(60%)Assume+that+the+following+sequences+of+instructions+are+exe-cuted+on+a+five-stage+pipelined+dat

时间: 2023-12-29 19:26:31 浏览: 64
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Unit2-Section+A+-+exercises.pdf

以下是一个用于找到方程ax^2 + bx + c = 0的实根的伪代码算法： ```python 1. 输入a, b, c 2. 计算delta = b^2 - 4ac 3. 如果delta < 0，则方程没有实根，结束 4. 如果delta = 0，则方程有一个实根，计算x = -b / (2a)，结束 5. 如果delta > 0，则方程有两个实根，计算x1 = (-b + sqrt(delta)) / (2a) 和 x2 = (-b - sqrt(delta)) / (2a)，结束 ``` 请注意，这只是一个伪代码算法，具体的实现可能会根据编程语言和环境的不同而有所不同。
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A circuit-switched network is a type of communication network where a dedicated communication path is established between two nodes for the duration of a communication session. In this network, the ...

Add a Library class with the following UML specification: +-------------------------------------------+ | Library | +-------------------------------------------+ | - name: String | | - users: ArrayList | +-------------------------------------------+ | + Library(String name) | | + addUser(IUser user): void | | + totalBorrowedBooks(): int | | + getBook(String name): int | | + moreBook(String name, int number): void | | + testLibrary(): void | +-------------------------------------------+ When a library is created, it has an arraylist of users (IUser) but the arraylist is empty (the arraylist does not contain any user). The addUser method takes a user (IUser) as argument and adds the user to the arraylist of users for the library. The totalBorrowedBooks method returns as result the total number of books borrowed by all users of the library (the result can be either positive or negative). The getBook method takes as argument the name of a user and returns as result the number of books currently borrowed by the user. If the library does not have a user with the given name, then the getBook method must throw an UnknownUserException with the message "User XXX unknown.", where XXX is replaced with the name of the user. Do not worry about multiple users having the same name. You can assume all user names are unique in the arraylist. The moreBook method takes as argument the name of a user and a number of books and changes the number of books currently borrowed by that user. If the library does not have a user with the given name, then the moreBook method must throw an UnknownUserException with the message "User XXX unknown.", where XXX is replaced with the name of the user. Do not worry about multiple users having the same name. Note: the moreBook method does not catch any exception, it only throws exceptions. Hint: use the equals method to compare strings, not the == operator which only works with constant strings. 写java文件

下面是实现上述UML规范的Java代码： java import java.util.ArrayList; public class Library { private String name; private ArrayList<IUser> users;... public Library(String name) { ...

continue to use Java language, Add a Library class with the following UML specification: +-------------------------------------------+ | Library | +-------------------------------------------+ | - name: String | | - users: ArrayList<IUser> | +-------------------------------------------+ | + Library(String name) | | + addUser(IUser user): void | | + totalBorrowedBooks(): int | | + getBook(String name): int | | + moreBook(String name, int number): void | | + testLibrary(): void | +-------------------------------------------+ When a library is created, it has an arraylist of users (IUser) but the arraylist is empty (the arraylist does not contain any user). The addUser method takes a user (IUser) as argument and adds the user to the arraylist of users for the library. The totalBorrowedBooks method returns as result the total number of books borrowed by all users of the library (the result can be either positive or negative). The getBook method takes as argument the name of a user and returns as result the number of books currently borrowed by the user. If the library does not have a user with the given name, then the getBook method must throw an UnknownUserException with the message "User XXX unknown.", where XXX is replaced with the name of the user. Do not worry about multiple users having the same name. You can assume all user names are unique in the arraylist. The moreBook method takes as argument the name of a user and a number of books and changes the number of books currently borrowed by that user. If the library does not have a user with the given name, then the moreBook method must throw an UnknownUserException with the message "User XXX unknown.", where XXX is replaced with the name of the user. Do not worry about multiple users having the same name. Note: the moreBook method does not catch any exception, it only throws exceptions. Hint: use the equals method to compare strings, not the == operator which only works with constant strings.

The testLibrary method in the Library class creates three users, adds them to the library, and tests the various methods of the library. The main method simply creates a Library object and calls its ...

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For the following C statement, write the corresponding RISC-V assembly code. Assume that the variables f, g, h, i, and j are assigned to registers x5, x6, x7, x28, and x29, respectively. Assume that the base address of the arrays A and B are in registers x10 and x11, respectively. B[8] = A[i−j];

C statement: c B[8] = A[i - j]; RISC-V assembly code: assembly subw x8, x28, x29 # 计算 i-j 并将结果放入 x8 寄存器 slli x8, x8, 2 # 将 x8 寄存器的值左移 2 位，即乘以 4 ...

用c++解决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.

这个问题涉及到一个货币兑换的问题，需要我们输入一些数据，然后判断是否能够增加货币的数量。 下面是用 C++ 解决这个问题的代码： c++ #include #include using namespace std; ...const int MAXN = 100 + 10;...

