Codeforces: MathJax в постах и комментариях

5 лет назад, # |

+36

$$$\begin{matrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{matrix}$$$

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TimonKnigge

5 лет назад, # |

← Rev. 2 →

+13

$$$\int_G f(g) \, \text{d}g = \frac{1}{|W|} \int_G \int_T \mathopen| \Delta(t) \mathclose|^2 f(gtg^{-1}) \, \text{d}t \, \text{d}g $$$

Looks very pretty :)

$$$1+2=3<4$$$

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oToToT

5 лет назад, # |

+57

$$$ \huge \mathcal{O}( N ) $$$

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5 лет назад, # |

+71

Testing

$$$\mathrm{f}[u]=b_u+\min\limits_{\substack{v\in anc_u\\\mathrm{dis}[u]-\mathrm{dis}[v]\le l_u}}(\mathrm{f}[v]-\mathrm{dis}[v]\times p_u)$$$

If $$$f(x)=\frac{1}{1-x^k}$$$, then $$$g(x)=(\ln f)(x)=\sum_{i=1}^{\infty}\frac{1}{i}x^{ik}$$$. Proof:

$$$\begin{aligned}g(x)&=\ln f(x)\\&=\int(\frac{\mathrm{d}}{\mathrm{d}x}\ln f)(x)\mathrm{d}x\\&=\int(\frac{f'(x)}{f(x)})\mathrm{d}x\\&=\int((1-x^k)f'(x))\mathrm{d}x\\&=\int((1-x^k)\sum_{i=1}^{\infty}k\cdot i\cdot x^{ki-1})\mathrm{d}x\\&=\int(\sum_{i=1}^{\infty}k\cdot i\cdot x^{ki-1}-\sum_{i=1}^{\infty}k\cdot i\cdot x^{ki-1}\cdot x^k)\mathrm{d}x\\&=\int(\sum_{i=1}^{\infty}k\cdot i\cdot x^{ki-1}-\sum_{i=1}^{\infty}k\cdot (i-1)\cdot x^{ki-1})\mathrm{d}x\\&=\int(\sum_{i=1}^{\infty}k\cdot x^{ki-1})\mathrm{d}x\\&=\sum_{i=1}^{\infty}\frac{1}{i}x^{ki}\end{aligned}$$$

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dmkz

5 лет назад, # ^ |

← Rev. 2 →

+73

I think, that you forgot to try \left( instead ( and \right) instead ). This allows to auto detect correct height of parentheses under summation, like this:

$$$\prod_{k=1}^{n-1} \left(\exp{\left(\dfrac{2 \cdot \pi \cdot i \cdot k}{n}\right)}-1\right) = \left(-1\right)^{n+1} \cdot n$$$

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5 лет назад, # ^ |

Yeah, you're right. But sometimes it's too complicated.

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_bom_bom_

5 лет назад, # |

Can explain me why backward compatibility is very important?

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Jhin

5 лет назад, # |

← Rev. 4 →

Deleted and screw uplaoding a photo

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aryaman

5 лет назад, # |

$$$\text{summitwei} \in \text{geniosity}$$$

Proof left as an exercise to the reader.

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fishy15

5 лет назад, # ^ |

$$$\text{fishy} \not\in \text{geniosity}$$$

I have already proved this B).

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nchn27

5 лет назад, # ^ |

hello, my platinum blue men

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aryaman

5 лет назад, # ^ |

nchn orz

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vibrant1

5 лет назад, # |

$$$a^2$$$

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Miraak

5 лет назад, # |

+42

$$$ы^2$$$

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algmyr

5 лет назад, # |

+227

Nice, Now I can draw beautiful owls in my posts. Thanks Mike!

$$$\underline{\widehat{\dbinom{\odot_\vee\odot}{{\raise-8pt"}\wr{\raise-8pt"}}}}$$$

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shanxizeng

5 лет назад, # ^ |

← Rev. 2 →

+36

LOL

$$$ \widetilde{ \dbinom{\odot_\vee\odot} {\underline{\ \smile\ }}} $$$ $$$ \widetilde{ \dbinom{\odot_\vee\odot} {\underline{\ -\ }}} $$$ $$$ \widetilde{ \dbinom{\odot_\vee\odot} {\underline{\ \frown\ }}} $$$

