c++中文C. Breach of Faith time limit per test2 seconds memory limit per test256 megabytes Breach of Faith - Supire feat.eili You and your team have worked tirelessly until you have a sequence a1,a2,…,a2n+1 of positive integers satisfying these properties. 1≤ai≤1018 for all 1≤i≤2n+1 . a1,a2,…,a2n+1 are pairwise distinct. a1=a2−a3+a4−a5+…+a2n−a2n+1 . However, the people you worked with sabotaged you because they wanted to publish this sequence first. They deleted one number from this sequence and shuffled the rest, leaving you with a sequence b1,b2,…,b2n . You have forgotten the sequence a and want to find a way to recover it. If there are many possible sequences, you can output any of them. It can be proven under the constraints of the problem that at least one sequence a exists. Input Each test contains multiple test cases. The first line contains the number of test cases t (1≤t≤104 ). The description of the test cases follows. The first line of each test case contains one integer n (1≤n≤2⋅105 ). The second line of each test case contains 2n distinct integers b1,b2,…,b2n (1≤bi≤109 ), denoting the sequence b . It is guaranteed that the sum of n over all test cases does not exceed 2⋅105 . Output For each test case, output 2n+1 distinct integers, denoting the sequence a (1≤ai≤1018 ). If there are multiple possible sequences, you can output any of them. The sequence a should satisfy the given conditions, and it should be possible to obtain b after deleting one element from a and shuffling the remaining elements. Example InputCopy 4 1 9 2 2 8 6 1 4 3 99 2 86 33 14 77 2 1 6 3 2 OutputCopy 7 9 2 1 8 4 6 9 86 99 2 77 69 14 33 4 6 1 2 3 1

从题目描述来看，这是一个关于序列恢复的问题，涉及数学推导和逻辑推理。我们先理解给定条件并尝试找到解决办法。 ### 题目分析： 1. **原始序列** \(a_1, a_2, \ldots, a_{2n+1}\) 满足两个关键性质： - 所有元素各不相同，并且范围在 [1, 10^18] 内。 - 序列满足公式：\[ a_1 = a_2 - a_3 + a_4 - a_5 + \cdots + a_{2n} - a_{2n+1} \] 2. **破坏后的序列** \(b\) 包含了原序列删除一个数之后剩下的所有其他数字，并被打乱顺序。 目标是从 \(b\) 中恢复出符合条件的 \(a\) 序列。我们知道一定存在这样的解法。 #### 解题思路： 为了找回缺失的那个数，我们可以考虑利用给定公式的特性去逆向求得它。 设被删掉的是第 \(k\) 位，则根据上述等式可以推出： - 如果 \( k=1 \)，那么新的第一项 \(a'_1\)=\(b[1]-b[2]+b[3]-...\pm b[2*n]\) - 否则 (\(k>1\))，此时将 \(b\) 看作去掉了一对加减操作中的某个值（如原本应该是“x-y”现在只剩下了另一个），因此我们需要调整剩余数值使得它们能构成正确的交错正负组合。 通过遍历检查每一个位置作为可能丢失的位置，计算对应的首项是否合理并且符合约束即可确定最终结果。 接下来是一个简单的算法框架用于解决问题： ```python def solve_breach_of_faith(t_cases): results = [] for _ in range(t_cases): # Read input data n = int(input()) b = list(map(int, input().split())) found_solution = False # Try every position as potentially missing value's index for i in range(len(b)+1): # Considering insertion point temp_sequence = b.copy() if i != len(temp_sequence): # Insert placeholder at ith place; will replace later based on calculations inserted_value = None adjusted_list_with_placeholder = ( [None]*(i) + [-sum((-1)**j * x for j,x in enumerate(b[:]))] + b[i:] ) expected_sum = sum( (-1)**idx * val for idx,val in enumerate([adjusted_list_with_placeholder[j] for j in range(i)]+[inserted_value]+[ adjusted_list_with_placeholder[j+i] for j in range(n)]) ) if min(adjusted_list_with_placeholder)>=1 and max(adjusted_list_with_placeholder)<=1e18: all_unique=True seen=set() for elem in adjusted_list_with_placeholder: if elem is not None: if elem in seen: all_unique=False break else: seen.add(elem) if all_unique and abs(expected_sum-inserted_value)<1e-9: result=[elem or inserted_value for elem in adjusted_list_with_placeholder] print(' '.join(str(x)for x in result)) found_solution = True break assert found_solution,"No valid solution was found." ``` 请注意这个示例代码并未完全优化也未经过全面测试，在实际提交前还需要进一步完善调试以确保其正确性和性能效率。 ### 示例解释 对于样例输入 `4\n1\n9 2` 的处理过程大致如下： 由于只有一个奇偶交替差分运算符 `-`, 我们可以直接猜测 9 或者 2 其中之一可能是缺失的那一位，然后验证哪个选择让新生成的第一项能够成立同时保持其余各项唯一非空合法范围内取值。 在这个例子中，假设缺少第一个元素时，容易发现选取 7 可以形成有效的 `(7, 9, 2)` 符合条件。 --- 希望以上解答对你有所帮助！