Suppose you are working for a fund management company. Knowing that you have learned machine learning, your boss has asked you to write “A Proposal To apply Machine Learning to Algorithmic Trading”. If your proposal is successful, your boss will allow you to lead a team. Therefore, you must explain clearly what you want to achieve with machine learning. What would the company be able to achieve if your proposal is supported? You should explain what support you need from the company. For example, what data must the company acquire? What expertise do you need in your team? To manage expectations, you should clearly state the limitations of your proposed project.

按照以下要求写出python代码，Suppose there are n assignments A= [a1, a2 …ai …an]that you need to complete before the deadline. An assignment ai= [durationi, deadlinei] need durationi days to complete and must be completed before or on deadlinei. You can only do one assignment at a time and start next assignment once the current assignment is completed. Assuming you start on day 1, implement an efficient algorithm to find the maximum number of assignments you can complete.

以下是Python代码： def max_assignments(A): # 按照deadlinei从小到大排序 A.sort(key=lambda x: x[1]) # 初始化当前时间和已完成的任务数 current_time = 0 completed = 0 for i in range(len(A)): # 如果当前时间加上任务需要的时间小于等于deadlinei，则可以完成该任务 if current_time + A[i][0] <= A[i][1]: current_time += A[i][0] completed += 1 return completed # 测试代码 A = [[2, 4], [3, 5], [1, 2], [4, 7], [5, 9]] print(max_assignments(A)) # 输出3

用python按以下要求写出代码，Suppose there are n assignments A= [a1, a2 …ai …an]that you need to complete before the deadline. An assignment ai= [durationi, deadlinei] need durationi days to complete and must be completed before or on deadlinei. You can only do one assignment at a time and start next assignment once the current assignment is completed. Assuming you start on day 1, implement an efficient algorithm to find the maximum number of assignments you can complete.

可以回答这个问题。以下是Python代码： def max_assignments(A): A = sorted(A, key=lambda x: x[1]) current_day = 1 count = 0 for duration, deadline in A: if current_day + duration - 1 <= deadline: count += 1 current_day += duration return count # 测试 A = [[2, 4], [3, 5], [1, 2], [4, 7], [1, 1]] print(max_assignments(A)) # 输出3，最多可以完成3个作业：[1, 2], [2, 4], [1, 1]

Suppose you have a lock that consists of 5 digits that are randomly chosen from the set {0,1,2,3,4,5,6,7,8,9}. Write a C++ program to demonstrate how to find the correct lock combination by using recursion and backtracking. Show step-by-step how to solve the problem. Draw three recursive level calls example for the recursion tree call with input values. Note that, the initial state of the key is {0,0,0,0,0} and the lock combination is randomly chosen {rand(0-9), rand(0-9), rand(0-9), rand(0-9), ), rand(0-9)}.

.Let color be the following structure: struct color int red;int green;int blue;};Suppose the resolution of the picture is 300 x 300 that is, you need to create a two-dimensional array of 300 x 300 pixels, each of which stores information in three channels: red, green, and blue. The color value is a random integer.

To create a two-dimensional array of 300 x 300 pixels, each storing information in the three color channels, we can use the following code in C++: struct color { int red; int green; int blue; }; const int width = 300; const int height = 300; color image[height][width]; // To initialize the color values with random integers between 0 and 255 for (int i = 0; i < height; i++) { for (int j = 0; j < width; j++) { image[i][j].red = rand() % 256; image[i][j].green = rand() % 256; image[i][j].blue = rand() % 256; } } This code defines a struct color that has three integer fields red, green, and blue. We then define two constants, width and height, to specify the dimensions of the image. The two-dimensional array image is then created with dimensions height and width, with each element of the array storing a color struct. To initialize the color values with random integers between 0 and 255, we use a nested loop to iterate over each element of the array and set its red, green, and blue fields to a random integer between 0 and 255 using the rand() function.

notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different permutation. check to see the result if we double it again! now you are suppose to check if there are more numbers with this property. that is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number.

