Denmark's two-world array. Apply the solution of tasks from two-world arrays, tasks for independent development. Tsіlepokladannya, setting the task

Golovna / Zahist

Laboratory work for schools and universities how to complete the task of completing two-world arrays of mine C (Ci). Redirection of the head of the laboratory work (click for help directing to the solution of the task, as well as є):

Addendum tasks:

Usability of current programs has been revised in CodeBlocks 16.01 (MinGW, Windows 10). You can give your solution to non-violent programs and help to improve the basics in the comments at the bottom of the page.

Download the file of the laboratory and all external code programs in the archive:

So you can change your decisions laboratory robots my C++ and many others:

No. 1: Denmark's two-world array. Find the minimum sum of elements in a row. The task is to complete without an additional one-dimensional array.

#include #include #include int main() ( int *arr; int x, y, i, j, sum = 0, min = 0, minstr = 0; srand(time(NULL)); x = 2 + rand() % 10; y = 2 + rand() % 6; arr = (int*)malloc(x*y*sizeof(int)); i

#2: Find the coordinate of the element closest to the middle value of all elements in the array.

#include #include #include #include int main() ( int *arr; int x, y, i, j, minx = 0, miny = 0; float ssum = 0; srand(time(NULL)); x = 1 + rand() % 10; y = 1 + rand() % 7; arr = (int*)malloc(x*y*sizeof(int)); ;i

No. 9: Danish two-world array from the paired number of columns. Remember the pillars of the left half of the massif with the pillars of the right half.

#include #include #include int main() ( int *arr; int x, y, i, j, d; srand(time(NULL)); x = 1 + rand() % 10; y = 2 * (1 + rand() % 3 ); arr = (int*)malloc(x*y*sizeof(int));printf("Array %d x %d: \n", x, y);

No. 14: For the skin row of a two-world array, the number of elements is determined, the rest of the number of those is more.

#include #include #include int main() ( int *arr; int x, y, i, j, a, count; srand(time(NULL)); printf("Enter a: "); scanf("%d", &a); x = 1 + rand() % 10; y = 1 + rand() % 7; arr = (int*)malloc(x*y*sizeof(int)); x, y), for (i = 0; i

#21: Danish two-world array b. Looking at yoga like a matrix, transpose b.

#include #include #include int main() ( int *arr, *arrT, x, y, i, j; srand(time(NULL)); x = 1 + rand() % 7; y = 1 + rand() % 7; arr = (int*)malloc(x*y*sizeof(int));printf("Matrix %d x %d: \n", x, y);

No. 1 (Additional): Denmark's two-world array. Arrange the rows while avoiding the first elements.

#include #include #include int main() ( int *arr, x, y, i, j, k, d, max = 0; srand(time(NULL)); x = 1 + rand() % 10; y = 1 + rand() %7;arr = (int*)malloc(x*y*sizeof(int));max*y)) max=j; for (k = 0; k

No. 2 (Additional): Significantly, chi є a square orthonormal matrix is given, tobto. such, in such a scalar supplement of the skin row of the lower rows, the cost is 0, and the scalar supplement of the skin row on itself is the cost of 1.

#include #include int main() ( int *arr; int x, i, j, k, sum = 0; printf("Enter size of the square matrix: "); scanf("%d", &x); arr = (int * )malloc(x*x*sizeof(int));

No. 3 (Additional): Danish two-dimensional square array. Know the sum of the elements of the skin diagonal, parallel to the side.

#include #include #include int main() ( int *arr; int x, y, i, j, sum; srand(time(NULL)); x = 2 + rand() % 6; arr = (int*)malloc(x*y* ) sizeof(int));printf("Array %d x %d: \n", x, x); for (i = 0; i

The black portion of the head of informatics for schoolchildren has been added. How many times can we look at a robot with two-world C++ arrays. Tsі zavdannya dosit tsіkavі. And the greater number of them was worthy of me.

