crossbow exercise

Problem DescriptionEnter an unsigned short integer k[hex.] and P[oct.], the high byte of K as the low byte of the result, and the high byte of p as the result of the high byte to form a new integer.InputK[hex.] and P[oct.]OutputThe operation gets the new integer n.Sample Input0XD9 01117Sample Output200HINTOutput is 16 binary1#include 2 intMain ()3 {4 5 intk,p;6 while(SCANF ("%x%o", k,p)! =EOF)7 {8 intNewint = (p0xff00)| (k0xff00) >>8;9printf"%x\n", newint);Ten }

‐‐‐| ')Print (' |‐‐‐1: Query contact information ‐‐‐| ')Print (' |‐‐‐2: Insert new contact ‐‐‐| ')Print (' |‐‐‐3: Delete existing contact ‐‐‐| ')Print (' |‐‐‐4: Exit Address Book Program ‐‐‐| ')contacts = Dict ()While 1:instr = Int (input (' \ n Please enter the relevant instruction code: '))If InStr = = 1:Name = input (' Please enter contact name: ')If name in Contacts:Print (name + ': ' + contacts[name])ElsePrint (' The name you entered is no longer in the address Book! ‘)If InStr = = 2:Name

in range (4): li.append (str (random.randint (0,9))) print ("". Join (LI))　　Generate four-bit verification codes with numbers and letters1, Method one Import random num_l = list (map (str,list (range))) #获取以0-9 String for the listing of elements del_l = lists (range (91,97)) Chr_ L = [Chr (i) if I not in del_l else "for I in Range (65,123)] #获取大小写的a-Z list, remove the middle 6 special characters num_l.extend (chr_l) #将字母和 The numbers are grouped together res_l = num_l res_str = ""

input integer n from keyboard, output 1+3+5+7+ ... The and of the first n items. (n/** \brief input integer n from keyboard, output 1+3+5+7+ ... The and of the first n items. (n*/#include#includeintMain () {intN//integer n for storing input inti =1;//as an intermediate variable, for counting intSum_n =0;//stores the first n items andprintf"Input integer n\n");//Output Input Reminderscanf"%d", n);//reading keyboard input while(I N) {sum_n+= (2*i-1) ; I++; } printf ("The first%d entries

Wikipedia's article as an unordered list.RunCommand line RunCopy that four rows listD:\>cd D:\Code\VimCode\Python_auto\RunD:\Code\VimCode\Python_auto\Run>python bulletPointAdder.pyScript RunStep Analysis:1. Clipboard copy and paste, using Text=pyperclip.paste (), this is not difficult, simple to use the Pyperclip library. Paste () and. Copy () on the line2. Detach the lines in the text and add an asterisk. line = Text.split (' \ n ').3. Link line after processing. Text= ' \ n '. Join (lines)Cod

Please find the maximum values in each decimal group, and then concatenate them together to form a new array.Tip: You can use a for loop to iterate through algebraic groups and arr[i] access each element of the array in a way.function Largestoffour (arr) { var l=New Array (arr.length); for (i=0;i) { l[i]=arr[i][0]; for (j=0;j) { if(arr[i][j]>l[i]) { l[i]=arr[ I][J];}} } return l;} Largestoffour ([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001,

(PID = =0) { //create a new session periodSetsid (); //set the current directory as the root directoryChDir"/"); //Set directory PermissionsUmask0); Close (Stderr_fileno); Close (Stdin_fileno); FILE*PFA =NULL; PFA= fopen ("/home/test/1/time.log","a"); if(PFA = =NULL) {Perror ("fopen () Err"); return-1; } //every 3 seconds intseconds =0; while(1) {seconds=3; Do{seconds=sleep (seconds); } while(Seconds >0); //Get current Time Chartimearr[ -] = {0

partition #我已创建sda3650) this.width=650; "title=" image "style=" border-top:0px;border-right:0px;background-image:none;border-bottom:0 px;padding-top:0px;padding-left:0px;border-left:0px;padding-right:0px; "border=" 0 "alt=" image "src=" http:// S3.51cto.com/wyfs02/m01/73/41/wkiom1x4rblzqmteaabjtjxvrrs509.jpg "height=" 117 "/>3) partition file system advanced format #mkfs. Ext4/dev/sda3650) this.width=650; "title=" image "style=" border-top:0px;border-right:0px;background-image:none;border-botto

array is sorted first, then the median is obtained, and the sum of the distance is output according to the requirements.Time complexity: O (NLG N)Space complexity: O (N)Code: Analysis: The value of n is very large, it is impossible to save the votes of everyone directly. Note that there are at most only 10000 votes, so the real get tickets will not exceed 10000 people, so it is good to run.One scheme is discretization (the word is somewhat high-end, but very simple), the stored array into

This exercise is from the Liaoche JS tutorial. The answer is self-written.function functions are described as follows:Turn a string 13579 into a Array -- [1, 3, 5, 7, 9] and then use reduce() it to write a function that converts the string to number.The function code is as follows:' Use strict ';function String2int (s) {var arr = [];The string is converted to an array, but each element of the arrays that this method gets is still a stringfor (var i =

