diff --git a/content/posts/leetcode.md b/content/posts/leetcode.md
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--- a/content/posts/leetcode.md
+++ b/content/posts/leetcode.md
@@ -22981,3 +22981,44 @@ public:
     }
 };
 ```
+## day340 2025-02-13
+### 3066. Minimum Operations to Exceed Threshold Value II 
+You are given a 0-indexed integer array nums, and an integer k.
+
+In one operation, you will:
+
+Take the two smallest integers x and y in nums.
+Remove x and y from nums.
+Add min(x, y) * 2 + max(x, y) anywhere in the array.
+Note that you can only apply the described operation if nums contains at least two elements.
+
+Return the minimum number of operations needed so that all elements of the array are greater than or equal to k.
+
+![0213vgK8sqTPCA3Q](https://testingcf.jsdelivr.net/gh/game-loader/picbase@master/uPic/0213vgK8sqTPCA3Q.png)
+
+### 题解
+本题注意到题目中会多次用到nums中的最小值，操作中的第三步min(x,y)*2+max(x,y)如果先拿到nums中的最小值，删掉这个最小值，再取出nums中的最小值则先取得的一定是更小的那个，后取得的是更大的那个。则此处可使用最小堆，弹出堆顶元素即为nums中的最小值，执行题目所述操作得到新的数字后再将新的数字插入堆中，如此反复直到堆顶元素大于等于k，返回记录的操作次数。
+
+### 代码
+```cpp 
+class Solution {
+public:
+    int minOperations(vector<int>& nums, int k) {
+        priority_queue<long long int,vector<long long int>,greater<long long int>> min_heap(nums.begin(), nums.end());
+        int opnum = 0;
+        long long int small = 0;
+        long long int big = 0;
+        while(min_heap.top() < k){
+            small = min_heap.top();
+            min_heap.pop();
+            big = min_heap.top();
+            min_heap.pop();
+            min_heap.push(small*2+big);
+            opnum++;
+        }
+        return opnum;
+    }
+};
+```
+### 总结
+注意到想构建优先级队列需要定义好三个属性，一是优先级队列中保存的数据类型，二是用来保存这些数据的容器类型，三则是优先级队列中的比较函数，定义好这三个抽象出来的属性就可以定义好一个优先级队列。