+1)1 ..., (X+1)8, (X+1)9, (X+2)0, ...\\After\ removing\ last\ digit,\ we\ get\ sequence\ x, ...,\ x,\ x+1, ...,\ x+1,\ x+2, ...\\What\ digits\ does\ x\ has\ to\ have?\\(X)y\ has\ a_1,\ so\ x\ must\ have\ a_1 \ {y}. (X)(y+1)\ has\ a_2,\ so\ x\ must\ have\ a_2 \ {y+1}\ and\ so\ on.\\By\ merging\ these\ requirements,\ we\ can\ get\ a\ new\ sequence\ of\ sets\ b_1,\ b_2, ...\\B_1\ declares\ required\ digits\ for\ x,\ b_2 -\ for (X+1)\ and\ so\ on.\\We\ repeat\ the\ same\ steps\ with\ our\ new\ sequence.\\What\ happens\ when\ we\ repeat\ these\ steps?\\We\ started\ with\ sequence\ length\ k,\ so\ after\ first\ step\ our\ new\ sequence\ length\ will\ be\ not\ bigger\ than [K/10]+1.\\So\ after\ log\ k\ steps\ our\ sequence\ will\ be\ of\ length 1\ or 2 -\ at\ this\ step\ we\ can\ produce\ the\ answer.\\Time\ complexity:\ o(K*log\ k)\\Solution 4 (42\ points):\\If\ the\ previous\ solution\ is\ programmed\ inefficiently,\ it\ is\ possible\ to\ acquire\ an\ o(K^2)\ solution.\\Time\ complexity:\ o(K^2).$