建议加强洛谷博客的LaTeX适应能力
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建议加强洛谷博客的LaTeX适应能力
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Chinese_zjc_楼主2020/9/29 11:11
$$
\begin{align}
sum&=(a+1)(b+1)\\
&=(a+1)(mid-a+1)\\
&=a\cdot mid-a^2+a+mid-a+1\\
&=-a^2+a\cdot mid+1+mid\\
&=-(a^2-a\cdot mid)+1+mid\\
&=-(a^2-a\cdot mid+\frac{mid^2}{4})+1+mid+\frac{mid^2}{4}\\
&=-(a-\frac{mid}{2})^2+1+mid+\frac{mid^2}{4}
\end{align}
$$

\begin{aligned}
sum&=(a+1)(b+1)\\
&=(a+1)(mid-a+1)\\
&=a\cdot mid-a^2+a+mid-a+1\\
&=-a^2+a\cdot mid+1+mid\\
&=-(a^2-a\cdot mid)+1+mid\\
&=-(a^2-a\cdot mid+\frac{mid^2}{4})+1+mid+\frac{mid^2}{4}\\
&=-(a-\frac{mid}{2})^2+1+mid+\frac{mid^2}{4}
\end{aligned}

这两个在 TyporaTypora 上是等价的,但是到了洛谷就只有第二个可以正常显示.

sum=(a+1)(b+1)=(a+1)(mid−a+1)=a⋅mid−a2+a+mid−a+1=−a2+a⋅mid+1+mid=−(a2−a⋅mid)+1+mid=−(a2−a⋅mid+mid24)+1+mid+mid24=−(a−mid2)2+1+mid+mid24\begin{aligned} sum&=(a+1)(b+1)\\ &=(a+1)(mid-a+1)\\ &=a\cdot mid-a^2+a+mid-a+1\\ &=-a^2+a\cdot mid+1+mid\\ &=-(a^2-a\cdot mid)+1+mid\\ &=-(a^2-a\cdot mid+\frac{mid^2}{4})+1+mid+\frac{mid^2}{4}\\ &=-(a-\frac{mid}{2})^2+1+mid+\frac{mid^2}{4} \end{aligned}

希望可以增强洛谷博客的适应能力(或许是洛谷用 KaTeXKaTeX 渲染的缘故?)

2020/9/29 11:11