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enoone

5 лет назад, # |

← Rev. 4 →

$$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$$

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dacin21

5 лет назад, # |

← Rev. 3 →

+117

$$$\require{AMScd}$$$ \begin{CD} T @>h>> A\\ @. @VV n V\\ M @<<s< K\\ @VV iV @.\\ K @>>e> !\\ \end{CD}

$$$ \frac{\color{purple}{4}}{\color{blue}{\Pi}} = \color{green}{1} + \cfrac{\color{red}{1}}{\color{orange}{2} + \cfrac{\color{red}{9}}{\color{orange}{2} + \cfrac{\color{red}{25}}{\color{orange}{2} + \cfrac{\color{red}{49}}{\color{orange}{2} + \dots}}}} $$$

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do_good_

5 лет назад, # ^ |

red coder maths r very strong

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FlyInTheSky

5 лет назад, # |

← Rev. 6 →

+24

$$$ \displaystyle \sum_{i=1}^n \sum_{j=1}^m d_1(\gcd(i, j)) \\ \displaystyle\sum_{d=1}^n \sum_{i=1}^n \sum_{j=1}^m d_1(d)[\gcd(i,j)=d])\\ \displaystyle\sum_{d=1}^n d_1(d) \sum_{i=1}^n \sum_{j=1}^m [\gcd(i,j)=d])\\ \displaystyle\sum_{d=1}^n d_1(d) \sum_{k=1}^{\lfloor \frac nd \rfloor} \mu(k) \lfloor \frac n{kd} \rfloor \displaystyle\lfloor \frac m{kd} \rfloor \\ $$$

Let $$$T=kd$$$,

$$$ \displaystyle\sum_{T=1}^n \lfloor \frac nT \rfloor \lfloor \frac mT \rfloor \sum_{d|T}d_1(d) \mu(\frac Td) $$$ $$$ \displaystyle dp(i)=\min\limits_{i < j \leq n, dp(j)\leq \operatorname{sum}(j - 1) - \operatorname{sum}(i - 1)}(\operatorname{sum}(j - 1) - \operatorname{sum}(i - 1)) $$$

Why the \sum became \prod?....

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shanxizeng

5 лет назад, # ^ |

Maybe you can use..

$$$\Sigma_{i=1}^n \Sigma_{j=1}^m d_1(\gcd(i, j))$$$

\Sigma_{i=1}^n \Sigma_{j=1}^m d_1(\gcd(i, j))

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FlyInTheSky

5 лет назад, # ^ |

← Rev. 2 →

It seems that it was fixed now, also thank you...

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shanxizeng

5 лет назад, # ^ |

I'm just crack a joke,if you use the \Sigma,it will be too small...

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5 лет назад, # |

+40

$$$a^{n}\equiv a^{\varphi(m)+(n \bmod \varphi(m))}\pmod{m}$$$

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ouuan

5 лет назад, # ^ |

+62

It is not always true when $$$n<\varphi(m)$$$. For example, $$$6^{2}\not\equiv 6^{4+2}\pmod8$$$.

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5 лет назад, # ^ |

+45

It's true for $$$n>\log_2 m$$$

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5 лет назад, # ^ |

+27

More accurately, it's true for $$$n\ge\max_{i=1}^k\alpha_i$$$, where $$$m=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$$$.

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5 лет назад, # ^ |

what's the theory behind the fomula plz

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5 лет назад, # ^ |

+21

Let $$$m=p_1^{d_1} \dots p_n^{d_n}$$$. Consider the periodicity of $$$a^n$$$ on each $$$p_k^{d_k}$$$ separately. By Chinese Remainder Theorem, pre-period of $$$a^n$$$ modulo $$$m$$$ will be equal to the maximum of pre-periods modulo $$$p_k^{d_k}$$$ and period itself will be least common multiple of periods modulo $$$p_k^{d_k}$$$.

On each $$$p_k^{d_k}$$$ pre-period will be either $$$0$$$ if $$$\gcd(a,p)=1$$$ or at most $$$d_k \leq \log_2 m$$$ otherwise. And period will always be the divisor of $$$\varphi(p_k^{d_k})$$$ which is in turn the divisor of $$$\varphi(m)$$$. Thus their least common multiple will be the divisor of $$$\varphi(m)$$$. This gives you the result above, given that $$$\varphi(m) \geq \log_2 m$$$.

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5 лет назад, # ^ |

thank you vvery much. seems that I have a lot of things to catch up with

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blinkshot

5 лет назад, # |

-10

I don't care. I want another codeforces round asap.

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VLADO

5 лет назад, # ^ |

You can have hundreds of virtual contests.

