0 7 9 4 0 0 8 0 4 5 0 1
0 2 4 0 0 0 1 6 2 8 6 0
0 1 1 1 1 0 8 5 6 8 0 7
0 5 1 0 0 0 1 3 8 1 0 1

Answer :
Implementation:
â€¢ Take a 12 digit code from user. Use integer array to store this 12
digit code.
â€¢ The array index start from 0 and goes to 11.
â€¢ Use a for() loop to take input from user.
/*for() loop to initilize value from 0-11*/
for(i=0;i<12;i++)
{
scanf("%d",&arr[i]); /*read input*/
}
â€¢ Use another for() loop to calculate the sum of digits in the odd-
numbered positions and even-numbered positions.
â€¢ Inside the for() a j variable is initialized with a value of 1. This
variable is incremented by a unit value for each iteration performed
by the loop.
â€¢ This j variable is used to find out the even and odd positions.
This is done by using an if() statement. If j is divisible by 2 then it
calculates the sum of even number else the same operation is done
for odd numbers.
for(i=0;i<11;i++)
{
j = j+1;
/*
*check the value of 'j' is divisible by 2 or not
*if() statement will executes if it is not divisible by 2,
*else else() part will execute

*/
if(j%2!=0)
{
/*sum of digits in the odd-numbered positions*/
odd_sum = odd_sum + arr[i];
}
else
{
/*sum of digits in the even-numbered positions*/
even_sum = even_sum + arr[i];
}
}
â€¢ Multiply the sum of odd-numbered position by 3.
odd_sum = (odd_sum*3);/*multiply by 3 to odd_sum*/
â€¢ Calculate the sum of even-numbered and odd-numbered
positions.
sum = odd_sum + even_sum;
â€¢ Find the last digit of sum, by using modulus operator(%).
last_digit = sum%10; /*to find the last digit*/
â€¢ If the last digit of sum is equals to 0 then print
â€œvalidatedâ€
±
, else subtract the last digit from 10 and compare
with the final digit of the 12 digit. If both are equal then print
â€œvalidateâ€
±
else print â€œerror in bar codeâ€
±
.

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- Summer '16
- boer
- Elementary arithmetic, Even and odd functions, Parity, Evenness of zero, sum of the digits