[BUG]:latex公式里的=号被识别为-号

[loowace]

Describe the bug

左边是typora的显示图被正确显示为=号，中间是源码，右边是yn的显示图

To Reproduce
用yn源码模式输入这些源码就能看到

10.5. SOFTWARE PIPELINING 755

$$
\begin{aligned}
& main() \{
\\
& \quad E' = \{e | e \text{ in } E, \delta_e = 0\}; \\
& \quad T_0 = \max \left( \max_j \left\lceil \frac{\sum_{n,i} RT_n(i,j)}{r_j} \right\rceil, \max_{c \text{ acyclein } G} \left\lceil \frac{\sum_{e \text{ in } c} d_e}{\delta_e} \right\rceil \right); \\
& \quad \text{for (} T = T_0, T_0 + 1, \ldots \text{ or until all SCC's in } G \text{ are scheduled)} \{
\\
& \quad \quad RT = \text{an empty reservation table with } T \text{ rows;} \\
& \quad \quad E^* = \text{AllPairsLongestPath}(G, T); \\
& \quad \quad \text{for (each SCC } C \text{ in } G \text{ in prioritized topological order)} \{
\\
& \quad \quad \quad \text{for (all } n \text{ in } C) \\
& \quad \quad \quad \quad s_0(n) = \max_{e=p\to n} \in E^*, p \text{ scheduled } (S(p) + d_e); \\
& \quad \quad \quad \quad \text{first = some } n \text{ such that } s_0(n) \text{ is a minimum;} \\
& \quad \quad \quad \quad s_0 = s_0(\text{first}); \\
& \quad \quad \quad \quad \text{for (} s = s_0; s < s_0 + T; s = s+1) \\
& \quad \quad \quad \quad \quad \text{if (SccScheduled (RT, T, C, first,s)) break;} \\
& \quad \quad \quad \quad \quad \text{if (C cannot be scheduled in RT) break;}
\\
& \quad \} \\
\} \\
\end{aligned}
$$

$$
\begin{aligned}
& \text{SccScheduled}(RT, T, c, \text{first}, s) \{
\\
& \quad RT' = RT; \\
& \quad \text{if (not NodeScheduled (RT', T, first, s)) return false;} \\
& \quad \text{for (each remaining } n \text{ in } c \text{ in prioritized} \\
& \quad \quad \text{topological order of edges in } E') \{
\\
& \quad \quad s_l = \max_{e=n'\to n} \in E^*, n' \text{ in } c, n' \text{ scheduled } S(n') + d_e - (\delta_e \times T); \\
& \quad \quad s_u = \min_{e=n\to n'} \in E^*, n' \text{ in } c, n' \text{ scheduled } S(n') - d_e + (\delta_e \times T); \\
& \quad \quad \text{for (} s = s_l; \leq \min(s_u, s_l + T - 1); s = s+1) \\
& \quad \quad \quad \text{if NodeScheduled(RT', T, n, s) break;} \\
& \quad \quad \quad \text{if (n cannot be scheduled in RT') return false;}
\\
\} \\
& \quad RT = RT'; \\
& \quad \text{return true;} \\
\} \\
\end{aligned}
$$

Figure 10.29: A software-pipelining algorithm for cyclic dependence graphs

-->

Expected behavior

Screenshots

Desktop (please complete the following information):

• Version [e.g. v3.23.0]
• OS: [e.g. macOS]
• Browser [e.g. chrome, safari]

Additional context