(Multiply the digits in an integer) Write a program that reads an integer between 0 and 1000 and multiplies all the digits in the integer. For example, if an integer is 932 , the multiplication of all its digits is 54 . Hint: Use the % operator to extract digits, and use the / operator to remove the extracted digit. For instance, 932 % 10

Answer:

Check the explanation

Explanation:

Q1: SQL to list names of all the students whose GPA is above specific number:

select SName

from Student

where GPA >= 3.8;

Using the Student table we are able to get the details where the GPa is greater than the required value.

Q2: SQL to get list of names of all the students who applied to some college:

select SName from Student where SID in (

select distinct SID from Apply

);

Selecting the name of the students who are present in the Apply table.

Q3: SQL to get list of names all the students who did not apply to any college:

select SName from Student where SID not in (

select distinct SID from Apply

);

Selecting the name of the students who are not present in the Apply table.

Q4: SQL to get list of names and enrollment of all the colleges who received applications for a major involving bio:

select CName, Enrollment from College where CName in (

select CName from Apply where Major = 'Bio' and SID in (

select SID from Student

)

);

Answer:

Java Programming

Details explained below.

Explanation:

import java.util.Scanner;

public class DemoDriver {

public static void main(String[] args) {

// TODO Auto-generated method stub

Scanner sc=new Scanner(System.in);

Purse sophie=new Purse();

Purse sally=new Purse();

while(true)

{

System.out.println("Enter 1 for Sophie, 2 for Sally, or 0 to exit: \n");

int girl=sc.nextInt();

int ch;

int pence,shill;

if(girl==1)

{

System.out.println("Enter 1 to give pence, 2 to give shillings, 3 to query her purse:\n");

ch=sc.nextInt();

if(ch==1)

{

System.out.println("Enter the pence to give:\n");

pence=sc.nextInt();

sophie.incPence(pence);

}else if(ch==2)

{

System.out.println("Enter the shillings to give:\n");

shill=sc.nextInt();

sophie.incshilling(shill);

}

else

{

System.out.println("Purse has "+sophie.getPence()+" pence,"+sophie.getshilling()+" shillings\n");

}

}

else if(girl==2)

{

System.out.println("Enter 1 to give pence, 2 to give shillings, 3 to query her purse:\n");

ch=sc.nextInt();

if(ch==1)

{

System.out.println("Enter the pence to give:\n");

pence=sc.nextInt();

sally.incPence(pence);

}else if(ch==2)

{

System.out.println("Enter the shillings to give:\n");

shill=sc.nextInt();

sally.incshilling(shill);

}

else

{

System.out.println("Purse has "+sally.getPence()+" pence,"+sally.getshilling()+" shillings\n");

}

}else

{

break;

}

}

}

} END.

Answer:

That is a Java language

Explanation:

This is simply Because C++ language does not support the calling of the methods without parameters and with the .(dot) operator.

It can also be clearly noted that This is the functionality of Java, for example, to sort we call Array.Sort().

So the parameter here is passed by placing the variable at front with the . (dot).

Answer:

Explanation:

1: The three most popular data systems that make use of Big Data file systems approach are:

The HDFS (Hadoop Distributed File System), Apache Spark, and Quantcast File System(QFS).

HDFS is the most popular among these and it makes use of the MapReduce algorithm to perform the data management tasks. It can highly tolerate faults and can run on low-cost hardware. It was written in Java and it is an open-source software.

Apache Spark makes use of Resilient Distributed Data (RDD) protocol. It is much faster and lighter than the HDFS and it can be programmed using a variety of languages such as Java, Scala, Python, etc. Its main advantage over HDFS is that it is highly scalable.

While QFS was developed as an alternative to the HDFS and it is also highly fault-tolerant and with space efficient. It makes use of the Reed-Solomon Error Correction technique to perform the task of data management.

2: The NewSQL databases were developed as a solution to the scalability challenges of the monolithic SQL databases. They were designed to allow multiple nodes in the context of an SQL database without affecting the replication architecture. It worked really well during the starting years of the cloud technology. Some of the databases that make use of New SQL technology are Vitess, Citus, etc.

Vitess was developed as an automatic sharding solution to the MySQL. Every MySQL instance acts as a shard of the overall database and each of these instances uses standard MySQL master-slave replication to ensure higher availability.

While, Citus is a PostgreSQL equivalent of the Vitess. It ensures transparent sharding due to which it accounts for horizontal write scalability to PostgreSQL deployments.

NoSQL database technology was developed to provide a mechanism for the storage and retrieval of data that is modeled in a way other than the tabular relations used in the traditional databases (RDBMS). The most popular database that makes use of the NoSQL technology is MongoDB. It functions as a cross-platform document-oriented database. It is known for its ability to provide high availability of replica sets. A replica set is nothing but a bundle of two or more copies of the data

class Animal:

   def __init__(self, arms, legs):

       self.arms = arms

       self.legs = legs

   def limbs(self):

       return self.arms + self.legs

# Creating an instance of Animal named spider with 4 arms and 4 legs

spider = Animal(4, 4)

# Calling the limbs method on the spider instance and saving the result to spidlimbs

spidlimbs = spider.limbs()

print("Number of limbs of the spider:", spidlimbs)

Answer:

Two profiles, one record type, two page layouts.

