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internbootcamp/bootcamp/ckuroniandimpossiblecalculation/ckuroniandimpossiblecalculation.py

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+"""# 
+
+### 谜题描述
+To become the king of Codeforces, Kuroni has to solve the following problem.
+
+He is given n numbers a_1, a_2, ..., a_n. Help Kuroni to calculate ∏_{1≤ i<j≤ n} |a_i - a_j|. As result can be very big, output it modulo m.
+
+If you are not familiar with short notation, ∏_{1≤ i<j≤ n} |a_i - a_j| is equal to |a_1 - a_2|⋅|a_1 - a_3|⋅ ... ⋅|a_1 - a_n|⋅|a_2 - a_3|⋅|a_2 - a_4|⋅ ... ⋅|a_2 - a_n| ⋅ ... ⋅ |a_{n-1} - a_n|. In other words, this is the product of |a_i - a_j| for all 1≤ i < j ≤ n.
+
+Input
+
+The first line contains two integers n, m (2≤ n ≤ 2⋅ 10^5, 1≤ m ≤ 1000) — number of numbers and modulo.
+
+The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9).
+
+Output
+
+Output the single number — ∏_{1≤ i<j≤ n} |a_i - a_j| mod m.
+
+Examples
+
+Input
+
+
+2 10
+8 5
+
+
+Output
+
+
+3
+
+Input
+
+
+3 12
+1 4 5
+
+
+Output
+
+
+0
+
+Input
+
+
+3 7
+1 4 9
+
+
+Output
+
+
+1
+
+Note
+
+In the first sample, |8 - 5| = 3 ≡ 3 mod 10.
+
+In the second sample, |1 - 4|⋅|1 - 5|⋅|4 - 5| = 3⋅ 4 ⋅ 1 = 12 ≡ 0 mod 12.
+
+In the third sample, |1 - 4|⋅|1 - 9|⋅|4 - 9| = 3 ⋅ 8 ⋅ 5 = 120 ≡ 1 mod 7.
+
+Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
+```python
+from collections import defaultdict as dd, Counter, OrderedDict as od, deque
+from itertools import combinations as comb, permutations as perm
+from functools import reduce
+from math import log, ceil, floor
+from fractions import gcd
+
+import os, sys
+
+#sys.stdin = open(\"pb.txt\", \"r\")
+raw_input = sys.stdin.readline
+
+int_input = lambda : int(raw_input())
+list_input = lambda : list(raw_input().strip())
+explode_input = lambda sep :raw_input().split(sep)
+format_input = lambda object : map(object, raw_input().split())
+
+
+n, m = format_input(int)
+a = format_input(int)
+
+if n > m:
+	print 0
+else:
+	cum = 1
+	n_itr1 = 0
+	while cum > 0 and n_itr1 < n:
+		for n_itr2 in xrange(n_itr1 + 1, n):
+			cum = (cum * abs(a[n_itr1] - a[n_itr2])) % m
+		n_itr1 += 1
+	print cum
+```
+
+
+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 random
+import re
+from collections import defaultdict
+from bootcamp import Basebootcamp
+
+class Ckuroniandimpossiblecalculationbootcamp(Basebootcamp):
+    def __init__(self, **params):
+        self.min_n = params.get('min_n', 2)
+        self.max_n = params.get('max_n', 20)
+        self.min_m = params.get('min_m', 1)
+        self.max_m = params.get('max_m', 1000)
+        self.a_min = params.get('a_min', 0)
+        self.a_max = params.get('a_max', 10**9)
+        self.n_gt_m_prob = params.get('n_gt_m_prob', 0.5)
+
+    def case_generator(self):
+        if random.random() < self.n_gt_m_prob:
+            # 确保m的取值范围适配当前max_n
+            m_upper = min(self.max_m, self.max_n-1)
+            if m_upper < self.min_m:
+                m = self.max_n  # 当无法满足则转为生成n <= m模式
+                n = random.randint(self.min_n, min(self.max_n, m))
+            else:
+                m = random.randint(self.min_m, m_upper)
+                n = random.randint(m+1, self.max_n)
+            a = [random.randint(self.a_min, self.a_max) for _ in range(n)]
+            return {
+                'n': n,
+                'm': m,
+                'a': a,
+                'expected': 0
+            }
+        else:
+            # 生成n <= m的案例
+            m = random.randint(self.min_m, self.max_m)
+            n = random.randint(self.min_n, min(self.max_n, m))
+            mods = defaultdict(int)
+            duplicates = False
+            a = []
+            for _ in range(n):
+                val = random.randint(self.a_min, self.a_max)
+                mod = val % m
+                if mods[mod] >= 1:
+                    duplicates = True
+                mods[mod] += 1
+                a.append(val)
+            
+            expected = 0 if duplicates else self._safe_compute(n, m, a)
+            return {
+                'n': n,
+                'm': m,
+                'a': a,
+                'expected': expected
+            }
+
+    def _safe_compute(self, n, m, a):
+        """安全计算模式（限制n不超过100）"""
+        if n > 100:
+            return 0
+        product = 1
+        for i in range(n):
+            for j in range(i+1, n):
+                product = (product * abs(a[i]-a[j])) % m
+        return product
+
+    @staticmethod
+    def prompt_func(question_case) -> str:
+        input_case = f"{question_case['n']} {question_case['m']}\n{' '.join(map(str, question_case['a']))}"
+        return f"""编程竞赛题目：
+计算所有两两绝对差值的乘积模m
+
+输入格式：
+第一行：n m（2≤n≤2e5，1≤m≤1000）
+第二行：a_1 a_2 ... a_n（0≤a_i≤1e9）
+
+示例说明：
+当n>m时结果为0，否则计算各对差值的乘积取模
+
+当前题目：
+输入：
+{input_case}
+
+请将最终答案用[answer]标签包裹，例如：[answer]0[/answer]"""
+
+    @staticmethod
+    def extract_output(output):
+        matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
+        if not matches:
+            return None
+        try:
+            return int(matches[-1].strip())
+        except:
+            return None
+
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        return solution == identity['expected']