Gyikoo/KT_QA huggingface dataset:
问题：Consider inserting a target element e into a BST. when it fails to be found during the search, which node does _hot correspond to? 对BST进行插入操作，对待插入的目标元素e进行查找后，若查找失败，_hot指向的节点为： {'A': 'The node to be inserted 待插入的节点', 'B': 'The parent of e after the insertion e被插入后的父亲', 'C': 'The left-child of e after the insertion e被插入后的左孩子', 'D': 'The root 根节点'}
答案：B
问题：When atempting to delete a node v of degree 2, what&#39;s the actual node being deleted? 当欲删除的节点v在BST中的度为2时，实际被删除的节点为： {'A': 'v&#39;s direct predecessor in the in-order sequence v在中序遍历下的直接前驱', 'B': 'v&#39;s direct successor in the pre-order sequence v在先序遍历下的直接后继', 'C': 'the last node in the left branch of v&#39;s right sub-tree v的右子树中左侧分支的最后一个节点', 'D': 'v&#39;s parent v的父亲'}
答案：C
问题：A BST has n nodes and a height of h. What is the worst-case time complexity for searching in it? 在含n个节点、高度为h的BST中进行查找的最坏时间复杂度为： {'A': '$O(h)$', 'B': '$O(n)$', 'C': '$O(nh)$', 'D': '$O(n/h)$'}
答案：A
问题：A BST has n nodes and a height of h. It is a balanced BST if:  含n个节点，高度为h的BST称为平衡二叉搜索树若它满足： {'A': '$n=O(h)$', 'B': '$h=O(n)$', 'C': '$h=O(log_2{n})$', 'D': '$n=O(log_2{h})$'}
答案：C
问题：Two equivalent balanced BST have identical: 两个等价的平衡二叉搜索树有相同的： {'A': 'pre-order sequence 先序遍历序列', 'B': 'in-order sequence 中序遍历序列', 'C': 'post-order sequence 后序遍历序列', 'D': 'hierachical sequence 层次遍历序列'}
答案：B
问题：What is the maximum number of imbalanced nodes after inserting a node in an AVL tree?  在AVL树中刚插入一个节点后失衡节点个数最多为 {'A': 'O(1)', 'B': 'O(lglgn)', 'C': 'O(lgn)', 'D': 'O(n)'}
答案：C
问题：What is the maximum number of imbalanced nodes after deleting a node in an AVL tree?  在AVL树中刚删除一个节点后失衡节点个数最多为 {'A': 'O(1)', 'B': 'O(lglgn)', 'C': 'O(lgn)', 'D': 'O(n)'}
答案：A
问题：The root of the AVL tree above has a balance factor of: 以上AVL树根节点的平衡因子为： {'A': '-1', 'B': '0', 'C': '1', 'D': '2'}
答案：A
问题：What is the minumum number of nodes in a AVL tree of height 3?高度为3的AVL树至少包含几个节点？[填空1]
答案：{"1": ["7"]}
问题：Inserting nodes into an AVL tree could lead to imbalance. After rebalancing the tree via rotation, the height of the sub-trees containing node g, p, and v  AVL树中插入节点引发失衡，经旋转调整后重新平衡，此时包含节点g,p,v的子树高度 {'A': 'decrease by 1 减小1', 'B': 'remain the same 不变', 'C': 'increase by 1 增加1', 'D': 'either remain the same or increase by 1 有可能不变也有可能增加1'}
答案：B
问题：AVL树中删除节点引发失衡，经旋转调整后重新平衡，此时包含节点g,p,v的子树高度 {'A': '减小1', 'B': '不变', 'C': '增加1', 'D': '有可能不变也有可能减小1'}
答案：D
问题：The _____ of an AVL tree is invariant under the 3+4 reconstruction.  经过3+4重构后的AVL树_____不变。 {'A': 'pre-order sequence 先序遍历序列', 'B': 'in-order sequence 中序遍历序列', 'C': 'post-order sequence 后序遍历序列', 'D': 'hierachical sequence 层次遍历序列'}
答案：B
问题：Which of the following is a BST? 以下哪项是二叉搜索树? {'A': '', 'B': '', 'C': '', 'D': ''}
答案：B
问题：What&#39;s the number of distinct BSTs containing nodes {1, 2, 3 ,4}?包含节点{1,2,3,4}的不同二叉搜索树有多少棵？[填空1]
答案：{"1": ["14"]}
问题：What is the 3rd element we compare to when searching for 14 in the BST above?在以上二叉搜索树中查找元素14，第3个和14发生比较的元素为：[填空1]
答案：{"1": ["11"]}