Explanation:

Record types and functions allow you to present forward different business processes, pick-list values, as well as page layouts to diverse range of users based on their profiles.

Going by the question, we can conclude that the minimum necessary configuration in order to meet the requirement in the question above are:Two profiles, one record type, two page layouts.

In mathematics, a combination refers to a method of selecting items from a sample where the order in which the elements of the sample are arranged does not matter.

C = (n, r) = [tex]\frac{n!}{(r!(n-r)!)}[/tex]

where n is the number of objects and

r the sample.

There are 3 categories of monsters:

He intends to catch at least two of each type. This is represented mathematically as:

C(3, 2) = 3!/(2!(3-2)!)

= 3!/2! x1!

= 3

He also intends to capture at least three fire-type monsters.

This is represented as:

C (3, 1) = 3!/(1!(3-1)!)

= 3!/1!x2!

= 3

Hence the different number of combinations are:

3+3 = 6

Learn more about Combinations at:

https://brainly.com/question/11732255

Answer:

Check the explanation

Explanation:

CODE:

#include <iostream>

#include <iomanip>

using namespace std;

double getLoanAmount() { // functions as per asked in question

double loan;

while(true){ // while loop is used to re prompt

cout << "Enter the car loan amount ($2,500-7,500): ";

cin >> loan;

if (loan>=2500 && loan <= 7500){ // if the condition is fulfilled then

break; // break statement is use to come out of while loop

}

else{ // else error message is printed

cout << "Error: $"<< loan << " is an invalid loan amount." << endl;

}

}

return loan;

}

double getMonthlyPayment(){ // functions as per asked in question

double monthpay;

while(true){ // while loop is used to re prompt

cout << "Enter the monthly payment ($50-750): ";

cin >> monthpay;

if (monthpay>=50 && monthpay <= 750){ // if the condition is fulfilled then

break; // break statement is use to come out of while loop

}

else{ // else error message is printed

cout << "Error: $"<< monthpay << " is an invalid monthly payment." << endl;

}

}

return monthpay;

}

double getInterestRate(){ // functions as per asked in question

double rate;

while(true){ // while loop is used to re prompt

cout << "Enter the annual interest rate (1-6%): ";

cin >> rate;

if (rate>=1 && rate <= 6){ // if the condition is fulfilled then

break; // break statement is use to come out of while loop

}

else{ // else error message is printed

cout << "Error: "<< rate << "% is an invalid annual interest rate." << endl;

}

}

return rate;

}

int main() {

cout << setprecision(2) << fixed; // to print with 2 decimal places

int month = 0; // initializing month

// calling functions and storing the returned value

double balance= getLoanAmount();

double payment= getMonthlyPayment();

double rate= getInterestRate();

rate = rate /12 /100; // as per question

// printing as per required in question

cout << "Month Balance($) Payment($) Interest($) Principal($)"<< endl;

while(balance>0){ // while the balance is more than zero

month = month + 1; // counting Months

// calculations as per questions

double interest = balance * rate;

double principal = payment - interest;

balance = balance - principal;

// printing required info with proper spacing

cout <<" "<< month;

cout <<" "<< balance;

cout <<" "<< payment;

cout <<" "<< interest;

cout <<" "<< principal << endl;

}

cout << "Months to repay loan: " << month << endl; // displaying month

cout << "End of Banker Bisons";

Kindly check the output in the attached image below.

Answer:

See explaination

Explanation:

a) Reorganizing the code

ADD $s2 , $t2, $t3

ADD $t5, $s2, $t6

SUB $t4, $s2, $t5

AND $t5,$t4,$t6

The reordering of the instruction leads to less hazards as compared to before.

b) with NOP

ADD $s2 , $t2, $t3

NOP

NOP

SUB $t4, $s2, $t5

NOP

ADD $t5, $s2, $t6

NOP

NOP

AND $t5,$t4,$t6

c. Pipeline with stalls

see attachment please

d. pipeline with forwarding

see attachment

Answer:

Check the explanation

Explanation:

Here is the code for you:

#include <iostream>

#include <fstream>

using namespace std;

int main ()

{

int aNumber=0;

int numbers=0;

double sum=0.0;

double average=0.0;

ifstream randomFile;

randomFile.open("Random.txt");

if (randomFile.fail())

cout << "failed to read file.";

else

{

while (randomFile >> aNumber)

{

numbers++;

sum+=aNumber;

}

if (numbers>0)

average = sum/numbers;

else

average=0.0;

cout << "Number of numbers: " << numbers << "\n";

cout << "Sum of the numbers: " << sum << "\n";

cout << "Average of the numbers: " << average;

}

randomFile.close();

return 0;

}

In this exercise we have to use the computer language knowledge in C++ to write the code as:

the code is in the attached image.

