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Anastasy [175]
3 years ago
10

Which of the following is a technique used by hackers to identify unsecured wireless network locations to other hackers?A. Blues

narfingB. War dialingC. War chalkingD. War driving
Computers and Technology
1 answer:
natka813 [3]3 years ago
5 0

Answer:Is called war driving

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in java how do i Write a program that reads a set of integers, and then prints the sum of the even and odd integers.
hammer [34]

Answer:

Here is the JAVA program:

import java.util.Scanner; //to take input from user

public class Main{

public static void main(String[] args) { //start of main function

Scanner input = new Scanner(System.in); // creates Scanner class object  

       int num, integer, odd = 0, even = 0; //declare variables

       System.out.print("Enter the number of integers: "); //prompts user to enter number of integers

       num = input.nextInt(); //reads value of num from user

       System.out.print("Enter the integers:\n"); //prompts user to enter the integers

       for (int i = 0; i < num; i++) { //iterates through each input integer

           integer = input.nextInt(); // reads each integer value

               if (integer % 2 == 0) //if integer value is completely divisible by 2

                   even += integer; //adds even integers

               else //if integer value is not completely divisible by 2

                   odd += integer;  } //adds odd integers

           System.out.print("Sum of Even Numbers: " + even); //prints the sum of even integers

           System.out.print("\nSum of Odd Numbers: " + odd);//prints the sum of odd integers

           }}

Explanation:

The program is explained in the comments mentioned with each line of the code. I will explain the logic of the program with the help of an example.

Suppose user wants to input 5 integers. So,

num = 5

Suppose the input integers are:

1, 2 , 3, 4, 5

for (int i = 0; i < num; i++)  is a for loop that has a variable i initialized to 0. The condition i<num is true because i=0 and num=5 so 0<5. Hence the statements inside body of loop execute.

At first iteration:

integer = input.nextInt(); statement reads the value of input integer. The first integer is 1

if (integer % 2 == 0) checks if integer is completely divisible by 2. The modulo operator is used which returns the remainder of the division and if this remainder is equal to 0 then it means that the integer is completely divisible by 2 and if the integer is completely divisible by 2 then this means the integer is even. 1%2 returns 1 so this means this condition evaluates to false and the integer is not even. So the else part executes:

odd += integer;   this becomes:

odd = odd + integer

odd = 1

Now the value of i is incremented to 1 so i=1

At second iteration:

integer = input.nextInt(); statement reads the value of input integer. The first integer is 2

if (integer % 2 == 0) checks if integer is completely divisible by 2. 2%2 returns 0 so this means this condition evaluates to true and the integer is even. So

even += integer;   this becomes:

even= even+ integer

even = 2

Now the value of i is incremented to 1 so i=2

At third iteration:

integer = input.nextInt(); statement reads the value of input integer. The first integer is 3

if (integer % 2 == 0) checks if integer is completely divisible by 2. 3%2 returns 1 so this means this condition evaluates to false and the integer is odd. So

odd += integer;   this becomes:

odd= odd + integer

odd= 1 + 3

odd = 4

Now the value of i is incremented to 1 so i=3

At fourth iteration:

integer = input.nextInt(); statement reads the value of input integer. The first integer is 4

if (integer % 2 == 0) checks if integer is completely divisible by 2. 4%2 returns 0 so this means this condition evaluates to true and the integer is even. So

even+= integer;   this becomes:

even = even + integer

even = 2 + 4

even = 6

Now the value of i is incremented to 1 so i=4

At fifth iteration:

integer = input.nextInt(); statement reads the value of input integer. The first integer is 5

if (integer % 2 == 0) checks if integer is completely divisible by 2. 5%2 returns 1 so this means this condition evaluates to false and the integer is odd. So

odd+= integer;   this becomes:

odd= odd+ integer

odd= 4 + 5

odd = 9

Now the value of i is incremented to 1 so i=5

When i=5 then i<num condition evaluate to false so the loop breaks. The next two print statement prints the values of even and odd. So

even = 6

odd = 9

Hence the output is:

Sum of Even Numbers: 6                                                                                                           Sum of Odd Numbers: 9  

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Programming CRe-type the code and fix any errors. The code should convert non-positive numbers to 1.
LUCKY_DIMON [66]

Answer:

Given

The above lines of code

Required

Rearrange.

The code is re-arrange d as follows;.

#include<iostream>

int main()

{

int userNum;

scanf("%d", &userNum);

if (userNum > 0)

{

printf("Positive.\n");

}

else

{

printf("Non-positive, converting to 1.\n");

userNum = 1;

printf("Final: %d\n", userNum);

}

return 0;

}

When rearranging lines of codes. one has to be mindful of the programming language, the syntax of the language and control structures in the code;

One should take note of the variable declarations and usage

See attachment for .cpp file

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