问题：What are the steps for deleting 16? 欲在以上二叉搜索树中删除节点16，可行的方案是： {'A': 'remove 16, make 25 the right child of 10 and 10 becomes the new root.将16摘除后令25为10的右子，从而10是新的根节点', 'B': 'execute a zig operation on 16 making it no longer the root, and then remove it directly以16为轴进行一次zig操作使之不再是根节点，再直接摘除16', 'C': 'exchange the key of node 16 and node 33, and remove the new node 16将节点16和节点33的关键码互换，再摘除新的节点16', 'D': 'exchange the key of node 16 and node 15, remove the new node 16 and make 13 the right child of 11将节点16和节点15的关键码互换，摘除新的节点16并令13为11的右子'}
答案：D
问题：A BST has n nodes and a height of h. Which of the following holds?二叉搜索树的高度h和节点个数n满足关系 {'A': 'h=O(1)', 'B': 'h=O(lgn)', 'C': 'h=O(n)', 'D': 'h=O(nlgn)'}
答案：C
问题：What is the worst-case time complexity for searching in it?在其上进行查找的最坏时间复杂度为 {'A': 'O(1)', 'B': 'O(lgn)', 'C': 'O(n)', 'D': 'O(nlgn)'}
答案：C
问题：Given that it&#39;s a balanced BST, which of the following holds?若已知它是平衡二叉搜索树，则n和h满足关系 {'A': 'h=O(1)', 'B': 'h=O(lgn)', 'C': 'h=O(n)', 'D': 'h=O(nlgn)'}
答案：B
问题：the complexity becomes 在其上进行查找的最坏时间复杂度为 {'A': 'O(1)', 'B': 'O(lgn)', 'C': 'O(n)', 'D': 'O(nlgn)'}
答案：B
问题：Given the BST above, what is the result after a zig operation on node 19?对以上二叉搜索树，以节点19为轴进行一次zig操作后得到的树为： {'A': '', 'B': '', 'C': '', 'D': ''}
答案：B
问题：What is the time complexity for searching in an AVL tree with n nodes?在包含n个节点的AVL树中进行查找的时间复杂度为 {'A': 'O(1)', 'B': 'O(lgn)', 'C': 'O(n)', 'D': 'O(nlgn)'}
答案：B
问题：What about inserting a node? 插入的时间复杂度为 {'A': 'O(1)', 'B': 'O(lgn)', 'C': 'O(n)', 'D': 'O(nlgn)'}
答案：B
问题：What about deleting a node? 删除的时间复杂度为 {'A': 'O(1)', 'B': 'O(lgn)', 'C': 'O(n)', 'D': 'O(nlgn)'}
答案：B
问题：After inserting a node into an AVL tree and rebalancing, the height of the rotated sub-treeAVL树中插入节点引发失衡，经旋转调整重新平衡后发生旋转的子树的高度 {'A': 'decrease by 1 减小1', 'B': 'remains the same or increase by 1不变或增加1', 'C': 'remains the same 不变', 'D': 'remains the same or decrease by 1 不变或减小1'}
答案：C
问题：What about the case when a node is deleted? 删除节点的情况呢？ {'A': 'decreaes by 1 减小1', 'B': 'remains the same or increase by 1 不变或增加1', 'C': 'remains the same 不变', 'D': 'remains the same or decrease by 1 不变或减小1'}
答案：D
问题：The AVL tree above is imblanced because node 13 was just inserted. Which one is the tree after rebalancing?以上AVL树因为刚插入节点13而失衡，利用课堂上所学重平衡算法，恢复平衡后的AVL树为： {'A': '', 'B': '', 'C': '', 'D': ''}
答案：A
问题：Choose the sub-tree after applying 3+4 reconstruction to the tree above.对以上子树进行3+4重构，得到的子树为： {'A': '', 'B': '', 'C': '', 'D': ''}
答案：B
问题：Which property of data access is taken advantage of by splay trees?伸展树利用了实际问题中数据访问的何种特点： {'A': 'massiveness 大量性', 'B': 'locality 局部性', 'C': 'holisticity 整体性', 'D': 'diversity 多样性'}
答案：B
问题：Which operation is executed for each accessed node in a splay tree?伸展树每次访问过某节点后都会把该节点： {'A': 'removing it 删除', 'B': 'moving it to the higer level 上移一层', 'C': 'moving it to the root 移动到根节点', 'D': 'accessing it again 再次访问该节点'}
答案：C
问题：In Tarjan&#39;s algorithm, how many layers are splayed together?Tarjan提出的伸展算法每几层一起伸展？ {'A': '1', 'B': '2', 'C': '3', 'D': '4'}
答案：B
