copywriting exercises

); theta = Theta-alpha (alpha_i) .*grad; end plot (0:49, Jtheta ( 1:50), char (Plotstyle (alpha_i)), ' LineWidth ', 2)% be sure to convert by Char function hold on if (1 = = Alpha (alpha_i))% The experiment found that Alpha was1 o'clock the effect is best, then the theta value after the iteration is the desired value theta_grad_descent = theta endendlegend (' 0.01 ', ' 0.03 ', ' 0.1 ', ' 0.3 ', ' 1 ', ' 1.3 '); Xlabel (' Number of iterations ') ylabel (' cost function ')% below is the pred

|grep ' \ (listen\) [[: Space:]]\?\+ '650) this.width=650; "title=" Image004.png "alt=" wkiol1xiviahojxbaabyjjrpda4905.jpg "src=" http://s3.51cto.com/ Wyfs02/m02/72/5e/wkiol1xiviahojxbaabyjjrpda4905.jpg "/>To extend the regular expression: # netstat-tan|egrep ' (LISTEN) [[: Space:]]?+ ' Add user bash,testbash,basher, and nologin user (nologin user's shell For /sbin/nologin); then find the user name in the/etc/passwd file with the shell name line; Regular Expressions: # grep ' ^\ (

P12 Exercises 1.21. $\begin{align*}\lim\limits_{n\rightarrow \infty} \frac{n^2 \arctan n}{1+n^2} = \frac{\pi}{2}\end{align*}$Card : Because$\begin{align*}\frac{n^2}{1+n^2} \end{align*}$So$\begin{align*}| \frac{n^2 \arctan n}{1+n^2}-\frac{\pi}{2}| \\= \frac{\pi}{2}-\frac{n^2 \arctan n}{1+n^2} \ \= \frac{\pi}{2}-\arctan n + \frac{\arctan n}{1+n^2},\end{align*}$And because $\arctan x $\begin{align*} | \frac{n^2 \arctan n}{1+n^2}-\frac{\pi}{2}| \\ = \tex

This exercise exercises the relationship between the car class and the factory class, and completes some output through calls between the methods.Note the parameters and the settings for the return value.1 PackageCom.cnblogs.java;2 3 Public classTestcar {4 Public Static voidMain (string[] args) {5Car c1=NewCar ();//instantiation of6 c1.info ();7C1.setname ("Bentley");8C1.setprice (1000000);9C1.wheel=4;Ten c1.info (); One AFAC f=NewFAC

Related exercises:1. The contents of the File1 file are: "1 2 3 4 5 6 7 8 9 10" Calculates the sum of all numbersEcho $[1+2+3+4+5+6+7+8+9+10]echo "1+2+3+4+5+6+7+8+9+10" | BcComputer Demo[[email protected] desktop]# echo $[1+2+3+4+5+6+7+8+9+10]55[Email protected] desktop]# echo "1+2+3+4+5+6+7+8+9+10" | Bc55[Email protected] desktop]#2, processing string "Xt.,l 1 jr#! $mn 2 c*/fe3 uz4", keep only the numbers and spacesecho "Xt.,l 1 jr#! $mn 2 c*/fe3 uz4

--about the primary foreign key exercises--adding and removing primary and foreign key syntax--deleting FOREIGN KEY syntax: ALTER TABLE name DROP constraint foreign KEY constraint name--Add foreign KEY syntax: ALTER TABLE NAME-constraint foreign KEY constraint name Forei GN key (column name) references Reference foreign key table (column name)--Delete primary KEY syntax: ALTER TABLE name DROP constraint PRIMARY KEY constraint name--Add PRIMARY KEY syn

Exercise 1.3Define a procedure thats three numbers as argument and return the sum of the square of both large number.Scheme implementation( define (square x) (* x x)) (define (Sum_of_square x y) (+ (square x) ( square y)) ( Define (bigger x y ) (if (> x y) x y)) (define (smaller x y) (if ( x y) x y) (define (Sum_square x y z) ( sum_of_square (bigger x y) (bigger (smaller x y) z)) c19/>C Language Implementation//some

# #习题1:List A = [11,22,24,29,30,32]1 inserting 28 to the end of the listA.append (28)2 inserting elements after element 29 57>>> a = [11,22,24,29,30,32]3>>> a.insert (a.index +1,57)>>> a[ 11, 22, 24, 29, 57, 30, 32]3 Change element 11 to 6A[0] = 6 a[a.index (11)] = 63 Deleting an Element 32del a[a.index (+)]del a[5]a.pop ()4 list from small to large sortA.sort () modify the original list directly# #习题2:List B = [1,2,3,4,5]1 output the following results in 2 ways:[1,2,3,4,5,6,7,8]Method One:B.app

[Lab4-1] Grasping routing exercises question 1:Fetching 172.18.0.0/16----172.29.0.0/16 routes with ACLsAnswer: 172.16.0.0 0.15.255.255, it looks perfect, but the box is actually a little wider.R1 (f0/0)-(f0/0) R2Interface address on R1:r1#sh IP int bInterface ip-address OK? Method Status Protocolfastethernet0/0 12.1.1.1 YES NVRAM upLoopback0 172.16.0.1 YES NVRAM upLoopback1 172.17.0.1 YES NVRAM upLoopback2 172.18.0.1 YES NVRAM upLoopback3 172.19.0.1 Y

