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internbootcamp/bootcamp/fslimeandsequenceseasyversion/fslimeandsequenceseasyversion.py

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+"""# 
+
+### 谜题描述
+Note that the only differences between easy and hard versions are the constraints on n and the time limit. You can make hacks only if all versions are solved.
+
+Slime is interested in sequences. He defined good positive integer sequences p of length n as follows:
+
+  * For each k>1 that presents in p, there should be at least one pair of indices i,j, such that 1 ≤ i < j ≤ n, p_i = k - 1 and p_j = k.
+
+
+
+For the given integer n, the set of all good sequences of length n is s_n. For the fixed integer k and the sequence p, let f_p(k) be the number of times that k appears in p. For each k from 1 to n, Slime wants to know the following value:
+
+$$$\left(∑_{p∈ s_n} f_p(k)\right)\ mod\ 998 244 353$$$
+
+Input
+
+The first line contains one integer n\ (1≤ n≤ 5000).
+
+Output
+
+Print n integers, the i-th of them should be equal to \left(∑_{p∈ s_n} f_p(i)\right)\ mod\ 998 244 353.
+
+Examples
+
+Input
+
+
+2
+
+
+Output
+
+
+3 1 
+
+
+Input
+
+
+3
+
+
+Output
+
+
+10 7 1 
+
+
+Input
+
+
+1
+
+
+Output
+
+
+1 
+
+Note
+
+In the first example, s=\{[1,1],[1,2]\}.
+
+In the second example, s=\{[1,1,1],[1,1,2],[1,2,1],[1,2,2],[2,1,2],[1,2,3]\}.
+
+In the third example, s=\{[1]\}.
+
+Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
+```python
+#include <bits/stdc++.h>
+using namespace std;
+inline bool cmin(int &x, int y) { return x > y ? x = y, 1 : 0; }
+inline bool cmax(int &x, int y) { return x < y ? x = y, 1 : 0; }
+const int _ = 5055, mod = 998244353;
+int N, fac[_], inv[_];
+int f[_][_];
+int ans[_];
+inline int dec(int x) { return x >= mod ? x - mod : x; }
+inline int Pow(int x, int y = mod - 2) {
+  int ans(1);
+  for (; y; y >>= 1, x = (long long)x * x % mod)
+    if (y & 1) ans = (long long)ans * x % mod;
+  return ans;
+}
+int main() {
+  cin >> N;
+  fac[0] = fac[1] = inv[0] = 1;
+  for (int i = 2, I = N; i <= I; ++i) fac[i] = 1ll * fac[i - 1] * i % mod;
+  inv[N] = Pow(fac[N]);
+  for (int i = N, I = 2; i >= I; --i) inv[i - 1] = 1ll * inv[i] * i % mod;
+  f[0][0] = 1;
+  for (int i = 1, I = N; i <= I; ++i)
+    for (int j = 1, I = i; j <= I; ++j)
+      f[i][j] =
+          (1ll * f[i - 1][j] * j + 1ll * f[i - 1][j - 1] * (i - j + 1)) % mod;
+  for (int i = 1, I = N; i <= I; ++i)
+    for (int j = 1, I = N; j <= I; ++j)
+      ans[j] = (ans[j] + 1ll * f[i][j] * inv[i]) % mod;
+  for (int i = 1, I = N; i <= I; ++i)
+    cout << 1ll * ans[i] * fac[N] % mod << \" \";
+  return 0;
+}
+```
+
+
+请完成上述谜题的训练场环境类实现，包括所有必要的方法。
+"""
+
+from bootcamp import Basebootcamp
+import re
+from bootcamp import Basebootcamp
+
+MOD = 998244353
+
+class Fslimeandsequenceseasyversionbootcamp(Basebootcamp):
+    def __init__(self, max_n=100):  # 默认限制测试用例规模
+        """
+        限制默认max_n以保证生成效率，用户可自行设置更大值
+        """
+        self.max_n = min(max_n, 5000)  # 强制不超过题目要求的5000上限
+    
+    def case_generator(self):
+        import random
+        n = random.randint(1, self.max_n)
+        expected = self.compute_answer(n)
+        return {'n': n, 'expected': expected}
+    
+    @staticmethod
+    def compute_answer(n):
+        mod = MOD
+        if n == 0:
+            return []
+        
+        # 预计算阶乘和逆阶乘
+        fac = [1]*(n+1)
+        for i in range(1, n+1):
+            fac[i] = fac[i-1] * i % mod
+            
+        inv_fac = [1]*(n+1)
+        inv_fac[n] = pow(fac[n], mod-2, mod)
+        for i in range(n-1, -1, -1):
+            inv_fac[i] = inv_fac[i+1] * (i+1) % mod
+        
+        # 使用滚动数组优化空间
+        ans = [0]*(n+2)
+        prev = [0]*(n+2)
+        prev[0] = 1
+        
+        for i in range(1, n+1):
+            curr = [0]*(n+2)
+            for j in range(1, i+1):
+                term1 = prev[j] * j % mod
+                term2 = prev[j-1] * (i-j+1) % mod
+                curr[j] = (term1 + term2) % mod
+            
+            # 实时累加结果，避免存储全表
+            for j in range(1, i+1):
+                ans[j] = (ans[j] + curr[j] * inv_fac[i]) % mod
+            
+            prev = curr  # 滚动更新
+        
+        # 最终调整
+        return [ans[j] * fac[n] % mod for j in range(1, n+1)]
+    
+    @staticmethod
+    def prompt_func(question_case):
+        n = question_case['n']
+        return f"""Given n = {n}, compute the sum of occurrences for each k (1-{n}) in all valid sequences. Enclose your answer in [answer][/answer] tags.
+
+Sample format:
+n=2 → [answer]3 1[/answer]
+n=3 → [answer]10 7 1[/answer]
+
+Rules:
+1. For any k>1 in sequence, must have earlier k-1
+2. Output n space-separated values mod 998244353
+3. Numbers must be in order from k=1 to k={n}
+
+Your answer for n={n}:"""
+    
+    @staticmethod
+    def extract_output(output):
+        matches = re.findall(r'\[answer\](.*?)\[\/answer\]', output, re.DOTALL)
+        if not matches:
+            return None
+        
+        last = matches[-1].strip()
+        try:
+            return list(map(int, last.split()))
+        except:
+            return None
+    
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        expected = identity['expected']
+        return solution == expected
+
+# 测试样例验证
+if __name__ == "__main__":
+    # 验证n=2的示例
+    bootcamp = Fslimeandsequenceseasyversionbootcamp()
+    test_case = {'n': 2}
+    assert bootcamp.compute_answer(2) == [3, 1], "Sample 2 failed"
+    
+    # 验证n=3的示例
+    test_case = {'n': 3}
+    assert bootcamp.compute_answer(3) == [10, 7, 1], "Sample 3 failed"
+    
+    print("All sample tests passed!")