问题：A splay tree degenerates into a list. What&#39;s its height after accessing the lowest node?在一棵退化成单链的伸展树中访问其最深的节点，经过伸展后树高大约为原先的： {'A': 'one third of the original height 三分之一', 'B': 'half of the original height 一半', 'C': 'the original height 不变', 'D': 'twice the original height 两倍'}
答案：B
问题：What&#39;s the amortized complexity for a single splay operation in Tarjan&#39;s algorithm?按照Tarjan的伸展算法，单次伸展操作的分摊复杂度为： {'A': 'O(1)', 'B': 'O(lgn)', 'C': 'O(n)', 'D': 'O(nlgn)'}
答案：B
问题：The node accessed is a right-child, and so is its parent. What should we do for a single splay operation?所访问的节点及其父亲都是右孩子，则双层伸展要执行的操作是： {'A': 'zig-zag', 'B': 'zag-zig', 'C': 'zig-zig', 'D': 'zag-zag'}
答案：D
问题：What&#39;s the amortized complexity in a sufficiently long sequence of accessing k nodes in a splay tree of size n?规模为n的伸展树中若所访问的节点只有k个，经过足够长时间的访问序列后，访问的分摊复杂度为：- 未答复 {'A': 'O(lgk)', 'B': 'O(klgn)', 'C': 'O(nlgk)', 'D': 'O(lgn)'}
答案：A
问题：Why does the memory becomes smaller and smaller?内存“越来越小”的原因是： {'A': 'With more and more processes running on one computer, the average memroy size for each process is getting smaller 机器上运行的程序越来越多，平均每个程序使用的内存变小', 'B': 'The real-world applications call for rocketing memory storage 实际应用对存储的需求急剧增长', 'C': 'The silicon used for producing chips is soon running out 用于制造内存芯片的硅资源消耗殆尽', 'D': 'The physical size of memory chips is getting smaller thanks to the advances in technology 随着工艺的进步，内存的体积变小，集成度变高'}
答案：B
问题：If it would take us a second for a single memory access, how long would it cost for a single access to the external storage?如果说访问一次内存需要1秒，则一次外存访问大概需要： {'A': 'a hour 1小时', 'B': 'a day 1天', 'C': 'a month 1月', 'D': 'a year 1年'}
答案：B
问题：B-trees are 树结构上的特点是： {'A': 'short and chubby, with each node having at most two children矮胖，每个节点至多两个孩子', 'B': 'short and chubby, with each node having potentially many children矮胖，每个节点可有多于两个孩子', 'C': 'tall and slim, with each node having at most two children瘦高，每个节点至多两个孩子', 'D': 'tall and slim, with each node having potentially many children瘦高，每个节点可有多于两个孩子'}
答案：B
问题：B-trees have small height so as to B树的层数少有助于： {'A': 'reduce the number of I/O operations 减少I/O次数', 'B': 'reduce the time consumption for a single I/O 减少每次I/O的时间', 'C': 'reduce the asymptotic time complexity for seaching 降低查找的渐进时间复杂度', 'D': 'reduce the storage requirement 节省存储空间'}
答案：A
问题：How many children are there for each node in a B-tree of order 4?4阶B树中每个节点的分支数为： {'A': '1~4', 'B': '2~5', 'C': '2~4', 'D': '3~5'}
答案：C
问题：What&#39;s the time complexity for searching in a B-tree of order 4 and size n?在存储了n个元素的4阶B树中查找，单个节点进行一次查找的时间复杂度为： {'A': 'O(1)', 'B': 'O(lgn)', 'C': 'O(n)', 'D': 'O(nlgn)'}
答案：A
问题：What&#39;s the return value for a failed search in a B-tree of order 4?B树查找算法若最终失败，返回值为： {'A': 'None', 'B': 'NULL', 'C': 'A pointer to the last examined node 指向最后一个所查找节点的指针', 'D': 'A pointer to the root 指向根节点的指针'}
答案：B
问题：What&#39;s the height of a B-tree of order 128 relative to a corresponding BBST?若B树的阶m=128，则它的高度大致是对应的BBST的： {'A': '1/5', 'B': '1/6', 'C': '1/7', 'D': '1/8'}
答案：B
