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167 lines
4.4 KiB
Python
Executable file
167 lines
4.4 KiB
Python
Executable file
"""#
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### 谜题描述
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Quite recently a creative student Lesha had a lecture on trees. After the lecture Lesha was inspired and came up with the tree of his own which he called a k-tree.
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A k-tree is an infinite rooted tree where:
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* each vertex has exactly k children;
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* each edge has some weight;
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* if we look at the edges that goes from some vertex to its children (exactly k edges), then their weights will equal 1, 2, 3, ..., k.
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The picture below shows a part of a 3-tree.
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<image>
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As soon as Dima, a good friend of Lesha, found out about the tree, he immediately wondered: \"How many paths of total weight n (the sum of all weights of the edges in the path) are there, starting from the root of a k-tree and also containing at least one edge of weight at least d?\".
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Help Dima find an answer to his question. As the number of ways can be rather large, print it modulo 1000000007 (109 + 7).
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Input
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A single line contains three space-separated integers: n, k and d (1 ≤ n, k ≤ 100; 1 ≤ d ≤ k).
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Output
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Print a single integer — the answer to the problem modulo 1000000007 (109 + 7).
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Examples
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Input
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3 3 2
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Output
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3
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Input
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3 3 3
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Output
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1
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Input
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4 3 2
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Output
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6
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Input
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4 5 2
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Output
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7
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Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
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```python
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MOD = 1000000007
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cache = {}
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def ways(curr_weight, target_weight, k, d, d_count):
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if curr_weight == target_weight and d_count > 0:
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return 1
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elif curr_weight > target_weight:
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return 0
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elif (curr_weight, d_count) not in cache:
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ans = 0
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for w in xrange(1, k+1):
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ans = (ans + ways(curr_weight+w, target_weight, k, d, d_count + int(w >= d))) % MOD
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cache[(curr_weight, d_count)] = ans
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return cache[(curr_weight, d_count)]
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n, k, d = tuple(map(int, raw_input().split()))
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print ways(0, n, k, d, 0)
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```
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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。
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"""
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from bootcamp import Basebootcamp
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import random
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import re
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from bootcamp import Basebootcamp
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MOD = 10**9 + 7
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class Cktreebootcamp(Basebootcamp):
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def __init__(self, min_n=1, max_n=100, min_k=1, max_k=100):
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self.min_n = min_n
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self.max_n = max_n
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self.min_k = min_k
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self.max_k = max_k
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def case_generator(self):
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# 确保生成的k >=1 且 <=100，d在合法范围内
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k = random.randint(max(1, self.min_k), min(100, self.max_k))
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d = random.randint(1, k)
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# 确保n不超过k的理论可能范围
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max_feasible_n = min(self.max_n, k * 10) # 合理限制最大n
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n = random.randint(max(1, self.min_n), max_feasible_n)
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return {'n': n, 'k': k, 'd': d}
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@staticmethod
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def prompt_func(question_case):
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n = question_case['n']
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k = question_case['k']
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d = question_case['d']
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prompt = f"""在k-tree中，每个节点有k个子节点，子节点边的权重分别为1到{k}。请计算从根出发总权重为{n}且至少包含一条权重≥{d}的路径数目（模10^9+7）。
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输入参数：n={n}, k={k}, d={d}
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答案格式：[answer]整数[/answer]"""
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return prompt
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@staticmethod
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def extract_output(output):
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# 增强正则匹配鲁棒性
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matches = re.findall(r'\[answer\]\s*(\d+)\s*\[/answer\]', output, re.IGNORECASE)
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return int(matches[-1]) % MOD if matches else None
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@classmethod
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def _verify_correction(cls, solution, identity):
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try:
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n, k, d = identity['n'], identity['k'], identity['d']
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correct = cls.calculate_answer(n, k, d)
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# 统一取模比较（处理负数和大数情况）
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return int(solution) % MOD == correct
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except:
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return False
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@staticmethod
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def calculate_answer(n, k, d):
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# 使用动态规划优化空间复杂度
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dp_total = [0] * (n + 1)
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dp_total[0] = 1
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for i in range(n + 1):
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for j in range(1, k + 1):
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if i + j <= n:
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dp_total[i + j] = (dp_total[i + j] + dp_total[i]) % MOD
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if d == 1:
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return dp_total[n] % MOD
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dp_no = [0] * (n + 1)
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dp_no[0] = 1
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for i in range(n + 1):
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for j in range(1, d):
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if i + j <= n:
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dp_no[i + j] = (dp_no[i + j] + dp_no[i]) % MOD
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return (dp_total[n] - dp_no[n]) % MOD
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