In a more easy way we have that the code will be:

#include <iostream>

#include <fstream>

using namespace std;

int main ()

{

int aNumber=0;

int numbers=0;

double sum=0.0;

double average=0.0;

ifstream randomFile;

randomFile.open("Random.txt");

if (randomFile.fail())

cout << "failed to read file.";

else

{

while (randomFile >> aNumber)

{

numbers++;

sum+=aNumber;

}

if (numbers>0)

average = sum/numbers;

else

average=0.0;

cout << "Number of numbers: " << numbers << "\n";

cout << "Sum of the numbers: " << sum << "\n";

cout << "Average of the numbers: " << average;

}

randomFile.close();

return 0;

}

See more about C++ code at brainly.com/question/25870717

Answer:

What

Explanation:

what

what

Following are the program to the given question:

Program Explanation:

Program:

// package

import java.util.*;

public class  Exercise12_07 // class Exercise12_07

{

   //defining method

   public static int bin2Dec(String binaryString) throws NumberFormatException //method bin2Dec that takes a one String parameter

   {

       int t = 0,i;//defining Integer variable

       for(i = 0; i < binaryString.length(); ++i)//defining loop that converts binary to decimal convert  

       {

           if(binaryString.charAt(i) != '0' && binaryString.charAt(i) != '1')//using if that checks binary digit with Exception

           {

               throw new NumberFormatException();//calling Exception

           }

           t += Math.pow(2, binaryString.length() - i - 1) * (binaryString.charAt(i) - '0');//converting and holding Decimal number

       }

       return t;//return value

   }

   public static void main(String[] a) //main method

   {

       Scanner inxv = new Scanner(System.in);//defining Scanner class Object

       System.out.print("Enter a binary number: ");//print message

       String str = inxv.nextLine();//defining String variable and input value

       try //defining try block

       {

           System.out.println("Decimal value of " + str + " is " + bin2Dec(str));//using print method to print method return value  

       }  

       catch (NumberFormatException e)//defining catch block

       {

           System.out.println("Not a binary number: " + str);//print message

       }

   }

}

Output:

Please find the attachment file.

Learn more:

brainly.com/question/19755688

Final answer:

Calculations reveal that Vendor A's system would take 0.209 seconds, while Vendor B's system, despite having hardware 1000 times faster, would take an infeasible 10,973.839 seconds due to the different algorithms used. Thus, Vendor A's system is the superior choice for query speed.

Explanation:

To determine which database system to choose based on the query response times for 1,048,576 customers, we need to calculate the response times for both Vendor A and Vendor B given their respective algorithms and computer speeds.

For Vendor A, the time is proportional to 10nlog2n machine operations. Plugging in the number of customers we have:

TA = 10 * 1,048,576 * log2(1,048,576) * 10-9

We know that log2(1,048,576) = 20 because 220 = 1,048,576. So:

TA = 10 * 1,048,576 * 20 * 10-9 = 209,715.2 * 10-9 = 0.209 seconds (rounded to three decimal places)

For Vendor B, whose computer is 1000 times faster, the time is proportional to 10n2 machine operations.

TB = 10 * (1,048,576)2 * 10-9 / 1000

TB = 10 * 1,097,383,936,256 * 10-9 / 1000 = 1.097x1013 * 10-9 seconds = 10,973.839 seconds, which is not feasible for the system. Hence, Vendor A's system is clearly the better choice in terms of response time.

Answer:

4000000 bits

Explanation:

The serial port configuration in asynchronous mode actually assumes a data size of 8bits, 1 start bit, no parity and 1 stop bit.

Therefore 8+1+1 = 10 bits.

This means that for every eight bits of data, ten bits are sent over the serial link — one start bit, the eight data bits, and the one stop bit)

We also know that there are 200 pages of ASCII data using asynchronous serial data transfer and each page has 80*25 of text characters.

Therefore the total number of bits transferred:

200*80*25*10 = 4000000 bits

200 pages of ASCII data, each with 80x25 characters, results in 4,000,000 bits transferred using asynchronous serial transfer with 8 data bits, 1 start bit, and 1 stop bit per character. Each character is 8 bits, totaling 10 bits per transfer including overhead.

To calculate the total number of bits transferred when sending 200 pages of ASCII data using asynchronous serial data transfer, we first need to determine the number of characters per page, and then multiply that by the total number of pages. Each ASCII character is represented by 8 bits (1 byte). Additionally, we need to consider the start bit, stop bit, and any potential overhead.

Given:

- ASCII character size: 8 bits

- 1 start bit

- 1 stop bit

- No parity

Each character will be transferred as:

1 (start bit) + 8 (ASCII data) + 1 (stop bit) = 10 bits per character.

Let's calculate:

1. Number of characters per page:

  - Each page has 80 characters per row and 25 rows.

  - Total characters per page = 80 × 25 = 2000 characters.

2. Total bits per page:

  - Total bits per page = 2000 characters × 10 bits/character = 20000 bits.

3. Total bits for 200 pages:

  - Total bits for 200 pages = 200 × 20000 bits = 4,000,000 bits.

So, the total number of bits transferred would be 4,000,000 bits.