Manager No. 1

Get the index of the first entry of the maximum element.
Input data format
The program takes into account an array of n and m, then rows of n by m numbers in the skin. n and m do not move 100.
Output data format
Enter two numbers: the row number and the column number, which have the largest element in the two-world array. If there are such elements of the kilka, then the one with the lower row number will be displayed, and if the row numbers are equal, the one with the lower row number.

sample input: 3 4 0 3 2 4 2 3 5 5 5 1 2 3 sample output: 1 2

#include using namespace std; int main() ( int n, m; cin >> n >> m; int a; // read for (int i = 0; i< n; i++) { for (int j = 0; j < m; j++) { cin >> a[i][j]; )) int max = a, max_i = 0, max_j = 0; for (int i = 0; i< n; i++) { for (int j = 0; j < m; j++) { if (a[i][j] >max) ( max = a[i][j]; max_i = i; max_j = j; ) ) ) cout<< max_i << " " << max_j; return 0; }

Manager No. 2

Given an unpaired number n, which does not exceed 15. Create a two-world array of n × n elements, filling it with the characters "." (Skin element array є a row of one character). Let's use the characters "*" to fill in the middle row of the array, the middle row of the array, the head diagonal and the side diagonal. As a result, "*" in the array is responsible for approving the image of the star. Display the array on the screen, splitting the elements of the array with spaces.

sample input: 5 sample output: * . * . * . * * * . * * * * * . * * * . * . * . *

#include using namespace std; int main() ( int n; cin >>< n; i++) { for (int j = 0; j < n; j++) { if (i == j || i == n - 1 - j || i == n / 2 || j == n / 2) a[i][j] = 1; else a[i][j] = 0; } } for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { if (a[i][j] == 0) cout << "." << " "; else cout << "*" << " "; } cout << endl; } return 0; }

Manager No. 3

You are given a number n, which does not exceed 100. Create an array with a size of n×n and store it after the next rule. The number 0 can be written on the head diagonal. The number 1 is written on the two diagonals that adjoin the head one. The number 2 is written on the two diagonals, etc.

sample input: 5 sample output: 0 1 2 3 4 1 0 1 2 3 2 1 0 1 2 3 2 1 0 1 4 3 2 1 0

#include #include using namespace std; int main() ( int n; cin >> n; int a; // for (int i = 0; i< n; i++) { for (int j = 0; j < n; j++) { a[i][j] = (int) abs(i - j); } } for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { cout << a[i][j] << " "; } cout << endl; } return 0; }

Manager No. 4

Danish two-world array and two numbers: i and j. Remember the array of columns with numbers i and j.
Input data format
The program takes into account the array n and m, so as not to change 100, then the elements of the array, then the numbers i j.
Output data format
View the result.

sample input: 0 1 sample output: 12 11 13 14 22 21 23 24 32 31 33 34

#include #include using namespace std; int main() ( int n, m, x, y, temp; cin >> n >>< n; i++) { for (int j = 0; j < m; j++) { cin >> a[i][j]; ) cin >> x >> y; // processing for (int i = 0; i< n; i++) { temp = a[i][x]; a[i][x] = a[i][y]; a[i][y] = temp; } // вывод for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { cout << a[i][j] << " "; } cout << endl; } return 0; }

Manager No. 5

You are given a number n, which does not exceed 10, and an array of size n × n. Flip, which array is symmetrical to the main diagonal. Enter the word "YES" as the array is symmetrical, and the word "NO" otherwise.

sample input: 3 0 1 2 1 2 3 2 3 4 sample output: YES

#include #include using namespace std; int main() ( int n; bool symmetric; cin >> n; int a; // for (int i = 0; i< n; i++) { for (int j = 0; j < n; j++) { cin >> a[i][j]; )) // processing symmetric = true; for (int i = 0; i< n; i++) { for (int j = 0; j < n; j++) { if (a[i][j] != a[j][i]) symmetric = false; } } // вывод if (symmetric) cout << "YES"; else cout << "NO"; return 0; }

Manager No. 6

Danish square two-world array of n×n size and the number k. Enter elements k-ї behind the head diagonal below the head diagonal (then if k = 1, then it is necessary to enter elements of the first diagonal, which lie lower behind the head, if k = 2, then another diagonal, etc.).
The value of k can be negative, for example, if k = −1, then it is necessary to enter the value of the first diagonal, which lies above the head. If k = 0, you must enter the elements of the head diagonal.
The program takes the number n as input, but do not change 10, then the array is n × n, then the number k.

