Sum of digits at Even and Odd Places in C - c

Question

I need to add the digits on the even and odd places in an integer. Say, Let number = 1234567. Sum of even place digits = 2+4+6 = 12 Sum of odd place digits = 1+3+5+7 = 16
The code I currently have is:
int returnsum(int num) {
while (num) {
rem = num % 10;
sum = sum + rem;
num = num / 10;
}
while (sum) {
a = sum % 10;
arr[i] = a:
sum = sum / 10;
i++;
}
for (i = 0; a[i]; i += 2) {
summ = summ + a[i];
}
return summ;
}
void main() {
int n, m, oddSum = 0, evenSum = 0;
printf("Please insert the number for the program:");
scanf("%d", & n);
while (n != 0) {
oddSum += n % 10;
n /= 10;
evenSum += n % 10;
n /= 10;
}
printf("Sum of digits in even places:%d\n", evenSum);
printf("Sum of digits in odd places:%d\n", oddSum);
}

Answer 1

Here is a solution for your problem:
void main()
{
int n,m,oddSum=0,evenSum=0;
printf("Please insert the number for the program:");
scanf("%d",&n);
int flag=0;
int counter=1;
while (n!=0) {
if(counter%2==0)
{
evenSum += n % 10;
n /= 10;
}
else
{
oddSum += n % 10;
n /= 10;
}
counter++;
}
if(counter%2==0)
{
int temp=oddSum;
oddSum=evenSum;
evenSum=temp;
}
printf("Sum of digits in even places:%d\n",evenSum);
printf("Sum of digits in odd places:%d\n",oddSum);
}

Answer 2

Okay, let's first assume the number has an even number of digits, so that the second-last and last are at odd and even positions respectively.
Then, the last digit can be retrieved with number % 10 and the second last with (number / 10) % 10.
So, knowing that, you can simply loop over the number, adding those values and dividing by a hundred until you get a number less than ten.
If that number is zero, then your assumption about the original having an even number of digits was correct and you can exit.
If it's a non-zero (1..9), then your assumption was wrong, you should swap the even and odd sums to date then add the final digit to the odd sum.
The pseudo-code goes something like this:
def getEvenOdd (num):
even = 0
odd = 0
while num > 9:
even = even + (num % 10)
odd = odd + ((num / 10) % 10)
num = num / 100
if num > 0:
temp = even
even = odd
odd = temp
odd = odd + num
return (even,odd)
print getEvenOdd(1234567)
And the output of that would be something like (12,16).
And it's no accident that pseudo-code looks like beginner's Python, since that language is the perfect pseudo-code language, provided you don't mess about with the more complex corners of it.

Answer 3

Check the below code:
#include <stdio.h>
int main(void){
char p[20];
int i=0,even=0,odd=0,n;
scanf("%d",&n);
sprintf(p,"%d",n);
while(p[i] != '\0')
{
if(i%2 == 0)
odd+= (p[i] - '0');
else
even+= (p[i] - '0');
i++;
}
printf("%d\n",even);
printf("%d\n",odd);
return 0;
}

Answer 4

To sum every other digit, simply divide by 100 in each iteration:
int sumodd(int num) {
int sum = 0, rem;
while(num) {
rem=num%10;
sum=sum+rem;
num=num/100);
}
return sum;
}
Since we already have a function that can sum every other digit, we can re-use it:
int sumeven(int num) {
return sumodd(num/10);
}

Answer 5

My five cents.:)
#include <stdio.h>
typedef struct pair
{
unsigned int odd;
unsigned int even;
} pair_t;
pair_t GetSums( unsigned int x )
{
const unsigned int Base = 10;
pair_t sums = { 0u, 0u };
size_t n = 0;
do
{
if ( ++n & 1 ) sums.odd += x % Base;
else sums.even += x % Base;
} while ( x /= Base );
if ( ( n & 1 ) == 0 )
{
unsigned int tmp = sums.odd;
sums.odd = sums.even;
sums.even = tmp;
}
return sums;
}
int main(void)
{
while ( 1 )
{
unsigned int x;
pair_t sums;
printf( "\nEnter a non-negative number (0 - exit): " );
if ( scanf( "%u", &x ) != 1 || x == 0 ) break;
sums = GetSums( x );
printf( "\nSum of digits in odd positions is %u\n", sums.odd );
printf( "Sum of digits in even positions is %u\n", sums.even );
}
return 0;
}
The output might look like
Enter a non-negative number (0 - exit): 1234567
Sum of digits in odd positions is 16
Sum of digits in even positions is 12
Enter a non-negative number (0 - exit): 123456
Sum of digits in odd positions is 9
Sum of digits in even positions is 12
Enter a non-negative number (0 - exit): 0