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301 lines
12 KiB
Python
Executable file
301 lines
12 KiB
Python
Executable file
"""#
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### 谜题描述
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After defeating a Blacklist Rival, you get a chance to draw 1 reward slip out of x hidden valid slips. Initially, x=3 and these hidden valid slips are Cash Slip, Impound Strike Release Marker and Pink Slip of Rival's Car. Initially, the probability of drawing these in a random guess are c, m, and p, respectively. There is also a volatility factor v. You can play any number of Rival Races as long as you don't draw a Pink Slip. Assume that you win each race and get a chance to draw a reward slip. In each draw, you draw one of the x valid items with their respective probabilities. Suppose you draw a particular item and its probability of drawing before the draw was a. Then,
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* If the item was a Pink Slip, the quest is over, and you will not play any more races.
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* Otherwise,
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1. If a≤ v, the probability of the item drawn becomes 0 and the item is no longer a valid item for all the further draws, reducing x by 1. Moreover, the reduced probability a is distributed equally among the other remaining valid items.
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2. If a > v, the probability of the item drawn reduces by v and the reduced probability is distributed equally among the other valid items.
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For example,
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* If (c,m,p)=(0.2,0.1,0.7) and v=0.1, after drawing Cash, the new probabilities will be (0.1,0.15,0.75).
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* If (c,m,p)=(0.1,0.2,0.7) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,0.25,0.75).
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* If (c,m,p)=(0.2,Invalid,0.8) and v=0.1, after drawing Cash, the new probabilities will be (0.1,Invalid,0.9).
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* If (c,m,p)=(0.1,Invalid,0.9) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,Invalid,1.0).
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You need the cars of Rivals. So, you need to find the expected number of races that you must play in order to draw a pink slip.
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Input
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The first line of input contains a single integer t (1≤ t≤ 10) — the number of test cases.
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The first and the only line of each test case contains four real numbers c, m, p and v (0 < c,m,p < 1, c+m+p=1, 0.1≤ v≤ 0.9).
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Additionally, it is guaranteed that each of c, m, p and v have at most 4 decimal places.
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Output
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For each test case, output a single line containing a single real number — the expected number of races that you must play in order to draw a Pink Slip.
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Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.
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Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.
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Example
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Input
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4
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0.2 0.2 0.6 0.2
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0.4 0.2 0.4 0.8
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0.4998 0.4998 0.0004 0.1666
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0.3125 0.6561 0.0314 0.2048
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Output
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1.532000000000
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1.860000000000
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5.005050776521
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4.260163673896
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Note
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For the first test case, the possible drawing sequences are:
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* P with a probability of 0.6;
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* CP with a probability of 0.2⋅ 0.7 = 0.14;
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* CMP with a probability of 0.2⋅ 0.3⋅ 0.9 = 0.054;
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* CMMP with a probability of 0.2⋅ 0.3⋅ 0.1⋅ 1 = 0.006;
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* MP with a probability of 0.2⋅ 0.7 = 0.14;
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* MCP with a probability of 0.2⋅ 0.3⋅ 0.9 = 0.054;
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* MCCP with a probability of 0.2⋅ 0.3⋅ 0.1⋅ 1 = 0.006.
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So, the expected number of races is equal to 1⋅ 0.6 + 2⋅ 0.14 + 3⋅ 0.054 + 4⋅ 0.006 + 2⋅ 0.14 + 3⋅ 0.054 + 4⋅ 0.006 = 1.532.
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For the second test case, the possible drawing sequences are:
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* P with a probability of 0.4;
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* CP with a probability of 0.4⋅ 0.6 = 0.24;
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* CMP with a probability of 0.4⋅ 0.4⋅ 1 = 0.16;
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* MP with a probability of 0.2⋅ 0.5 = 0.1;
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* MCP with a probability of 0.2⋅ 0.5⋅ 1 = 0.1.
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So, the expected number of races is equal to 1⋅ 0.4 + 2⋅ 0.24 + 3⋅ 0.16 + 2⋅ 0.1 + 3⋅ 0.1 = 1.86.
