add: contest weekly 375

math/combinatorics/combination.md

@@ -1879,6 +1879,126 @@ public:
 
 * * *
 
+> [!NOTE] **[LeetCode 2963. 统计好分割方案的数目](https://leetcode.cn/problems/count-the-number-of-good-partitions/)**
+> 
+> 题意: TODO
+
+> [!TIP] **思路**
+> 
+> 区间合并 + 组合数 (后者也可转化成 $2^{m-1}$ 思考)
+
+<details>
+<summary>详细代码</summary>
+<!-- tabs:start -->
+
+##### **C++**
+
+```cpp
+using LL = long long;
+const static int N = 1e5 + 10, MOD = 1e9 + 7;
+
+int qmi(int a, int k, int p) {
+    int ret = 1;
+    while (k) {
+        if (k & 1)
+            ret = (LL)ret * a % p;
+        a = (LL)a * a % p;
+        k >>= 1;
+    }
+    return ret;
+}
+
+int fact[N], infact[N];
+bool inited = false;
+
+void init() {
+    if (inited)
+        return;
+    inited = true;
+
+    fact[0] = infact[0] = 1;
+    for (int i = 1; i < N; ++ i ) {
+        fact[i] = (LL)fact[i - 1] * i % MOD;
+        infact[i] = (LL)infact[i - 1] * qmi(i, MOD - 2, MOD) % MOD;
+    }
+}
+
+int comb(int a, int b) {
+    return (LL)fact[a] * infact[b] % MOD * infact[a - b] % MOD;
+}
+
+class Solution {
+public:
+    using PII = pair<int, int>;
+
+    void merge(vector<PII> & segs) {
+        vector<PII> res;
+        sort(segs.begin(), segs.end());
+        int st = -2e9, ed = -2e9;
+        for (auto seg : segs)
+            if (ed < seg.first) {
+                if (st != -2e9)
+                    res.push_back({st, ed});
+                st = seg.first, ed = seg.second;
+            } else
+                ed = max(ed, seg.second);
+        if (st != -2e9)
+            res.push_back({st, ed});
+        segs = res;
+    }
+
+    int numberOfGoodPartitions(vector<int>& nums) {
+        init();
+
+        int n = nums.size();
+
+        unordered_set<int> S;
+        unordered_map<int, int> l, r;
+        {
+            // 先找到 最左侧/最右侧 出现的位置
+            for (int i = 0; i < n; ++ i ) {
+                int x = nums[i];
+                if (S.count(x)) {
+                    l[x] = min(l[x], i), r[x] = max(r[x], i);
+                } else {
+                    l[x] = r[x] = i;
+                    S.insert(x);
+                }
+            }
+        }
+        vector<PII> xs;
+        {
+            // 根据出现过的元素划分基本区间
+            for (auto [k, v] : l)
+                xs.push_back({v, r[k]});
+            // 区间合并
+            merge(xs);
+        }
+
+        // 留下的都是可以任意合并的区间
+        // 最坏情况下 m=1e5
+        int m = xs.size(), res = 0;
+        for (int i = 0; i < m; ++ i )
+            // C{m-1}{i}
+            res = (res + comb(m - 1, i)) % MOD;
+        return res;
+    }
+};
+```
+
+##### **Python**
+
+```python
+
+```
+
+<!-- tabs:end -->
+</details>
+
+<br>
+
+* * *
+
 ### 排列数
 
 > [!NOTE] **[AcWing 1309. 车的放置](https://www.acwing.com/problem/content/1311/)** [TAG]