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156 lines
5.1 KiB
Python
Executable file
156 lines
5.1 KiB
Python
Executable file
"""#
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### 谜题描述
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In this task Anna and Maria play the following game. Initially they have a checkered piece of paper with a painted n × m rectangle (only the border, no filling). Anna and Maria move in turns and Anna starts. During each move one should paint inside the last-painted rectangle a new lesser rectangle (along the grid lines). The new rectangle should have no common points with the previous one. Note that when we paint a rectangle, we always paint only the border, the rectangles aren't filled.
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Nobody wins the game — Anna and Maria simply play until they have done k moves in total. Count the number of different ways to play this game.
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Input
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The first and only line contains three integers: n, m, k (1 ≤ n, m, k ≤ 1000).
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Output
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Print the single number — the number of the ways to play the game. As this number can be very big, print the value modulo 1000000007 (109 + 7).
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Examples
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Input
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3 3 1
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Output
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1
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Input
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4 4 1
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Output
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9
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Input
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6 7 2
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Output
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75
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Note
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Two ways to play the game are considered different if the final pictures are different. In other words, if one way contains a rectangle that is not contained in the other way.
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In the first sample Anna, who performs her first and only move, has only one possible action plan — insert a 1 × 1 square inside the given 3 × 3 square.
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In the second sample Anna has as much as 9 variants: 4 ways to paint a 1 × 1 square, 2 ways to insert a 1 × 2 rectangle vertically, 2 more ways to insert it horizontally and one more way is to insert a 2 × 2 square.
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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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n, m, k = map(int, raw_input().strip().split())
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def fac(x):
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ret = 1
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for i in xrange(1, x + 1):
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ret *= i
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return ret
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def c(n, r):
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return fac(n) / (fac(n - r) * fac(r)) if r <= n else 0
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print c(n - 1, k * 2) * c(m - 1, k * 2) % (10 ** 9 + 7)
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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 re
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import random
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from bootcamp import Basebootcamp
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MOD = 10**9 + 7
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class Egameswithrectanglebootcamp(Basebootcamp):
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def __init__(self, n_min=3, n_max=20, m_min=3, m_max=20, k_min=1, k_max=5):
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# 确保参数的合理性
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self.n_min = max(n_min, 3)
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self.n_max = max(n_max, self.n_min)
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self.m_min = max(m_min, 3)
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self.m_max = max(m_max, self.m_min)
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# 动态调整k的取值范围
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max_possible_k = min((self.n_max-1)//2, (self.m_max-1)//2)
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self.k_min = max(k_min, 1)
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self.k_max = min(k_max, max_possible_k)
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if self.k_max < self.k_min:
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raise ValueError("Invalid parameter range: Unable to generate valid k")
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def case_generator(self):
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"""生成保证有解的合法测试用例"""
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while True:
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# 计算当前参数允许的k范围
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max_k = min((self.n_max-1)//2, (self.m_max-1)//2)
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k = random.randint(self.k_min, min(self.k_max, max_k))
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# 动态调整维数生成范围
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min_n = 2*k + 1
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min_m = 2*k + 1
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valid_n_range = (max(self.n_min, min_n), self.n_max)
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valid_m_range = (max(self.m_min, min_m), self.m_max)
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if valid_n_range[0] > valid_n_range[1] or valid_m_range[0] > valid_m_range[1]:
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continue # 跳过无效范围
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n = random.randint(*valid_n_range)
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m = random.randint(*valid_m_range)
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return {'n': n, 'm': m, 'k': k}
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@staticmethod
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def prompt_func(question_case):
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n, m, k = question_case['n'], question_case['m'], question_case['k']
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return f"""Anna和Maria正在进行矩形嵌套游戏。初始有一个{n}x{m}的矩形边框，需进行正好{k}步操作。计算不同的游戏方式总数（模10^9+7）。
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游戏规则：
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1. 每个新矩形必须完全在前一个矩形内部且不共享任何边界点
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2. 每个矩形必须在两个维度上都至少缩小2个单位
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3. 不同顺序的矩形添加视为不同方案
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4. 最终答案需模10^9+7
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请将最终答案包裹在[answer]标签内，例如：[answer]42[/answer]。"""
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@staticmethod
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def extract_output(output):
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matches = re.findall(
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r'\[answer\]\s*(\d+)\s*\[/answer\]',
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output,
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re.IGNORECASE | re.DOTALL
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)
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return matches[-1] if matches else None
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@classmethod
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def _verify_correction(cls, solution, identity):
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n, m, k = identity['n'], identity['m'], identity['k']
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t = 2 * k
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def comb_mod(n_, r):
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"""使用动态规划计算组合数模MOD"""
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if r < 0 or r > n_:
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return 0
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C = [0] * (r + 1)
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C[0] = 1
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for i in range(1, n_ + 1):
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for j in range(min(i, r), 0, -1):
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C[j] = (C[j] + C[j-1]) % MOD
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return C[r]
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try:
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expected = (comb_mod(n-1, t) * comb_mod(m-1, t)) % MOD
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return int(solution.strip()) == expected
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except (ValueError, TypeError):
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return False
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