One dimensional arrays

[intersted2405]

Hello! How do I make a program in C++,in a one-dimensional array consisting of n integer elements, calculate:
-number of paired elements standing in even places in the array,
-the sum of the elements of the array, located after the first maximum element in the array.
-Compress the array, removing from it all the elements whose value coincides with n (the number of elements in the array). Fill in the blanks at the end of the array with zeros.Does not always produce even numbers and I don’t know how to do the last

#include <iostream>
#include <math.h>
using namespace std;

int main()
{
	int i, n, imax = 0, a = 0, b = 0;
	double max = 0, crazy = 0, sum = 0, c = 0;
	double sort = 1;
	cout << "Enter the number of elements in the array:";
	cin >> n;
	cout << "Enter array elements:" << endl;
	double* p = new double[n];
	double* p1 = new double[n];
	for (i = 0; i < n; i++)
		cin >> p[i];
	for (i = 0; i < n; i++) 
	{
		if ((i % 2) == false) 
		{
			p1[a] = p[i];
			a++;
		}
	}
	while(sort) 
	{
		sort = 0;
		for (i = 0; i < a - 1; i++) 
		{
			if (p1[i] > p1[i + 1]) 
			{
				c = p1[i];
				p1[i] = p1[i + 1];
				p1[i + 1] = c;
				sort = 1;
			}
		}

		
	}
	for (i = 0; i < a; i++)
		cout << p1[i] << " ";
	cout << endl;
	
	
}

[seeplus]

number of paired elements standing in even places in the array,

Can you explain please. What is 'paired elements standing in even places'??

[intersted2405]

instead of the word "paired", there are even numbers

[markyrocks]

It looks like you're checking if the element position is even but not checking if the contents of said position is even.

[intersted2405]

yes, I don't know how to do it yet

[markyrocks]

It would look something like...

if (i % 2== 0 && p[i] % 2 ==0) {
// do stuff
}

[intersted2405]

do you know how to do it?.Compress the array, removing from it all the elements whose value coincides with n (the number of elements in the array). Fill in the blanks at the end of the array with zeros.

[intersted2405]

still don't understand how to do it,the sum of the elements of the array, located after the first maximum element in the array

[markyrocks]

   double* p1 = new double[n];
ZeroMemory(p1,n*sizeof(double));

Sets all the elements to Zero b4 any work is done. If you want to delete the elements then that's a different story. The size is in bytes so that should be correct.

[intersted2405]

can you write the entire code? For me to understand,and do you know how to do the second?

[markyrocks]

Lol no. You ain't never going to learn that way.

theres usually already function available for most array manipulation pre-written in the algorithm.

#include <algorithm>
using namespace std;
auto max=max_element(begin(p1),end(p1));

As a side note everything im writing is untested and not guaranteed to be error free.

[Ganado]

It's <algorithm> in standard C++, not <algorithm.h>.

[markyrocks]

Edited.

[seeplus]

Perhaps something like:

#include <iostream>
#include <numeric>
#include <limits>
using namespace std;

int main()
{
	size_t n {}, even {}, maxpos {};
	int maxno {numeric_limits<int>::min()};

	cout << "Enter the number of elements in the array: ";
	cin >> n;

	int* p {new int[n] {}};

	cout << "Enter array elements:\n";
	for (size_t i = 0; i < n; ++i) {
		cin >> p[i];
		even += ((p[i] % 2 == 0) && (i % 2 == 0));
		if (p[i] > maxno) {
			maxno = p[i];
			maxpos = i;
		}
	}

	cout << "There are " << even << " paired evens\n";

	const int sum {accumulate(p + maxpos + 1, p + n, 0)};
	cout << "The sum of the numbers after the maximum element is " << sum << '\n';

	// Remove elements == number of elements and set to 0
	for (size_t i = 0, nn = n; i < nn; ++i)
		if (p[i] == n) {
			p[i] = p[--nn];
			p[nn] = 0;
		}

	for (size_t i = 0; i < n; ++i)
		cout << p[i] << ' ';

	cout << '\n';

	delete[] p;
}