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Sergio [31]
2 years ago
8

Suppose an array with six rows and eight columns is stored in row major order starting at address 20 (base 10). If each entry in

the array requires only one memory cell, what is the address of the entry in the third row and fourth column?
Computers and Technology
1 answer:
Ronch [10]2 years ago
3 0

Answer:

39

Explanation:

Since each of the address occupies only 1 memory cell and the 2-D array is row-major 2-D array.So the elements will be filled row wise.First the first row will be fully filled after that second row and so on.Since we want the address of the element at third row and fourth column.

we can generalize this :

address of the element at ith row and jth column=s + ( c * ( i - 1 ) + ( j - 1 ) ).

s=Starting address.

c=Number of columns in the 2-D array.

address=20+(8*(3-1)+(4-1))

=20+(8*2+3)

=20+16+3

=39

Or you can make a matrix of six rows and eight columns and put first cell with 20.Start filling the elements row wise one by one and look what is the count of 3rd row and 4th column.

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Which software programs should students avoid using to create and submit course work? (Choose all that apply.
mel-nik [20]

Answer:

The answer to this question is given below in the explanation section.

Explanation:

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4 0
2 years ago
Consider an array inarr containing atleast two non-zero unique positive integers. Identify and print, outnum, the number of uniq
Delvig [45]

Answer:

Program.java  

import java.util.Scanner;  

public class Program {  

   public static boolean isPalindrome(String str){

       int start = 0, end = str.length()-1;

       while (start < end){

           if(str.charAt(start) != str.charAt(end)){

               return false;

           }

           start++;

           end--;

       }

       return true;

   }  

   public static int countPalindromePairs(String[] inarr){

       int count = 0;

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

           for(int j=i+1; j<inarr.length; j++){

               StringBuilder sb = new StringBuilder();

               sb.append(inarr[i]).append(inarr[j]);

               if(isPalindrome(sb.toString())){

                   count++;

               }

           }

       }

       return count == 0 ? -1 : count;

   }

   public static void main(String[] args) {

       Scanner sc = new Scanner(System.in);

       String line = sc.next();

       String[] inarr = line.split(",");

       int count = countPalindromePairs(inarr);

       System.out.println("RESULT: "+count);

   }

}

Explanation:

OUTPUT:

3 0
2 years ago
Suppose you are given three data items that indicate a starting index in a list, an array of characters, and an array of link in
Artemon [7]

Answer:

see explaination

Explanation:

INCLUDE Irvine32.inc

;declare the variables

SPACE = 32

startIndex = 1

.data

statement BYTE "The correct sequence of array: ",0

; declare the chars('H','A','C','E','B','D','F','G')

chars BYTE 48h,41h,43h,45h,42h,44h,46h,47h

sizeOfArray = ($ - chars)

links DWORD 0,4,5,6,2,3,7,0

outputArray BYTE sizeOfArray DUP(?)

.code

main PROC

; assign starting index

mov eax,startIndex

; assign offset of array chars

mov ebx,OFFSET chars

; assign the size of the array

mov ecx, sizeOfArray

; assign the offset of array links

mov edi,OFFSET links

; assign the offset of new array

mov esi,OFFSET outputArray

; call the traverseLink procedure

call traverseLink

; display the characters of new array

mov edx,OFFSET statement

call writeString

; assign the offset of new array

mov esi,OFFSET outputArray

; assign the size of the array

mov ecx, sizeOfArray

; call the procedure printArray to print the ordered array

call printArray

exit

main ENDP

traverseLink PROC

forLoop:

; get the character at index eax

mov dl,[ebx+eax]

; assign it to new array

mov [esi] , dl

; get the value at the next index

mov eax,[edi+eax*4]

; goto next location of new array

inc esi

; repeat until ecx=0

loop forLoop

ret

traverseLink ENDP

printArray PROC

forLoop:

; get the character at index eax

mov al,[esi]

; print the character

call WriteChar

; increament esi to get next character

inc esi

; print the space character

mov al,SPACE

call WriteChar

loop forLoop

call Crlf

ret

printArray ENDP

END main

7 0
2 years ago
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