问题：What&#39;s the &#39;overflow&#39; for a B-tree?B树的上溢是指： {'A': 'The property of a B-tree is violated due to removing a key 删除关键码后违反了B树的性质', 'B': 'The property of a B-tree is violated due to inserting a key 插入新的关键码后违反了B树的性质', 'C': 'The property of a B-tree is violated due to a split 分裂后违反了B树的性质', 'D': 'A B-tree becomes too high B树的高度过高'}
答案：B
问题：Which consequence follows as a B-tree becomes higher?B树高度的增加一定伴随着： {'A': 'Each node has a larger number of keys 每个节点所存放的关键码数量增加', 'B': 'Each node has a smaller number of keys 每个节点所存放的关键码数量减少', 'C': 'Splitting to the root 分裂到根', 'D': 'Splitting to the leaves 分裂到叶'}
答案：C
问题：The underflow of a B-tree happens whenB树的下溢发生于： {'A': 'the height of the tree decreases B树高度减少', 'B': 'the property of the B-tree is violated due to inserting a key 插入关键码后违反了B树的性质', 'C': 'the property of the B-tree is violated due to deleting a key 删除关键码后违反了B树的性质', 'D': 'inserting a key to a leaf node 在叶节点插入关键码'}
答案：C
问题：The height of a B-tree decreases only whenB树高度的减少只会发生于 {'A': 'merging two children of the root 根节点的两个孩子合并', 'B': 'removing the root 根节点被删除', 'C': 'rotating the root 根节点发生旋转', 'D': 'there are multiple keys in the root 根节点有多个关键码'}
答案：A
问题：A node in a red-balck tree could be红黑树节点的颜色有- 未答 {'A': 'red 红', 'B': 'black 黑', 'C': 'yellow 黄', 'D': 'blue 蓝', 'E': 'green 绿', 'F': 'purple 紫'}
答案：A, B
问题：What&#39;s unique about red-black trees compared to AVL trees? 红黑树相比于AVL树的特点是： {'A': 'The balance factor of each node falls within [-1, 1] 每个节点的平衡因子的绝对值不超过1', 'B': 'It is a balanced binary search tree 是平衡二叉搜索树', 'C': 'The search time is O(lgn) 支持O(lgn)时间的查找', 'D': 'The topology changes no more than O(1) after each insertion/deletion每次插入/删除后拓扑结构的变化不超过O(1)'}
答案：D
问题：The root of a red-black tree is 红黑树的根节点的颜色是 {'A': 'red 红', 'B': 'black 黑'}
答案：B
问题：The external nodes are 外部节点的颜色是 {'A': 'red 红', 'B': 'black 黑'}
答案：B
问题：A red-black tree is equivalent to a B-tree of order __红黑树等价于[填空1]阶B树:
答案：{"1": ["4"]}
问题：In a red-black tree of size n, the black height wouldn&#39;t exceed __规模为n的红黑树，黑高度不超过： {'A': 'O(1)', 'B': 'O(lgn)', 'C': 'O(n)', 'D': 'O(nlgn)'}
答案：B
问题：The height wouldn&#39;t exceed __ 高度不超过： {'A': 'O(1)', 'B': 'O(lgn)', 'C': 'O(n)', 'D': 'O(nlgn)'}
答案：B
问题：What is a &#39;double red&#39; in a red-black tree? 在红黑树中，何为双红缺陷： {'A': 'the root is red 根节点为红色', 'B': 'both the root and the external nodes are red 根节点和外部节点都为红色', 'C': 'both a parent and its child are red 相邻的两个父子节点都为红色', 'D': 'there are two red nodes in the tree 树中有两个红色节点'}
答案：C
问题：How does the topology changes when fixing a double red and the uncle node u is red.当叔父节点u为红色时，修正双红缺陷导致的红黑树拓扑结构的变化为： {'A': 'no change 没有变化', 'B': 'it changes no more than O(1) 有变化，但是不超过O(1)', 'C': 'it changes no more than O(lgn) 有变化，但是不超过O(lgn)', 'D': 'it changes no more than O(n) 有变化，但是不超过O(n)'}
答案：A
问题：Splay trees have large worst-case time complexity for a single operation, however, they reduce the number of I/O by exploiting hierachical storage伸展树虽然单次操作的最坏时间复杂度比较大，但是可以利用存储器的层次结构降低I/O的次数 {'true': '', 'false': ''}
答案：false
问题：Splay trees are difficult to implement 伸展树相较于AVL树的缺点是它实现起来较为复杂 {'true': '', 'false': ''}
答案：false
问题：Splay trees have larger amortized complexity than AVL trees for insertion伸展树插入操作的分摊复杂度比AVL树大 {'true': '', 'false': ''}
