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internbootcamp/bootcamp/cengineerartem/cengineerartem.py

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+"""# 
+
+### 谜题描述
+Artem is building a new robot. He has a matrix a consisting of n rows and m columns. The cell located on the i-th row from the top and the j-th column from the left has a value a_{i,j} written in it. 
+
+If two adjacent cells contain the same value, the robot will break. A matrix is called good if no two adjacent cells contain the same value, where two cells are called adjacent if they share a side. 
+
+Artem wants to increment the values in some cells by one to make a good.
+
+More formally, find a good matrix b that satisfies the following condition — 
+
+  * For all valid (i,j), either b_{i,j} = a_{i,j} or b_{i,j} = a_{i,j}+1. 
+
+
+
+For the constraints of this problem, it can be shown that such a matrix b always exists. If there are several such tables, you can output any of them. Please note that you do not have to minimize the number of increments.
+
+Input
+
+Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10). Description of the test cases follows.
+
+The first line of each test case contains two integers n, m (1 ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of rows and columns, respectively.
+
+The following n lines each contain m integers. The j-th integer in the i-th line is a_{i,j} (1 ≤ a_{i,j} ≤ 10^9).
+
+Output
+
+For each case, output n lines each containing m integers. The j-th integer in the i-th line is b_{i,j}.
+
+Example
+
+Input
+
+
+3
+3 2
+1 2
+4 5
+7 8
+2 2
+1 1
+3 3
+2 2
+1 3
+2 2
+
+
+Output
+
+
+1 2
+5 6
+7 8
+2 1
+4 3
+2 4
+3 2
+
+Note
+
+In all the cases, you can verify that no two adjacent cells have the same value and that b is the same as a with some values incremented by one. 
+
+Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
+```python
+from sys import stdin
+t = int(stdin.readline())
+for _ in xrange(t):
+    n,m = map(int,stdin.readline().split())
+    for i in xrange(n):
+        a = map(int,stdin.readline().split())
+        for j in xrange(m):
+            x = a[j]
+            if (i+j)%2 != x%2:
+                x+=1
+            print x,
+        print \"\"
+```
+
+
+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
+from bootcamp import Basebootcamp
+import random
+import re
+
+class Cengineerartembootcamp(Basebootcamp):
+    def __init__(self, min_n=1, max_n=100, min_m=1, max_m=100, seed=None):
+        super().__init__()
+        self.min_n = min_n
+        self.max_n = max_n
+        self.min_m = min_m
+        self.max_m = max_m
+        self.rng = random.Random(seed)
+    
+    def case_generator(self):
+        n = self.rng.randint(self.min_n, self.max_n)
+        m = self.rng.randint(self.min_m, self.max_m)
+        
+        # 确保生成的b矩阵满足奇偶棋盘模式
+        b = []
+        for i in range(n):
+            row = []
+            for j in range(m):
+                base_value = self.rng.randint(1, 10**3)  # 缩小数值范围便于测试
+                parity = (i + j) % 2
+                if (base_value % 2) != parity:
+                    base_value += 1
+                row.append(base_value)
+            b.append(row)
+        
+        # 生成对应的a矩阵（确保值不低于1）
+        a = []
+        for i in range(n):
+            row_a = []
+            for j in range(m):
+                if self.rng.choice([True, False]) and b[i][j] > 1:
+                    row_a.append(b[i][j] - 1)
+                else:
+                    row_a.append(b[i][j])
+            a.append(row_a)
+        
+        return {
+            'n': n,
+            'm': m,
+            'a': a,
+            'expected_parity_pattern': [[(i+j)%2 for j in range(m)] for i in range(n)]  # 添加校验模式
+        }
+    
+    @staticmethod
+    def prompt_func(question_case) -> str:
+        matrix_str = '\n'.join(' '.join(map(str, row)) for row in question_case['a'])
+        return f"""Artem需要调整一个矩阵使其相邻单元格不重复。每个单元格可以选择保持原值或+1。
+
+当前矩阵（{question_case['n']}行×{question_case['m']}列）：
+{matrix_str}
+
+请输出修改后的矩阵，确保：
+1. 相邻单元格（上下左右）的值不同
+2. 每个单元格只能是原值或原值+1
+3. 将最终答案用[answer]标签包裹，例如：
+[answer]
+1 2
+3 4
+[/answer]"""
+
+    @staticmethod
+    def extract_output(output):
+        # 支持多答案块和大小写标签
+        pattern = re.compile(r'\[answer\](.*?)\[/?answer\]', re.DOTALL | re.IGNORECASE)
+        matches = pattern.findall(output)
+        if not matches:
+            return None
+        
+        # 处理最后一个答案块
+        last_answer = matches[-1].strip()
+        matrix = []
+        for line in last_answer.split('\n'):
+            line = line.strip()
+            if line:
+                try:
+                    matrix.append( list(map(int, line.split())) )
+                except ValueError:
+                    continue
+        return matrix if matrix else None
+    
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        # 维度校验
+        if len(solution) != identity['n'] or any(len(row)!=identity['m'] for row in solution):
+            return False
+        
+        a = identity['a']
+        # 值域校验
+        for i in range(identity['n']):
+            for j in range(identity['m']):
+                if solution[i][j] not in {a[i][j], a[i][j]+1}:
+                    return False
+        
+        # 相邻校验
+        directions = [(-1,0),(1,0),(0,-1),(0,1)]
+        for i in range(identity['n']):
+            for j in range(identity['m']):
+                for dx, dy in directions:
+                    x, y = i+dx, j+dy
+                    if 0 <= x < identity['n'] and 0 <= y < identity['m']:
+                        if solution[i][j] == solution[x][y]:
+                            return False
+        return True