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im0qianqian

5 лет назад, # |

← Rev. 2 →

+22

WoW, it's cool! Thank you~

It was finally realized. A question about using LaTeX in blogs

$$$ \begin{bmatrix} F_i \\ F_{i-1} \\ (i+1)^3 \\ (i+1)^2 \\ i+1 \\ 1 \end{bmatrix} \Longleftarrow \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 3 & 3 & 1 \\ 0 & 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix} \times \begin{bmatrix} F_{i-1} \\ F_{i-2} \\ i^3 \\ i^2 \\ i \\ 1 \end{bmatrix} $$$

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ouuan

5 лет назад, # |

+44

The most convenient thing is that you can copy the formula by $$$\text{Right Click}\rightarrow\text{Show Math As}\rightarrow\text{TeX Commands}$$$.

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5 лет назад, # |

WAIT! Why all \sums became \prods? $$$\underset{\text{actually \sum}}{\underline{\sum}}\neq\underset{\text{real \prod}}{\underline{\prod}}$$$

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5 лет назад, # ^ |

Fixed!

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dacin21

5 лет назад, # |

← Rev. 2 →

+38

Looks like I found the following feature: If you use \newcommand or \renewcommand, it will affect all comments below yours. Apologies to all the people between my two comments that got confused by this.

Stackexchange had the same issue, but there some people could at least edit the answers/comments and remove the malicious code. Their fix (mentioned in an answer to the linked post) was to wrap posts in \begingroup and \endgroup. Another fix mentioned there was to use \resetstack. MikeMirzayanov could you look into that?

Edit: It looks like \newcommand and \renewcommand now get stripped from mathjax parts of comments. This should deal with people replacing $$$\pi$$$ with 40 owls for now. Nevertheless, I would love to see support for a \newcommand whose scope is just the comment it was written in. (But I see that implementing and testing that might take some time.)

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wucstdio

5 лет назад, # |

← Rev. 3 →

+12

$$$\begin{align*}\sum_{i=1}^n\sum_{j=1}^nij\gcd\left(i,j\right)&=\sum_{d=1}^nd\sum_{i=1}^n\sum_{j=1}^nij[\gcd\left(i,j\right)==d]\\&=\sum_{d=1}^nd^3\sum_{i=1}^{\lfloor\dfrac nd\rfloor}\sum_{j=1}^{\lfloor\dfrac nd\rfloor}ij[\gcd\left(i,j\right)==1]\\&=\sum_{d=1}^nd^3\sum_{i=1}^{\lfloor\dfrac nd \rfloor}\sum_{j=1}^{\lfloor\dfrac nd\rfloor}ij\sum_{t|\gcd\left(i,j\right)}\mu\left(t\right)\\&=\sum_{d=1}^nd^3\sum_{t=1}^{\lfloor\dfrac nd \rfloor}\mu\left(t\right)t^2\sum_{i=1}^{\lfloor\dfrac n{td}\rfloor}\sum_{j=1}^{\lfloor\dfrac n{td}\rfloor}ij\\&=\sum_{d=1}^nd^3\sum_{t=1}^{\lfloor\dfrac nd \rfloor}\mu\left(t\right)t^2\left(1+2+3+\cdots+\lfloor\dfrac n{td}\rfloor\right)^2\\&=\sum_{d=1}^nd^3\sum_{d|T}^n\mu\left(\dfrac Td\right)\left(\dfrac Td\right)^2sum^2\left(\lfloor\dfrac nT\rfloor\right)\\&=\sum_{T=1}^nsum^2\lfloor\dfrac nT\rfloor T^2\sum_{d|T}d\mu\left(\dfrac Td\right)\\&=\sum_{T=1}^nsum^2\lfloor\dfrac nT\rfloor T^2\phi\left(T\right)\end{align*}$$$

It is a great joy to solve mathematical problems using $$$LaTeX$$$ !

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....................

5 лет назад, # ^ |

-10

this is disgusting, genuinely

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_wisdom

5 лет назад, # |

+20

$$$gcd(Fib_{a},Fib_{b})=Fib_{gcd(a,b)}$$$

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TimonKnigge

5 лет назад, # ^ |

← Rev. 2 →

More generally, $$$F_n \equiv 0 \mod F_m$$$ if $$$n \equiv 0 \mod m$$$.

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5 лет назад, # ^ |

-8

Is this fomula true? What's this theroem called?

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DeeaX99

5 лет назад, # |

← Rev. 2 →

+21

\begin{equation*} e^{\pi i} + 1 = 0 \end{equation*}

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migeater

5 лет назад, # |

Well, Let's see some tables

$$$ \begin{array}{c|lcr} n & \text{Left} & \text{Center} & \text{Right} \\ \hline 1 & 0.24 & 1 & 125 \\ 2 & -1 & 189 & -8 \\ 3 & -20 & 2000 & 1+10i \end{array} $$$

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akidiekatze

5 лет назад, # |

-14

1. 1. it doesnt work

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mmello

5 лет назад, # |

← Rev. 6 →

-9

Testing

$$$\sin \pi = 0$$$

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mmello

5 лет назад, # ^ |

← Rev. 4 →

Testing a second time

$$$\sin \pi = 0$$$

Yep, dacin21's feature is still not fixed.

