"""# ### 谜题描述 Ksenia has an array a consisting of n positive integers a_1, a_2, …, a_n. In one operation she can do the following: * choose three distinct indices i, j, k, and then * change all of a_i, a_j, a_k to a_i ⊕ a_j ⊕ a_k simultaneously, where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). She wants to make all a_i equal in at most n operations, or to determine that it is impossible to do so. She wouldn't ask for your help, but please, help her! Input The first line contains one integer n (3 ≤ n ≤ 10^5) — the length of a. The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — elements of a. Output Print YES or NO in the first line depending on whether it is possible to make all elements equal in at most n operations. If it is possible, print an integer m (0 ≤ m ≤ n), which denotes the number of operations you do. In each of the next m lines, print three distinct integers i, j, k, representing one operation. If there are many such operation sequences possible, print any. Note that you do not have to minimize the number of operations. Examples Input 5 4 2 1 7 2 Output YES 1 1 3 4 Input 4 10 4 49 22 Output NO Note In the first example, the array becomes [4 ⊕ 1 ⊕ 7, 2, 4 ⊕ 1 ⊕ 7, 4 ⊕ 1 ⊕ 7, 2] = [2, 2, 2, 2, 2]. Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution. ```python from __future__ import print_function # for PyPy2 from collections import Counter, OrderedDict from itertools import permutations as perm from collections import deque from sys import stdin from bisect import * from heapq import * import math g = lambda : stdin.readline().strip() gl = lambda : g().split() gil = lambda : [int(var) for var in gl()] gfl = lambda : [float(var) for var in gl()] gcl = lambda : list(g()) gbs = lambda : [int(var) for var in g()] mod = int(1e9)+7 inf = float(\"inf\") n, = gil() a = gil() xor = 0 ans = [] if ~n&1 : n-=1 for v in a: xor ^= v if xor : print(\"NO\") else: print(\"YES\") print(n-2) i = 3 while i<= n: # print(i-2, i-1, i) ans.append(str(i-2)+\" \"+str(i-1)+\" \"+str(i)+\"\n\") i += 2 i=n-2 while i > 1: # print(i-2, i-1, i) ans.append(str(i-2)+\" \"+str(i-1)+\" \"+str(i)+\"\n\") i -= 2 print(\"\".join(ans)) ``` 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 bootcamp import Basebootcamp class Dpowerfulkseniabootcamp(Basebootcamp): def __init__(self, min_n=3, max_n=10, possible=None): self.min_n = max(3, min_n) self.max_n = max(self.min_n, max_n) self.possible = possible def case_generator(self): if self.possible is not None: should_generate_possible = self.possible else: should_generate_possible = random.choice([True, False]) if should_generate_possible: # 确定可行的生成类型 generate_odd = False possible_odd = [x for x in range(self.min_n, self.max_n+1) if x%2 == 1] possible_even = [x for x in range(max(4, self.min_n), self.max_n+1) if x%2 == 0] # 动态选择生成类型 if possible_odd and possible_even: generate_odd = random.choice([True, False]) elif possible_odd: generate_odd = True elif possible_even: generate_odd = False else: raise ValueError("No valid n in given range") if generate_odd: n = random.choice(possible_odd) elements = [random.randint(1, 10) for _ in range(n)] return {'n': n, 'a': elements} else: n = random.choice(possible_even) elements = [random.randint(1, 10) for _ in range(n-1)] total_xor = 0 for num in elements: total_xor ^= num elements.append(total_xor) return {'n': n, 'a': elements} else: # 生成不可能的偶数案例 possible_even = [x for x in range(max(4, self.min_n), self.max_n+1) if x%2 == 0] if not possible_even: raise ValueError("No even n in given range for impossible case") n = random.choice(possible_even) elements = [random.randint(1, 10) for _ in range(n-1)] total_xor = 0 for num in elements: total_xor ^= num elements.append(total_xor ^ random.randint(1, 10)) return {'n': n, 'a': elements} @staticmethod def prompt_func(question_case): n = question_case['n'] a = question_case['a'] return ( f"Problem: Given array of {n} integers: {', '.join(map(str, a))}\n" "Operation: Choose 3 distinct indices and set all to their XOR\n" "Task: Determine if achievable in ≤n operations\n" "Answer format:\n" "[answer]\n" "YES/NO\n" "m\n" "i j k\n" "...\n" "[/answer]\n" "Note: Indices must be 1-based" ) @staticmethod def extract_output(output): answer_blocks = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL) if not answer_blocks: return None return answer_blocks[-1].strip() @classmethod def _verify_correction(cls, solution, identity): try: lines = [l.strip() for l in solution.split('\n') if l.strip()] if not lines: return False # 验证首行有效性 first_line = lines[0].upper() if first_line not in {'YES', 'NO'}: return False # 获取原始数据 n = identity['n'] a = identity['a'].copy() total_xor = 0 for num in a: total_xor ^= num # 校验理论正确性 expected = 'YES' if (n%2 == 1) or (total_xor == 0) else 'NO' if first_line != expected: return False # NO情况直接返回正确 if expected == 'NO': return True # 验证操作步骤 if len(lines) < 2: return False # 缺失操作数行 try: m = int(lines[1]) if m < 0 or m > n: return False except ValueError: return False # 验证每个操作 operations = [] arr = a.copy() for line in lines[2:2+m]: if not line: continue parts = line.split() if len(parts) != 3: return False try: i, j, k = map(int, parts) if len({i, j, k}) != 3 or any(x < 1 or x > n for x in (i, j, k)): return False # 转换为0-based索引 i -= 1 j -= 1 k -= 1 # 执行操作 xor_val = arr[i] ^ arr[j] ^ arr[k] arr[i] = arr[j] = arr[k] = xor_val except (ValueError, IndexError): return False # 检查最终统一性 return all(x == arr[0] for x in arr) except Exception: return False