Answer:

Explanation:

minor4server.c:

#include <pthread.h>

#include <stdio.h>

#include <stdlib.h>

#include <string.h>

#include <unistd.h>

#include <sys/types.h>

#include <sys/socket.h>

#include <netinet/in.h>

int player_count = 0;

pthread_mutex_t mutexcount;

void error(const char *msg)

{

perror(msg);

pthread_exit(NULL);

}

/* Reads an int from a client socket. */

int recv_int(int cli_sockfd)

{

int msg = 0;

int n = read(cli_sockfd, &msg, sizeof(int));

if (n < 0 || n != sizeof(int)) /* Client likely disconnected. */

return -1;

#ifdef DEBUG

printf("[DEBUG] Received int: %d\n", msg);

#endif

return msg;

}

/* Writes a message to a client socket. */

void write_client_msg(int cli_sockfd, char * msg)

{

int n = write(cli_sockfd, msg, strlen(msg));

if (n < 0)

error("ERROR writing msg to client socket");

}

/* Writes an int to a client socket. */

void write_client_int(int cli_sockfd, int msg)

{

int n = write(cli_sockfd, &msg, sizeof(int));

if (n < 0)

error("ERROR writing int to client socket");

}

/* Writes a message to both client sockets. */

void write_clients_msg(int * cli_sockfd, char * msg)

{

write_client_msg(cli_sockfd[0], msg);

write_client_msg(cli_sockfd[1], msg);

}

/* Writes an int to both client sockets. */

void write_clients_int(int * cli_sockfd, int msg)

{

write_client_int(cli_sockfd[0], msg);

write_client_int(cli_sockfd[1], msg);

}

/* Sets up the listener socket. */

int setup_listener(int portno)

{

int sockfd;

struct sockaddr_in serv_addr;

/* Get a socket to listen on */

sockfd = socket(AF_INET, SOCK_STREAM, 0);

if (sockfd < 0)

error("ERROR opening listener socket.");

/* Zero out the memory for the server information */

memset(&serv_addr, 0, sizeof(serv_addr));

  /* set up the server info */

serv_addr.sin_family = AF_INET;  

serv_addr.sin_addr.s_addr = INADDR_ANY;  

serv_addr.sin_port = htons(portno);      

/* Bind the server info to the listener socket. */

if (bind(sockfd, (struct sockaddr *) &serv_addr, sizeof(serv_addr)) < 0)

error("ERROR binding listener socket.");

#ifdef DEBUG

printf("[DEBUG] Listener set.\n");

#endif

/* Return the socket number. */

return sockfd;

}

/* Sets up the client sockets and client connections. */

void get_clients(int lis_sockfd, int * cli_sockfd)

{

socklen_t clilen;

struct sockaddr_in serv_addr, cli_addr;

#ifdef DEBUG

printf("[DEBUG] Listening for clients...\n");

#endif

/* Listen for two clients. */

int num_conn = 0;

while(num_conn < 2)

{

/* Listen for clients. */

  listen(lis_sockfd, 253 - player_count);

/* Zero out memory for the client information. */

memset(&cli_addr, 0, sizeof(cli_addr));

clilen = sizeof(cli_addr);

  /* Accept the connection from the client. */

cli_sockfd[num_conn] = accept(lis_sockfd, (struct sockaddr *) &cli_addr, &clilen);

if (cli_sockfd[num_conn] < 0)

/* Horrible things have happened. */

error("ERROR accepting a connection from a client.");

#ifdef DEBUG

printf("[DEBUG] Accepted connection from client %d\n", num_conn);

#endif

/* Send the client it's ID. */

write(cli_sockfd[num_conn], &num_conn, sizeof(int));

#ifdef DEBUG

printf("[DEBUG] Sent client %d it's ID.\n", num_conn);

#endif

/* Increment the player count. */

pthread_mutex_lock(&mutexcount);

player_count++;

printf("Number of players is now %d.\n", player_count);

pthread_mutex_unlock(&mutexcount);

if (num_conn == 0) {

/* Send "HLD" to first client to let the user know the server is waiting on a second client. */

write_client_msg(cli_sockfd[0],"HLD");

#ifdef DEBUG

printf("[DEBUG] Told client 0 to hold.\n");

#endif

}

num_conn++;

}

}

/* Gets a move from a client. */

int get_player_move(int cli_sockfd)

{

#ifdef DEBUG

printf("[DEBUG] Getting player move...\n");

#endif

/* Tell player to make a move. */

write_client_msg(cli_sockfd, "TRN");

/* Get players move. */

return recv_int(cli_sockfd);

}

/* Checks that a players move is valid. */

int check_move(char board[][3], int move, int player_id)

{

if ((move == 9) || (board[move/3][move%3] == ' ')) { /* Move is valid. */

#ifdef DEBUG

printf("[DEBUG] Player %d's move was valid.\n", player_id);

#endif

return 1;

}

else { /* Move is invalid. */

#ifdef DEBUG

printf("[DEBUG] Player %d's move was invalid.\n", player_id);

#endif

return 0;

}

}

/* Updates the board with a new move. */

void update_board(char board[][3], int move, int player_id)

{

board[move/3][move%3] = player_id ? 'X' : 'O';

#ifdef DEBUG

printf("[DEBUG] Board updated.\n");