Sample Input 1: 4 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 1 Sample Output 1: 5 1 6 Sample Input 2: 4 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 -2 Sample Output 2: 3 8

#include using namespace std; int main() ( int n, k; cin >> n; int a[n][n]; // for (int i = 0; i< n; i++) { for (int j = 0; j < n; j++) { cin >> a[i][j]; ) cin >> k; // processing that view for (int i = 0; i< n; i++) { for (int j = 0; j < n; j++) { if (i - j - k == 0) cout << a[i][j] << " "; } } return 0; }

Manager No. 7

Danish two-world array with a size of n × m (n and m do not exceed 1000). A symmetrical youmou like a head diagonal array is called a transposition to it. Vіn maє m×n: the rows of the outer array become the columns of the transposed one, the columns of the outer array become the rows of the transposed one.
For this array, try to transpose the array and bring it to the screen.

sample input: 3 4 11 12 13 14 21 22 23 24 31 32 33 34 sample output: 11 21 31 12 22 32 13 23 33 14 24 34

#include using namespace std; int main() ( int n, m, x, y, temp; cin >> n >>< n; i++) { for (int j = 0; j < m; j++) { cin >> < n; i++) { for (int j = 0; j < m; j++) { b[j][i] = a[i][j]; } } // вывод for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { cout << b[i][j] << " "; } cout << endl; } return 0; }

Manager No. 8

At the cinema, there are n rows of m each for the skin one (n and m do not exceed 20). The two-world array collects information about the ticket sales, the number 1 means that the ticket for the place has already been sold, the number 0 means that the ticket is free. Nadіyshov ask for sales tickets for the next month in one row. Wiznachte, chi can vikonati ask.
Input data format
The program takes the input of numbers n and m. Give n rows to replace m numbers (0 or 1), separated by spaces. Then we are given a number k.
Output data format
The program is responsible for entering the number of the row, in which there are є k free months to go to bed. For such rows of kilka, enter the number of the smallest row. There is no need for a row, enter the number 0.

sample input: 3 4 0 1 0 1 1 0 0 1 1 1 1 1 2 sample output: 2

#include using namespace std; int main() ( int n, m, k, r = 0; cin >> n >> m; int a[n][m]; // for (int i = 0; i< n; i++) { for (int j = 0; j < m; j++) { cin >> a[i][j]; ) cin >> k; // processing for (int i = 0; i< n; i++) { int near_free = 0; for (int j = 0; j < m; j++) { if (a[i][j] == 0) { near_free++; if (near_free == k) { r = i + 1; break; } } else near_free = 0; } if (near_free == k) break; } // вывод cout << r; return 0; }

Manager No. 9

Given a rectangular array with a size of n×m. Rotate yoga by 90 degrees after the year arrow, writing the result in a new array with a size of m × n.
Input data format
Enter two numbers n and m, not to exceed 100, then an array with a size of n×m.
Output data format
View the array as you see it. Divide the numbers with one gap when visualized.

sample input: 3 4 11 12 13 14 21 22 23 24 31 32 33 34 sample output: 31 21 11 32 22 12 33 23 13 34 24 14

#include using namespace std; int main() ( int n, m; cin >> n >> m; int a[n][m]; int b[m][n]; // for (int i = 0; i< n; i++) { for (int j = 0; j < m; j++) { cin >> a[i][j]; ) ) // processing for (int i = 0; i< n; i++) { for (int j = 0; j < m; j++) { b[j] = a[i][j]; } } // вывод for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { cout << b[i][j] << " "; } cout << endl; } return 0; }

Manager No. 10

After the given numbers n and m, fill in a two-world array with n×m numbers from 1 to n×m with a “snake”, as shown in the butt.
Input data format
Two numbers n and m are entered, without changing them 20.
Output data format

sample input: 3 5 sample output: 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15

