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Sunny_sXe [5.5K]
3 years ago
6

Помогите решить на с++

Computers and Technology
2 answers:
gulaghasi [49]3 years ago
8 0

Answer:

what?

Explanation:

Ugo [173]3 years ago
8 0
Yup i totally understand this
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________ sets up a point-to-point connection between two computer systems over an Internet Protocol (IP) network. A. Point-to-Po
nlexa [21]

Answer:

The correct answer is PPTP.

Explanation:

PPTP is the protocol that is used to create a connection which is referred to as tunnel and it transports data between two systems over the internet while also securing it with encryption. I hope this answer helps.

4 0
3 years ago
Consider the following method.
Akimi4 [234]

Answer:

The output of the given code as follows:

Program:

public class Main //define class

{

   public static void printSomething (int num, boolean val) //define method.

{

   num--; //decrease value by 1.

   System.out.print(val); //print value.

   System.out.print(num); //print value.

   

}

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

{

   printSomething(1, true); //calling function

   printSomething(2, true); //calling function

}

}

Output:

true 0

true 1

Explanation:

In the question, it is given there is a method that is "printSomething" is defined. This method accepts two parameters that are num and val. where num is an integer variable and val is boolean variable that holds two values only that are true or false. inside the method, we decrease the value of the num variable by 1 and then we print the value of val and num variable. This function does not return any value because we use return type void.

The function is defined in the same class so, we call the function two times that can be described as:

  • In first time calling we pass 1 and true value in function so it will "print true 0".
  • In second time calling we pass 2 and true value in function so it will "print true 1".

3 0
3 years ago
Create a do-while loop that asks the user to enter two numbers. The numbers should be added and the sum displayed. The loop shou
Gala2k [10]

Answer:

import java.util.Scanner;

public class TestClock {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       int sum = 0;

       char op;

       do{

           System.out.println("Enter two numbers");

           int num1= in.nextInt();

           int num2 = in.nextInt();

           sum = sum+num1+num2;

           System.out.println("Do you wish to perform another operation, Y/N");

           op =in.next().charAt(0);

       }while(op =='Y'||op=='y');

       System.out.println("sum "+sum);

   }

}

Explanation:

The code is implemented in Java progrmming language

  1. create an integer variable sum
  2. create a character variable op
  3. create a do while loop to request user to enter num1 and num2
  4. Request the user to also enter a char Y/N (yes or no repectively)
  5. While the char is Y keep adding up the sum and prompt the user to enter two new numbers
  6. Else break out of the loop and print the sum
6 0
3 years ago
30 points) Suppose you are given a string containing only the characters ( and ). In this problem, you will write a function to
navik [9.2K]

Answer:

The screenshot is attached below

Explanation:

The below program check balence of paranthesis using stack.

if the input charecter is ( then push to stack

if the input charecter is ) then pop top element and compair both elements for match.

Program:

public class Main

{

static class stack // create class for stack

{

int top=-1; // initialy stack is empty so top = -1

char items[] = new char[50]; // Create an array for stack to store stack elements

 

void push(char a) // Function push element to stack

{

if (top == 49) // chack stack is full ?

{

System.out.println("Stack full");

}

else

{

items[++top] = a; // increment the array index and store the element

}

}

 

char pop() // function to return the elements from stack

{

if (top == -1) // check the stack is empty

{

System.out.println("Empty stack");

return '\0';

}

else

{

char element = items[top]; // return the top element

top--; // Decrement the array index by one

return element;

}

}

 

boolean isEmpty() // Check the stack is empty or not

{

return (top == -1) ? true : false;

}

}

static boolean isMatch(char character1, char character2) // Check the input charecters are matching pairs or not

{

if (character1 == '(' && character2 == ')') // If they match return true

return true;

else

return false;

}

 

static boolean isBalanced(String parantheses) // check the input string is balenced or not

{

char[] exp = new char[parantheses.length()]; // Create a charecter array

for (int i = 0; i < parantheses.length(); i++) { // Convert the string parantheses to charecter array

exp[i] = parantheses.charAt(i);

}

stack st=new stack(); // Declare an empty character stack    

for(int i=0;i<exp.length;i++)

{  

if (exp[i] == '('))   // if input charecter is '(' push to stack

st.push(exp[i]);

if (exp[i] == ')') // if input charecter is ')' pop top element from stack

{

if (st.isEmpty())

{

return false;

}

 

else if ( !isMatch(st.pop(), exp[i]) ) // Call isMatch function

{

return false;

}

}

 

}

 

if (st.isEmpty())

return true; //balanced

else

{

return false; //not balanced

}

}

 

 

public static void main(String[] args)

{

 

System.out.println(isBalanced("()()()")); // Output true if string is balenced

 

}

 

}

Screenshot:

5 0
3 years ago
Letm1, m2,···mnbe distinct numbers on the number line, in the increasing order. Your goalis to color all of them blue. You have
siniylev [52]

Answer:

Following are the algorithm to this question:

y = 0 //  initialize variable y that assigns the value  0

p = 1 // initialize value 1 in the variable p which also known as starting position

init num = 1//define variable num that assign value 1

for j = 1 to n: //defining loop

y = m[j] - m[p]

if (y > 10) //defining if block

num++;  //increment num variable

p=i; //holding loop value in p variable

y= 0//assign value 0 in y variable

Explanation:

Following are the runtime analysis of the above-given algorithm:

The above-provided algorithm is greedy, but if it doesn't exceed the scope, it operates by greedily choosing its next object. Therefore the algorithm selects the fewest number of pens.

Running time:

This algorithm merely iterates once over all the points. The run-time is therefore O(n).

7 0
3 years ago
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