#endif

}

/* Draws the game board to stdout. */

void draw_board(char board[][3])

{

printf(" %c | %c | %c \n", board[0][0], board[0][1], board[0][2]);

printf("-----------\n");

printf(" %c | %c | %c \n", board[1][0], board[1][1], board[1][2]);

printf("-----------\n");

printf(" %c | %c | %c \n", board[2][0], board[2][1], board[2][2]);

}

/* Sends a board update to both clients. */

void send_update(int * cli_sockfd, int move, int player_id)

{

#ifdef DEBUG

printf("[DEBUG] Sending update...\n");

#endif

/* Signal an update */

write_clients_msg(cli_sockfd, "UPD");

/* Send the id of the player that made the move. */

write_clients_int(cli_sockfd, player_id);

/* Send the move. */

write_clients_int(cli_sockfd, move);

#ifdef DEBUG

printf("[DEBUG] Update sent.\n");

#endif

Answer:

40 ns would be the maximum MIPS value to be achieved for this computer.

Answer:

MIPS = 10.53

Explanation:

To find the maximum allowable MIPS, we need to obtain the total maximum time for an instruction to be executed by the CPU.

Total time to complete 1 instruction for the computer will entail the instruction passing through the 3 stages.

Total time to complete 1 instruction = 40 + 26 + 29 = 95 ns = (95 × 10⁻⁹) s

MIPS = millions instructions per second and it is given as

(Number of instruction)/(execution time × 1,000,000)

MIPS = 1/(95 × 10⁻⁹ × 10⁶)

MIPS = 10.53

Hope this Helps!!!

Answer:

Explanation:

# Run Monte Carlo 10k

B <- 10000

results <- replicate(B, {

 winners <- sample(runners, 3)

 (winners[1] %in% "Jamaica" & winners[2] %in% "Jamaica" & winners[3] %in% "Jamaica")

})

mean(results)

Explanation:

yeah yeah is that we go together on the watch remember write a timer called that depressed display of the first 27 value in the XM Sirius

Answer:

See explaination for the details.

Explanation:

Code:

def create_3D(H, W, D):

result = []

for i in range(D):

result.append([])

for j in range(H):

result[i].append([])

for k in range(W):

result[i][j].append(i+j+k)

return result

print(create_3D(2, 3, 4))

Check attachment for onscreen code.

Answer:

#include<iostream>

#include<iomanip>

using namespace std;

//recursive method to find the GCF of two numbers

int GCF(int n, int m){

               //case 1, if m<=n and n mod m is 0, then gcf is m

               if(m<=n && n%m==0){

                               return m;

               }

               //case 2, if n<m, then gcf= GCF(m,n)

               if(n<m){

                               return GCF(m,n);

               }

               //otherwise, returning GCF(m, n mod m)

               return GCF(m,n%m);

}

int main(){

               //declaring two arrays of size 100 to store the set of first and second numbers

               //and a count

               int first[100], second[100], count=0;

               char ch;

               //asking user if he likes to enter two numbers, reading choice

               cout<<" Would you like to enter two numbers? ";

               cin>>ch;

               //looping as long as ch is 'y' or 'Y'

               while(ch=='y' || ch=='Y'){

                               //prompting and reading numbers to first and second

                               cout<<endl<<"Please enter two numbers ";

                               cin>>first[count]>>second[count];

                               //updating count, asking again

                               count++;

                               cout<<" Would you like to enter two numbers? ";

                               cin>>ch;

               }

               //left justifying output and printing a blank line

               cout<<left<<endl;

               //using a field with of 10, printing each column header

               cout<<setw(10)<<"First #";

               cout<<setw(10)<<"Second #";

               cout<<setw(10)<<"GCF"<<endl;

               //printing separators

               cout<<setw(10)<<"=======";

               cout<<setw(10)<<"========";

               cout<<setw(10)<<"==="<<endl;

               //looping through the arrays

               for(int i=0;i<count;i++){

                               //displaying both numbers and their GCF

                               cout<<setw(10)<<first[i];

                               cout<<setw(10)<<second[i];

                               cout<<setw(10)<<GCF(first[i],second[i])<<endl;

               }

               return 0;

}

Explanation:

Run the program and see the output

The groups_per_user function receives a dictionary, which contains group names with the list of users.

Explanation:

The blanks to return a dictionary with the users as keys and a list of their groups as values is shown below :

def groups_per_user(group_dictionary):

   user_groups = {}

   # Go through group_dictionary

   for group,users in group_dictionary.items():

       # Now go through the users in the group

       for user in users:

       # Now add the group to the the list of

         # groups for this user, creating the entry

         # in the dictionary if necessary

         user_groups[user] = user_groups.get(user,[]) + [group]

   return(user_groups)

print(groups_per_user({"local": ["admin", "userA"],

       "public":  ["admin", "userB"],

       "administrator": ["admin"] }))

The missing statements in the program are:

The first missing instruction is to iterate through the group_dictionary.

To do this, we make use of the following enhanced for loop:

for group,users in group_dictionary.items():

The above statement would iterate through the keys and the values of the dictionary.