#include using namespace std; int main() ( int n, m, c = 0; cin >> n >> m; int a[n][m]; // for (int i = 0; i< n; i++) { for (int j = 0; j < m; j++) { c++; if (i%2 == 0) a[i][j] = c; else a[i] = c; } } // вывод for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (a[i][j] % 10 == a[i][j]) cout << " "; else if (a[i][j] % 100 == a[i][j]) cout << " "; else if (a[i][j] % 1000 == a[i][j]) cout << " "; cout << a[i][j]; } cout << endl; } return 0; }

Manager No. 11

After the given numbers n and m, fill in a two-dimensional array with n × m numbers like 1 to n × m “diagonals”, as shown in the example.
Input data format

Output data format
To see an array, leading to the sighting of the skin element is equal to 4 symbols.

sample input: 3 5 sample output: 1 2 4 7 10 3 5 8 11 13 6 9 12 14 15

#include using namespace std; int main() ( int n, m, pos = 0, row = 0; cin >> n >> m; int a[n][m]; // processing int start_row = 0; int number = 1; for ( int min_row = 0;min_row< n; min_row++) { if (min_row >0) start_row = pos - 1; else start_row = 0; for(pos=start_row; pos< m; pos++) { row = min_row; for (int col = pos; col >= 0; col--) ( if (row< n) { a = number; number++; row++; } else break; } } } // вывод for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (a[i][j] % 10 == a[i][j]) cout << " "; else if (a[i][j] % 100 == a[i][j]) cout << " "; else if (a[i][j] % 1000 == a[i][j]) cout << " "; cout << a[i][j]; } cout << endl; } return 0; }

Manager No. 12

Numbers n and m are given. Fill in the array with a size of n × m in check order: the keys of one color are filled with zeros, and the elements of the other color are filled with numbers of the natural row downwards, left-handed, right-handed. The number 1 is written at the upper left corner.
Input data format
Enter two numbers n and m, so as not to change 100.
Output data format
To see an array, leading to the sighting of the skin element is equal to 4 symbols.

sample input: 3 5 sample output: 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8

#include using namespace std; int main() ( int n, m, sm; cin >> n >> m; int a[n][m]; // processing int number = 1; for (int i = 0; i< n; i++) { for (int j = 0; j < m; j++) { a[i][j] = 0; } } for (int i = 0; i < n; i++) { if (i % 2 == 1) sm = 1; else sm = 0; for (int j = sm; j < m; j++) { a[i][j] = number; number++; j++; } } for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (a[i][j] % 10 == a[i][j]) cout << " "; else if (a[i][j] % 100 == a[i][j]) cout << " "; else if (a[i][j] % 1000 == a[i][j]) cout << " "; cout << a[i][j]; } cout << endl; } return 0; }

Manager No. 13

For given numbers n and m, fill in a two-dimensional array with a size of n × m with numbers from 1 to n × m in a spiral, which comes out of the left upper kut and twisted behind the annual arrow, as shown in the butt.
Input data format
Enter two numbers n and m, so as not to change 100.
Output data format
To see an array, leading to the sighting of the skin element is equal to 4 symbols.

sample input: 4 5 sample output: 1 2 3 4 5 14 15 16 17 6 13 20 19 18 7 12 11 10 9 8

#include using namespace std; int main() ( int n, m; cin >> n >> m; int a; for (int i = 0; i<= n + 1; i++) { for (int j = 0; j <= m + 1; j++) { a[i][j] = -1; } } for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { a[i][j] = 0; } } int num = 0, row = 1, col = 0; while (num < n * m) { while (a == 0) { col++; num++; a = num; } while (a == 0) { row++; num++; a = num; } while (a == 0) { col--; num++; a = num; } while (a == 0) { row--; num++; a = num; } } for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { if (a[i][j] % 10 == a[i][j]) cout << " "; else if (a[i][j] % 100 == a[i][j]) cout << " "; else if (a[i][j] % 1000 == a[i][j]) cout << " "; cout << a[i][j]; } cout << endl; } return 0; }

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butt 3

Denmark is a large two-world array, size n x m,know the smallest element of the array and the number of the row in which to know the vin.

