Update 1486. 数组异或操作.md

杨世超 · 杨世超 · commit 664351ce67e7 · 2022-04-08T15:34:05.000+08:00
diff --git a/Solutions/1486. 数组异或操作.md b/Solutions/1486. 数组异或操作.md
@@ -35,11 +35,11 @@
 
 就变为了计算前 n 项序列的异或值。假设我们定义一个函数 sumXor(x) 用于计算前 n 项数的异或结果，通过观察可得出：
 
-$sumXor(x) = \left\{ \begin{array} \ x, & x = 4i, k \in Z \\ (x-1) \oplus x, & x = 4i+1, k \in Z \\ (x-2) \oplus (x-1) \oplus x, & x = 4i+2, k \in Z \\ (x-3) \oplus (x-2) \oplus (x-3) \oplus x, & x = 4i+3, k \in Z \end{array} \right.$
+$sumXor(x) = \left\{ \begin{array} \ x, & x = 4i, k \in Z \cr (x-1) \oplus x, & x = 4i+1, k \in Z \cr (x-2) \oplus (x-1) \oplus x, & x = 4i+2, k \in Z \cr (x-3) \oplus (x-2) \oplus (x-3) \oplus x, & x = 4i+3, k \in Z \end{array} \right.$
 
 继续化简得：
 
-$sumXor(x) = \left\{ \begin{array} \ x, & x = 4i, k \in Z \\ 1, & x = 4i+1, k \in Z \\ x+1, & x = 4i+2, k \in Z \\ 0, & x = 4i+3, k \in Z \end{array} \right.$
+$sumXor(x) = \left\{ \begin{array} \ x, & x = 4i, k \in Z \cr 1, & x = 4i+1, k \in Z \cr x+1, & x = 4i+2, k \in Z \cr 0, & x = 4i+3, k \in Z \end{array} \right.$
 
 则最终结果为 $sumXor(s-1) \oplus sumXor(s+n-1) * 2 + e$。
 
