Answer:
The correct answer is letter "A": The message is broken into independent blocks which are encrypted.
Explanation:
There are five (5) main forms of encryption: <em>Electronic Code Book (ECB), Cipher Block Chaining (CBC), Cipher Feedback (CFB), Output Feedback (OFB), and Output Feedback (OFB)</em>. Electronic Code Book (ECB) is the simplest of all of them. Using this method to encrypt information implies dividing a message into two parts to encrypt each block independently. ECB does not hide patterns effectively because the blocks are encrypted using identical plaintexts.
Answer:
PART ONE
- import java.util.Scanner;
- public class CountToLimit {
- public static void main(String[] args) {
- Scanner scnr = new Scanner(System.in);
- int countLimit = 0;
- int printVal = 0;
- // Get user input
- System.out.println("Enter Count Limit");
- countLimit = scnr.nextInt();
- do {
- System.out.print(printVal + " ");
- printVal = printVal + 1;
- } while ( printVal<=countLimit );
- System.out.println("");
- return;
- }
- }
PART TWO
- import java.util.Scanner;
- public class NumberPrompt {
- public static void main (String [] args) {
- Scanner scnr = new Scanner(System.in);
- System.out.print("Your number < 100: ");
- int userInput = scnr.nextInt();
- do {
- System.out.print("Your number < 100: ");
- userInput = scnr.nextInt();
- }while (userInput>=100);
- System.out.println("Your number < 100 is: " + userInput);
- return;
- }
- }
Explanation:
In Part one of the question, The condition for the do...while loop had to be stated this is stated on line 14
In part 2, A do....while loop that will repeatedly prompt user to enter a number less than 100 is created. from line 7 to line 10
Answer:
Technician B.
Explanation:
The claim of technician B that some vehicle manufacturers use a stepped ECT circuit inside the PCM to broaden the accuracy of the sensor is correct.
EDSAC is the first computer ever
Answer:
See explaination for the program code
Explanation:
The code below
Pseudo-code:
//each item ai is used at most once
isSubsetSum(A[],n,t)//takes array of items of size n, and sum t
{
boolean subset[n+1][t+1];//creating a boolean mtraix
for i=1 to n+1
subset[i][1] = true; //initially setting all first column values as true
for i = 2 to t+1
subset[1][i] = false; //initialy setting all first row values as false
for i=2 to n
{
for j=2 to t
{
if(j<A[i-1])
subset[i][j] = subset[i-1][j];
if (j >= A[i-1])
subset[i][j] = subset[i-1][j] ||
subset[i - 1][j-set[i-1]];
}
}
//returns true if there is a subset with given sum t
//other wise returns false
return subset[n][t];
}
Recurrence relation:
T(n) =T(n-1)+ t//here t is runtime of inner loop, and innner loop will run n times
T(1)=1
solving recurrence:
T(n)=T(n-1)+t
T(n)=T(n-2)+t+t
T(n)=T(n-2)+2t
T(n)=T(n-3)+3t
,,
,
T(n)=T(n-n-1)+(n-1)t
T(n)=T(1)+(n-1)t
T(n)=1+(n-1)t = O(nt)
//so complexity is :O(nt)//where n is number of element, t is given sum
