diff --git a/src/main/java/leetcode/editor/cn/FourXy4Wx.java b/src/main/java/leetcode/editor/cn/FourXy4Wx.java
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+//小力将 N 个零件的报价存于数组 `nums`。小力预算为 `target`，假定小力仅购买两个零件，要求购买零件的花费不超过预算，请问他有多少种采购方案。
+//
+//
+//注意：答案需要以 `1e9 + 7 (1000000007)` 为底取模，如：计算初始结果为：`1000000008`，请返回 `1`
+//
+//
+//**示例 1：**
+//>输入：`nums = [2,5,3,5], target = 6`
+//>
+//>输出：`1`
+//>
+//>解释：预算内仅能购买 nums[0] 与 nums[2]。
+//
+//**示例 2：**
+//>输入：`nums = [2,2,1,9], target = 10`
+//>
+//>输出：`4`
+//>
+//>解释：符合预算的采购方案如下：
+//>nums[0] + nums[1] = 4
+//>nums[0] + nums[2] = 3
+//>nums[1] + nums[2] = 3
+//>nums[2] + nums[3] = 10
+//
+//**提示：**
+//- `2 <= nums.length <= 10^5`
+//- `1 <= nums[i], target <= 10^5`
+// 👍 25 👎 0
+
+package leetcode.editor.cn;
+//LCP 28:采购方案
+public class FourXy4Wx{
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new FourXy4Wx().new Solution();
+    }
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+class Solution {
+    public int purchasePlans(int[] nums, int target) {
+        int[] sort = new int[target];
+        long[] count = new long[target];
+        for (int num : nums) {
+            if (num < target) {
+                sort[num] += 1;
+            }
+        }
+        long sum = 0;
+        for (int i = 1; i < target; i++) {
+            sum += sort[i];
+            count[i] = sum;
+        }
+        long result = 0;
+        for (int num : nums) {
+            if (target > num) {
+                result += num <= target - num ? count[target - num] - 1 : count[target - num];
+            }
+        }
+        return (int) (result / 2 % (Math.pow(10, 9) + 7));
+    }
+}
+//leetcode submit region end(Prohibit modification and deletion)
+
+}
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diff --git a/src/main/java/leetcode/editor/cn/FourXy4Wx.md b/src/main/java/leetcode/editor/cn/FourXy4Wx.md
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+小力将 N 个零件的报价存于数组 `nums`。小力预算为 `target`，假定小力仅购买两个零件，要求购买零件的花费不超过预算，请问他有多少种采购方案。
+
+注意：答案需要以 `1e9 + 7 (1000000007)` 为底取模，如：计算初始结果为：`1000000008`，请返回 `1`
+
+
+**示例 1：**
+>输入：`nums = [2,5,3,5], target = 6`
+>
+>输出：`1`
+>
+>解释：预算内仅能购买 nums[0] 与 nums[2]。
+
+**示例 2：**
+>输入：`nums = [2,2,1,9], target = 10`
+>
+>输出：`4`
+>
+>解释：符合预算的采购方案如下：
+>nums[0] + nums[1] = 4
+>nums[0] + nums[2] = 3
+>nums[1] + nums[2] = 3
+>nums[2] + nums[3] = 10
+
+**提示：**
+- `2 <= nums.length <= 10^5`
+- `1 <= nums[i], target <= 10^5`
+<div><li>👍 25</li><li>👎 0</li></div>
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