add: biweekly 120

math/theory.md

@@ -1,5 +1,62 @@
 ## 习题
 
+### 多边形构造
+
+> [!NOTE] **[LeetCode 2971. 找到最大周长的多边形](https://leetcode.cn/problems/find-polygon-with-the-largest-perimeter/)**
+> 
+> 题意: TODO
+
+> [!TIP] **思路**
+> 
+> 如果有 k (k >= 3) 个 正 数 a1,a2,a3, ... ak 满足 a1 <= a2 <= a3 <= ... <= ak 且 a1 + a2 + a3 + ... + ak-1 > ak ，那么 一定 存在一个 k 条边的多边形，每条边的长度分别为 a1, a2, a3, ... ak
+> 
+> 排序贪心即可
+
+<details>
+<summary>详细代码</summary>
+<!-- tabs:start -->
+
+##### **C++**
+
+```cpp
+class Solution {
+public:
+    using LL = long long;
+    const static int N = 1e5 + 10;
+
+    LL s[N];
+
+    long long largestPerimeter(vector<int>& nums) {
+        sort(nums.begin(), nums.end());
+        int n = nums.size();
+        {
+            s[0] = 0;
+            for (int i = 1; i <= n; ++ i )
+                s[i] = s[i - 1] + nums[i - 1];
+        }
+
+        // ATTENTION: 贪心即可 【不需要双指针】
+        for (int i = n; i >= 1; -- i )
+            if (s[i - 1] > nums[i - 1])
+                return s[i];
+        return -1;
+    }
+};
+```
+
+##### **Python**
+
+```python
+
+```
+
+<!-- tabs:end -->
+</details>
+
+<br>
+
+* * *
+
 ### 拉格朗日四平方和
 
 > [!NOTE] **[LeetCode 279. 完全平方数](https://leetcode-cn.com/problems/perfect-squares/)**

topic/bf-improve.md

@@ -408,6 +408,92 @@ public:
 
 * * *
 
+> [!NOTE] **[LeetCode 2973. 树中每个节点放置的金币数目](https://leetcode.cn/problems/find-number-of-coins-to-place-in-tree-nodes/)** [TAG]
+> 
+> 题意: TODO
+
+> [!TIP] **思路**
+> 
+> 数学分析 优化数据范围
+
+<details>
+<summary>详细代码</summary>
+<!-- tabs:start -->
+
+##### **C++**
+
+```cpp
+class Solution {
+public:
+    using LL = long long;
+    const static int N = 2e4 + 10, M = 4e4 + 10;
+
+    int h[N], e[M], ne[M], idx;
+    void init() {
+        memset(h, -1, sizeof h);
+        idx = 0;
+    }
+    void add(int a, int b) {
+        e[idx] = b, ne[idx] = h[a], h[a] = idx ++ ;
+    }
+
+    int n;
+    vector<int> cs;
+    vector<LL> res;
+
+    vector<int> dfs(int u, int fa) {
+        vector<int> ret = {cs[u]};
+        for (int i = h[u]; ~i; i = ne[i]) {
+            int j = e[i];
+            if (j == fa)
+                continue;
+            auto t = dfs(j, u);
+            ret.insert(ret.end(), t.begin(), t.end());
+        }
+
+        sort(ret.begin(), ret.end());
+        int m = ret.size();
+        // 求最值
+        if (m >= 3) {
+            res[u] = max({(LL)ret[m - 3] * ret[m - 2] * ret[m - 1],
+                        (LL)ret[0] * ret[1] * ret[m - 1], 0ll});
+        }
+        // ATTENTION 收缩不必要的区间
+        if (m > 5)
+            ret = {ret[0], ret[1], ret[m - 3], ret[m - 2], ret[m - 1]};
+        return ret;
+    }
+
+    vector<long long> placedCoins(vector<vector<int>>& edges, vector<int>& cost) {
+        init();
+        {
+            for (auto & e : edges)
+                add(e[0], e[1]), add(e[1], e[0]);
+        }
+
+        this-> cs = cost;
+        this->n = cs.size();
+
+        res = vector<LL>(n, 1);
+        dfs(0, -1);
+        return res;
+    }
+};
+```
+
+##### **Python**
+
+```python
+
+```
+
+<!-- tabs:end -->
+</details>
+
+<br>
+
+* * *
+
 ### 双指针
 
 > [!NOTE] **[LeetCode 1521. 找到最接近目标值的函数值](https://leetcode-cn.com/problems/find-a-value-of-a-mysterious-function-closest-to-target/)** [TAG]
@@ -985,6 +1071,77 @@ public:
 
 * * *
 
+> [!NOTE] **[LeetCode 2972. 统计移除递增子数组的数目 II](https://leetcode.cn/problems/count-the-number-of-incremovable-subarrays-ii/)** [TAG]
+> 
+> 题意: TODO
+
+> [!TIP] **思路**
+> 
+> 思维 + 双指针
+> 
+> 注意不要漏掉【只保留左侧 l+2】的计数值
+
+<details>
+<summary>详细代码</summary>
+<!-- tabs:start -->
+
+##### **C++**
+
+```cpp
+class Solution {
+public:
+    // 移除一段连续序列 使得左右两侧合并为LIS(严格递增)
+    // 求有多少个序列移除办法
+    // 
+    // 显然要先从左右两侧收缩区间 找到最小的不满足LIS的位置
+    // 随后分情况讨论 + 枚举右端点
+
+    using LL = long long;
+
+    int n, l, r;
+
+    long long incremovableSubarrayCount(vector<int>& nums) {
+        this->n = nums.size();
+        l = 0, r = n - 1;
+        {
+            while (l + 1 < n && nums[l + 1] > nums[l])
+                l ++ ;
+            while (r > 0 && nums[r - 1] < nums[r])
+                r -- ;
+        }
+        if (l >= r)
+            return (LL)(n + 1) * n / 2;
+
+        // ATTENTION: case 1 如果只保留左侧(完全移除右侧)
+        LL res = l + 2;
+
+        // case 2 只保留右侧位置
+        // 枚举右侧的起始位置 则左侧终止位置需要满足特定条件
+        for (int i = 0, j = r; j < n; ++ j ) {
+            // 找到第一个不符合的位置
+            while (i <= l && nums[i] < nums[j])
+                i ++ ;
+
+            res += (i + 1);
+        }
+        return res;
+    }
+};
+```
+
+##### **Python**
+
+```python
+
+```
+
+<!-- tabs:end -->
+</details>
+
+<br>
+
+* * *
+
 ### 边遍历边维护
 
 > [!NOTE] **[LeetCode 2763. 所有子数组中不平衡数字之和](https://leetcode.cn/problems/sum-of-imbalance-numbers-of-all-subarrays/)**