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267 lines
8.3 KiB
Python
Executable file
267 lines
8.3 KiB
Python
Executable file
"""#
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### 谜题描述
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There is an old tradition of keeping 4 boxes of candies in the house in Cyberland. The numbers of candies are special if their arithmetic mean, their median and their range are all equal. By definition, for a set {x1, x2, x3, x4} (x1 ≤ x2 ≤ x3 ≤ x4) arithmetic mean is <image>, median is <image> and range is x4 - x1. The arithmetic mean and median are not necessary integer. It is well-known that if those three numbers are same, boxes will create a \"debugging field\" and codes in the field will have no bugs.
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For example, 1, 1, 3, 3 is the example of 4 numbers meeting the condition because their mean, median and range are all equal to 2.
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Jeff has 4 special boxes of candies. However, something bad has happened! Some of the boxes could have been lost and now there are only n (0 ≤ n ≤ 4) boxes remaining. The i-th remaining box contains ai candies.
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Now Jeff wants to know: is there a possible way to find the number of candies of the 4 - n missing boxes, meeting the condition above (the mean, median and range are equal)?
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Input
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The first line of input contains an only integer n (0 ≤ n ≤ 4).
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The next n lines contain integers ai, denoting the number of candies in the i-th box (1 ≤ ai ≤ 500).
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Output
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In the first output line, print \"YES\" if a solution exists, or print \"NO\" if there is no solution.
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If a solution exists, you should output 4 - n more lines, each line containing an integer b, denoting the number of candies in a missing box.
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All your numbers b must satisfy inequality 1 ≤ b ≤ 106. It is guaranteed that if there exists a positive integer solution, you can always find such b's meeting the condition. If there are multiple answers, you are allowed to print any of them.
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Given numbers ai may follow in any order in the input, not necessary in non-decreasing.
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ai may have stood at any positions in the original set, not necessary on lowest n first positions.
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Examples
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Input
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2
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1
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1
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Output
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YES
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3
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3
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Input
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3
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1
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1
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1
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Output
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NO
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Input
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4
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1
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2
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2
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3
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Output
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YES
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Note
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For the first sample, the numbers of candies in 4 boxes can be 1, 1, 3, 3. The arithmetic mean, the median and the range of them are all 2.
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For the second sample, it's impossible to find the missing number of candies.
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In the third example no box has been lost and numbers satisfy the condition.
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You may output b in any order.
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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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#!/usr/bin/env python
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#-*- coding:utf-8 -*-
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import sys
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import math
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import random
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import operator
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from fractions import Fraction, gcd
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from decimal import Decimal, getcontext
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from itertools import product, permutations, combinations
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getcontext().prec = 100
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MOD = 10**9 + 7
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INF = float(\"+inf\")
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n = int(raw_input())
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arr = [int(raw_input()) for _ in range(n)]
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def check(arr):
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x = arr
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v1 = sum(x) / 4.0
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v2 = (x[1] + x[2]) / 2.0
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v3 = x[3] - x[0]
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return v1 == v2 == v3
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if len(arr) == 4:
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print \"YES\" if check(sorted(arr)) else \"NO\"
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quit()
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for x1 in xrange(1, 501):
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for x2 in xrange(1, 501):
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val = 2 * x1
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x4 = 3 * x1
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x3 = 2 * val - x2
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res = [x1, x2, x3, x4]
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if not (x1 <= x2 <= x3 <= x4):
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continue
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if not check(res):
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continue
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for v in arr:
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if v not in res:
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break
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res.remove(v)
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else:
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print \"YES\"
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# print res
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for c in res:
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print c
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quit()
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print \"NO\"
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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 Bcandyboxesbootcamp(Basebootcamp):
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def __init__(self, solution_prob=0.5, max_x1=166, max_retry=100):
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self.solution_prob = solution_prob
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self.max_x1 = max_x1