答案：false
问题：Splay trees have larger worst-case time complexity than AVL trees for searching伸展树单次查找操作的最坏时间复杂度比AVL树大 {'true': '', 'false': ''}
答案：true
问题：This is a splay tree. What is the result after accessing node v and performing a two-layer splay?上图是伸展树，其中节点v刚被访问过，双层伸展后的结果是： {'A': '', 'B': '', 'C': '', 'D': ''}
答案：D
问题：This is a splay tree. What&#39;s the result after inserting a node 1 and performing a two-layer splay? 在以上伸展树中插入节点1并经过双层伸展后的结果是 {'A': '', 'B': '', 'C': '', 'D': ''}
答案：C
问题：Which statement regarding (2, 4)-trees is incorrect? 关于(2, 4)-树，下列命题不正确的是： {'A': '(2, 4)-trees are equivalent to red-black trees (2,4)-树等价于红黑树', 'B': 'Each internal node has no more than 4 branches 每个内部节点的分支数不超过4', 'C': 'Each internal node has no less than 2 branches 每个内部节点的分支数不小于2', 'D': 'Each node has exactly 3 keys except for the root 除了根节点外，每个内部节点都恰好包含3个关键码'}
答案：D
问题：An overflow occurs after inserting node 52 into a (3,6)-tree, what&#39;s the result after splitting?上图是(3,6)-树中刚插入节点52后的情形，可以看出发生了上溢，分裂后的结果为： {'A': '', 'B': '', 'C': '', 'D': ''}
答案：A
问题：An underflow occurs after removing a node from a (3,6)-tree, what&#39;s the result after the adjustment?上图是(3,6)-树中刚删除某节点后的情形，可以看出发生了下溢，调整后的结果为： {'A': '', 'B': '', 'C': '', 'D': ''}
答案：B
问题：Which statement regarding red-black trees is wrong? 以下关于红黑树的说法，错误的是：- 未答复 {'A': 'A red-black tree of size n has a balck height of O(lgn), but the height is not necessarily O(lgn)含n个节点的红黑树，其黑高度为O(lgn)，但是总的高度却未必是O(lgn)', 'B': 'There cannot be two consecutive red nodes in a path from a external node to the root从红黑树的任一外部节点上溯到根节点，沿途不可能经过连续两个红色节点', 'C': 'The black height cannot be smaller than half the height红黑树的黑高度一定不小于总高度的一半', 'D': 'x (black) and y (black) are two children of a black node. The black height of the sub-trees x and y must equal.红黑树中的黑色节点u有黑色左孩子x和黑色右孩子y,则x与y的黑高度一定相等'}
答案：A
问题：What&#39;s the result after fixing the double red? 对上图中红黑树的节点x进行双红修正的结果是： {'A': '', 'B': '', 'C': '', 'D': ''}
答案：C
问题：Wich property could be violated after removing a node in a red-black tree using the algorithm for BST node removal?在红黑树中直接按照常规的BST删除节点算法删除一个节点，关于红黑树结构的四条性质是否有可能被破坏？1、the root is balck 树根必为黑色 {'A': 'could be violated 有可能会被破坏', 'B': 'won&#39;t be violated 不会被破坏'}
答案：A
问题：（接上题）2、the external nodes are black 外部节点必均为黑色 {'A': 'could be violated 有可能会被破坏', 'B': 'won&#39;t be violated 不会被破坏'}
答案：B
问题：（接上题）3、the children of a red node are black 红色节点的孩子必为黑 {'A': 'could be violated 有可能会被破坏', 'B': 'won&#39;t be violated 不会被破坏'}
答案：A
问题：（接上题）4、there are equal number of black nodes in each path from a external node to the root从根到外部节点的不同路径途中黑色节点数目相等 {'A': 'could be violated 有可能会被破坏', 'B': 'won&#39;t be violated 不会被破坏'}
答案：A
问题：We know many BBSTs up until now. 至此，我们接触了以下几种平衡二叉搜索树AVL trees AVL树Splay trees 伸展树B-trees B-树Red-black trees 红黑树kd-trres (see PA3 and the lecture note) kd-树（见PA3以及讲义）Please choose a BBST for each of thw following scenarios针对下列应用的特点，请你选取最合适的平衡二叉搜索树accessing massive amount of data (which cannot fit into the memory)对大规模的数据（不能全部存放于内存中）的存取 {'A': 'AVL trees AVL树', 'B': 'Splay trees 伸展树', 'C': 'B-trees B-树', 'D': 'Red-black trees 红黑树', 'E': 'kd-trres kd-树'}
