write a program in Java to initialize a SDA of 10 numbers and find the average of the odd numbers
pls help me
Answers
SOLUTION.
Here comes my approach for the problem.
import java.util.*;
public class Array{
public static void main(){
Scanner sc=new Scanner(System.in);
System.out.println("Enter 10 elements. ");
int a[]=new int[10];
int c=0;
float sum=0;
for(int i=0;i<10;i++){
System.out.print("> ");
a[i]=sc.nextInt();
if(a[i]%2!=0){
sum+=a[i];
c++;
}
}
sc.close();
sum/=(float)c;
System.out.println("Average of odd numbers: "+sum);
}
}
LOGIC.
- Accept the numbers.
- Initialise sum=0, count=0. count stores the number of odd numbers present in array.
- If number is not divisible by 2, add the number to sum and increment the value of count by 1.
- Divide sum by count and store it in same variable.
- Display the value of the variable 'sum'.
See attachment for output.
Answer:
Program:-
import java.util.*;
public class Main
{
public static void main(String args[])
{
Scanner in=new Scanner(System.in);
int a[]=new int[10];
int sum=0,c=0;
double avg=0;
System.out.println("Enter the elements of the array:");
for(int i=0;i<a.length;i++)
{
a[i]=in.nextInt();
}
System.out.println("Given Array:"+Arrays.toString(a));
for(int i=0;i<a.length;i++)
{
if(a[i]%2!=0)
{
sum=sum+a[i];
c++;
}
}
avg=sum/(double)c;
System.out.println("The average of odd elements="+avg);
}
}
Logic:-
- Enter the elements of the array
- If an element is not divisible by 2 i.e an odd element, so c is incremented. Variable c is the number of odd elements that is present in the array.
- Finally , avg=sum/(double) c (explicit conversion in order to get the correct average as Java treats '/' as only the quotient not in decimal places so explicit conversion need to be there.
- Finally print the average.
Output is attached.