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self.max_retry = max_retry
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def case_generator(self):
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if random.random() < self.solution_prob:
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# 生成可解案例：四数满足特定结构
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x1 = random.randint(1, self.max_x1)
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x2 = random.randint(x1, 2 * x1)
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x3 = 4 * x1 - x2
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x4 = 3 * x1
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quadruple = sorted([x1, x2, x3, x4]) # 保证有序结构
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n = random.randint(0, 4)
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ai = random.sample(quadruple, n) if n else []
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return {'n': n, 'ai': ai}
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else:
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# 生成不可解案例，保证n与ai长度严格匹配
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n = random.choice([2, 3, 4])
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for _ in range(self.max_retry):
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ai = [random.randint(1, 500) for _ in range(n)]
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if not self._has_solution(n, ai):
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return {'n': n, 'ai': ai}
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# 回退策略：根据n生成已验证的不可解案例
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if n == 2:
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ai = [1, 5] # 无法形成有效四元组
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elif n == 3:
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ai = [1, 1, 1] # 三个相同数不可解
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else:
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ai = [1, 2, 3, 4] # 四数不符合条件
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return {'n': n, 'ai': ai}
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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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ai = question_case['ai']
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problem = f"""Jeff has 4 special boxes of candies where the arithmetic mean, median, and range must all be equal. Currently, {n} box(es) remain with these candy counts: {', '.join(map(str, ai)) if n else 'None'}.
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Determine if the missing boxes can be filled (1-1,000,000 candies each) to satisfy the condition. If possible, write "YES" followed by the missing numbers; otherwise, write "NO".
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Format your answer as:
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[answer]
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YES|NO
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[missing numbers (if YES)]
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[/answer]"""
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return problem
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@staticmethod
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def extract_output(output):
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# 提取最后一个answer块，确保捕获最终答案
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matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
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if not matches:
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return None
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content = matches[-1].strip()
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lines = [line.strip() for line in content.split('\n') if line.strip()]
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if not lines or lines[0].upper() not in ('YES', 'NO'):
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return None
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result = {'answer': lines[0].upper(), 'missing': []}
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if result['answer'] == 'YES' and len(lines) >= 2:
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for line in lines[1:]:
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if not line.isdigit():
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return None
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num = int(line)
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if not (1 <= num <= 10**6):
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return None
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result['missing'].append(num)
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return result
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@classmethod
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def _verify_correction(cls, solution, identity):
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answer = solution.get('answer')
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missing = solution.get('missing', [])
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n = identity['n']
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ai = identity['ai']
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if answer not in ('YES', 'NO'):
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return False
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if answer == 'YES':
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# 验证数量与数值范围
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if len(missing) != 4 - n or any(not (1 <= num <= 10**6) for num in missing):
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return False
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combined = sorted(ai + missing)
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# 检查四数条件
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mean = sum(combined) / 4
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median = (combined[1] + combined[2]) / 2
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range_val = combined[-1] - combined[0]
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return abs(mean - median) < 1e-9 and abs(median - range_val) < 1e-9
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else:
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# 检查确实无解
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return not cls._has_solution(n, ai)
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@classmethod
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def _has_solution(cls, n, ai):
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if n == 4:
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sorted_ai = sorted(ai)
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mean = sum(sorted_ai) / 4
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median = (sorted_ai[1] + sorted_ai[2]) / 2
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range_val = sorted_ai[-1] - sorted_ai[0]
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return abs(mean - median) < 1e-9 and abs(median - range_val) < 1e-9
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# 遍历所有可能四元组结构
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for x1 in range(1, 167):
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for x2 in range(x1, 2 * x1 + 1):
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x3 = 4 * x1 - x2
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x4 = 3 * x1
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quad = sorted([x1, x2, x3, x4])
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temp = quad.copy()
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try:
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for num in ai:
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temp.remove(num)
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except ValueError:
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continue # ai包含不在四元组中的数
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if len(temp) == 4 - n:
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return True
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return False
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