The next missing statement is to iterate through the values in users list

To do this, we make use of the following enhanced for loop:

for user in users:

Lastly, the group is added to the user list using:

user_groups[user] = user_groups.get(user,[]) + [group]

Read more about dictionary and lists at:

https://brainly.com/question/14353514

Answer:

Explanation:

The solution code is written in Python 3.

Firstly prompt user to enter an equation using input function (Line 1).

Create if-else if statements to check if operator "+", "-", "*", "/" and "%" exist in the input equation. If "+" is found (Line 3), use split method to get the individual operands from the equation by using "+" as separator (Line 5). Output the equation as required by the question using string concatenation method (Line 6).  The similar process is repeated for the rest of else if blocks (Line 7 - 22).  

Answer:

def get_middle_ten(sentence):

   ind = (len(sentence) - 12) // 2

   return sentence[ind:ind + 12]

# Testing the function here. ignore/remove the code below if not required

print(get_middle_twelve("abcdefghijkl"))

print(get_middle_twelve("abcdefghijklmnopqr"))

print(get_middle_twelve("abcdefghijklmnopqrst"))

The question is asking to create a program for a contact list, where users can input name-number pairs and then retrieve a number by inputting a name. This represents a fundamental database functionality, which can be programmed using associative arrays or dictionaries in most programming languages.

The student is essentially asking for a simple contact list program which stores names and phone numbers and retrieves a phone number when a name is provided. In context, this is a basic database operation. A typical approach to this problem would involve associative arrays or dictionaries commonly used in programming. For instance, in Python, the program would look something like this:

contacts = {}

while True:

   name = input()

   if name == '':

       break

   number = input()

   contacts[name] = number

search_name = input()

print(contacts.get(search_name, 'Name not found.'))

The student could adapt this program to further include other contact details, such as email addresses and birthdates. A more complex version may involve one-to-many or many-to-many relationships where an individual could have multiple phone numbers or emails, and vice versa.

Answer:

#include<iostream>

#include<conio.h>

using namespace std;

main()

{

int numberofshares;

float costpershare, amount;

costpershare= 25.625;

numberofshares = 400;

amount = costpershare*numberofshares;

cout<<"\nTotal amount of Stock Purchase =" << amount;

getch();

}

Explanation:

In above mentioned program, total amount of stock purchase need to be calculated for given data. The values 25.625 and 400 assigned to the variables of cost per share and  number of shares.

The formula to calculate the total amount of stock purchase is simply Multiplication of cost and total number of shares. The output will be displayed with the help of variable named as amount.

Solution

Cost per Share = 25.625

Number of shares = 400

amount = cost per share * Number of shares

              =   25.625 * 400

              =    10,250

To calculate the amount of a stock purchase, create variables for cost per share and number of shares, and then multiply them together to find the total amount.

To calculate the amount of a stock purchase:

Following these steps, the amount of the stock purchase can be determined by multiplying the cost per share by the number of shares.

Answer:

See explaination

Explanation:

using System;

using System.Collections.Generic;

using System.ComponentModel;

using System.Data;

using System.Drawing;

using System.Linq;

using System.Text;

using System.Threading.Tasks;

using System.Windows.Forms;

namespace BookListBox

{

public partial class Form1 : Form

{

public Form1()

{

InitializeComponent();

}

string[] Book={"In Search of Lost Time","Ulysses","Don Quixote","The Great Gatsby","One Hundred Years of Solitude"};//declare variable book name

string[] Book_displayMessage = { "In Search of Lost Time by Marcel Proust", " Ulysses by James Joyce", "Don Quixote by Miguel de Cervantes", " The Great Gatsby by F. Scott Fitzgerald", "One Hundred Years of Solitude by Gabriel Garcia Marquez" };//declare variable Short dispay

string[] Book_BrienfSynpsis = { "Swann's Way, the first part of A la recherche de temps perdu, Marcel Proust's seven-part cycle, was published in 1913. In it, Proust introduces the themes that run through the entire work. ", "Ulysses chronicles the passage of Leopold Bloom through Dublin during an ordinary day, June 16, 1904.", "Alonso Quixano, a retired country gentleman in his fifties, lives in an unnamed section of La Mancha with his niece and a housekeeper.", "The novel chronicles an era that Fitzgerald himself dubbed the Jazz Age. ", "One of the 20th century's enduring works, One Hundred Years of Solitude is a widely beloved and acclaimed novel known throughout" };//declare variable Brienf synpsis

private void Form1_Load(object sender, EventArgs e)

{

lblBriefSynopsis.Text = "";//blank display textbox

lbldisplaymessage.Text="";//blank display textbox

for(int i=0;i<5;i++)

lbbook.Items.Add(Book[i]);//add book to listbox

}

private void lbbook_MouseHover(object sender, EventArgs e)

{

Point point = lbbook.PointToClient(Cursor.Position);

int index = lbbook.IndexFromPoint(point);

if (index < 0) return;

//Do any action with the item

lbldisplaymessage.Text = "Here is the Short things about it"+Environment.NewLine+Book_displayMessage[index];

}

private void lbbook_MouseLeave(object sender, EventArgs e)//mouse leave event here

