Given an array of integers an). the array is already not write c++ statements to fill the array eye write c++ statements that print to the screen how many values in the how many values are odd. in addition, the statements output to the scre the array that are even. s

#include<iostream>

using namespace std;

int main () {

   int n;

   int odd=0,even=0;

  cout<<"Enter the number of element store in the array: ";

  cin>>n;

  int arr[n];

  for(int i=0;i<n;i++){

     cin>>arr[i];

  }

  for(int i=0;i<n;i++){

       if(arr[i]%2==1){

           odd++;

       }else{

           even++;

       }

  }

  cout<<"The number of odd values is: "<<odd<<endl;

  cout<<"The number of even values is: "<<even<<endl;

}

Explanation:

Include the library iostream for input/output.

Create the main function and declare the variables.

Then, print the message on the screen and store the value enter by the user for the size of the array. After that, take a for loop and store the values in the array enter by the user.

then, take a for loop for traversing to the array and check the condition for odd and even by using if-else statement.

for odd: If odd values divided by 2 it gives 1 remainder.

for even: If even values divided by 2 it gives zero remainder.

if condition true, then update the counter.

finally, print the result.

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