答案：C
问题：（接上题）Easy implementation and O(lgn) complexity 需要易于实现，而且各接口的分摊复杂度为O(lgn) {'A': 'AVL trees AVL树', 'B': 'Splay trees 伸展树', 'C': 'B-trees B-树', 'D': 'Red-black trees', 'E': 'kd-trres kd-树'}
答案：B
问题：（接上题）Problems related to geometry 处理和几何有关的问题 {'A': 'AVL trees AVL树', 'B': 'Splay trees 伸展树', 'C': 'B-trees B-树', 'D': 'Red-black trees 红黑树', 'E': 'kd-trres kd-树'}
答案：E
问题：（接上题）Implementing a version control system 扩充后可支持对历史版本的访问 {'A': 'AVL trees AVL树', 'B': 'Splay trees 伸展树', 'C': 'B-trees B-树', 'D': 'Red-black trees 红黑树', 'E': 'kd-trres kd-树'}
答案：D
问题：If Baidu determines its own phone number in the above manner, the number is百度(baidu)如果用以上方式确定自己的电话号码，则号码是[填空1]
答案：{"1": ["22438"]}
问题：The actual storage location of the key is:关键码key实际存放的位置是： {'A': 'key', 'B': 'key - 1', 'C': 'hash(key)', 'D': 'hash(hash(key))'}
答案：C
问题：If the hash function is h(x) = x % 20, then the conflicts in the keys 25, 85, 15, 20 are:散列函数是 h(x) = x % 20，则关键码25, 85, 15, 20中会发生冲突的是： {'A': '25和85', 'B': '25和15', 'C': '15和85', 'D': '20和15'}
答案：A
问题：In the hash, if the number of keys exceeds the actual space used, is it possible to avoid conflict:在散列中，关键码个数超过实际使用的空间时，有没有可能不发生冲突： {'A': 'possible可能', 'B': 'impossible不可能'}
答案：B
问题：The hash function h(x) = x % M, the M should be:散列函数 h(x) = x % M 中的M应该取： {'A': '1', 'B': 'Power of 2 2的幂', 'C': 'A prime number 素数', 'D': 'A composite number 合数'}
答案：C
问题：MAD is:MAD是: {'A': 'h(x)=1', 'B': 'h(x)=x', 'C': 'h(x)=(a * x + b) % M', 'D': 'h(x)=MAD'}
答案：C
问题：The key type for the polynomial method is:多项式法适用的关键码类型为：- {'A': 'Integer整数', 'B': 'Float浮点数', 'C': 'Char字符', 'D': 'String字符串'}
答案：D
问题：Key to improving your programming skills[填空1]
答案：{"1": ["Learning Tsinghua Data Structures & Algorithms"]}
问题：Which kind of data structure can be used to solve the problem of multiple slots method:用哪种数据结构可以解决多槽位法的不足： {'A': 'Vector向量', 'B': 'List列表', 'C': 'Stack栈', 'D': 'Queue队列'}
答案：B
问题：In a linear probing, in the event of a conflict, turn to probe:在线性试探中，一旦发生冲突，转而试探： {'A': 'Predecessor of current position当前位置的前驱', 'B': 'Succcessor of current position当前位置的后继', 'C': 'Current position当前位置', 'D': 'The next element in the list of current position当前位置的列表中的下一个元素'}
答案：B
问题：Using the quadratic probing method, the positions of the previous probings are 0, 1, 4, 9, and the position of the next probing is: (assuming the bucket array is large enough)使用平方试探法，前几次试探的位置是0, 1, 4, 9，则下一次试探的位置是：（假设桶数组足够大） {'A': '9', 'B': '10', 'C': '15', 'D': '16'}
答案：D
问题：When the length of the table is prime, in order to make the quadratic probing always successful, the load factor needs to be less than:当表的长度为素数时，为了使平方试探总是成功，装填因子需要少于： {'A': '10%', 'B': '50%', 'C': '80%', 'D': '100%'}
答案：B
问题：The range of values of N elements to be sorted is [1, M]. The time complexity of counting sort is:N个待排序元素的取值范围是[1, M]，计数排序的时间复杂度为： {'A': 'O(NlgN)', 'B': 'O(N)', 'C': 'O(M)', 'D': 'O(M+N)'}
答案：D