Please write a simply code in python, the requirement is as followed: The Tower of Hanoi is a mathematical game or puzzle. It consists of three rods, and a number of disks of different sizes which can slide onto any rod. The puzzle starts with the disks in a neat stack in ascending order of size on one rod, the smallest at the top, thus making a conical shape. The following figure shows the initial state of the Tower of Hanoi with 5 disks.The objective of the puzzle is to move the entire stack to another rod, obeying the following simple rules: 1. Only one disk can be moved at a time. 2. Each move consists of taking the upper disk from one of the stacks and placing it on top of another stack i.e. a disk can only be moved if it is the uppermost disk on a stack. 3. No disk may be placed on top of a smaller disk. Assume that initially all the disks are placed on rod A. Write a non-recursive Python function to print out the steps to move all the disks from rod A to rod C via rod B (Hint: a recursive algorithm can be converted into a non-recursive algorithm using stack). The header of the function is: def HanoiTower(n) Here n represents the number of disks. example: when n = 3, the program will output: A -> C 换行 A -> B 换行 C -> B 换行 A -> C 换行 B -> A 换行 B -> C 换行 A -> C

这是一个使用非递归算法实现的汉诺塔问题的 Python 代码： python def HanoiTower(n): stack = [] stack.append((n, 'A', 'B', 'C')) while len(stack) > 0: n, from_rod, aux_rod, to_rod = stack.pop() ...

写一段Java实现Input In the first line you are given the number n of billionaires (1 ≤ n ≤ 10000). The following n lines contain information about these people: their names, cities where they were staying at the beginning of the period, and their fortunes. In the next line you are given the number m of days in the period for which you have the information (1 ≤ m ≤ 50000) and the number k of travels of the billionaires (0 ≤ k ≤ 50000). The following k lines contain the list of travels in the following format: the number of the day (from 1 to m−1), the name of the person, and the city of destination. You may assume that billionaires depart late at night and arrive to the destination city on the next day's morning. They cannot make more than one travel each day. The numbers of days in the list are not decreasing. All names of people and cities consist of at most 20 English letters; you must take into consideration the case of the symbols. The fortunes are in the range from 1 to 100 billions (one billion is a thousand million). Output In each line of the output give the name of a city and, after a space, the number of days during which this city was the first with respect to the sum of fortunes of the billionaires staying there. Leave out those cities for which there were no such days. The cities must be sorted alphabetically (with the usual symbol order: ABC...Zabc...z).

以下是Java实现代码： java import java.util.*; public class Main { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); int n = scanner.nextInt();...

SSTACK+SEGMENT+STACK+DW+32+DUP(?)+SSTACK+ENDS+CODE+SEGMENT+ASSUME+CS:CODE+START:+MOV+DX,+0646H+MOV+A

CODE SEGMENT是定义了一个代码段，ASSUME CS:CODE表示将CS寄存器与CODE段相关联。MOV指令是将数据从一个位置移动到另一个位置，DX和AL是寄存器，数字是内存地址或立即数。OUT指令是将数据输出到端口，DELAY是一个...

We will write a procedure every-nth that accepts two arguments: num is an integer and list-of-sents is a list of a certain number of lists. every-nth selects the num-th elements from each list in list-of-sents and returns the list of selected elements. Assume that all lists in listof-sents have more than num elements.

Here's a possible implementation of the every-nth procedure in Python: python def every_nth(num, list_of_sents): result = [] for sentence in list_of_sents: result.append(sentence[num-1]) ...

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C statement: c f = g + (h - 5); RISC-V assembly code: assembly addw x5, x6, x7 # 将 g 和 h 相加并将结果放入 x5 寄存器 addi x7, x7, -5 # 将 h 寄存器中的值减去 5，结果放入 x7 寄存器 ...

WishingBone plans to decorate his hexagon-shaped room with unique flooring patterns. First he divides it into triangles, and colors them with different colors. Now that he has got a satisfactory pattern, he uses small rugs to cover the floor. The rugs sold at shops are diamond-shaped; exact the shape of two adjacent triangles. He is not willing to cut any of them. Different colored rugs are labeled different price. And he does care about the price - he cares about the most expensive rugs that he should use to finish the covering. Input There are multiple tests, for each test: The first line is an integer specifying the side of his room (1 <= n <= 50). The following 2n lines with the pattern of the room, with upper-case letters denoting different colors. So there are at most 26 colors. Since the room is considerably large and sometimes needs pillars to support the ceiling, some triangles are marked with '.' rather than a letter. Such triangles are not to be covered. The following line contains one integer m (1 <= m <= 351) - number of rugs in the shop. The following m lines contains one type of rug (with two upper-case letters) and its price p (1 <= p <= 100000). There will not be duplicate items listed, that is, if AB is listed, BA will not appear. You may assume the number of each type to be infinite. Output One line for each test, the price of the most expensive rugs used. Of course, you have to minimize this value. If there is no solution to the problem, print a 0.

这是一道图论和最小权匹配问题，需要使用匈牙利算法来解决。 首先，将六边形的每个三角形看作一个节点，将相邻的三角形之间的边看作图中的边。这样就将问题转化为了求解这个图的最小权完美匹配问题。...

We will write a procedure every-nth that accepts two arguments: num is an integer and list-of-sents is a list of a certain number of lists. every-nth selects the num-th elements from each list in list-of-sents and returns the list of selected elements. Assume that all lists in listof-sents have more than num elements(piease use racket))

At each step, it selects the num-th element from the first list in lst using list-ref, adds it to the accumulator acc using cons, and recurses on the rest of the list using cdr. When the ...
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