Write a shell program to find the sum of odd and even numbers from the given set of numbers?

Answer 1

In order to write a program, we have to have an algorithm. In this case, we can use two temporary variables to store up the sums: evenSum = 0 oddSum = 0 for i = 0 to array.size do: if isEven(i): evenSum = evenSum + array[i] else: oddSum = oddSum + array[i]

Answer 2

echo enter n value

read n

sumodd=0

sumeven=0

i=0

while [ $i -ne $n ]

do

echo "Enter Number"

read num

if [ `expr $num % 2` -ne 0 ]

then

sumodd=`expr $sumodd + $num`

sumeven=`expr $sumeven+$num`

fi

i=`expr $i + 1`

done

echo Sum of odd numbers = $sumodd

echo Sum of even numbers = $sumeven

Answer 3

#include<stdio.h>

main()

{

int a,b,sum;

print f("enter two numbers");

scan f("%d%d",&a&b);

sum=a+b

print f(the sum of a and b is %d,sum);

}

Answer 4

== == #include<stdio.h> #include<conio.h> void main() { int a,b,c; clrscr(); a=20,b=30; c=a+b; printf("the sum of two number is %d",c); getch(); }/*the program is closed here*/ output of the program is the sum of two numbers is 50

Answer 5

/*C program to find the entered number is even of odd*/ #include #include void main() { int n; clrscr(); printf("Enter the number\n"); scanf("%d",&n); if((n%2)==0) printf("%d is even number",n); else printf("%d is odd number",n); getch(); }

Answer 6

public void sumEvenOdd(int[] input)

{

int sumEven = 0;

int sumOdd = 0;

for (int i = 0; i < input.length; i++)

{

if (input[i] % 2 == 0)

{

sumEven = sumEven + input[i];

}

else

{

sumOdd = sumOdd + input[i];

}

}

//Do what you want here to sumEven and sumOdd like printing it.

}

Answer 7

int t=27; // assuming youre looking for the first 27 even numbers

int n=1; int s=0;

while(n<=t){

s=s+(n*2);

n++;

}

printf("%dn", s);