注意，数字123456789是一个由数字1到9组成的9位数，没有重复。将它翻倍，我们将得到246913578，这恰好是另一个由数字1到9组成的9位数，只是排列不同。现在你应该检查如果我们再次将其翻倍的结果是什么！现在你应该检查是否还有更多具有此属性的数字。也就是说，将一个有k位数字翻倍，你需要判断结果数字是否只由原始数字的排列组成。

Suppose there are five feature A, B,C, D, E with constraints B , 3,C , 2, A , B, B , C,C < D, A = D, E < A, E < B, E < C, E < D, B , D Define the loss function as val({A = a, . . . , E = e})) = the number of violated constraints Apply greedy descent, write down a sequence of solutions obtained.

首先，我们可以通过绘制限制条件图来更好地理解问题： E --- B --- D | | | | C --- A |_____| 在这个限制条件图中，箭头表示限制条件，例如 "B > C" 表示从 B 到 C 有一条箭头，表示 B 的值必须大于 C 的值。接下来，我们可以使用贪心下降算法来求解此问题。首先，我们可以随机生成一个初始解，例如 A=1, B=2, C=3, D=1, E=0。然后，我们对每个特征进行局部搜索，以找到可以改进的解。对于特征 A，由于它必须等于 D，因此我们可以将 A 的值改为 1，这样就满足了限制条件。对于特征 B，由于它必须大于 E 和 C，因此我们可以将 B 的值改为 3，这样就满足了限制条件。对于特征 C，由于它必须小于 B 和 D，因此我们可以将 C 的值改为 1，这样就满足了限制条件。对于特征 D，由于它必须大于 C，因此我们可以将 D 的值改为 2，这样就满足了限制条件。对于特征 E，由于它必须小于 A、B、C 和 D，因此我们无法对其进行改进。最终，我们得到的最优解为 A=1, B=3, C=1, D=2, E=0，该解中没有违反任何限制条件，因此损失函数的值为 0。

Recall that an exponential random variable 𝑋~exp (𝜆) has mean 𝜆 1 and pdf given by: 𝑓(𝑥) = 𝜆𝑒 −𝜆𝑥 on 𝑥 ≥ 0. (a) Compute 𝑃(𝑋 ≥ 𝑥). (b) Suppose that 𝑋1 and 𝑋2 are independent exponential random variables with mean 𝜆 1 . Let 𝑇 = min (𝑋1 ,𝑋2 ). Find the cdf of 𝑇. (Hint: what is 𝑃(𝑇 ≥ 𝑡)?)

(a) 𝑃(𝑋 ≥ 𝑥) = ∫𝑥∞ 𝑓(𝑥) d𝑥 = ∫𝑥∞ 𝜆𝑒^−𝜆𝑥 d𝑥 = e^−𝜆𝑥 (b) 𝑃(𝑇 ≥ 𝑡) is the probability that both 𝑋1 and 𝑋2 are greater than or equal to 𝑡, which is given by: 𝑃(𝑇 ≥ 𝑡) = 𝑃(𝑋1 ≥ 𝑡)𝑃(𝑋2 ≥ 𝑡) = e^−𝜆1𝑡 * e^−𝜆1𝑡 = e^(−2𝜆1𝑡) To find the cdf of 𝑇, we can use the fact that the cdf is the integral of the pdf: 𝐹𝑇(𝑡) = 𝑃(𝑇 ≤ 𝑡) = 1 − 𝑃(𝑇 ≥ 𝑡) = 1 − e^(−2𝜆1𝑡) Therefore, the cdf of 𝑇 is 𝐹𝑇(𝑡) = 1 − e^(−2𝜆1𝑡).

Suppose we have T(n)≤ c =O(1)for all≤3,andforeveryn≥4,we have ，T(n)<=T(n/4)+T(0.75*n)+cn,Use Mathematical Induction to prove that T (n) = O(n log n) for all n ≥ 4.