问题：S is the space of all possible entries, A is the space of all available addresses (|A| &lt; |S|), h is a hash function, then:S为所有可能词条的空间，A为所有可用地址的空间（|A| &lt; |S|），h是散列函数，则： {'A': 'h maps from S to A, it must be a surjectiveh从S映射到A，一定是满射', 'B': 'h maps from S to A, it cannot be a injective从S映射到A，不可能是单射', 'C': 'h maps from A to S, it cannot be a surjectiveh从A映射到S，不可能是满射', 'D': 'h maps from A to S, it must be a injectiveh从A映射到S，一定是单射'}
答案：B
问题：In the hash table, what are the characteristics of a good hash function h?在散列表中，一个好的散列函数h需要具有哪些特点？It is a injective是单射 {'true': '', 'false': ''}
答案：false
问题：（接上题）Range covers as many addresses as possible值域尽可能覆盖更多的地址 {'true': '', 'false': ''}
答案：true
问题：（接上题）Every time we calculate the value of h in the same entry, we get different results每次计算h在同一个词条的取值，得到的结果均不同 {'true': '', 'false': ''}
答案：false
问题：It can efficiently calculate the function value of an given entry给定词条，能够高效地计算对应的函数值 {'true': '', 'false': ''}
答案：true
问题：（接上题）For different entries, the distribution of function values is as uniform as possible to avoid aggregation对于不同词条，函数值的分布尽可能均匀，避免聚集 {'true': '', 'false': ''}
答案：true
问题：（接上题）Unable to find a pair of conflicting entries within a reasonable calculating time无法在合理的计算时间内找到一对发生冲突的词条 {'true': '', 'false': ''}
答案：false
问题：Consider the set of keys S = {0, 8, 16, 24, 32, 40, 48, 56, 64}考虑key的集合S = {0, 8, 16, 24, 32, 40, 48, 56, 64}Hash function constructed by division method用除余法构造的散列函数h1(key) = key % 12h2(key) = key % 11How many elements does the range of S mapped by h1 has?h1将S映射到的值域有几个元素？[填空1]How many elements does the range of S mapped by h2 has?h2将S映射到的值域有几个元素？[填空2]
答案：{"1": ["3"], "2": ["9"]}
问题：Regarding the method of resolving conflicts, which of the following statement is correct:关于排解冲突的方法，以下说法正确的是： {'A': 'Using separate chaining method to resolve conflicts, all entries are actually stored inside the bucket array用独立链法排解冲突，所有词条的实际存放位置均在桶数组内部', 'B': 'Using open addressing method to resolve conflicts, the actual storage location of an entry is not necessarily the corresponding hash function value用开放定址排解冲突，词条的实际存放位置不一定是对应的散列函数值', 'C': 'Using open addressing to resolve conflicts, entries are stored in a list用开放定址排解冲突，词条被存放在列表中', 'D': 'As long as the hash function is designed properly, it does not necessarily need to resolve conflicts只要散列函数设计得当，不一定需要排解冲突的策略'}
答案：B
问题：The current state of the bucket array of size 11 is A = {*, *, *, *, *, 0, 15, 26, *, 5, 9}, where * means empty bucket规模为11的桶数组当前状态为 A = { *, *, *, *, *, 0, 15, 26, *, 5, 9}，其中*表示空桶The hash function is h(key) = (3 * key + 5) % 11散列函数为h(key) = (3 * key + 5) % 11Using Open addressing + Linear probing to solve conflicts用开放定址+线性试探排解冲突Insert the entry 4, its actual storage location is插入词条4, 它的实际存放位置是- 未答复 {'A': 'A[4]', 'B': 'A[6]', 'C': 'A[7]', 'D': 'A[8]'}
答案：D
问题：The current state of the bucket array of size 11 is A = {*, *, *, *, *, 0, 15, 26, *, 5, 9}, where * means empty bucket规模为11的桶数组当前状态为 A = { *, *, *, *, *, 0, 15, 26, *, 5, 9}，其中*表示空桶The hash function is h(key) = (3 * key + 5) % 11散列函数为h(key) = (3 * key + 5) % 11Using Open addressing +quadraticprobing to solve conflicts用开放定址+平方试探排解冲突Insert the entry 4, its actual storage location is插入词条4, 它的实际存放位置是 {'A': 'A[4]', 'B': 'A[6]', 'C': 'A[7]', 'D': 'A[8]'}
