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184 lines
6.3 KiB
Python
Executable file
184 lines
6.3 KiB
Python
Executable file
"""#
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### 谜题描述
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In the probability theory the following paradox called Benford's law is known: \"In many lists of random numbers taken from real sources, numbers starting with digit 1 occur much more often than numbers starting with any other digit\" (that's the simplest form of the law).
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Having read about it on Codeforces, the Hedgehog got intrigued by the statement and wishes to thoroughly explore it. He finds the following similar problem interesting in particular: there are N random variables, the i-th of which can take any integer value from some segment [Li;Ri] (all numbers from this segment are equiprobable). It means that the value of the i-th quantity can be equal to any integer number from a given interval [Li;Ri] with probability 1 / (Ri - Li + 1).
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The Hedgehog wants to know the probability of the event that the first digits of at least K% of those values will be equal to one. In other words, let us consider some set of fixed values of these random variables and leave only the first digit (the MSD — most significant digit) of each value. Then let's count how many times the digit 1 is encountered and if it is encountered in at least K per cent of those N values, than such set of values will be called a good one. You have to find the probability that a set of values of the given random variables will be a good one.
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Input
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The first line contains number N which is the number of random variables (1 ≤ N ≤ 1000). Then follow N lines containing pairs of numbers Li, Ri, each of whom is a description of a random variable. It is guaranteed that 1 ≤ Li ≤ Ri ≤ 1018.
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The last line contains an integer K (0 ≤ K ≤ 100).
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All the numbers in the input file are integers.
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Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).
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Output
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Print the required probability. Print the fractional number with such a precision that the relative or absolute error of the result won't exceed 10 - 9.
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Examples
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Input
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1
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1 2
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50
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Output
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0.500000000000000
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Input
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2
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1 2
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9 11
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50
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Output
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0.833333333333333
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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 = input()
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T = [1]+[0]*N
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for i in range(N):
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L,R = map(int,raw_input().split())
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c = 1
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n = 0
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while c<=10**18:
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n += max(0,min(R,2*c-1)-max(L,c)+1)
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c *= 10
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p = float(n)/(R-L+1)
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B = T
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T = [B[0]*(1-p)]+[0]*N
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for j in range(1,N+1):
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T[j] = B[j]*(1-p)+B[j-1]*p
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K = input()
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print sum(T[(N*K+99)/100:])
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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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from bootcamp import Basebootcamp
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import math
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import re
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import random
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class Cfirstdigitlawbootcamp(Basebootcamp):
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def __init__(self, max_N=10, **params):
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super().__init__(**params)
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self.max_N = max_N # 控制问题规模以便测试
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def case_generator(self):
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N = random.randint(1, self.max_N)
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variables = []
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for _ in range(N):
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# 生成更丰富的数值区间分布
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if random.random() < 0.3:
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# 特殊形式：生成以1开头的大范围数值
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exp = random.randint(0, 17)
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Li = 10**exp
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Ri = 10**exp * 2 - 1
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else:
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# 通用生成逻辑
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exp = random.randint(0, 18)
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Li = random.randint(10**exp // 10, 10**exp - 1) if exp > 0 else 1
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max_exp_ri = random.randint(exp, 18)
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max_ri = min(10**max_exp_ri - 1, 10**18)
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Ri = random.randint(Li, max_ri)
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variables.append({"L": Li, "R": Ri})
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K = random.randint(0, 100)
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return {
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"N": N,
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"variables": variables,
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"K": K
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}
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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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variables = question_case['variables']
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K = question_case['K']
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required_num = (N * K + 99) // 100
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problem = f"""根据本福特定律研究问题，请计算{N}个独立随机变量满足条件的概率：
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每个变量的取值范围如下："""
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for idx, var in enumerate(variables, 1):
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problem += f"\n变量{idx}: [{var['L']}, {var['R']}]，其中每个整数等概率出现"
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problem += f"""
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要求计算至少{required_num}个变量的最高位为1的概率（即至少达到{K}%的比例）。
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请输出精确到小数点后15位的概率值，并确保绝对/相对误差≤1e-9。答案请用[answer]和[/answer]标签包裹。
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示例：
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[answer]0.123456789012345[/answer]"""
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return problem
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@staticmethod
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def extract_output(output):
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matches = re.findall(r'\[answer\]([\d.eE+-]+)\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 float(matches[-1].strip().replace(',', ''))
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except ValueError:
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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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try:
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N = identity['N']
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variables = identity['variables']
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K = identity['K']
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# 动态规划数组初始化
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dp = [1.0] + [0.0] * N
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for var in variables:
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L, R = var['L'], var['R']
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total = R - L + 1
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# 计算有效数目
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valid_count = 0
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c = 1
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while c <= 1e18:
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lower_bound = max(L, c)
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upper_bound = min(R, 2 * c - 1)
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valid_count += max(0, upper_bound - lower_bound + 1)
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c *= 10
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# 更新概率分布
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p = valid_count / total
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new_dp = [0.0] * (N + 1)
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new_dp[0] = dp[0] * (1 - p)
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for j in range(1, N + 1):
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new_dp[j] = dp[j] * (1 - p) + dp[j-1] * p
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dp = new_dp
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# 计算阈值
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required = (N * K + 99) // 100
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threshold = max(0, min(required, N))
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correct_prob = sum(dp[threshold:]) if threshold <= N else 0.0
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# 浮点数精确校验
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return math.isclose(solution, correct_prob, rel_tol=1e-9, abs_tol=1e-12)
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except:
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return False
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