ACM Exercises (v)1.//ASCLL code Value Ordering/* #include using namespace Std;void Chushi (char a[3]){for (int i=0;i{cin>>a[i];}}void Paixu (char a[3]){for (int i1=0;i1for (int i2=i1;i2{if (A[i1]>a[i2]){Char TEMP=A[I1];A[I1]=A[I2];A[i2]=temp;}}cout}int main (){int N;cin>>n;Char a[3];for (int m=0;m{Chushi (a);Paixu (a);}return 0;}*/2.//Odd-even separation/* #include using namespace Std;void Jishu (int n){int i=0;for (i;i{if (i%2==1){cout}}cout}void Ors

summed up the beauty of programming above on the list of problems, there is not the right place, welcome to shoot bricks, and other programming beauty to read, back to brush other problems encountered linked list and then add ~The catalogue is as follows (click to expand the above directory to the topic of interest):/************************************************************************//* Linked List Classic exercises 1. Removing nodes from the Hea

TopicCode/* ---------------------------------------* Date: 2015-07-05* sjf0115* Title: Object Manager * Source: Huawei Machine Test Exercises----------------------- ------------------*/#include #include "ObjMgt.h"#include using namespace STD;structobject{unsigned intKey1;unsigned intKey2;unsigned intKey3; Object (unsigned intAunsigned intBunsigned intc) {key1 = A; Key2 = b; Key3 = C; } Object () {}}; vectorContainer/ *-----------------------

TopicDescribe:找出1至n之间同构数的个数。同构数是这样一组数：它出现在平方数的右边。例如：5是25右边的数，25是625右边的数，5和25都是同构数。Detailed Description:Interface descriptionPrototype:intSearchSameConstructNum(int n);Input parameters:int n：查找1至n之间的全部同构数return value:int：1至n之间同构数的个数Practice Stage:Code/* ---------------------------------------* Date: 2015-07-05* sjf0115* title: Find the number of isomorphic numbers * Source: Huawei Machine Test Exercises-------------------- ---------------------*/#inclu

TopicDescription: Topic title:求某二进制数中1的个数。给定一个unsigned int型的正整数，求其二进制表示中“1”的个数，要求算法的执行效率尽可能地高。Detailed Description:Prototype:int GetCount(unsigned int num)Input parameters: num 给定的正整数Output parameters (the memory area pointed to by the pointer is guaranteed to be valid):无return value:返回1的个数举例：输入13，则对应的二进制是1101，那么1的个数为3个。则：返回3。Practice Stage:初级 Code/* ---------------------------------------* Date: 2015-07-03* sjf0115* title: Number of 1 in a binary number * Source: Huawei Machine Test

usingSystem;usingSystem.Collections;usingSystem.Collections.Generic;usingSystem.Linq;usingSystem.Text;usingSystem.Text.RegularExpressions;namespaceC Programming Exercises { Public enumqqstate {OnLine=1,//The enumeration type has a default initial value of 0 and assigns a value of 1 to match the console output. OffLine, Leave, Busy, QMe,}classProgram {Static voidMain (string[] args) { //prompts the user to select an online status, we

Topic描述: 详细描述：N个人围坐在一个圆桌上，顺时针报数，报数的初始值为第一个人设置。当有成员报出的数字为7的倍数或数字中包含7，则该人退出圆桌，而后由下一个人开始重新继续该游戏。实现以下接口： 1、设定输入原始的圆桌游戏的人数。以最先开始报数的人编号为1，顺时针排序。 2、设定第一个人的初始值，获取按照规则退出圆桌的人的编号。 3、结束游戏。样例:比如初始化为4人的游戏：第1轮初始值为1，退出为3；第2轮初始值为4，退出为4号；第三轮初始值为16，退出为2号；第四轮初始值为6，退出为1号；练习阶段: 中级 Code/* ---------------------------------------* Date: 2015-06-31* sjf0115* title: Roundtable game * Source: Huawei Machine Test Exercises----------------------- ------------------*/#inclu

// 1. the exercises in this chapter are purely for the sake of existence... Typedef struct phone {char * quhao; char * jiaohuantai; char * zhanhaoma;}; typedef struct call {char * date; char * time; phone * shiyong; phone * self; phone * hujiao ;}; 2. typedef struct cash {float retail_price; float sale; float sales_tax; float licensing_fee ;}; typedef struct daikuan {float hour; float sales_tax; float licensing_ment; float interval; float loan_duratio

Exercise 1.45 is a summary of many previous exercises on fixed points. This topic reviews the fixed point search method we used in Section 1.3.3. When looking for the fixed point of Y-> x/y, this transformation itself does not converge, an average damping is required. For the transformation y-> X/(y ^ 2), it can also be reduced by an average damping. However, one average damping is not enough for the four equations. That is to say, for a transforma

This is a self-made exercise and may be incorrect. You are welcome to discuss and discuss various optimization and reconstruction solutions.Based on the reader's feedback or code review, I will update and supplement the answer or related content of this article, which will be added to the comments of this blog.I try to ensure that the answer code for each question is complete, not just functions or classes. Open the Python 2.7 IDLE and copy the complete code to debug and run it.Welcome to Balian

about the drivers and responsibilities. First of all, everyone should be sure that everyone is responsible for the rise and fall of the team. The good and bad of a person involve the entire team. When everyone reaches a consensus, for example, let everyone build their own workflows and plans first, it must be simple and practical. It should be set based on the performance gap. Of course, the reward system should be required, and the rewards should be rewarded to raise the responsibility. 5.E