Assume the postorder of binary tree T is RELATIONDB, the in order of T is REILTABOND,then the preorder of T will be

时间: 2024-06-02 15:10:11 浏览: 104
To determine the preorder of T, we need to first identify the root of the binary tree. In an inorder traversal of a binary tree, the nodes to the left of the root correspond to the left subtree of the root, and the nodes to the right of the root correspond to the right subtree of the root. From the given inorder sequence REILTABOND, we can see that the root of the binary tree T is "L". Next, we need to determine the left and right subtrees of the root. Since the postorder traversal of T is RELATIONDB, we know that the last element "B" is the root of the right subtree of "L". To find the root of the left subtree of "L", we need to look for the last element in the postorder sequence that appears before "B" in the inorder sequence. From the inorder sequence REILTABOND, we can see that the last element before "B" is "T". Therefore, "T" is the root of the left subtree of "L". Now we can construct the preorder traversal of T by starting at the root "L" and recursively traversing the left and right subtrees in preorder. The preorder traversal of T is: - Root: L - Left subtree: T, R, E, I, L, A - Right subtree: B, O, N, D Therefore, the preorder of T is LTRIELABOND.
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Write a program to simulate a process of translation from a logical address to physical address. Assumptions 1. Assume the file la.txt includes the sequence of generated addresses from CPU. 2. Use a part of memory as backing store that store data for a process. 3. The backing store size is 128 bytes 4. The size of process p is 128 bytes. 5. The contents of p is included in a file pdata.bin which is a binary file. 6. Use a part of memory as RAM. The size of physical memory is 256 bytes, from 0 to 255. All the physical memory is available, allocating starting from beginning in sequence. That is, allocate frame 0 first, then frame 1, then frame 2…. 7. The size of a frame is 32 bytes, i.e., 5 bits for the offset in a frame, total number of frames is 8. 8. At beginning, no page table is available for process p. Requirements Write a program to 1. Setup a simulating backing store in memory. Read the data from pdata.bin to this backing store. 2. Initialize a page table for process p, set the frame number to be -1 for each page, indicating that the page is not loaded into memory yet. 3. Read logical addresses one by one from la.txt. 4. For each logical address, a) if its page has been loaded into physical memory, simply find the frame number in the page table, then generate physical address, find and print out the physical address and data inside this address. b) if the page is used for the first time, i.e., in page table, its frame number is -1,then the page that contains this address should be loaded into a free frame in physical memory (RAM). Then update the page table by adding the frame number to the right index in the page table. Then repeat 4a). Refer to Figure 1 for the relationships and how physical memory, backing store, and CPU are simulated.写一个c文件

用c++解决Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations only with these currencies. There can be several points specializing in the same pair of currencies. Each point has its own exchange rates, exchange rate of A to B is the quantity of B you get for 1A. Also each exchange point has some commission, the sum you have to pay for your exchange operation. Commission is always collected in source currency. For example, if you want to exchange 100 US Dollars into Russian Rubles at the exchange point, where the exchange rate is 29.75, and the commission is 0.39 you will get (100 - 0.39) * 29.75 = 2963.3975RUR. You surely know that there are N different currencies you can deal with in our city. Let us assign unique integer number from 1 to N to each currency. Then each exchange point can be described with 6 numbers: integer A and B - numbers of currencies it exchanges, and real RAB, CAB, RBA and CBA - exchange rates and commissions when exchanging A to B and B to A respectively. Nick has some money in currency S and wonders if he can somehow, after some exchange operations, increase his capital. Of course, he wants to have his money in currency S in the end. Help him to answer this difficult question. Nick must always have non-negative sum of money while making his operations. Input The first line contains four numbers: N - the number of currencies, M - the number of exchange points, S - the number of currency Nick has and V - the quantity of currency units he has. The following M lines contain 6 numbers each - the description of the corresponding exchange point - in specified above order. Numbers are separated by one or more spaces. 1 ≤ S ≤ N ≤ 100, 1 ≤ M ≤ 100, V is real number, 0 ≤ V ≤ 103. For each point exchange rates and commissions are real, given with at most two digits after the decimal point, 10-2 ≤ rate ≤ 102, 0 ≤ commission ≤ 102. Let us call some sequence of the exchange operations simple if no exchange point is used more than once in this sequence. You may assume that ratio of the numeric values of the sums at the end and at the beginning of any simple sequence of the exchange operations will be less than 104. Output If Nick can increase his wealth, output YES, in other case output NO.

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