Problem Descriptionthis day the teacher as a routine to small Hao out a topic: The teacher to small Hao a string, let small Hao will the string in reverse order output. InputThe first line includes a T, which indicates how many sets of test data are:Next T line, each line includes a string (length not exceeding 65535).OutputFor each test data output it's reverse string.Sample Input1oaHoaiX EvoL ISample OutputI Love Xiaohao1#include 2#include string.h>3 intMain ()4 {5 intL,i,n;6 Chara[655

) - buy_lists.append (Goods_list[num]) to buy_price.append (Goods_price[num]) +elif num = =2and Money >=Goods_price[num]: -Money-=int(Goods_price[num]) thePrint"you have purchased%s"% Goods_list[num],'%d elements remaining'%Money ) * buy_lists.append (Goods_list[num]) $ buy_price.append (Goods_price[num])Panax Notoginsengelif num = =3and Money >=Goods_price[num]: -Money-=int(Goods_price[num]) thePrint"you have purchased%s"% Goods_list[num],'%d elements remaining'%Money ) + buy_lists.append (Goo

#first, mathematical functions are introducedImportMath#create a class of pointsclassPoint ():#the coordinates of the initialization point (x, y) def __init__(self,x=0,y=0): self.x=x self.y=y#gets the x-axis coordinates of a point defGetX (self):returnself.x#get the y-coordinate of a point defGetY (self):returnSelf.y#define a class of linesclassLine ():#two points of the initialization line def __init__(SELF,P1,P2): self.x= P1.getx ()-p2.getx () self.y= P1.gety ()-p2.gety ()#Pythagor

)/(Verctormagnitudetest (c) *verctormagnitudetest (b));//with custom dot product and modulus functions m_ ptemp.x = PTEMP2.X-PTEMP1.X;M_PTEMP.Y = Ptemp2.y-ptemp1.y;m_ptemp.z = Ptemp2.z-ptemp1.z;return m_pTemp;} The angle between the a vector and the B vector on the plane formed by vector A and vector b//ARSin (A and B fork multiplication of the die/(A-mode *b die)) function VectorZuoye7 (A:VECTOR3,B:VECTOR3): float{m_ftemp = Mathf.asin (Verctormagnitudetest (V Ectorchatest (b))/(Verctormagnitude

password")if (username== "or pwd=="):return 0Elseresult = Login (USERNAME,PWD)if (result==0):Print ("Account password cannot be empty, please login again")Return Logintip ()Elif (result==-1):Print ("Password is incorrect, please login again")Return Logintip ()Elif (result==1):Print ("Login Successful")ReturnIsbool = TrueWhile Isbool:Answer = input ("Do you want to log in or create a new account?") Please answer n/l ")if (answer== ' N '):Addtip ()Elif (answer== ' L '):Logintip ()ElsePrint ("Inpu

Tag:uppercase number ++str string statapiabc practice /*public class test1{public static void Main (String[]args) {String s = "abcdeEFHDKEI38475"; Char a[] = S.tochararray (); int lower = 0,upper = 0,other = 0; for (int i=0; iKey idea: Extract each character in the string and compare it. See the Java API documentation for details. Https://docs.oracle.com/javase/8/docs/api/index.htmlJava Exercise: Specify the number of upperc

in the warm burning relay, so there is the nearest neighbor algorithm, with the greedy algorithm, with the insertion algorithm, with the Christofides algorithm, with the LKH algorithm, with a linear programming algorithm. ” Vint Cerf and many other world-class computer scientists recommend "Publisher Weekly" Science and other magazines praise the Tide A brief history of time as a humorous p/np problem A feast on computation, Mathematics and logic Watercress 7.4 min"Thi

:") Span style= "FONT-FAMILY:CALIBRI; Font-size:14pt ">if (Num==num[::-1]): nbsp;print "is a palindrome number! " else: print" is not a palindrome number " Ten, beg greatest common divisor and least common multiplemin_num = min (num1, num2)for I in range (1, Min_num + 1):if (num1% i = = 0 and num2% i = = 0):Gys = iGBS = NUM1 * Num2/gysprint "%d and%d greatest common divisor:%d"% (NUM1, num2, Gys)print "%d and%d Max public:%d"% (NUM1, num2, GBS)Xi. guessing number gamesImport Rand

Original question:Write a program to copy its input to its output, replacing each string of one or more blanks by a single blank.The first type:This is most commonly used to set a inspace as a Boolean variable that flags whether the current input is in a character, or outside of a character#include The second kind: own thought, although the effect is similar, but the mentality is differentNumber as the flag bit, record the amount of space#include The third type:Use a while loop to "take" multipl

" "" Self.trans (Self.tree); def test (self): "" "test data, read from left to right and read from right to left" "" Tempnode = Self.tree while tempnode.ge Tleftnode () is not None: #找到最左节点 Tempnode = Tempnode.getleftnode () #print Tempnode.getdat A () #从左向右读 while Tempnode.getrightnode () are not none:print tempnode.getdata () Tempnod E = Tempnode.getrightnode () print Tempnode.getdata() #从右向左读 while tempnode are not none:print tempnode.getdata () Tempnode = Tempnode.get Leftnode () if __name__