Block diagram:

Program code:
var a: array of integer;
i, j, m, n, min, k: integer;
Begin
Write(' How many rows?'); readln(n);
Write(' How many columns?'); readln(m);
For i:=1 to n do
For j:=1 to m do
begin
write('a[',i,',',j,']='); readln(a); (Introduction of a two-world array)
end;
min:=a; (Minimum element)
k:=1; (row number)
For i:=1 to n do
For j:=1 to m do
If a< min then
begin
min:=a; k:=i; (Search for the minimum and "remembering" the row number)
end;
Writeln('smallest number ',min,' iterates over ', k , ' in a row');
end.

manager

. Find out the sum and total of all the elements in the array.
Danish large two-world array, size n x m. Know the amount of money and dobutok guys elementіv.
Danish large two-world array, size n x m. Find the sum and number of elements that are multiples of 3 and 5.
Danish large two-world array, size n x m. Know the number of negative elements greater than -9.
Danish large two-world array, size n x m.
Danish large two-world array, size n x m
Danish large two-world array, size n x m
Danish large two-world array, size n x m. Find the arithmetic mean of all elements in the array.
Danish large two-world array, size n x m. Z'yasuvati, like the number is growing in a certain row earlier - more positively and more negatively.
Danish large two-world array, size n x m. Z'yasuvati, in a row there is a sequence of growing numbers.
Danish large two-world array, size n x m. Vivestiyoga elements, іndexi є є steps dvіyki (1, 2, 4, 8, 16, ...).
Danish large two-world array, size n x m. Find the number of multiple elements 7.
Danish large two-world array, size n x m. Bring on the screen the elements, like the squares of any number.
Danish large two-world array, size n x m. Know the numbers of unpaired elements to stand on paired places.
Danish large two-world array, size n x m. Know the maximum and minimum. Remember them with the moons.
Danish large two-world array, size n x m. Replace all the elements on your squares.
Danish large two-world array, size n x m. Replace all elements with opposite values.
Danish large two-world array, size n x m. Remember the first and the last elements with the help of the moons.
Danish large two-world array, size n x m. Form a new array, based on the protilegish vіdpovіdnih elements.
Danish large two-world array, size n x m. Show on the screen those elements, which have some leftovers in the form of rozpodіlu on m dorivnyuє k.
The results of the control work of 10 students are requested. Decide on the number of unsatisfactory, positive, good, authoritative assessments. Enter the average mark, won by the students for the control work.
Enter grades for N students from K subjects. It is important to show the number of students on the screen, as they did not take away the desired "5".
The group learns N students, the students took chotiri credits for their sleep. Indicate the number of unsuccessful students and the average score of the group.
Danish large two-world array, size n x m. Calculate the sum of numbers, ordinal numbers and those of fibonacci numbers.
Danish large two-world array, size n x m. Fold the essential elements.

Danish two-world array 5x5. Organize the introduction of elements, seeing the result as a vicon. Change after pauses the colors of the vicons, deselect the tasks in the same way to the point of clearing the elements.

Job options:

10.

11. Know more elements, how to know more for the main diagonal. Replace with the known number of all unpaired elements.

12. Find the sum of elements that are in unpaired positions (the sum of indices (i+j) for Х ij is an unpaired number). Substitute the known number of all negative elements.

13. Find out how to get all the elements that are on the main diagonal. Replace with the known number of the elements of the fifth row and that of the other column.

14. Find the arithmetic mean of the elements on the head diagonal and the arithmetic mean of the side diagonal. Remember the elements on the diagonals.

15. Know the arithmetic mean of positive elements, as they are higher than the head diagonal. Substitute the known number of all negative elements.

16. Find the sum of unpaired elements. Replace with a known number all the elements that are found more in the secondary diagonal.

17. Find additional elements of the i-row and j-stow (i, j- type in front from the keyboard). Replace with a known number all the elements below in the side diagonal.

18. Find additional elements that are in pair positions (the sum of indices (i+j) for X ij is an unpaired number). Replace with a known number all the elements of the head diagonal.

19. Find the arithmetic mean of unpaired elements. Substitute the known number of all the elements of the secondary diagonal.

20. To know the sum of all the pair elements, as if they were trying more side diagonals. Replace with the known number of the elements of the fourth row.

21. Know more elements, how to know more for the main diagonal. Replace with the known number of all unpaired elements.

22. Find the sum of elements that are in unpaired positions (the sum of indices (i+j) for Х ij is an unpaired number). Substitute the known number of all negative elements.