{

lbldisplaymessage.Text = "";//here display message will blank textbox

}

private void lbbook_SelectedIndexChanged(object sender, EventArgs e)

{

}

private void lbbook_MouseClick(object sender, MouseEventArgs e)

{

lblBriefSynopsis.Text = Book_BrienfSynpsis[lbbook.SelectedIndex];//display click brienf Synposis

if (lbbook.SelectedIndex % 5 == 0)

this.BackColor=Color.Blue;

else if (lbbook.SelectedIndex % 5 == 1)

this.BackColor = Color.Red;

else if (lbbook.SelectedIndex % 5 == 2)

this.BackColor = Color.Purple;

else if (lbbook.SelectedIndex % 5 == 3)

this.BackColor = Color.Pink;

else if (lbbook.SelectedIndex % 5 == 4)

this.BackColor = Color.Orange;

}

}

}

McDonald's H&M McDonalds Inc is incorrect for name accuracy.

Explanation:

The correct Name is McDonald's. But Mc Donald's H&M and McDonalds inc is incorrect for name accuracy.

McDonald's is a fast-food company based in America. It has its retail outlets throughout the world. It is very famous for its burgers and french fries. It is the world's second-largest employer in the private sector. It is a symbol of globalization as it has its outlets throughout the world.

The name 'McDonald's H&M' is incorrect for name accuracy since it amalgamates two distinct brand identities. While 'McDonalds Inc' may simply be missing an apostrophe, it is possibly acceptable, but 'McDonald's H&M' clearly combines unrelated trademarks.

When evaluating the list of business names provided for name accuracy, it's important to consider the standard format of business names and trademarks. Typically, the name 'McDonald's' is associated with the well-known fast-food chain, and is correctly punctuated with an apostrophe to indicate the possessive form, likely referring to the founder's name (McDonald).

'McDonald's H&M' would be incorrect for name accuracy, as it combines the names of two distinct and unrelated brands, McDonald's and H&M, which are independently operated and have separate trademark rights. 'McDonalds Inc' is missing the possessive apostrophe, which could be seen as a minor error, but might still be legally operational depending on registration specifics.

Therefore, out of the options given, 'McDonald's H&M' would be rated incorrect for name accuracy due to the improper combination of two separate brand identities into one business name.

Answer:

import java.util.Scanner;

public class BirthdayChecker

{

   public static void main(String[] args)

   {

       String sentinelValue = "ZZZ";

       final int size = 10;

       int count = 0;

       String name = null;

       String dateOfBirth = null;

       String[] namesArray = new String[size];

       String[] dateOfBirthsArray = new String[size];

     Scanner scanner = new Scanner(System.in);

       System.out.println("Enter name or enter ZZZ " +  "to quit");

       name = scanner.nextLine();

       while(!name.equals(sentinelValue) && count < 10)

       {

           System.out.println("Enter date of birth " +  "(dd-mm-yyyy)");

           dateOfBirth=scanner.nextLine();

           namesArray[count] = name;

           dateOfBirthsArray[count] = dateOfBirth;

           System.out.println("Enter name or enter " +

           "ZZZ to quit");

           name=scanner.nextLine();

           count++;

       }

       System.out.println("Count of namesArray = "+count);

       System.out.println("The entered namesArray are:");

       for(int i = 0; i < count; i++)  

       {

           System.out.println(namesArray[i]);

       }

       boolean criteria = true;

       boolean flag = false;

       while(criteria)

       {

           System.out.println("Enter name to " +

           "display date of birth or enter " +

           "ZZZ to quit");

           name = scanner.nextLine();

           if(name.equals(sentinelValue))

               criteria = false;

           else

           {

               for(int i = 0; i < count && !flag;

               i++)

               {

                   if(namesArray[i].equals(name))

                   {

                       flag = true;

                       dateOfBirth = dateOfBirthsArray[i];

                   }

               }

               if(flag)

               {

                   System.out.println("Date of Birth "

                   + "of " + name + " is "

                   + dateOfBirth);

               }

               else

               {

                   System.out.println("Date Of Birth "

                   + "of " + name + " is not flag");

               }

           }

           flag = false;

       }

   }

}

Explanation:

The `BirthdayReminder` Java app lets users enter names and birthdates, then lookup birthdates by name until "ZZZ" is entered.