首先，我们假设 T(n) ≤ cn，其中c为一个常数；接下来，假设我们知道T(n) ≤ cn，其中c为一个常数，且T(k) ≤ ck，其中k ≤ n，且k > 4。根据我们的假设，我们有：T(n) ≤ T(n/4) + T(0.75n)，由此可以得出：T(n) ≤ c(n/4) + c(0.75n)，于是，T(n)≤ cn。所以，我们可以得出结论：T(n) = O(n log n)，其中n ≥ 4。

Use a.any() or a.all()

The difference between a.any() and a.all() in Python is that a.any() returns True if any element in the array a is true, while a.all() returns True only if all elements in the array a are true. For example, suppose we have an array a = [True, False, True]. Then a.any() would return True because there is at least one True element in the array, while a.all() would return False because not all elements in the array are True. Here's some sample code to demonstrate this: import numpy as np a = np.array([True, False, True]) print(a.any()) # Output: True print(a.all()) # Output: False

Consider a datagram network using 32-bit host addresses. Suppose a router has four links, numbered 0 through 3, and packets are to be forwarded to the link interfaces as follows: Destination Address Range Link Interface 11100000 00000000 00000000 00000000 through 11100000 00111111 11111111 11111111 0 11100000 01000000 00000000 00000000 through 11100000 01000000 11111111 11111111 1 11100000 01000001 00000000 00000000 through 11100001 01111111 11111111 11111111 2 Otherwise 3 1) Provide a forwarding table that has five entries, uses longest prefix matching, and forwards packets to the correct link interfaces. (6 points) Prefix Match Link Interface

Prefix Match Link Interface 11100000 00000000 00000000 00000000 through 11100000 00111111 11111111 11111111 0 11100000 01000000 00000000 00000000 through 11100000 01000000 11111111 11111111 1 11100000 01000001 00000000 00000000 through 11100001 01111111 11111111 11111111 2 Default (otherwise) 3

用C++编写程序，实现以下问题2、题目ID Codes（POJ1146） Time Limit: 1000MS Memory Limit: 10000K 描述： It is 2084 and the year of Big Brother has finally arrived, albeit a century late. In order to exercise greater control over its citizens and thereby to counter a chronic breakdown in law and order, the Government decides on a radical measure--all citizens are to have a tiny microcomputer surgically implanted in their left wrists. This computer will contains all sorts of personal information as well as a transmitter which will allow people's movements to be logged and monitored by a central computer. (A desirable side effect of this process is that it will shorten the dole queue for plastic surgeons.) An essential component of each computer will be a unique identification code, consisting of up to 50 characters drawn from the 26 lower case letters. The set of characters for any given code is chosen somewhat haphazardly. The complicated way in which the code is imprinted into the chip makes it much easier for the manufacturer to produce codes which are rearrangements of other codes than to produce new codes with a different selection of letters. Thus, once a set of letters has been chosen all possible codes derivable from it are used before changing the set. For example, suppose it is decided that a code will contain exactly 3 occurrences of a', 2 of b' and 1 of c', then three of the allowable 60 codes under these conditions are: abaabc abaacb ababac These three codes are listed from top to bottom in alphabetic order. Among all codes generated with this set of characters, these codes appear consecutively in this order. Write a program to assist in the issuing of these identification codes. Your program will accept a sequence of no more than 50 lower case letters (which may contain repeated characters) and print the successor code if one exists or the message No Successor' if the given code is the last in the sequence for that set of characters. 输入： Input will consist of a series of lines each containing a string representing a code. The entire file will be terminated by a line consisting of a single #. 输出： Output will consist of one line for each code read containing the successor code or the words 'No Successor'. 样例输入 abaacb cbbaa # 样例输出 ababac No Successor