23. Find out how to get all the elements that are on the main diagonal. Replace with the known number of the elements of the fifth row and that of the other column.

24. Find the arithmetic mean of the elements on the head diagonal and the arithmetic mean of the side diagonal. Remember the elements on the diagonals.

25. Know the arithmetic mean of positive elements, as they are higher than the head diagonal. Substitute the known number of all negative elements.

For example: Find the arithmetic mean of the elements on the head diagonal

After a pause:

Control nutrition

1. What is the text mode and how does it work in the sound mode?

2. What is it like?

3. How to move the cursor at the given window coordinate?

4. How do I change the color of the font/background?

Entry

Modern technical attachments, designated for control and automation, widely use double schemes and double number system. For the analysis and synthesis of combinatorial and subsequent circuits, the theory of algebra of logic is developed.

The material of the lectures was presented in three sections. The first one gives the basic understanding of the theory of the algebra of logic, looks at the different forms of the manifestation of Boolean functions. Another division of assignments to power minimization is the appearance of Boolean functions, which is the basis of the abstract synthesis of discrete additions. The third section is about ways to implement Boolean functions based on relay circuits, which is the basis for the structural synthesis of discrete extensions.

When writing the abstract of the lectures, the following literature was scored: divided 1.1 -; for division 1.2 -; for division 1.3 -; for division 2.1 -; for division 2.2 -; for distribution 3.1, 3.2 -.

Task for two-world arrays

1) Danish two-world array of integers. Fold the program:

a) displaying on the screen an element stashed at the upper right corner of the array;

b) displaying on the screen an element, stashed in the lower left corner of the array;

c) displaying on the screen any element of another row array;

d) displaying on the screen any third element of the array;

e) displaying any element of the array on the screen.

2) Danish dvіrny tsіlіsny array. Remember the missions:

a) elements, stitched at the upper right and lower left corners;

b) elements, ruffled in the lower right and upper left corners;

3) Enter the number of the row, on which the element of the main diagonal of the two-dimensional array of integers is placed. Display the value of that element on the screen.

4) Display on the screen (in one row):

a) all elements of the head diagonal array

b) all the elements of the head diagonal array, starting from the element, stitched in the lower right fold.

5) Change the values of all elements in the head diagonal of the whole array to zero.

6) Replace the value of all elements of the side diagonal of the whole array with a value equal to 100.

7) Designate:

a) the sum of the elements of the head diagonal of the array;

b) the sum of the elements on the side diagonal of the array;

c) the arithmetic mean of the elements in the head diagonal of the array;

d) the arithmetic mean of the elements of the side diagonal of the array;

e) the minimum (maximum) element of the head diagonal of the array;

f) the maximum (minimum) element of the side diagonal of the speech array;

g) coordinates of the first maximum element of the main diagonal;

h) coordinates of the first minimal element of the main diagonal.

a) display on the screen all the elements of the fifth row array;

b) display on the screen all elements of the third row array, starting from the remaining element of the third row;

c) display all elements of the s-th column array on the screen;

d) replace the value of all elements of another row in the array by the number 5;

e) replace the value of all elements of the fifth column by the number 10.

f) determine the maximum (minimum) element of the third column;

9) Danish double array of integers. Fold the program:

a) a rozrahunka to the creation of two be-any elements in a be-any row of an array

b) the sum of all elements, whether in a row or array;

10) Danish two-world array of integers. Designate:

a) a) for some stovptsі the array of elements is less, for the first one it remains;

b) b) in some row, the twir elements are larger, in the other row, the third.

11) Danish two-world array with a size of 10 x 10. Display a part of the array on the screen:

a) I’ve stitched it behind the head diagonal;

b) ruffled below the head diagonal;

c) expanded more side diagonals;

d) folded lower in the side diagonal;

12) Danish two-world array of integers:

a) in skin yoga, to know the maximum (minimum) element;

b) coordinates of the maximum (minimum) element;

c) the sum of unpaired elements at the skin stalk (row);

d) the number of elements that are multiples of A or Art.

e) know the row of the maximum sum of elements;

e) to know the composition of the minimum sum of elements.