init-commit

internbootcamp/bootcamp/eleavingthebar/eleavingthebar.py

@@ -0,0 +1,316 @@
+"""# 
+
+### 谜题描述
+For a vector \vec{v} = (x, y), define |v| = √{x^2 + y^2}.
+
+Allen had a bit too much to drink at the bar, which is at the origin. There are n vectors \vec{v_1}, \vec{v_2}, ⋅⋅⋅, \vec{v_n}. Allen will make n moves. As Allen's sense of direction is impaired, during the i-th move he will either move in the direction \vec{v_i} or -\vec{v_i}. In other words, if his position is currently p = (x, y), he will either move to p + \vec{v_i} or p - \vec{v_i}.
+
+Allen doesn't want to wander too far from home (which happens to also be the bar). You need to help him figure out a sequence of moves (a sequence of signs for the vectors) such that his final position p satisfies |p| ≤ 1.5 ⋅ 10^6 so that he can stay safe.
+
+Input
+
+The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of moves.
+
+Each of the following lines contains two space-separated integers x_i and y_i, meaning that \vec{v_i} = (x_i, y_i). We have that |v_i| ≤ 10^6 for all i.
+
+Output
+
+Output a single line containing n integers c_1, c_2, ⋅⋅⋅, c_n, each of which is either 1 or -1. Your solution is correct if the value of p = ∑_{i = 1}^n c_i \vec{v_i}, satisfies |p| ≤ 1.5 ⋅ 10^6.
+
+It can be shown that a solution always exists under the given constraints.
+
+Examples
+
+Input
+
+3
+999999 0
+0 999999
+999999 0
+
+
+Output
+
+1 1 -1 
+
+
+Input
+
+1
+-824590 246031
+
+
+Output
+
+1 
+
+
+Input
+
+8
+-67761 603277
+640586 -396671
+46147 -122580
+569609 -2112
+400 914208
+131792 309779
+-850150 -486293
+5272 721899
+
+
+Output
+
+1 1 1 1 1 1 1 -1 
+
+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, stdout
+
+n = input()
+inp = stdin.readlines()
+
+if n == 1:
+    print 1
+    exit()
+
+a = [0 for i in xrange(3 * n)]
+
+child = [[-1, -1] for i in xrange(len(a))]
+sign = [[1, 1] for i in xrange(len(a))]
+
+i = 0
+for line in inp:
+    a[i] = tuple(map(int, line.split()))
+    i += 1
+
+r = 1000000
+R = r * r
+
+idx = n
+free = [i for i in xrange(n)]
+
+while len(free) >= 3:
+    pos = []
+    for it in xrange(3): pos.append(free.pop())
+
+    found = 0
+    for i in xrange(3):
+        for j in xrange(i + 1, 3):
+            # |v[i] + v[j]| <= r 
+            pt = tuple([a[pos[i]][0] + a[pos[j]][0], a[pos[i]][1] + a[pos[j]][1]])
+
+            if pt[0] * pt[0] + pt[1] * pt[1] <= R:
+                child[idx][0] = pos[i]
+                child[idx][1] = pos[j]
+
+                sign[idx][0] = 1
+                sign[idx][1] = 1
+
+                a[idx] = pt
+                free.append(idx)
+
+                free.append(pos[i ^ j ^ 3])
+
+                idx += 1
+                found = 1
+                break
+
+            # |v[i] - v[j]| <= r
+            pt = tuple([a[pos[i]][0] - a[pos[j]][0], a[pos[i]][1] - a[pos[j]][1]])
+
+            if pt[0] * pt[0] + pt[1] * pt[1] <= R:
+                child[idx][0] = pos[i]
+                child[idx][1] = pos[j]
+
+                sign[idx][0] = 1
+                sign[idx][1] = -1
+
+                a[idx] = pt
+                free.append(idx)
+
+                free.append(pos[i ^ j ^ 3])
+
+                idx += 1
+                found = 1
+                break
+
+        if found: break
+
+pos = []
+for i in xrange(2): pos.append(free.pop())
+
+pt = tuple([a[pos[0]][0] + a[pos[1]][0], a[pos[0]][1] + a[pos[1]][1]])
+
+if pt[0] * pt[0] + pt[1] * pt[1] <= 2 * R:
+    child[idx][0] = pos[0]
+    child[idx][1] = pos[1]
+
+    sign[idx][0] = 1
+    sign[idx][1] = 1
+
+else:
+    child[idx][0] = pos[0]
+    child[idx][1] = pos[1]
+
+    sign[idx][0] = 1
+    sign[idx][1] = -1
+
+ans = [1 for i in xrange(3 * n)]
+
+stk = [idx]
+ptr = 0
+
+while ptr < len(stk):
+    u = stk[ptr]
+    v1, v2 = child[u]
+
+    if v1 != -1 and v2 != -1:
+        stk.append(v1)
+        stk.append(v2)
+
+        ans[v1] = sign[u][0] * ans[u]
+        ans[v2] = sign[u][1] * ans[u]
+
+    ptr += 1
