By using DE Morgan's law simplify the logical expression not ((style>=8) and (bribe>5)) to the one that contains no parentheses .

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not ((style>=8) and (bribe>5)) Using De Morgan's law, we can rewrite the above expression as: not (style>=8) or not (bribe>5) Simplifying further, we can remove the parentheses and write: not style<8 or not bribe>5
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John is going on a fishing trip. He has h hours available (1 <= h <= 16), and there are n lakes in the area (2 <= n <= 25) all reachable along a single, one-way road. John starts at lake 1, but he can finish at any lake he wants. He can only travel from one lake to the next one, but he does not have to stop at any lake unless he wishes to. For each i = 1,...,n - 1, the number of 5-minute intervals it takes to travel from lake i to lake i + 1 is denoted ti (0 < ti <=192). For example, t3 = 4 means that it takes 20 minutes to travel from lake 3 to lake 4. To help plan his fishing trip, John has gathered some information about the lakes. For each lake i, the number of fish expected to be caught in the initial 5 minutes, denoted fi( fi >= 0 ), is known. Each 5 minutes of fishing decreases the number of fish expected to be caught in the next 5-minute interval by a constant rate of di (di >= 0). If the number of fish expected to be caught in an interval is less than or equal to di , there will be no more fish left in the lake in the next interval. To simplify the planning, John assumes that no one else will be fishing at the lakes to affect the number of fish he expects to catch. Write a program to help John plan his fishing trip to maximize the number of fish expected to be caught. The number of minutes spent at each lake must be a multiple of 5.这道题的分析

1. 状态定义：设f(i, j)表示从起点到i时，已经用了j个5分钟的时间，能抓到的最大鱼数。 2. 状态转移：因为从i到j只能走一次，所以可以考虑枚举上一个节点k，转移方程为：f(i, j) = max{f(k, j-ti) + max(0, fi-(j-...

De Morgan s Law

De Morgan's laws are two rules in Boolean algebra that relate to the logical operators "AND" and "OR". The laws are as follows: 1. The complement of the conjunction of two statements is equivalent to...

请用java实现7. Implement the class MixedNumber that you designed in the previous project. Use the operations in Fraction whenever possible. For example, to add two mixed numbers, convert them to fractions, add the fractions by using Fraction’s add operation, and then convert the resulting fraction to mixed form. Use analogous techniques for the other arithmetic operations. Handling the sign of a mixed number can be a messy problem if you are not careful. Mathematically, it makes sense for the sign of the integer part to match the sign of the fraction. But if you have a negative fraction, for example, the toString method for the mixed number could give you the string "−5 −1/2", instead of "−5 1/2", which is what you would normally expect. Here is a possible solution that will greatly simplify computations. Represent the sign of a mixed number as a character data field. Once this sign is set, make the integer and fractional parts positive. When a mixed number is created, if the given integer part is not zero, take the sign of the integer part as the sign of the mixed number and ignore the signs of the fraction’s numerator and denominator. However, if the given integer part is zero, take the sign of the given fraction as the sign of the mixed number.

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De Morgan's Law

De Morgan's Law is a fundamental principle in mathematics and logic that states that the negation of a conjunction (AND) or a disjunction (OR) is equivalent to the disjunction or conjunction, ...