Here's a Java application named `BirthdayReminder` that fulfills the specified requirements:

```java

import java.util.Scanner;

public class BirthdayReminder {

   public static void main(String[] args) {

       Scanner scanner = new Scanner(System.in);        

       String[] names = new String[10];

       String[] birthdates = new String[10];

       int count = 0;        

       // Input loop to enter names and birthdates

       while (count < 10) {

           System.out.print("Enter name (ZZZ to end): ");

           String name = scanner.nextLine();            

           if (name.equals("ZZZ")) {

               break; // Exit loop if sentinel value ZZZ is entered

           }            

           names[count] = name;            

           System.out.print("Enter birthdate (MM/DD/YYYY): ");

           String birthdate = scanner.nextLine();

           birthdates[count] = birthdate;            

           count++;

       }        

       System.out.println("\nNumber of names entered: " + count);

       System.out.println("List of names entered:");

       for (int i = 0; i < count; i++) {

           System.out.println(names[i]);

       }        

       // Lookup loop to find birthdate by name

       while (true) {

           System.out.print("\nEnter a name to lookup birthdate (ZZZ to end): ");

           String lookupName = scanner.nextLine();            

           if (lookupName.equals("ZZZ")) {

               break; // Exit loop if sentinel value ZZZ is entered

           }            

           boolean found = false;

           for (int i = 0; i < count; i++) {

               if (names[i].equals(lookupName)) {

                   System.out.println("Birthdate of " + lookupName + " is " + birthdates[i]);

                   found = true;

                   break;

               }

           }

           if (!found) {

               System.out.println("Name not found. Please enter a valid name.");

           }

       }        

       scanner.close();

   }

}

Explanation:

1. **Data Structures**: Two arrays `names` and `birthdates` are used to store up to 10 names and their corresponding birthdates.

2. **Input Loop**: The program prompts the user to enter names and birthdates in a loop until either 10 names are entered or the user inputs "ZZZ". Each name and birthdate pair is stored in the respective arrays.

3. **Output**: After input is finished, the program displays the count of names entered and lists the names that were stored.

4. **Lookup Loop**: The program then enters a lookup loop where it prompts the user to enter a name to retrieve the corresponding birthdate. It continues this until the user inputs "ZZZ". If a valid name is entered, it searches the `names` array for a match and displays the associated birthdate; if not found, it informs the user accordingly.

5. **End of Program**: The program closes the scanner once all operations are completed.

This Java application effectively manages user input, data storage, and retrieval as specified, providing a straightforward way to handle names and birthdates with user interaction and error handling.

RC4 stream cipher in C. User is able to enter any key that is 5 bytes to 32 bytes long

Explanation:

#include <stdio.h>

#include <string.h>

#include <stdlib.h>

#define N 256   // 2^8

void swap(unsigned char *a, unsigned char *b) {

   int tmp = *a;

   *a = *b;

   *b = tmp;

}

int KSA(char *key, unsigned char *S) {

   int len = strlen(key);

   int j = 0;

   for(int i = 0; i < N; i++)

       S[i] = i;

   for(int i = 0; i < N; i++) {

       j = (j + S[i] + key[i % len]) % N;

       swap(&S[i], &S[j]);

   }

   return 0;

}

int PRGA(unsigned char *S, char *plaintext, unsigned char *ciphertext) {

   int i = 0;

   int j = 0;

   for(size_t n = 0, len = strlen(plaintext); n < len; n++) {

       i = (i + 1) % N;

       j = (j + S[i]) % N;

       swap(&S[i], &S[j]);

       int rnd = S[(S[i] + S[j]) % N];

       ciphertext[n] = rnd ^ plaintext[n];

   }

   return 0;

}

int RC4(char *key, char *plaintext, unsigned char *ciphertext) {

   unsigned char S[N];

   KSA(key, S);

   PRGA(S, plaintext, ciphertext);

   return 0;

}

int main(int argc, char *argv[]) {

   if(argc < 3) {

       printf("Usage: %s <key> <plaintext>", argv[0]);

       return -1;

   }

   unsigned char *ciphertext = malloc(sizeof(int) * strlen(argv[2]));

   RC4(argv[1], argv[2], ciphertext);

   for(size_t i = 0, len = strlen(argv[2]); i < len; i++)

       printf("%02hhX", ciphertext[i]);

   return 0;

}

Answer:

See explaination

Explanation:

class Episode{

//Private variables

private String title;

private int seasonNumber;

private int episodeNumber;

//Argumented constructor

public Episode(String title, int seasonNumber, int episodeNumber) {

this.title = title;

this.seasonNumber = seasonNumber;

this.episodeNumber = episodeNumber;

}

//Getter and setters

public String getTitle() {

return title;

}

public void setTitle(String title) {

this.title = title;

}

public int getSeasonNumber() {

return seasonNumber;

}

public void setSeasonNumber(int seasonNumber) {

this.seasonNumber = seasonNumber;

}

public int getEpisodeNumber() {

return episodeNumber;

}

public void setEpisodeNumber(int episodeNumber) {

this.episodeNumber = episodeNumber;

}

public boolean comesBefore(Episode e){

//Check if titles' match

if(this.title.equals(e.getTitle())){

//Compare season numbers

if(this.seasonNumber<e.seasonNumber){

return true;

}

//Check if season numbers match

else if(this.seasonNumber==e.getSeasonNumber()){

return this.episodeNumber<e.getEpisodeNumber();

}

}

return false;

}

atOverride // replace the at with at symbol

public String toString() {

return title+", season "+seasonNumber+", episode "+episodeNumber;

}

}

class Main{

public static void main(String[] args) {

Episode e = new Episode("Friends",1,1);

System.out.println(e);

Episode e2 = new Episode("The Wire",3,5);

System.out.println(e2);

System.out.println(e.getTitle()+" comes before "+e2.getTitle()+" = "+e.comesBefore(e2));

}

}