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cxkno1

5 лет назад, # |

$$$2019 \ good \ luck \ and \ high \ rating $$$ $$$2019 \ become \ master:D$$$

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SYCstudio

5 лет назад, # ^ |

Wow, I find a person use the same picture as me! orz.

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riela

5 лет назад, # |

$$$-\dfrac{\hslash^2}{2m} \, \dfrac{\mathrm{d}^2 \psi}{\mathrm{d} x^2}$$$

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riela

5 лет назад, # |

$$$-\dfrac{\hslash^2}{2m} \, \dfrac{\mathrm{d}^2 \psi}{\mathrm{d} x^2}$$$

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5 лет назад, # ^ |

Oh boi we should forbid posting Hamiltonians on codeforces...

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TimonKnigge

5 лет назад, # ^ |

$$$\begin{pmatrix}1&0\\0&-1\\\end{pmatrix}$$$

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riela

5 лет назад, # ^ |

+14

strange suggestion, coming from a string theorist...

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....................

5 лет назад, # |

-27

and what does this have to do on a programming website? this website is for competitive programming, not math.

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retired.old.cake

5 лет назад, # ^ |

Relax competitive programming is just maths + coding

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....................

5 лет назад, # ^ |

-14

do you realise what you're saying? just because some random website called codeforces decides to mess around with us by giving us terrible, useless problems full of math doesn't mean the whole thing called COMPETITIVE PROGRAMMING is like this.

read before you comment because people who don't know better will take what you say as being true.

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DKozl50

5 лет назад, # |

+28

\begin{CD} Me\\ @VV V\\ @>>> \\ solving\ problems @VVV @.\\ @<<< \\ @VVV\\ playing\ with\ LaTeX \end{CD}

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algmyr

5 лет назад, # ^ |

$$$\LaTeX$$$

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Chasingdreams

5 лет назад, # |

It's fantastic! $$$f(x)=g(y)+1$$$

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Retired_xryjr233

5 лет назад, # |

Great!It will make codeforces better. :D

$$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$$

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QAQAutoMaton

5 лет назад, # |

+30

$$$F(\omega_n^m)=F_0(\omega_{\frac{n}{2}}^m)+\omega_n^mF_1(\omega_{\frac{n}{2}}^m)$$$ $$$F(\omega_n^{m+\frac{n}{2}})=F_0(\omega_{\frac{n}{2}}^m)+\omega_n^{m+\frac{n}{2}}F_1(\omega_{\frac{n}{2}}^m)=F_0(\omega_{\frac{n}{2}}^m)-\omega_n^{m}F_1(\omega_{\frac{n}{2}}^m)$$$

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XLor

5 лет назад, # |

$$$ c_i = \sum_{\gcd(j,k)=i} a_j \cdot b_k $$$

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Suraj_IOI

5 лет назад, # |

$$$2\sum_{n=1}^n n^3$$$

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StudyingFather

5 лет назад, # |

← Rev. 2 →

$$$ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $$$ $$$ \LaTeX $$$

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yPolarnova

5 лет назад, # |

$$$\begin{aligned}\left[ mx\right] =\sum ^{m}_{j=1}\left[ x+\dfrac {j-1}{m}\right] \left( m\in \mathbb{N} ^{+}\right) \end{aligned}$$$

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Hamza_Abusnaineh

5 лет назад, # |

That's what you love :D AhmadLoiy

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wangrq

5 лет назад, # |

← Rev. 4 →

$$$a_n=\sum_{k=0}^{n}\binom{n}{k}b_k\iff b_n=\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}a_k$$$ $$$\int_a^bf(x)\text{d}x\approx\frac{b-a}{6}(f(a)+4f(\frac{a+b}{2})+f(b))$$$

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Bilguunzen

5 лет назад, # |

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babin

5 лет назад, # |

Is it rated ?

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belal_bh

5 лет назад, # |

Does it support Markdown and MathJax in the same post/comment ?

Here I'm trying to test both of this:

Markdown & Latex Example

Codeforces is awesome!! Now let's try to write a matrix: Inline $$$\begin{bmatrix}a & b\ c & d\ e & f \end{bmatrix}$$$ and

Now in new line matrix:`

$$$\begin{bmatrix}a & b\\ c & d\\ e & f \end{bmatrix}$$$

Seems it doesn't work in inline form.

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