/*Write a C/C++ application that defines two matrices of integer values. The matrices’ dimensions and their elements are read from the standard input. Write the functions that display:
a) the positions of the odd elements in each matrix;
b) the sum of all even elements in both matrices;
c) the sum of the two initial matrices;
Author: Adorjan Istvan
Date:30/11/2017*/
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <conio.h>

using namespace std;
int position(int*[]); // function for the position of the odd numbers
int sum(int*[]); // function for the sum of all even numbers
int matrixsum(int*[], int*[]); // function for the sum of two matrices
int print(int*[]); // function to print this two matrix

void main()
{
	int a[3][4] = { 0,1,2,3,4,5,6,7,8,9,10,11 };
	int b[3][4] = { 12,13,14,15,16,17,18,19,20,21,22,23 };
	int *p, *o;
	p = a[0];
	o = b[0];
	printf("\n Our first matrix looks like: \n");
	print(p);
	printf("\n Our second matrix looks like: \n");
	print(o);
	printf("\n Let's find the odd elements int the first matrix: ");
	position(p);
	printf("\n\n Let's find the odd elements int the second matrix: ");
	position(o);
	printf("\n\n The sum of the even numbers in both matrices is equal with: %d\n", sum(p) + sum(o));
	printf("\n The sum of this two matrices is equal with: \n");
	matrixsum(p, o);

	_getch();
}
int position(int* p)
{
	int i, j;

	for (i = 0; i<3; ++i)
	{
		for (j = 0; j<4; ++j)
		{
			if (*(*(p + i) + j) % 2 != 0)//Checking odd elements.
			{
				printf("\n %d row %d collum, which is: %d", i + 1, j + 1, *(*(p + i) + j));
			}
		}
	}
	return 0;
}
int sum(int* p)
{
	int sum = 0;
	int i, j;
	for (i = 0; i<3; ++i)
	{
		for (j = 0; j<4; ++j)
		{
			if (*(*(p + i) + j) % 2 == 0)//Checking odd elements.
			{
				sum += *(*(p + i) + j);
			}
		}
	}
	return sum;
}
int matrixsum(int* p, int* o)
{
	int i, j;
	int sum[3][4];
	int* f;
	f = sum[0];
	for (i = 0; i < 3; i++) {
		for (j = 0; j < 4; j++) {
			*f = *(*(p + i) + j) + *(*(o + i) + j); // Calculating the sum of the elements.
			printf("%d\t", (*(f + i) + j));
		}
		printf("\n");
	}
	return 0;
}
int print(int* p[])
{
	int i, j;
	for (i = 0; i < 3; i++) {
		for (j = 0; j < 4; j++) {
			printf("%d\t", *(*(p + i) + j));
		}
		printf("\n");
	}
	return 0;
}