Calculate sum of odd and even in an array in C++

Calculate sum of odd and even in an array in C++

In this tutorial, we will discuss how to use the  C++ program to calculate the sum of odd and even numbers in an array.

In this article, we are going to learn  how to  calculate the sum of odd and even numbers in the C++ programming language

 

What is an even or odd number?

When any integer ends in 0,2,4,6,8 and it can be divided by two with the remainder of zero, it is called as an even number.

Example of even numbers – 34,-64,78,788

When any integer ends in 0,1,3,5,7,9 and it cannot be divided without a remainder, it is called as an odd number.

Example for odd numbers – 33,-69,75,785

 

Here, we can use a modular operator to find odd or even number in an array

if n%2==0,  n is an even number – if the number is even, the remainder is zero.

if n%2==1,  n is an odd number – if the number is odd, the remainder is one.

 

Calculating the sum of odd and even numbers using “for loop”

Program 1

This program allows the user to calculate the sum of odd and even numbers in the given array using “for loop”.

Program 1

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

int main()
{
    int arr[6]={50,45,40,35,30,25};
    int i,oddSum=0,evenSum=0;

    for(i=0; i<6; i++){
        if(arr[i]%2==0){
        evenSum=evenSum+arr[i];
        }
    else{
             oddSum=oddSum+arr[i];
        }
    }
    cout<<"The sum of odd numbers are:"<<oddSum;
    cout<<"\nThe sum of even numbers are:"<<evenSum;
    getch();
    return 0;
}

When the above code is executed, it produces the following results

The sum of odd numbers are: 105
The sum of even numbers are: 120

Methodology:

First, define an array with elements.

Next, declare and initialize two variables to find sum as oddSum=0, evenSum=0

Then, use the “for loop” to take the elements one by one from the array.

The “if statement” finds a number and then if the number is even, it is added to evenSum. If the number is not even, it is dealt with by  “else statement”.

The “else statement”  finds odd numbers and adds to oddSum.

Finally, the sums of odd numbers and even numbers are displayed.

 

Program 2

This program allows the user to choose any value for array length and input values into the array.  The program will calculate the sum of odd and even numbers from the array using “for loop”.

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

int main()
{

    int oddSum=0,evenSum=0;
int i,size;
    cout<<"Enter the size of the array: ";
    cin>>size;
int arr[size];
    cout<<"Enter the Array elements: ";

    for(i=0; i<size; i++){
            cin>>arr[i];
    }
    for(i=0; i<size; i++){
        if(arr[i]%2==0){
        evenSum=evenSum+arr[i];
        }
    else{
             oddSum=oddSum+arr[i];
        }
    }
    cout<<"The sum of odd numbers are:"<<oddSum;
    cout<<"\nThe sum of even numbers are:"<<evenSum;
    getch();
    return 0;
}

When the above code is executed, it produces the following results

Enter the size of the array: 5
Enter the Array elements: 6
7
8
9
10

The sum of odd numbers are: 16
The sum of even numbers are: 24

 

Method:

To begin with , declares and initializes three variables: oddSum=0, evenSum=0 and size.

Next, receives input from the user regarding the array length.

Then, defines an array without including the elements.

Afterwards, the elements for an array is received from the user.

Then, using the “for loop”, the elements are added one by one to the array.

Then, using the second “for loop”, elements are picked one by one from the array to determine oddness and evenness.

Next, “if statements” are used to find a number and then if it is even, it is added to evenSum.If the number is not even, it is dealt with by  “else statement”.

After that, the “else statement” adds to oddSum.

Finally, the sums of odd numbers and even numbers are displayed.

 

Calculating the sum of odd and even numbers using “while loop”

Program 1

This program allows the user to calculate the sum of odd and even numbers in the given array using “while loop”.

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

int main()
{
    int arr[6]={45,40,35,30,25,20};
    int i,oddSum=0,evenSum=0;

    i=0;
    while(i<6){
        if(arr[i]%2==0){
        evenSum=evenSum+arr[i];

        }
    else{
             oddSum=oddSum+arr[i];
        }
        i++;
    }
    cout<<"The sum of odd numbers are:"<<oddSum;
    cout<<"\nThe sum of even numbers are:"<<evenSum;
    getch();
    return 0;
}

When the above code is executed, it produces the following results

The sum of odd numbers are: 105
The sum of even numbers are: 90

Methodology:

First, define an array with elements.

Next, declare and initialize two variables to find sum as oddSum=0, evenSum=0

Then, use the “while loop” to take the elements one by one from the array.

The “if statement” finds a number and then if the number is even, it is added to evenSum. If the number is not even, it is dealt with by  “else statement”.

The “else statement”  finds odd numbers and adds to oddSum.

Finally, the sums of odd numbers and even numbers are displayed.

 

Program 2

This program allows the user to choose any value for array length and input values into the array.  The program is calculating the sum of odd and even numbers from the array using “while loop”.

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

int main()
{

    int oddSum=0,evenSum=0;
int i,size;
    cout<<"Enter the size of the array: ";
    cin>>size;
int arr[size];
    cout<<"Enter the Array elements: ";

    i=0;
    while(i<size){
            cin>>arr[i];
             i++;
    }
    i=0;
    while(i<size){
        if(arr[i]%2==0){
        evenSum=evenSum+arr[i];
        }
    else{
             oddSum=oddSum+arr[i];
        }
         i++;
    }
    cout<<"The sum of odd numbers are:"<<oddSum;
    cout<<"\nThe sum of even numbers are:"<<evenSum;
    getch();
    return 0;
}

When the above code is executed, it produces the following results

Enter the size of the array: 6
Enter the array elements: 15
24
33
12
24
35
The sum of odd numbers are:83
The sum of even numbers are:60

Method:

To begin with , declares and initializes three variables: oddSum=0, evenSum=0 and size.

Next, receives input from the user regarding the array length.

Then, defines an array without including the elements.

Afterwards, the elements for an array is received from the user.

Then, using the “while loop”, the elements are added one by one to the array.

Then, using the second “while loop”, elements are picked one by one from the array to determine oddness and evenness.

Next, “if statements” are used to find a number and then if it is even, it is added to evenSum.If the number is not even, it is dealt with by  “else statement”.

After that, the “else statement” adds to oddSum.

Finally, the sums of odd numbers and even numbers are displayed.

 

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