diff --git a/website/content/ChapterFour/1000~1099/1009.Complement-of-Base-10-Integer.md b/website/content/ChapterFour/1000~1099/1009.Complement-of-Base-10-Integer.md
new file mode 100644
index 00000000..b448d6c2
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+++ b/website/content/ChapterFour/1000~1099/1009.Complement-of-Base-10-Integer.md
@@ -0,0 +1,74 @@
+# [1009. Complement of Base 10 Integer](https://leetcode.com/problems/complement-of-base-10-integer/)
+
+
+## 题目
+
+The **complement** of an integer is the integer you get when you flip all the `0`'s to `1`'s and all the `1`'s to `0`'s in its binary representation.
+
+- For example, The integer `5` is `"101"` in binary and its **complement** is `"010"` which is the integer `2`.
+
+Given an integer `n`, return *its complement*.
+
+**Example 1:**
+
+```
+Input: n = 5
+Output: 2
+Explanation: 5 is "101" in binary, with complement "010" in binary, which is 2 in base-10.
+
+```
+
+**Example 2:**
+
+```
+Input: n = 7
+Output: 0
+Explanation: 7 is "111" in binary, with complement "000" in binary, which is 0 in base-10.
+
+```
+
+**Example 3:**
+
+```
+Input: n = 10
+Output: 5
+Explanation: 10 is "1010" in binary, with complement "0101" in binary, which is 5 in base-10.
+
+```
+
+**Constraints:**
+
+- `0 <= n < 109`
+
+## 题目大意
+
+每个非负整数 N 都有其二进制表示。例如， 5 可以被表示为二进制 "101"，11 可以用二进制 "1011" 表示，依此类推。注意，除 N = 0 外，任何二进制表示中都不含前导零。
+
+二进制的反码表示是将每个 1 改为 0 且每个 0 变为 1。例如，二进制数 "101" 的二进制反码为 "010"。
+
+给你一个十进制数 N，请你返回其二进制表示的反码所对应的十进制整数。
+
+## 解题思路
+
+- 简单题。求一个十进制数的反码，只需要让该数和全 1 的数进行异或计算即可。所以本题重点在如何构造 mask 上。
+
+## 代码
+
+```go
+package leetcode
+
+func bitwiseComplement(n int) int {
+ mask := 1
+ for mask < n {
+ mask = (mask << 1) + 1
+ }
+ return mask ^ n
+}
+```
+
+
+----------------------------------------------
+
diff --git a/website/content/ChapterFour/1000~1099/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60.md b/website/content/ChapterFour/1000~1099/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60.md
new file mode 100644
index 00000000..b8edb0eb
--- /dev/null
+++ b/website/content/ChapterFour/1000~1099/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60.md
@@ -0,0 +1,70 @@
+# [1010. Pairs of Songs With Total Durations Divisible by 60](https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/)
+
+
+## 题目
+
+You are given a list of songs where the ith song has a duration of `time[i]` seconds.
+
+Return *the number of pairs of songs for which their total duration in seconds is divisible by* `60`. Formally, we want the number of indices `i`, `j` such that `i < j` with `(time[i] + time[j]) % 60 == 0`.
+
+**Example 1:**
+
+```
+Input: time = [30,20,150,100,40]
+Output: 3
+Explanation: Three pairs have a total duration divisible by 60:
+(time[0] = 30, time[2] = 150): total duration 180
+(time[1] = 20, time[3] = 100): total duration 120
+(time[1] = 20, time[4] = 40): total duration 60
+
+```
+
+**Example 2:**
+
+```
+Input: time = [60,60,60]
+Output: 3
+Explanation: All three pairs have a total duration of 120, which is divisible by 60.
+
+```
+
+**Constraints:**
+
+- `1 <= time.length <= 6 * 104`
+- `1 <= time[i] <= 500`
+
+## 题目大意
+
+在歌曲列表中，第 i 首歌曲的持续时间为 time[i] 秒。
+
+返回其总持续时间（以秒为单位）可被 60 整除的歌曲对的数量。形式上，我们希望下标数字 i 和 j 满足 i < j 且有 (time[i] + time[j]) % 60 == 0。
+
+## 解题思路
+
+- 简单题。先将数组每个元素对 60 取余，将它们都转换到 [0,59] 之间。然后在数组中找两两元素之和等于 60 的数对。可以在 0-30 之内对半查找符合条件的数对。对 0 和 30 单独计算。因为多个 0 相加，余数还为 0 。2 个 30 相加之和为 60。
+
+## 代码
+
+```go
+func numPairsDivisibleBy60(time []int) int {
+ counts := make([]int, 60)
+ for _, v := range time {
+ v %= 60
+ counts[v]++
+ }
+ res := 0
+ for i := 1; i < len(counts)/2; i++ {
+ res += counts[i] * counts[60-i]
+ }
+ res += (counts[0] * (counts[0] - 1)) / 2
+ res += (counts[30] * (counts[30] - 1)) / 2
+ return res
+}
+```
+
+
+----------------------------------------------
+
diff --git a/website/content/ChapterFour/1000~1099/1011.Capacity-To-Ship-Packages-Within-D-Days.md b/website/content/ChapterFour/1000~1099/1011.Capacity-To-Ship-Packages-Within-D-Days.md
index 8cd28a6c..c812630a 100755
--- a/website/content/ChapterFour/1000~1099/1011.Capacity-To-Ship-Packages-Within-D-Days.md
+++ b/website/content/ChapterFour/1000~1099/1011.Capacity-To-Ship-Packages-Within-D-Days.md
@@ -117,6 +117,6 @@ func calSum(mid, m int, nums []int) bool {
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diff --git a/website/content/ChapterFour/1200~1299/1295.Find-Numbers-with-Even-Number-of-Digits.md b/website/content/ChapterFour/1200~1299/1295.Find-Numbers-with-Even-Number-of-Digits.md
index 17d3b6ba..f9e0d8f3 100644
--- a/website/content/ChapterFour/1200~1299/1295.Find-Numbers-with-Even-Number-of-Digits.md
+++ b/website/content/ChapterFour/1200~1299/1295.Find-Numbers-with-Even-Number-of-Digits.md
@@ -66,5 +66,5 @@ func findNumbers(nums []int) int {
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