以下是使用C++编写的程序，可以解决该问题： c++ #include <iostream> #include <algorithm> #include <cstring> using namespace std; int main() { char str[55]; while(cin >> str && str[0] != '#') { if(next_permutation(str, str + strlen(str))) cout << str << endl; else cout << "No Successor" << endl; } return 0; } 首先，我们定义了一个字符数组 str 来存储输入的字符串。然后，在一个循环中不断读入字符串，直到读入单独的一个 '#' 字符为止。对于每个字符串，我们可以使用 next_permutation 函数来找到它的下一个排列。如果找到了下一个排列，则输出它；否则输出 "No Successor"。需要注意的是，我们必须将 str 的长度作为第二个参数传递给 next_permutation 函数，否则它将无法正确地处理含有重复字符的字符串。

1） Inheritance：Suppose there is a base class Animal that has a name property and an eat() method. Now there are two derived classes Cat and Dog, both of which inherit from the Animal class. The Cat class has a catchMouse() method and the Dog class has a bark() method. Write the main() function that does the following: create a Cat object and a Dog object; assign values to their name properties, respectively; call the eat() and catchMouse() methods of the Cat object; call the eat() and bark() methods of the Dog object. 2）Polymorphism： Based on 1), add a virtual function speak() in the base class Animal to represent the behavior of animals making sounds. The two derived classes Dog and Cat implement the speak() function respectively, and output "Wow Woof Woof" and "Meow Meow Meow" respectively. Please define a pointer to the Animal object in the main function, let it point to the Dog object and the Cat object, and call their speak() function. Submit your completed program as a single C++ source file.

1) Here's the code for the inheritance example: c++ #include <iostream> #include <string> class Animal { public: std::string name; void eat() { std::cout << name << " is eating." << std::endl; } }; class Cat : public Animal { public: void catchMouse() { std::cout << name << " is catching a mouse." << std::endl; } }; class Dog : public Animal { public: void bark() { std::cout << name << " is barking." << std::endl; } }; int main() { Cat cat; cat.name = "Tom"; cat.eat(); cat.catchMouse(); Dog dog; dog.name = "Fido"; dog.eat(); dog.bark(); return 0; } 2) Here's the updated code for the polymorphism example: c++ #include <iostream> #include <string> class Animal { public: std::string name; void eat() { std::cout << name << " is eating." << std::endl; } virtual void speak() { std::cout << name << " is making a sound." << std::endl; } }; class Cat : public Animal { public: void catchMouse() { std::cout << name << " is catching a mouse." << std::endl; } void speak() override { std::cout << "Meow Meow Meow" << std::endl; } }; class Dog : public Animal { public: void bark() { std::cout << name << " is barking." << std::endl; } void speak() override { std::cout << "Wow Woof Woof" << std::endl; } }; int main() { Animal* animalPtr; Cat cat; cat.name = "Tom"; animalPtr = &cat; animalPtr->speak(); Dog dog; dog.name = "Fido"; animalPtr = &dog; animalPtr->speak(); return 0; } Note that the speak() function is marked as virtual in the base class Animal, which means that it can be overridden by derived classes. In the Cat and Dog classes, we override the speak() function to output the appropriate sound for each animal. In the main() function, we define a pointer to an Animal object, and we use it to point to both a Cat object and a Dog object. We then call the speak() function through the pointer, and each animal outputs its own sound.

Recall that an exponential random variable 𝑋~exp (𝜆) has mean 𝜆 1 and pdf given by: 𝑓(𝑥) = 𝜆𝑒 −𝜆𝑥 on 𝑥 ≥ 0. (b) Suppose that 𝑋1 and 𝑋2 are independent exponential random variables with mean 𝜆 1 . Let 𝑇 = min (𝑋1 ,𝑋2 ). Find the cdf of 𝑇. (Hint: what is 𝑃(𝑇 ≥ 𝑡)?)教我怎么做

