diff --git a/Algorithms/1017. Convert to Base -2/1017. Convert to Base -2.go b/Algorithms/1017. Convert to Base -2/1017. Convert to Base -2.go
new file mode 100644
index 00000000..4821baa1
--- /dev/null
+++ b/Algorithms/1017. Convert to Base -2/1017. Convert to Base -2.go	
@@ -0,0 +1,20 @@
+package leetcode
+
+import "strconv"
+
+func baseNeg2(N int) string {
+	if N == 0 {
+		return "0"
+	}
+	res := ""
+	for N != 0 {
+		remainder := N % (-2)
+		N = N / (-2)
+		if remainder < 0 {
+			remainder += 2
+			N++
+		}
+		res = strconv.Itoa(remainder) + res
+	}
+	return res
+}
diff --git a/Algorithms/1017. Convert to Base -2/1017. Convert to Base -2_test.go b/Algorithms/1017. Convert to Base -2/1017. Convert to Base -2_test.go
new file mode 100644
index 00000000..f016c590
--- /dev/null
+++ b/Algorithms/1017. Convert to Base -2/1017. Convert to Base -2_test.go	
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1017 struct {
+	para1017
+	ans1017
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1017 struct {
+	one int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1017 struct {
+	one string
+}
+
+func Test_Problem1017(t *testing.T) {
+
+	qs := []question1017{
+
+		question1017{
+			para1017{2},
+			ans1017{"110"},
+		},
+
+		question1017{
+			para1017{3},
+			ans1017{"111"},
+		},
+
+		question1017{
+			para1017{4},
+			ans1017{"110"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1017------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1017, q.para1017
+		fmt.Printf("【input】:%v       【output】:%v\n", p, baseNeg2(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/1017. Convert to Base -2/README.md b/Algorithms/1017. Convert to Base -2/README.md
new file mode 100755
index 00000000..7c4bbfdd
--- /dev/null
+++ b/Algorithms/1017. Convert to Base -2/README.md	
@@ -0,0 +1,47 @@
+# [1017. Convert to Base -2](https://leetcode.com/problems/convert-to-base-2/)
+
+
+## 题目:
+
+Given a number `N`, return a string consisting of `"0"`s and `"1"`s that represents its value in base **`-2`** (negative two).
+
+The returned string must have no leading zeroes, unless the string is `"0"`.
+
+**Example 1:**
+
+    Input: 2
+    Output: "110"
+    Explantion: (-2) ^ 2 + (-2) ^ 1 = 2
+
+**Example 2:**
+
+    Input: 3
+    Output: "111"
+    Explantion: (-2) ^ 2 + (-2) ^ 1 + (-2) ^ 0 = 3
+
+**Example 3:**
+
+    Input: 4
+    Output: "100"
+    Explantion: (-2) ^ 2 = 4
+
+**Note:**
+
+1. `0 <= N <= 10^9`
+
+
+## 题目大意
+
+给出数字 N，返回由若干 "0" 和 "1"组成的字符串，该字符串为 N 的负二进制（base -2）表示。除非字符串就是 "0"，否则返回的字符串中不能含有前导零。
+
+提示：
+
+- 0 <= N <= 10^9
+
+
+
+## 解题思路
+
+- 给出一个十进制的数，要求转换成 -2 进制的数
+- 这一题仿造十进制转二进制的思路，短除法即可。
+