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- int* p: int a[3]{1, 2, 3}; p = a; What is the value of *(p+2)?A struct to store a single playing card is as follows: struct card { char suit; char kind; }; The valid suits are 'S', 'H', 'D' and 'C', while the valid kinds are 'A', '2', '3', '4', '5', '6', '7', '8', '9', 'T', 'J', 'Q', and 'K'. A poker hand can be stored as an array of five struct cards. A flush is a hand that contains five cards, all of the same suit. Write a function that returns 1, if the poker hand passed to it has five cards of the same suit, and 0 otherwise. int isFlush(struct card hand[]); int isFlush(struct card hand[]) {var adventurersName = ["George", " Tim", " Sarah", " Mike", " Edward"];var adventurersKilled = 3;var survivors;var leader = "Captain Thomas King";var numberOfAdventurers = adventurersName.length; survivors = numberOfAdventurers - adventurersKilled; console.log("Welcome to The God Among Us\n");console.log("A group of adventurers began their search for the mystical god said to live among us. In charge of the squad was " + leader + " who was famous for his past exploits. Along the way, the group of comrades were attacked by the god's loyal followers. The adventurers fought with bravado and strength under the tutelage of "+ leader + " the followers were defeated but they still suffered great losses. After a headcount of the remaining squad, "+ adventurersKilled +" were found to be dead which left only " + survivors + " remaining survivors.\n"); console.log("Current Statistics :\n");console.log("Total Adventurers = " + numberOfAdventurers);console.log("Total Killed = " +…
- C++ A robot is initially located at position (0; 0) in a grid [?5; 5] [?5; 5]. The robot can move randomly in any of the directions: up, down, left, right. The robot can only move one step at a time. For each move, print the direction of the move and the current position of the robot. If the robot makes a circle, which means it moves back to the original place, print "Back to the origin!" to the console and stop the program. If it reaches the boundary of the grid, print \Hit the boundary!" to the console and stop the program. A successful run of your code may look like:Down (0,-1)Down (0,-2)Up (0,-1)Left (-1,-1)Left (-2,-1)Up (-2,0)Left (-3,0)Left (-4,0)Left (-5,0)Hit the boundary! or Left (-1,0)Down (-1,-1)Right (0,-1)Up (0,0)Back to the origin! About: This program is to give you practice using the control ow, the random number generator, and output formatting. You may use <iomanip> to format your output. You may NOT use #include "stdafx.h".Doctor -signature:String -doctorID:int - medicine:Arraylist +Doctor(signature:String,doctorID:int) +PrescribeMedicine():void + salary () +checkRecords():void Medicine Pharmacist -medName:String -startTime:int -dosage :int -endTime:int -date_prescribed:int - medicine:Arraylist +Medicine(medName:String,-dosage :int,date_prescribed:int) +Pharmacist (startTime:int,endTime:int) +checkForConflict():double +confirm_prescription():String +getStartTime():int +getEndTime():int +setStartTime(time:int):void +setEndTime(time1:int).voidString Pair // Problem Description // One person hands over the list of digits to Mr. String, But Mr. String understands only strings. Within strings also he understands only vowels. Mr. String needs your help to find the total number of pairs which add up to a certain digit D. // The rules to calculate digit D are as follow // Take all digits and convert them into their textual representation // Next, sum up the number of vowels i.e. {a, e, i, o, u} from all textual representation // This sum is digit D // Now, once digit D is known find out all unordered pairs of numbers in input whose sum is equal to D. Refer example section for better understanding. // Constraints // 1 <= N <= 100 // 1 <= value of each element in second line of input <= 100 // Number 100, if and when it appears in input should be converted to textual representation as hundred and not as one hundred. Hence number…
- int a[ 10] = {1,5,3,8,4,2,6,9,7}; int *p; p = &a[2]; p[0] = 10; p[1] = 15; What are the values of a? Select one: O a. {10,15,3,8,4,2,6,9,7} O b. {1,15,10,8,4,2,6,9,7} O c. {1,5,10,15,4,2,6,9,7} O d. {1,10,15,8,4,2,6,9,7}T/F 3) A Java main method uses the parameter (String[ ] variable) so that a user can run the program and supply“command-line” parameters. Since the parameter is a String array, however, the user does not have to supply anyparameters.Complete the code: string cars[5] = {"Volvo", "BMW", "Ford", "Mazda", "Honda"}; for(int i = 0; i < - ; i++) { cout << cars[i]< "\n"; } %3D
- Data Structure & Algorithum java program Do the following: 1) Add a constructor to the class "LList" that creates a list from a given array of objects.2) Add a method "addAll" to the "LList" class that adds an array of items to the end of the list. The header of the method is as follows, where "T" is the generic type of the objects in the list. 3) Write a Test/Driver program that thoroughly tests all the methods in the class "LList".c++ coding language I need help with part B and C please. If you are unable to do both, then PLEASE prioritize part C. I am really stuck and really can use the help. This is the code for c that was provided in order to guide me: const int N =31; // N parking spaces bool parking[N]; // the garage void EmptyTheLot(bool parking[], int N) { for(int i=0; i<N; i++) p[i]=false; // empty space } // returns -1 if no space found, //otherwise it returns 0<=i<N for a valid space. int FindSpace(int PlateNumber, bool parking[], int N) { // ????? } main() { EmptyTheLot(parking, N); // start with an empty parking garage. // get plate numbers and fill lot. }dictionaries = [] dictionaries.append({"First":"Bob", "Last":"Jones"}) dictionaries.append({"First":"Harpreet", "Last":"Kaur"}) dictionaries.append({"First":"Mohamad", "Last":"Argani"}) for i in range(0, len(dictionaries)): # Condition that's print the full name when Condition is True if(dictionaries[i]['First']=='Bob')or(dictionaries[i]['Last']=='Kaur'): print(dictionaries[i]['First']+" "+dictionaries[i]['Last']) ********************************** please modify the code to use a while loop instead of a for loop