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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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PREC = 0.00000005
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def floatEq(a, b):
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return abs(a-b) < PREC
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def f(c, m, p, v, n):
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r = (n+1)*p
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if c is not None and (c < v or floatEq(c, v)):
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if m is not None:
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r += c*f(None, m+c/2.0, p+c/2.0, v, n+1)
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else:
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r += c*f(None, None, p+c, v, n+1)
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elif c is not None and c > v:
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if m is not None:
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r += c*f(c-v, m+v/2.0, p+v/2.0, v, n+1)
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else:
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r += c*f(c-v, None, p+v, v, n+1)
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if m is not None and (m < v or floatEq(m, v)):
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if c is not None:
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r += m*f(c+m/2.0, None, p+m/2.0, v, n+1)
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else:
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r += m*f(None, None, p+m, v, n+1)
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elif m is not None and m > v:
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if c is not None:
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r += m*f(c+v/2.0, m-v, p+v/2.0, v, n+1)
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else:
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r += m*f(None, m-v, p+v, v, n+1)
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return r
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I=lambda: map(float, raw_input().split())
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t = input()
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for _ in xrange(t):
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c, m, p, v = I()
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print f(c, m, p, v, 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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class Cneedforpinkslipsbootcamp(Basebootcamp):
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def __init__(self, **params):
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self.params = params
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super().__init__(**params)
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def case_generator(self):
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# 随机选择生成类型：普通案例、含极值p、初始值触发无效条件
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case_type = random.choice(['normal', 'extreme_p', 'low_prob'])
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if case_type == 'normal':
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# 标准生成方式
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a = random.randint(1, 9998)
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max_b = 10000 - a - 1 # 保证至少留1给d
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b = random.randint(1, max_b)
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d = 10000 - a - b
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c = round(a / 10000, 4)
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m = round(b / 10000, 4)
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p = round(d / 10000, 4)
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v = round(random.randint(1000, 9000) / 10000, 4)
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elif case_type == 'extreme_p':
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# 生成极小的p值（类似第三个测试案例）
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p = round(random.choice([0.0001, 0.0002, 0.0003, 0.0004]), 4)
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remaining = round(1 - p, 4)
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c = round(remaining * random.uniform(0.4, 0.6), 4)
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m = round(remaining - c, 4)
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v = round(random.uniform(0.15, 0.25), 4)
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elif case_type == 'low_prob':
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# 生成会立即触发无效条件的案例
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v_val = round(random.uniform(0.1, 0.3), 4)
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target = random.choice(['c', 'm'])
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if target == 'c':
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c = round(random.uniform(0.01, v_val), 4)
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remaining = 1 - c
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m = round(remaining * random.uniform(0.3, 0.7), 4)
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p = round(remaining - m, 4)
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else:
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m = round(random.uniform(0.01, v_val), 4)
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remaining = 1 - m
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c = round(remaining * random.uniform(0.3, 0.7), 4)
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p = round(remaining - c, 4)
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v = v_val
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# 校验并修正浮点精度
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total = round(c + m + p, 4)
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if total != 1.0:
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p = round(1 - c - m, 4)
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return {
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'c': c,
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'm': m,
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'p': p,
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'v': v
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}
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@staticmethod
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def prompt_func(question_case):
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c = question_case['c']
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m = question_case['m']
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p = question_case['p']
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v = question_case['v']
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prompt = f"""你刚刚击败了一个黑名单对手，获得了一个抽取奖励纸条的机会。初始时有三个有效纸条：现金纸条（Cash Slip，初始概率为{c:.4f}）、扣押解除标记（Impound Strike Release Marker，初始概率为{m:.4f}）和对手车辆的粉色纸条（Pink Slip，初始概率为{p:.4f}）。每次比赛获胜后，你可以抽取一张纸条。只要未抽到粉色纸条，你就可以继续参赛。每次抽取后，根据以下规则调整剩余纸条的概率：
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1. 如果抽到的是粉色纸条，游戏结束，不再进行任何比赛。
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2. 如果抽到的是其他纸条（如现金或解除标记）：
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- 若该纸条当前的抽取概率a小于等于波动因子v（当前v={v:.4f}），则该纸条变为无效（概率变为0），其原有概率平均分配给剩余的合法纸条。
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- 若该纸条的概率a大于v，则其概率减少v，减少的部分平均分配给其他合法纸条。
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你的任务是计算需要进行的比赛的期望次数（即抽到粉色纸条之前的总抽取次数）。
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输入数据为四个实数：{c:.4f} {m:.4f} {p:.4f} {v:.4f}。请严格根据上述规则计算期望值，并确保答案的绝对或相对误差不超过1e-6。答案必须包含足够多的小数位（例如，类似1.532000000000的格式）。
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请将计算出的期望值放在[answer]和[/answer]的标签内。例如：[answer]1.532000000000[/answer]。"""
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return prompt
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@staticmethod
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def extract_output(output):
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matches = re.findall(r'\[answer\]\s*([\d.]+)\s*\[/answer\]', output, re.IGNORECASE)
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if not matches:
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return None
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try:
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return float(matches[-1].strip())
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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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expected = cls._calculate_expected(identity['c'], identity['m'], identity['p'], identity['v'])
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if solution is None:
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return False
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try:
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solution_float = float(solution)
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except:
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return False
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# 双重验证机制
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abs_error = abs(solution_float - expected)
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if abs_error < 1e-6:
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return True
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rel_error = abs_error / max(1.0, abs(expected))
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return rel_error <= 1e-6
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@staticmethod
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def _calculate_expected(c_val, m_val, p_val, v_val):
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PREC = 1e-8
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def float_eq(a, b):
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return abs(a - b) < PREC
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memo = {}
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def f(c, m, p, n):
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key = (c if c is None else round(c,8),
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m if m is None else round(m,8),
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n)
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if key in memo:
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return memo[key]
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total = (n+1)*p
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remaining = []
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if c is not None:
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remaining.append(('c', c))
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if m is not None:
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remaining.append(('m', m))
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for item, value in remaining:
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if item == 'c':
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current_c = value
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if current_c < v_val or float_eq(current_c, v_val):
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# Case 1: becomes invalid
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if m is not None:
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delta = current_c / 2
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contribution = current_c * f(None, m+delta, p+delta, n+1)
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else:
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contribution = current_c * f(None, None, p+current_c, n+1)
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else:
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# Case 2: reduce by v
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new_c = current_c - v_val
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if m is not None:
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delta = v_val / 2
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contribution = current_c * f(new_c, m+delta, p+delta, n+1)
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else:
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contribution = current_c * f(new_c, None, p+v_val, n+1)
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total += contribution
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elif item == 'm':
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current_m = value
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if current_m < v_val or float_eq(current_m, v_val):
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# Case 1: becomes invalid
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if c is not None:
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delta = current_m / 2
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contribution = current_m * f(c+delta, None, p+delta, n+1)
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else:
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contribution = current_m * f(None, None, p+current_m, n+1)
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else:
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# Case 2: reduce by v
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new_m = current_m - v_val
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if c is not None:
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delta = v_val / 2
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contribution = current_m * f(c+delta, new_m, p+delta, n+1)
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else:
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contribution = current_m * f(None, new_m, p+v_val, n+1)
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total += contribution
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memo[key] = total
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return total
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c = c_val if c_val > 1e-8 else None
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m = m_val if m_val > 1e-8 else None
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p = p_val
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return f(c, m, p, 0)
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