首先，我们可以使用下面的公式计算 $P(T \geq t)$： $$P(T \geq t) = P(X_1 \geq t, X_2 \geq t)$$ 因为 $X_1$ 和 $X_2$ 是独立的，所以我们可以将概率分解为： $$P(X_1 \geq t, X_2 \geq t) = P(X_1 \geq t) \cdot P(X_2 \geq t)$$ 由于 $X_1$ 和 $X_2$ 都是指数分布，所以： $$P(X_i \geq t) = e^{-\lambda_i t}, \ i = 1,2$$ 因此，我们有： $$P(T \geq t) = P(X_1 \geq t) \cdot P(X_2 \geq t) = e^{-\lambda_1 t} \cdot e^{-\lambda_2 t} = e^{-(\lambda_1 + \lambda_2) t}$$ 现在我们可以使用 $P(T \geq t)$ 来计算 $T$ 的累积分布函数 (CDF)： $$F_T(t) = P(T \leq t) = 1 - P(T \geq t) = 1 - e^{-(\lambda_1 + \lambda_2) t}$$ 因此，$T$ 的 CDF 为 $F_T(t) = 1 - e^{-(\lambda_1 + \lambda_2) t}$。

Inheritance：Suppose there is a base class Animal that has a name property and an eat() method. Now there are two derived classes Cat and Dog, both of which inherit from the Animal class. The Cat class has a catchMouse() method and the Dog class has a bark() method. Write the main() function that does the following: create a Cat object and a Dog object; assign values to their name properties, respectively; call the eat() and catchMouse() methods of the Cat object; call the eat() and bark() methods of the Dog object. 2）Polymorphism： Based on 1), add a virtual function speak() in the base class Animal to represent the behavior of animals making sounds. The two derived classes Dog and Cat implement the speak() function respectively, and output "Wow Woof Woof" and "Meow Meow Meow" respectively. Please define a pointer to the Animal object in the main function, let it point to the Dog object and the Cat object, and call their speak() function.

1) Inheritance: c++ #include <iostream> #include <string> class Animal { protected: std::string name; public: Animal(std::string n) : name(n) {} void eat() { std::cout << name << " is eating\n"; } }; class Cat : public Animal { public: Cat(std::string n) : Animal(n) {} void catchMouse() { std::cout << name << " is catching a mouse\n"; } }; class Dog : public Animal { public: Dog(std::string n) : Animal(n) {} void bark() { std::cout << name << " is barking\n"; } }; int main() { Cat cat("Kitty"); cat.eat(); cat.catchMouse(); Dog dog("Buddy"); dog.eat(); dog.bark(); return 0; } Output: Kitty is eating Kitty is catching a mouse Buddy is eating Buddy is barking 2) Polymorphism: c++ #include <iostream> #include <string> class Animal { protected: std::string name; public: Animal(std::string n) : name(n) {} void eat() { std::cout << name << " is eating\n"; } virtual void speak() {} }; class Cat : public Animal { public: Cat(std::string n) : Animal(n) {} void catchMouse() { std::cout << name << " is catching a mouse\n"; } void speak() override { std::cout << "Meow Meow Meow\n"; } }; class Dog : public Animal { public: Dog(std::string n) : Animal(n) {} void bark() { std::cout << name << " is barking\n"; } void speak() override { std::cout << "Wow Woof Woof\n"; } }; int main() { Animal* ptr; Cat cat("Kitty"); Dog dog("Buddy"); ptr = &cat; ptr->speak(); ptr = &dog; ptr->speak(); return 0; } Output: Meow Meow Meow Wow Woof Woof

Suppose the scores for a certain exam are normally distributed with a mean of 80 and a standard deviation of 4. Find the z-score for an exam score of 87.

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Suppose the scores for a certain exam are normally distributed with a mean of 80 and a standard deviation of 4. Find the z-score for an exam score of 87.

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Suppose there are five feature A, B,C, D, E with constraints B , 3,C , 2, A , B, B , C,C < D, A = D, E < A, E < B, E < C, E < D, B , D Define the loss function as val({A = a, . . . , E = e})) = the number of violated constraints Apply greedy descent, write down a sequence of solutions obtained.

Suppose we have T(n)≤ c =O(1)for all≤3,andforeveryn≥4,we have ，T(n)<=T(n/4)+T(0.75*n)+cn,Use Mathematical Induction to prove that T (n) = O(n log n) for all n ≥ 4.

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