To draw a 4x4 array, create a grid consisting of 4 rows and 4 columns. Start by drawing four horizontal lines parallel to each other, then draw four vertical lines intersecting them to form a grid. Each cell in the grid represents a position in the array, totaling 16 cells. You can label each cell with coordinates, such as (1,1) for the top-left cell and (4,4) for the bottom-right cell.
To draw a 5x7 array, start by creating 5 rows and 7 columns. You can represent each row as a horizontal line and each column as a vertical line, forming a grid. Fill each cell of the grid to visualize the array, where each cell corresponds to a single unit. This creates a total of 35 cells, representing the product of 5 and 7.
I suggest using an array with as many elements as the longest row you need. To keep it simple, keep two copies of the array, and calculate each element of the "new" array as the sum of the corresponding element, plus the previous element, of the "old" array. Then copy the information back for the next step.
This type of sorting can b performd by simply transferring all the matrix elements in a single dimension array of 1X16 size and then sorting this array and then transferring the elements back to 4X4 matrix. You can also treat the 4x4 matrix as a simple array using pointers and, thus, not need to transfer from matrix to array and back. Example, using ellipses (...) to simulate indentation for clarity... int matrix[4][4] = {...some values...} int *element; int flag = 1; while (flag == 1) { /* simple bubble sort */ ... flag = 0; ... /* loop from first element to next to last element */ ... for (element = &matrix[0][0]; element < &matrix[3][3]; element ++) { ... ... if (*element > *(element + 1)) { ... ... ... flag = 1; ... ... ... *element ^= *(element + 1); /* exclusive or swap */ ... ... ... *(element + 1) ^= *element; ... ... ... *element ^= *(element + 1); ... ... } ... } }
A single dimensional array is an array of items. A two-dimensional array is an array of arrays of items.
To find the highest value in an array, start with the first element as the initial maximum. Iterate through each element in the array, comparing it to the current maximum. If an element is greater than the current maximum, update the maximum to this element. After checking all elements, the current maximum will be the highest value in the array.
you use lines and arrows and draw through numbers
it is 3 squared
like this
25
15 is the answer The + will be used as the "dots" for the array Array:
The answer will depend on what the problem is: some can be solved using an array but for others, arrays are a complete waste of time.
you have to pick if you want to draw a 1*5 or a 5*1. then you have eaither put from up to down way or a across way thankyou
To draw a 5x7 array, start by creating 5 rows and 7 columns. You can represent each row as a horizontal line and each column as a vertical line, forming a grid. Fill each cell of the grid to visualize the array, where each cell corresponds to a single unit. This creates a total of 35 cells, representing the product of 5 and 7.
I suggest using an array with as many elements as the longest row you need. To keep it simple, keep two copies of the array, and calculate each element of the "new" array as the sum of the corresponding element, plus the previous element, of the "old" array. Then copy the information back for the next step.
9 sets of five pennies
One example is making 7 cans of juice in each of six rows: ••••••• ••••••• ••••••• ••••••• ••••••• •••••••
This type of sorting can b performd by simply transferring all the matrix elements in a single dimension array of 1X16 size and then sorting this array and then transferring the elements back to 4X4 matrix. You can also treat the 4x4 matrix as a simple array using pointers and, thus, not need to transfer from matrix to array and back. Example, using ellipses (...) to simulate indentation for clarity... int matrix[4][4] = {...some values...} int *element; int flag = 1; while (flag == 1) { /* simple bubble sort */ ... flag = 0; ... /* loop from first element to next to last element */ ... for (element = &matrix[0][0]; element < &matrix[3][3]; element ++) { ... ... if (*element > *(element + 1)) { ... ... ... flag = 1; ... ... ... *element ^= *(element + 1); /* exclusive or swap */ ... ... ... *(element + 1) ^= *element; ... ... ... *element ^= *(element + 1); ... ... } ... } }