continue to use Java language, Add a class Borrower that extends User. The constructor of the Borrower class takes a name and a number of books borrowed by the borrower. If the number of books given as argument is strictly less than zero, then the constructor must throw a NotALenderException with the message “A new borrower cannot lend books.”. The borrower class does not have any instance variable. The moreBook method of the Borrower class increases the number of books borrowed by the borrower by the number of books given as argument to the method (so the books borrowed by the borrower becomes more positive!) For example, if a borrower currently borrows 10 books and moreBook(2) is called then the borrower borrows 12 books. It is fine for the moreBook method to be given a negative value as argument, which means the borrower then just returned some books. For example, if a borrower currently borrows 10 books and moreBook(-2) is called then the borrower borrows 8 books. However, a borrower cannot lend books, so the number of books borrowed by the borrower must always be positive or zero, never negative. If the argument given to the moreBook method is too negative and would change the book variable into a negative value, then the number of books borrowed by the borrower must not change and the moreBook method must throw a NotALenderException with the message “A borrower cannot lend XXX book(s).”, where XXX is replaced with the result of -(book + number). For example, if a borrower currently borrows 10 books and moreBook(-12) is called then the borrower still borrows 10 books and the method throws a NotALenderException with the message “A borrower cannot lend 2 book(s).”. Note: to simplify the project, do not worry about the setBook method. Change other classes and interfaces as necessary

The moreBook method of the Borrower class increases or decreases the number of books borrowed by the borrower by the number of books given as argument to the method. If the argument given to the ...

#include <iostream> using namespace std; int n, m; inline int gcd(int a, int b) { if(!b) return a; return gcd(b, a % b); } struct Frac { int fz, fm; double val; void simplify() { register int g = gcd(fz, fm); fz /= g; fm /= g; val = (double)fz / (double)fm; } void init(int fm_, int fz_) { fm = fm_; fz = fz_; simplify(); } Frac plus(Frac a) { register int gfm = a.fm * fm; register int gfz = a.fz * fm + fz * a.fm; Frac ans; ans.init(gfm, gfz); ans.simplify(); return ans; } Frac times(int a) { register int gfm = fm; register int gfz = fz * a; Frac ans; ans.init(gfm, gfz); ans.simplify(); return ans; } }; inline int Solve(int cnt, int last, Frac sum) { if(cnt == n) { if(sum.fm - sum.fz == 1 && sum.fm >= last) { return 1; } return 0; } if(sum.val >= 1) { return 0; } Frac lst; lst.init(last, 1); Frac add = lst.times(n - cnt + 1); Frac t = sum.plus(add); if(t.val == 1) return 1; if(t.val < 1) { return 0; } lst.init(m, 1); add = lst.times(n - cnt); t = sum.plus(add); if(t.val >= 1) { return 0; } int count = 0; for(register int i = last; i <= m; ++i) { Frac temp; temp.init(i, 1); count += Solve(cnt + 1, i, sum.plus(temp)); } return count; } int main() { cin >> n >> m; cout << Solve(1, 2, {0, 1, 0.00}) << endl; return 0; } 优化上述代码

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不使用LINQ查询和操作集合 改进代码 namespace SandwichCalories { class Program { static void Main(string[] args) { // sandwich ingredients and their associated calories Dictionary<string, int> ingredients = new Dictionary<string, int>() { { "Bread", 100 }, { "Ham", 150 }, { "Lettuce", 10 }, { "Tomato", 20 }, { "Mayonnaise", 50 }, { "Cheese", 120 } }; // prompt user for calorie range Console.Write("Enter minimum calories: "); int min_calories = int.Parse(Console.ReadLine()); Console.Write("Enter maximum calories: "); int max_calories = int.Parse(Console.ReadLine()); // calculate the minimum and maximum calories for the sandwich int min_sandwich_calories = 2 * ingredients["Bread"] + ingredients.Values.Min() * 2; int max_sandwich_calories = 2 * ingredients["Bread"] + ingredients.Values.Max() * 2; // check if the calorie range is valid if (max_calories < min_sandwich_calories) { Console.WriteLine("Sorry, it is impossible to create a sandwich within the given calorie range."); } else { // create the sandwich List<string> sandwich = new List<string> { "Bread", "Bread" }; int sandwich_calories = 2 * ingredients["Bread"]; while (sandwich_calories < min_calories) { // add random ingredient string ingredient = ingredients.Keys.ElementAt(new Random().Next(ingredients.Count)); sandwich.Add(ingredient); sandwich_calories += ingredients[ingredient]; } while (sandwich_calories <= max_calories) { // add random ingredient string ingredient = ingredients.Keys.ElementAt(new Random().Next(ingredients.Count)); // check if the ingredient is the same as the previous one if (sandwich.Count >= 3 && ingredient == sandwich[sandwich.Count - 2]) { continue; } sandwich.Add(ingredient); sandwich_calories += ingredients[ingredient]; // check if the sandwich is already at the maximum calorie limit if (sandwich_calories == max_sandwich_calories) { break; } } // add the last slice of bread sandwich.Add("Bread"); // print the sandwich and its total calories Console.WriteLine("Your sandwich: " + string.Join(", ", sandwich)); Console.WriteLine("Total calories: " + sandwich_calories); } } } }

