Add solution 1018

leetcode/1018.Binary-Prefix-Divisible-By-5/1018. Binary Prefix Divisible By 5.go

@@ -0,0 +1,10 @@
+package leetcode
+
+func prefixesDivBy5(a []int) []bool {
+	res, num := make([]bool, len(a)), 0
+	for i, v := range a {
+		num = (num<<1 | v) % 5
+		res[i] = num == 0
+	}
+	return res
+}

leetcode/1018.Binary-Prefix-Divisible-By-5/1018. Binary Prefix Divisible By 5_test.go

@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1018 struct {
+	para1018
+	ans1018
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1018 struct {
+	a []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1018 struct {
+	one []bool
+}
+
+func Test_Problem1018(t *testing.T) {
+
+	qs := []question1018{
+
+		{
+			para1018{[]int{0, 1, 1}},
+			ans1018{[]bool{true, false, false}},
+		},
+
+		{
+			para1018{[]int{1, 1, 1}},
+			ans1018{[]bool{false, false, false}},
+		},
+
+		{
+			para1018{[]int{0, 1, 1, 1, 1, 1}},
+			ans1018{[]bool{true, false, false, false, true, false}},
+		},
+
+		{
+			para1018{[]int{1, 1, 1, 0, 1}},
+			ans1018{[]bool{false, false, false, false, false}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1018------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1018, q.para1018
+		fmt.Printf("【input】:%v       【output】:%v\n", p, prefixesDivBy5(p.a))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/1018.Binary-Prefix-Divisible-By-5/README.md

@@ -0,0 +1,70 @@
+# [1018. Binary Prefix Divisible By 5](https://leetcode.com/problems/binary-prefix-divisible-by-5/)
+
+
+## 题目
+
+Given an array `A` of `0`s and `1`s, consider `N_i`: the i-th subarray from `A[0]` to `A[i]` interpreted as a binary number (from most-significant-bit to least-significant-bit.)
+
+Return a list of booleans `answer`, where `answer[i]` is `true` if and only if `N_i` is divisible by 5.
+
+**Example 1:**
+
+```
+Input: [0,1,1]
+Output: [true,false,false]
+Explanation: 
+The input numbers in binary are 0, 01, 011; which are 0, 1, and 3 in base-10.  Only the first number is divisible by 5, so answer[0] is true.
+
+```
+
+**Example 2:**
+
+```
+Input: [1,1,1]
+Output: [false,false,false]
+
+```
+
+**Example 3:**
+
+```
+Input: [0,1,1,1,1,1]
+Output: [true,false,false,false,true,false]
+
+```
+
+**Example 4:**
+
+```
+Input: [1,1,1,0,1]
+Output: [false,false,false,false,false]
+
+```
+
+**Note:**
+
+1. `1 <= A.length <= 30000`
+2. `A[i]` is `0` or `1`
+
+## 题目大意
+
+给定由若干 0 和 1 组成的数组 A。我们定义 N_i：从 A[0] 到 A[i] 的第 i 个子数组被解释为一个二进制数（从最高有效位到最低有效位）。返回布尔值列表 answer，只有当 N_i 可以被 5 整除时，答案 answer[i] 为 true，否则为 false。
+
+## 解题思路
+
+- 简单题。每扫描数组中的一个数字，累计转换成二进制数对 5 取余，如果余数为 0，则存入 true，否则存入 false。
+
+## 代码
+
+```go
+package leetcode
+
+func prefixesDivBy5(a []int) []bool {
+	res, num := make([]bool, len(a)), 0
+	for i, v := range a {
+		num = (num<<1 | v) % 5
+		res[i] = num == 0
+	}
+	return res
+}
+```

website/content/ChapterFour/1018.Binary-Prefix-Divisible-By-5.md

@@ -0,0 +1,70 @@
+# [1018. Binary Prefix Divisible By 5](https://leetcode.com/problems/binary-prefix-divisible-by-5/)
+
+
+## 题目
+
+Given an array `A` of `0`s and `1`s, consider `N_i`: the i-th subarray from `A[0]` to `A[i]` interpreted as a binary number (from most-significant-bit to least-significant-bit.)
+
+Return a list of booleans `answer`, where `answer[i]` is `true` if and only if `N_i` is divisible by 5.
+
+**Example 1:**
+
+```
+Input: [0,1,1]
+Output: [true,false,false]
+Explanation: 
+The input numbers in binary are 0, 01, 011; which are 0, 1, and 3 in base-10.  Only the first number is divisible by 5, so answer[0] is true.
+
+```
+
+**Example 2:**
+
+```
+Input: [1,1,1]
+Output: [false,false,false]
+
+```
+
+**Example 3:**
+
+```
+Input: [0,1,1,1,1,1]
+Output: [true,false,false,false,true,false]
+
+```
+
+**Example 4:**
+
+```
+Input: [1,1,1,0,1]
+Output: [false,false,false,false,false]
+
+```
+
+**Note:**
+
+1. `1 <= A.length <= 30000`
+2. `A[i]` is `0` or `1`
+
+## 题目大意
+
+给定由若干 0 和 1 组成的数组 A。我们定义 N_i：从 A[0] 到 A[i] 的第 i 个子数组被解释为一个二进制数（从最高有效位到最低有效位）。返回布尔值列表 answer，只有当 N_i 可以被 5 整除时，答案 answer[i] 为 true，否则为 false。
+
+## 解题思路
+
+- 简单题。每扫描数组中的一个数字，累计转换成二进制数对 5 取余，如果余数为 0，则存入 true，否则存入 false。
+
+## 代码
+
+```go
+package leetcode
+
+func prefixesDivBy5(a []int) []bool {
+	res, num := make([]bool, len(a)), 0
+	for i, v := range a {
+		num = (num<<1 | v) % 5
+		res[i] = num == 0
+	}
+	return res
+}
+```

website/content/menu/index.md

@@ -483,6 +483,7 @@ headless: true
     - [1005.Maximize-Sum-Of-Array-After-K-Negations]({{< relref "/ChapterFour/1005.Maximize-Sum-Of-Array-After-K-Negations.md" >}})
     - [1011.Capacity-To-Ship-Packages-Within-D-Days]({{< relref "/ChapterFour/1011.Capacity-To-Ship-Packages-Within-D-Days.md" >}})
     - [1017.Convert-to-Base--2]({{< relref "/ChapterFour/1017.Convert-to-Base--2.md" >}})
+    - [1018.Binary-Prefix-Divisible-By-5]({{< relref "/ChapterFour/1018.Binary-Prefix-Divisible-By-5.md" >}})
     - [1019.Next-Greater-Node-In-Linked-List]({{< relref "/ChapterFour/1019.Next-Greater-Node-In-Linked-List.md" >}})
     - [1020.Number-of-Enclaves]({{< relref "/ChapterFour/1020.Number-of-Enclaves.md" >}})
     - [1021.Remove-Outermost-Parentheses]({{< relref "/ChapterFour/1021.Remove-Outermost-Parentheses.md" >}})