add: contest biweekly 117

OpenKikCoc · Dec 1, 2023 · 176f873 · 176f873
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diff --git a/math/combinatorics/inclusion-exclusion-principle.md b/math/combinatorics/inclusion-exclusion-principle.md
@@ -1285,4 +1285,115 @@ int main() {
 
 <br>
 
+* * *
+
+> [!NOTE] **[LeetCode 2930. 重新排列后包含指定子字符串的字符串数目](https://leetcode.cn/problems/number-of-strings-which-can-be-rearranged-to-contain-substring/)**
+> 
+> 题意: TODO
+
+> [!TIP] **思路**
+> 
+> DP 或 容斥
+
+<details>
+<summary>详细代码</summary>
+<!-- tabs:start -->
+
+##### **C++ 容斥**
+
+```cpp
+class Solution {
+public:
+    // 初步思想: 直接 dp
+    // 
+    // 进阶 考虑容斥 直接排除"不好"的即可
+    using LL = long long;
+    const static int N = 1e5 + 10, MOD = 1e9 + 7;
+
+    // 容斥
+    LL qpow(LL a, LL b) {
+        LL ret = 1;
+        while (b) {
+            if (b & 1)
+                ret = ret * a % MOD;
+            a = a * a % MOD;
+            b >>= 1;
+        }
+        return ret;
+    }
+    int stringCount(int n) {
+        // 反向考虑
+        return ((qpow(26, n)                                    // 所有字符串
+            - qpow(25, n - 1) * (25 * 3 + n) % MOD              // 不含至少某一种 (中间有重复统计)
+            + qpow(24, n - 1) * (24 * 3 + n * 2) % MOD          // 不含至少其中两种
+            - qpow(23, n - 1) * (23 + n)) % MOD + MOD) % MOD;   // 都不含
+    }
+};
+```
+
+##### **C++ DP**
+
+```cpp
+class Solution {
+public:
+    // 初步思想: 直接 dp
+    // 
+    // 进阶 考虑容斥 直接排除"不好"的即可
+    using LL = long long;
+    const static int N = 1e5 + 10, MOD = 1e9 + 7;
+
+    LL f[N][2][3][2];   // 考虑前 i 个字符: l 个数 0/>=1 e个数0/1/>=2 ...
+
+    void madd(LL & a, LL b) {
+        a = (a + b) % MOD;
+    }
+
+    int stringCount(int n) {
+        memset(f, 0, sizeof f);
+        f[0][0][0][0] = 1;
+
+        // 正推更好实现
+        for (int i = 0; i < n; ++ i ) {
+            for (int x = 0; x < 2; ++ x )
+                for (int y = 0; y < 3; ++ y )
+                    for (int z = 0; z < 2; ++ z ) {
+                        LL t = f[i][x][y][z];
+                        
+                        madd(f[i + 1][x][y][z], t * 23 % MOD);
+                        madd(f[i + 1][min(x + 1, 1)][y][z], t);
+                        madd(f[i + 1][x][min(y + 1, 2)][z], t);
+                        madd(f[i + 1][x][y][min(z + 1, 1)], t);
+                    }
+        }
+
+        // 逆推
+        for (int i = 1; i <= n; ++ i ) {
+            for (int x = 0; x < 2; ++ x )
+                for (int y = 0; y < 3; ++ y )
+                    for (int z = 0; z < 2; ++ z ) {
+                        LL & t = f[i][x][y][z];
+                        
+                        madd(t, f[i - 1][x][y][z] * 23 % MOD);
+                        madd(t, f[i - 1][max(x - 1, 0)][y][z]);
+                        madd(t, f[i - 1][x][max(y - 1, 0)][z]);
+                        madd(t, f[i - 1][x][y][max(z - 1, 0)]);
+                    }
+        }
+
+        return f[n][1][2][1];
+    }
+};
+```
+
+##### **Python**
+
+```python
+
+```
+
+<!-- tabs:end -->
+</details>
+
+<br>
+
 * * *