One possible way to improve the code without using LINQ and manipulating collections is to simplify the while loop that adds random ingredients to the sandwich. Instead of generating a random index ...

解释代码#include <stdio.h> #include <stdlib.h> #include <unistd.h> #include <string.h> #include <sys/socket.h> #include <netinet/ip.h> #define PORT 9090 /* Changing this size will change the layout of the stack. * We have added 2 dummy arrays: in main() and myprintf(). * Instructors can change this value each year, so students * won't be able to use the solutions from the past. * Suggested value: between 0 and 300 */ #ifndef DUMMY_SIZE #define DUMMY_SIZE 100 #endif char *secret = "A secret message\n"; unsigned int target = 0x11223344; void myprintf(char *msg) { uintptr_t framep; // Copy the ebp value into framep, and print it out asm("movl %%ebp, %0" : "=r"(framep)); printf("The ebp value inside myprintf() is: 0x%.8x\n", framep); /* Change the size of the dummy array to randomize the parameters for this lab. Need to use the array at least once */ char dummy[DUMMY_SIZE]; memset(dummy, 0, DUMMY_SIZE); // This line has a format-string vulnerability printf(msg); printf("The value of the 'target' variable (after): 0x%.8x\n", target); } /* This function provides some helpful information. It is meant to * simplify the lab tasks. In practice, attackers need to figure * out the information by themselves. */ void helper() { printf("The address of the secret: 0x%.8x\n", (unsigned) secret); printf("The address of the 'target' variable: 0x%.8x\n", (unsigned) &target); printf("The value of the 'target' variable (before): 0x%.8x\n", target); }

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Simplify the expression (2i − 2)(1 − 3i)/(2i + 1). Derive your answer step-by-step.

答案: 步骤1：将分子中的2i和1-3i相乘： 2i × (1 - 3i) = 2i - 6i² 步骤2：将...(-12i³ + 4i² - 2i + 6i²) / (4i² - 2i - 1) 步骤5：将分子和分母中的因数相加： (-12i³ + 6i² - 2i - 1) / (4i² - 2i - 1)

