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internbootcamp/bootcamp/cperfecttriples/cperfecttriples.py

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+"""# 
+
+### 谜题描述
+Consider the infinite sequence s of positive integers, created by repeating the following steps:
+
+  1. Find the lexicographically smallest triple of positive integers (a, b, c) such that 
+    * a ⊕ b ⊕ c = 0, where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). 
+    * a, b, c are not in s. 
+Here triple of integers (a_1, b_1, c_1) is considered to be lexicographically smaller than triple (a_2, b_2, c_2) if sequence [a_1, b_1, c_1] is lexicographically smaller than sequence [a_2, b_2, c_2]. 
+  2. Append a, b, c to s in this order. 
+  3. Go back to the first step. 
+
+
+
+You have integer n. Find the n-th element of s.
+
+You have to answer t independent test cases.
+
+A sequence a is lexicographically smaller than a sequence b if in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
+
+Input
+
+The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
+
+Each of the next t lines contains a single integer n (1≤ n ≤ 10^{16}) — the position of the element you want to know.
+
+Output
+
+In each of the t lines, output the answer to the corresponding test case.
+
+Example
+
+Input
+
+
+9
+1
+2
+3
+4
+5
+6
+7
+8
+9
+
+
+Output
+
+
+1
+2
+3
+4
+8
+12
+5
+10
+15
+
+Note
+
+The first elements of s are 1, 2, 3, 4, 8, 12, 5, 10, 15, ... 
+
+Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
+```python
+from datetime import datetime
+from random import randint
+import sys
+
+__author__ = 'ratmir'
+
+[t] = [int(x) for x in sys.stdin.readline().split()]
+
+
+def getFirstInTriple(count):
+    st2 = 1
+    while st2 < 3 * count:
+        st2 <<= 2
+    st2 >>= 2
+    firstInTriple = st2 + count - ((st2 - 1) / 3) - 1
+    return firstInTriple
+
+
+def getValue(position):
+    tripleIndex = 1 + (position - 1) / 3
+    firstInTriple = getFirstInTriple(tripleIndex)
+    if position % 3 == 1:
+        return firstInTriple
+    if position % 3 == 2:
+        value = 1
+        res = 0
+        while firstInTriple > 0:
+            x = firstInTriple & 3
+            if x == 1:
+                res += value << 1
+            if x == 2:
+                res += 3 * value
+            if x == 3:
+                res += value
+            value <<= 2
+            firstInTriple >>= 2
+        return res
+    if position % 3 == 0:
+        value = 1
+        res = 0
+        while firstInTriple > 0:
+            x = firstInTriple & 3
+            if x == 1:
+                res += 3 * value
+            if x == 3:
+                res += value << 1
+            if x == 2:
+                res += value
+            value <<= 2
+            firstInTriple >>= 2
+        return res
+
+output = \"\"
+
+# start_time = datetime.now()
+#
+# for j in range(0,10):
+#     for i in range(0, 100000):
+#         getValue(9999999999900001 + i)
+#
+# print(datetime.now() - start_time)
+
+for i in range(0, t):
+    position = int(sys.stdin.readline())
+    output += str(getValue(position)) + \"\n\"
+
+print output
+```
+
+
+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
+from bootcamp import Basebootcamp
+import random
+import re
+
+class Cperfecttriplesbootcamp(Basebootcamp):
+    def __init__(self, min_n=1, max_n=10**6):  # 默认最大值调整为1e6
+        self.min_n = min_n
+        self.max_n = max_n
+    
+    def case_generator(self):
+        # 生成策略：25%小值，25%中等值，50%参数范围
+        rand = random.random()
+        if rand < 0.25:
+            n = random.randint(1, 10)        # 基础测试样例区
+        elif rand < 0.5:
+            n = random.randint(100, 10**4)   # 中等规模测试区
+        else:
+            n = random.randint(self.min_n, self.max_n)
+        return {'n': n}
+    
+    @staticmethod
+    def prompt_func(question_case):
+        n = question_case['n']
+        return f"""Given n = {n}, compute the n-th element in the XOR triple sequence. Rules:
+1. Sequence is built by adding lex smallest (a,b,c) with a^b^c=0
+2. Each triple's elements are appended in order
+3. Sequence starts with 1,2,3,4,8,12,5,10,15...
+
+Put your final answer within [answer] tags like: [answer]42[/answer]"""
+    
+    @staticmethod
+    def extract_output(output):
+        matches = re.findall(r'\[answer\]\s*(\d+)\s*\[/answer\]', output, re.IGNORECASE)
+        return int(matches[-1]) if matches else None
+
+    @classmethod
+    def _get_st2(cls, count):
+        """ 优化st2计算：通过位长度快速定位起始点 """
+        if count == 0:
+            return 0
+        target = 3 * count
+        bit_len = target.bit_length()
+        exponent = (bit_len + 1) // 2  # 4^exponent初始估算
+        st2 = 1 << (2 * exponent)
+        
+        # 精确调整
+        while st2 > target:
+            exponent -= 1
+            st2 >>= 2
+        while st2 * 4 <= target:
+            st2 <<= 2
+        return st2
+
+    @classmethod
+    def _getFirstInTriple(cls, count):
+        st2 = cls._get_st2(count)
+        return st2 + count - (st2 - 1) // 3 - 1
+
+    @classmethod
+    def _getValue(cls, position):
+        # 保持原算法结构，优化计算效率
+        triple_index = (position + 2) // 3
+        first = cls._getFirstInTriple(triple_index)
+        
+        mod = position % 3
+        if mod == 1:
+            return first
+        
+        # 公共计算逻辑提取
+        res = 0
+        value = 1
+        f = first
+        while f > 0:
+            x = f & 3
+            if mod == 2:
+                res += (value << 1) if x == 1 else (3*value if x ==2 else value)
+            else:
+                res += (3*value) if x ==1 else (value<<1 if x==3 else value)
+            value <<= 2
+            f >>= 2
+        return res
+
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        try:
+            return solution == cls._getValue(identity['n'])
+        except:
+            return False