C++ code to Calculate average of odd and even in an array
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C++ code to Calculate average of odd and even in an array
C++ code to Calculate average of odd and even in an array
In this tutorial, we will discuss a concept of the C++ code to calculate the average of odd and even numbers in an array.
In this article, we are going to learn how to calculate the Average of odd and even numbers of an array 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.
C++ program to Calculate the average of odd and even numbers using for loop
Program 1
This program allows the user to calculate the average of odd and even numbers in the given array using “for loop”.
#include <iostream> #include <conio.h> using namespace std; int main() { int arr[6]={50,45,40,35,30,25}; //1 int i,oddSum=0,evenSum=0,oddCount=0,evenCount=0; //2 double avgOdd, avgEven; //3 for(i=0; i<6; i++){ //4 if(arr[i]%2==0){ evenSum=evenSum+arr[i]; //5 evenCount++; } else{ //6 oddSum=oddSum+arr[i]; oddCount++; } } double avgOdd=oddSum/oddCount; //7 double avgEven=evenSum/evenCount; cout<<"The average of odd numbers are:"<<avgOdd; //8 cout<<"\nThe average of even numbers are:"<<avgEven; getch(); return 0; }
When the above code is executed, it produces the following results
The average of odd numbers are:35 The average of even numbers are:40
Method:
- First, define an array with elements.
- Next, declare and initialize four variables to find sum as oddSum=0, evenSum=0, oddCount=0, evenCount=0;
- Then, declare two variables to find average named as avgOdd, avgEven;
- Thereafter, 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.
- Then, find the average of odd and even numbers
- Finally, the average of odd numbers and even numbers are displayed.
Program 2
This program allows the user to choose any value for array length(Number of array elements) and input values into the array. The program will calculate the Average of odd and even numbers from the list using “for loop”.
#include <iostream> #include <conio.h> using namespace std; int main() { int i,oddSum=0,evenSum=0,oddCount=0,evenCount=0,arr_size; //1 cout<<"Enter the size of the array: "; cin>>arr_size; //2 int arr[arr_size]; //3 cout<<"Enter the Array elements: "; //4 for(i=0; i<arr_size; i++){ //5 cin>>arr[i]; } for(i=0; i<arr_size; i++){ //6 if(arr[i]%2==0){ //7 evenSum=evenSum+arr[i]; oddCount++; } else{ oddSum=oddSum+arr[i]; //8 evenCount++; } } double avgOdd=oddSum/oddCount; //9 double avgEven=evenSum/evenCount; cout<<"The average of odd numbers are:"<<avgOdd; //10 cout<<"\nThe average of even numbers are:"<<avgEven; 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: 30 29 28 27 26 25 The average of odd numbers are: 27 The average of even numbers are: 28
Method:
- To begin with , declare and initialize two variables to find sum as oddSum=0, evenSum=0, oddCount=0, evenCount=0;
- 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.IIf the number is not even, it is dealt with by “else statement”.
- After that, the “else statement” find odd number and adds to oddSum.
- Then, find the average of odd and even numbers
- Finally, the average of odd numbers and even numbers are displayed.
C++ code to Calculate the average of odd and even numbers using while loop
Program 1
This program allows the user to calculate the average 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]={60,65,70,75,80,85}; //1 int i,oddSum=0,evenSum=0,oddCount=0,evenCount=0; /2 double avgOdd,avgEven; //3 i=0; while(i<6){ //4 if(arr[i]%2==0){ //5 evenSum=evenSum+arr[i]; evenCount++; } else{ //6 oddSum=oddSum+arr[i]; oddCount++; } i++; } double avgOdd=oddSum/oddCount; //7 double avgEven=evenSum/evenCount; cout<<"The average of odd numbers are:"<<avgOdd; //8 cout<<"\nThe average of even numbers are:"<<avgEven; getch(); return 0; }
When the above code is executed, it produces the following results
The average of odd numbers are: 75 The average of even numbers are:70
Method:
- First, define an array with elements.
- Next, declare and initialize four variables to find sum as oddSum=0, evenSum=0, oddCount=0, evenCount=0;
- Then, declare and two variables to find average as avgOdd, avgEven;
- Thereafter, 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.
- Then, find the average of odd and even numbers
- Finally, the average of odd numbers and even numbers are displayed.
Program 2
This program allows the user to choose any value for array length(Number of array elements) and input values into the array. The program will calculate the Average of odd and even numbers from the list using “while loop”.
#include <iostream> #include <conio.h> using namespace std; int main() { int i,oddSum=0,evenSum=0,oddCount=0,evenCount=0,arr_size; //1 cout<<"Enter the size of the array: "; //2 cin>>arr_size; int arr[arr_size]; //3 cout<<"Enter the Array elements: ";//4 i=0; while(i<arr_size){ //5 cin>>arr[i]; i++; } i=0; while(i<arr_size){ //6 if(arr[i]%2==0){ //7 evenSum=evenSum+arr[i]; oddCount++; } else{ //8 oddSum=oddSum+arr[i]; evenCount++; } i++; } double avgOdd=oddSum/oddCount; //9 double avgEven=evenSum/evenCount; cout<<"The average of odd numbers are:"<<avgOdd; //10 cout<<"\nThe average of even numbers are:"<<avgEven; 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: 100 99 98 97 96 95 The average of odd numbers are:97 The average of even numbers are:98
Method:
- To begin with , declare and initialize two variables to find sum as oddSum=0, evenSum=0, oddCount=0, evenCount=0;
- 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.
- Then, find the average of odd and even numbers
- Finally, the average of odd numbers and even numbers are displayed.
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