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184 lines
4.9 KiB
Python
Executable file
184 lines
4.9 KiB
Python
Executable file
"""#
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### 谜题描述
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For a vector \vec{v} = (x, y), define |v| = √{x^2 + y^2}.
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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}.
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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.
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Input
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The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of moves.
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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.
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Output
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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.
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It can be shown that a solution always exists under the given constraints.
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Examples
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Input
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3
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999999 0
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0 999999
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999999 0
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Output
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1 1 -1
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Input
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1
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-824590 246031
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Output
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1
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Input
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8
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-67761 603277
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640586 -396671
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46147 -122580
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569609 -2112
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400 914208
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131792 309779
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-850150 -486293
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5272 721899
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Output
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1 1 1 1 1 1 1 -1
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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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# randomized greedy solution
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from sys import stdin, stdout
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from random import seed, shuffle
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seed(0x5ad5eed)
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n = input()
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inp = stdin.readlines()
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pts = [0 for i in xrange(n)]
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pos = [i for i in xrange(n)]
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ans = [0 for i in xrange(n)]
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i = 0
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for line in inp:
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pts[i] = tuple(map(int, line.split()))
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i += 1
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R = 1500000 ** 2
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while True:
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X, Y = 0, 0
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for i in pos:
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x1, y1 = X + pts[i][0], Y + pts[i][1]
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x2, y2 = X - pts[i][0], Y - pts[i][1]
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if (x1 * x1 + y1 * y1 < x2 * x2 + y2 * y2):
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ans[i] = 1
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X, Y = x1, y1
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else:
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ans[i] = -1
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X, Y = x2, y2
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if X * X + Y * Y <= R: break
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shuffle(pos)
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out = [str(ans[i]) for i in xrange(n)]
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stdout.write(' '.join(out))
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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 math
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import re
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from bootcamp import Basebootcamp
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class Cleavingthebarbootcamp(Basebootcamp):
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def __init__(self, min_n=1, max_n=1000, max_vector_length=10**6):
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self.min_n = min_n
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self.max_n = max_n
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self.max_vector_length = max_vector_length
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def case_generator(self):
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n = random.randint(self.min_n, self.max_n)
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vectors = []
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max_len = self.max_vector_length
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for _ in range(n):
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# 随机选择轴对齐或任意方向
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if random.random() < 0.3: # 30%概率生成轴对齐向量
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axis = random.choice(['x', 'y'])
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sign = random.choice([1, -1])
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r = random.randint(0, max_len)
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vec = (sign*r, 0) if axis == 'x' else (0, sign*r)
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else: # 70%概率生成任意方向向量
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while True:
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x = random.randint(-max_len, max_len)
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y = random.randint(-max_len, max_len)
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if x**2 + y**2 <= max_len**2:
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break
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vec = (x, y)
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vectors.append(vec)
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return {'n': n, 'vectors': vectors}
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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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vectors = question_case['vectors']
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vectors_str = '\n'.join(f"{x} {y}" for x, y in vectors)
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return f"""Allen醉后需要从酒吧原点出发完成{n}次移动，每次沿±向量方向移动。请选择移动方向使得最终位置距离原点不超过1.5×10^6。
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输入格式：
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第一行为n
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接下来n行每行两个整数x_i y_i
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输入：
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{n}
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{vectors_str}
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输出要求：
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一行n个1/-1，1表示沿原向量方向，-1表示反向
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答案请包裹在[answer]标签内，例如：[answer]1 -1 1 ...[/answer]"""
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@staticmethod
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def extract_output(output):
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matches = re.findall(r'\[answer\]\s*([1\s-]+)\s*\[/answer\]', output)
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if not matches:
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return None
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try:
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return list(map(int, matches[-1].strip().split()))
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except:
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return None
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@classmethod
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def _verify_correction(cls, solution, identity):
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if len(solution) != identity['n']:
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return False
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x_total, y_total = 0, 0
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for sign, (dx, dy) in zip(solution, identity['vectors']):
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x_total += sign * dx
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y_total += sign * dy
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return math.hypot(x_total, y_total) <= 1.5e6
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