diff --git a/solution/1500-1599/1524.Number of Sub-arrays With Odd Sum/README.md b/solution/1500-1599/1524.Number of Sub-arrays With Odd Sum/README.md
index 97fbd0514a02f..ade0242a435e8 100644
--- a/solution/1500-1599/1524.Number of Sub-arrays With Odd Sum/README.md	
+++ b/solution/1500-1599/1524.Number of Sub-arrays With Odd Sum/README.md	
@@ -80,15 +80,15 @@ tags:
 
 ### 方法一：前缀和 + 计数器
 
-我们定义一个长度为 $2$ 的数组 $cnt$ 作为计数器，其中 $cnt[0]$ 和 $cnt[1]$ 分别表示前缀和为偶数和奇数的子数组的个数。初始时 $cnt[0] = 1$，而 $cnt[1] = 0$。
+我们定义一个长度为 $2$ 的数组 $\textit{cnt}$ 作为计数器，其中 $\textit{cnt}[0]$ 和 $\textit{cnt}[1]$ 分别表示前缀和为偶数和奇数的子数组的个数。初始时 $\textit{cnt}[0] = 1$，而 $\textit{cnt}[1] = 0$。
 
 接下来，我们维护当前的前缀和 $s$，初始时 $s = 0$。
 
-遍历数组 $arr$，对于遍历到的每个元素 $x$，我们将 $s$ 的值加上 $x$，然后根据 $s$ 的奇偶性，将 $cnt[s \mod 2 \oplus 1]$ 的值累加到答案中，然后我们将 $cnt[s \mod 2]$ 的值加 $1$。
+遍历数组 $\textit{arr}$，对于遍历到的每个元素 $x$，我们将 $s$ 的值加上 $x$，然后根据 $s$ 的奇偶性，将 $\textit{cnt}[s \mod 2 \oplus 1]$ 的值累加到答案中，然后我们将 $\textit{cnt}[s \mod 2]$ 的值加 $1$。
 
 遍历结束后，我们即可得到答案。注意答案的取模运算。
 
-时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组 $arr$ 的长度。
+时间复杂度 $O(n)$，其中 $n$ 为数组 $\textit{arr}$ 的长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -177,6 +177,25 @@ function numOfSubarrays(arr: number[]): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn num_of_subarrays(arr: Vec<i32>) -> i32 {
+        const MOD: i32 = 1_000_000_007;
+        let mut cnt = [1, 0];
+        let mut ans = 0;
+        let mut s = 0;
+        for &x in arr.iter() {
+            s += x;
+            ans = (ans + cnt[((s & 1) ^ 1) as usize]) % MOD;
+            cnt[(s & 1) as usize] += 1;
+        }
+        ans
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->
diff --git a/solution/1500-1599/1524.Number of Sub-arrays With Odd Sum/README_EN.md b/solution/1500-1599/1524.Number of Sub-arrays With Odd Sum/README_EN.md
index 5cf6739dc03d8..901fd189ef2d8 100644
--- a/solution/1500-1599/1524.Number of Sub-arrays With Odd Sum/README_EN.md	
+++ b/solution/1500-1599/1524.Number of Sub-arrays With Odd Sum/README_EN.md	
@@ -67,7 +67,17 @@ All sub-arrays have even sum and the answer is 0.
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Prefix Sum + Counter
+
+We define an array $\textit{cnt}$ of length 2 as a counter, where $\textit{cnt}[0]$ and $\textit{cnt}[1]$ represent the number of subarrays with even and odd prefix sums, respectively. Initially, $\textit{cnt}[0] = 1$ and $\textit{cnt}[1] = 0$.
+
+Next, we maintain the current prefix sum $s$, initially $s = 0$.
+
+Traverse the array $\textit{arr}$, for each element $x$ encountered, add the value of $x$ to $s$, then based on the parity of $s$, add the value of $\textit{cnt}[s \mod 2 \oplus 1]$ to the answer, and then increment the value of $\textit{cnt}[s \mod 2]$ by 1.
+
+After the traversal, we get the answer. Note the modulo operation for the answer.
+
+Time complexity is $O(n)$, where $n$ is the length of the array $\textit{arr}$. Space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -156,6 +166,25 @@ function numOfSubarrays(arr: number[]): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn num_of_subarrays(arr: Vec<i32>) -> i32 {
+        const MOD: i32 = 1_000_000_007;
+        let mut cnt = [1, 0];
+        let mut ans = 0;
+        let mut s = 0;
+        for &x in arr.iter() {
+            s += x;
+            ans = (ans + cnt[((s & 1) ^ 1) as usize]) % MOD;
+            cnt[(s & 1) as usize] += 1;
+        }
+        ans
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->
diff --git a/solution/1500-1599/1524.Number of Sub-arrays With Odd Sum/Solution.rs b/solution/1500-1599/1524.Number of Sub-arrays With Odd Sum/Solution.rs
new file mode 100644
index 0000000000000..72c387ec66a85
--- /dev/null
+++ b/solution/1500-1599/1524.Number of Sub-arrays With Odd Sum/Solution.rs	
@@ -0,0 +1,14 @@
+impl Solution {
+    pub fn num_of_subarrays(arr: Vec<i32>) -> i32 {
+        const MOD: i32 = 1_000_000_007;
+        let mut cnt = [1, 0];
+        let mut ans = 0;
+        let mut s = 0;
+        for &x in arr.iter() {
+            s += x;
+            ans = (ans + cnt[((s & 1) ^ 1) as usize]) % MOD;
+            cnt[(s & 1) as usize] += 1;
+        }
+        ans
+    }
+}