From Proposition 1, we plug ri,O = li(μ)τi into (39) and rewrite problem (38) as maximize ri,O 􏰗ai − μ li (μ) − Yi(t)g [li(μ)]􏰘 ri,O li (μ)hi (41a) March 2, 2021 DRAFT maximize ˆr O subject to 0 ≤ ri,O ≤ Qi(t), (41b) 0, ifa − μ −Y(t)g[li(μ)] <0, subject to where the optimal solution is r∗ i,O Accordingly, we have τ∗ = r∗ ii,Oi i1 of μ in (32) as 1−􏰀i∈M1 τi∗. Then, we obtain the optimal dual variable μ∗ through the ellipsoid method (bi-section search in this case) over the range [0,∆], where ∆ is a sufficiently large value, until a prescribed precision requirement is met. Given the optimal μ∗, we denote the optimal ratio obtained from (40) as li (μ∗) 􏰝 r∗ /τ∗, i,O i ∀i ∈ M1. Notice that the optimal solution 􏰕τi∗, r∗ , ∀i ∈ M1􏰖 of the dual problem may not be i,O primal feasible. Therefore, to find a primal optimal solution to (31), we substitute τi = ri,O/li (μ∗) into (31) and simplify the problem as = i li(μ) i li(μ)hi (42) otherwise. Qi (t), /l (μ). After obtaining τ∗, ∀i ∈ M , we calculate the subgradient 􏰁 􏰗ai − Yi(t)g [li(μ∗)]􏰘 ri,O (43a) i ∈ M 1 h i l i ( μ ∗ ) 􏰁 ri,O ≤ 1, ri,O ≤ Qi(t), ∀i ∈ M1. (43b) i∈M1 li(μ∗) The above problem is a simple linear programming (LP) that can be easily solved. With a bit abuse of notation, we denote the optimal solution of (43) as ˆr∗ = 􏰕r∗ , ∀i ∈ M 􏰖 and retrieve 20 the optimal solution to (31) as τ∗=r∗ /l(μ∗),e∗ =τi∗g[li(μ∗)],∀i∈M. (44) i i,O i i,O hili(μ∗) 1 Denote τˆ∗ = {τi∗,∀i ∈ M1} and ˆe∗O = 􏰕e∗i,O,∀i ∈ M1􏰖. As {τˆ∗,ˆe∗O,ˆr∗O,μ∗} satisfies the KKT conditions, {τˆ∗,ˆe∗O,ˆr∗O} is an optimal solution to (31). By combining the optimal solutions in (30) and (44), we obtain an optimal solution of (P4). We summarize the pseudo-code of the O i,O 1 algorithm to solve (P4) in Algorithm 2.,翻译并解释li和hi是什么

在这段文字中，li和hi分别代表优化问题中的变量。li是一个非负数，它表示第i个用户在当前时刻能够分配的最大带宽。hi是一个非负数，它表示第i个用户的利润，即用户在当前时刻通过分配带宽所能获得的收益。...

用Python语言翻译如下代码#include <stdio.h> #include <stdlib.h> #include <string.h> char str_plaincode[100];//存放p_text.txt文件中读到的明文 int str_ciphertext[100];//加密为密文存放到c_text.txt文件中 char test[100];//存放从c_text.txt文件中读到的密文解密得到的明文 int Keys[16];//由会话密钥生成的16轮对称加密使用的密钥 int j = 6;//OFB模式移位寄存器移j位 int IV = 117;//OFB模式移位寄存器初始值 int K;//通过私钥解密得到的会话密钥 int K_;//随机数生成函数生成会话密钥 #define p 7879 #define q 8971 #define e 19751321 #define d 31060661 long long ciphertext;//存放加密会话密钥后得到的密文 long long Test;//解密后得到的会话密钥暂存 long long n = (long long)p * q; int b[32] = { 0 }; //得到简化模次方计算所需的bi和k int get_b_return_k(int h) { int i = 0; for (int j = 0; j < 32; j++) { b[j] = 0; } long long x = h; while (x) { b[i] = (x & 1); x = x >> 1; i++; } i--; return i; } //简化模次方计算 long long simplify1(int k) { long long dd = 1; for (int j = k; j >= 0; j--) { dd = (dd * dd) % n; if (b[j] == 1) { dd = (dd * K_) % n; } } return dd; } //简化模次方计算 long long simplify2(int k) { long long dd = 1; for (int j = k; j >= 0; j--) { dd = (dd * dd) % n; if (b[j] == 1) { dd = (dd * ciphertext) % n; } } return dd; } //加密会话密钥 void encrypt_key(int k) { ciphertext =simplify1(k); printf("发送方用公钥加密会话密钥为：%d\n", ciphertext); } //解密会话密钥 int decode_key(int k) { Test= simplify2(k); return Test; }

x >>= 1 i += 1 i -= 1 return i # 简化模次方计算 def simplify1(k): dd = 1 for j in range(k, -1, -1): dd = (dd * dd) % n if b[j] == 1: dd = (dd * K_) % n return dd # 简化模次方计算 def ...

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