答案：A
问题：The size of the hash table is a prime number. Using open addressing + quadratic probing to resolve conflicts. To ensure that new entries can be inserted, the load factor of the hash table cannot exceed (please fill in decimal).散列表的规模是素数，用开放定址+平方试探法排解冲突，若要保证新的词条能够顺利插入，散列表的装填因子不能超过（请填十进制小数）[填空1]
答案：{"1": ["0.5"]}
问题：Sort n integers ranged in [0, M] by counting sort, the time complexity is用计数排序对n个[0, M)内的整数进行排序，时间复杂度为 {'A': 'O(n+M)', 'B': 'O(min{n, M})', 'C': 'O(nlgn)', 'D': 'O(n^2)'}
答案：A
问题：Sort the integers {10, 4, 2, 9, 3, 1, 2, 2, 4, 9, 8, 5, 9, 10, 7, 6, 9} in [0, 11) by the counting sort. The Accum[] table is:对[0, 11)中的整数{ 10, 4, 2, 9, 3, 1, 2, 2, 4, 9, 8, 5, 9, 10, 7, 6, 9 }用计数排序，得到的accum[]表为： {'A': '{0, 1, 3, 1, 2, 1, 1, 1, 1, 4, 2}', 'B': '{ 10, 4, 2, 9, 3, 1, 2, 2, 4, 9, 8, 5, 9, 10, 7, 6, 9 }', 'C': '{0, 1, 4, 5, 7, 8, 9, 10, 11, 15, 17}', 'D': '{0, 1, 4, 7, 9, 12, 13, 14, 15, 16, 17}'}
答案：C
问题：The access mode of elements in the priority queue is:优先级队列中元素的访问模式是：- 未答复 {'A': 'Smaller rank one is firstly visited.秩小者先访问', 'B': 'Bigger rank one is firstly visited.秩大者先访问', 'C': 'Higher priority one is firstly visited.优先级高者先访问', 'D': 'Front-located element is firstly visited位置靠前者先访问'}
答案：C
问题：The role of delMax() in the priority queue is:优先级队列的delMax()的作用是： {'A': 'Delete the largest element.删除最大的元素', 'B': 'Delete the element with highest priority.删除优先级最大的元素', 'C': 'Delete the first element.删除首元素', 'D': 'Delete the last element.删除末元素'}
答案：B
问题：Why do we usually not use BBST to implement priority queue?为什么我们通常不用BBST实现优先级队列？ {'A': 'Time efficiency is not high enough时间效率不够高', 'B': 'BBST is too complex for simple functions in priority queues对于实现优先级队列的简单功能来说，BBST过于复杂', 'C': 'BBST is not what we are going to learn in this chapter.我们本章要学的不是BBST', 'D': 'The vector is more efficient向量的效率更高'}
答案：B
问题：One way to implement priority queues is:优先级队列的一种实现方式是：Logically, equivalent to逻辑上，等同于 {'A': 'vector向量', 'B': 'stack栈', 'C': 'True binary tree真二叉树', 'D': 'Complete binary tree完全二叉树'}
答案：D
问题：（接上题）Physically, equivalent to物理上，等同于 {'A': 'vector向量', 'B': 'stack栈', 'C': 'True binary tree真二叉树', 'D': 'Complete binary tree完全二叉树'}
答案：A
问题：The parent node&#39;s priority in a complete binary heap完全二叉堆中父节点的优先级 {'A': 'Not smaller than its children不小于它的孩子', 'B': 'Not equal to their children不等于它的孩子', 'C': 'There is no necessary relationship with its children on size和它的孩子没有必然的大小关系', 'D': 'Equal to its children.等于它的孩子'}
答案：A
问题：The method for inserting elements in a full binary heap is在完全二叉堆中插入元素的方法是 {'A': 'Insert into the bottom, percolateUp插入到底层，上滤', 'B': 'Insert into the root, percolateDown插入到根节点，下滤', 'C': 'Insert into the bottom directly直接插入到底层', 'D': 'Insert into the root directly直接插入到根节点'}
答案：A
问题：The time complexity of inserting elements in a complete binary heap of size n is:规模为n的完全二叉堆中插入元素的时间复杂度为：- 未答复 {'A': 'O(nlgn)', 'B': 'O(n)', 'C': 'O(lgn)', 'D': 'O(1)'}
答案：C