+
+out = [str(ans[i]) for i in xrange(n)]
+stdout.write(' '.join(out))
+```
+
+
+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 math
+import random
+import re
+from bootcamp import Basebootcamp
+
+class Eleavingthebarbootcamp(Basebootcamp):
+    def __init__(self, min_n=1, max_n=100, vector_max_magnitude=1e6):
+        """
+        初始化谜题参数
+        
+        参数:
+            min_n (int): 最小移动次数，默认为1
+            max_n (int): 最大移动次数，默认为100
+            vector_max_magnitude (float): 单步移动向量的最大模长，默认为1e6
+        """
+        self.min_n = min_n
+        self.max_n = max_n
+        self.vector_max_magnitude = vector_max_magnitude
+    
+    def case_generator(self):
+        """
+        生成合法谜题实例，保证存在解
+        
+        返回:
+            dict: 包含n和向量列表的可序列化字典
+        """
+        n = random.randint(self.min_n, self.max_n)
+        vectors = []
+        max_m = int(self.vector_max_magnitude)
+        max_sq = self.vector_max_magnitude ** 2
+        
+        for _ in range(n):
+            x = random.randint(-max_m, max_m)
+            x_sq = x ** 2
+            remaining = max_sq - x_sq
+            if remaining < 0:  # 当x绝对值超过最大值时（理论上不可能）
+                y = 0
+            else:
+                max_y = int(math.sqrt(remaining))
+                y = random.randint(-max_y, max_y)
+            vectors.append((x, y))
+        
+        return {
+            'n': n,
+            'vectors': vectors
+        }
+    
+    @staticmethod
+    def prompt_func(question_case) -> str:
+        """
+        生成符合格式的系统提示
+        
+        参数:
+            question_case: case_generator生成的谜题实例
+            
+        返回:
+            str: 包含完整问题描述的字符串
+        """
+        n = question_case['n']
+        vectors = question_case['vectors']
+        prompt_lines = [
+            "Allen喝醉了，正在从位于坐标原点的酒吧出发回家。他需要进行一系列移动，每次可以选择沿着给定方向或反方向移动。",
+            "你需要帮助他选择每个移动方向的正负号，使得最终位置与原点的距离不超过1.5×10^6。",
+            "\n移动规则说明:",
+            "1. 共进行n次移动，每次对应一个二维向量",
+            "2. 第i次移动时，Allen可以选择移动方向为c_i·(x_i, y_i)，其中c_i只能是1（正方向）或-1（反方向）",
+            "3. 最终位置为所有移动向量的矢量和",
+            f"\n当前移动参数 (n={n}):",
+            f"{n}"
+        ]
+        prompt_lines.extend(f"{x} {y}" for x, y in vectors)
+        prompt_lines.append("\n请输出包含n个1或-1的序列表示移动方向，格式要求：")
+        prompt_lines.append("- 数字之间用空格分隔")
+        prompt_lines.append("- 将最终答案放置在[answer]和[/answer]标记之间")
+        prompt_lines.append("例如：[answer]1 -1 1[/answer]")
+        
+        return '\n'.join(prompt_lines)
+    
+    @staticmethod
+    def extract_output(output):
+        """
+        从模型输出中提取最后一个符合格式的答案
+        
+        参数:
+            output: 模型的完整输出文本
+            
+        返回:
+            list[int] | None: 提取的符号序列（未经验证）
+        """
+        pattern = r'\[answer\](.*?)\[/answer\]'
+        matches = re.findall(pattern, output, flags=re.DOTALL)
+        if not matches:
+            return None
+        
+        last_answer = matches[-1].strip()
+        try:
+            solution = list(map(int, last_answer.split()))
+            if all(c in {1, -1} for c in solution):
+                return solution
+        except ValueError:
+            pass
+        return None
+    
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        """
+        验证答案的正确性
+        
+        参数:
+            solution: 提取的符号序列
+            identity: 谜题实例
+            
+        返回:
+            bool: 是否满足最终位置约束
+        """
+        n = identity['n']
+        vectors = identity['vectors']
+        
+        # 基础校验
+        if len(solution) != n or not all(c in {1, -1} for c in solution):
+            return False
+        
+        # 计算最终位置
+        total_x, total_y = 0, 0
+        for c, (x, y) in zip(solution, vectors):
+            total_x += c * x
+            total_y += c * y
+        
+        # 检查模长约束（平方运算避免浮点误差）
+        max_allowed_sq = (1.5e6) ** 2
+        total_sq = total_x ** 2 + total_y ** 2
+        return total_sq <= max_allowed_sq