Thread: Output issues

Hey guys, I wanted to create a program that will read an integer and display the sum of the digits of the integer, if the sum of the digits of the integer is more than 9 then the sum of the digits of that number will be calculated as well. This is done until the answer becomes an integer that is between 1 to 9. Don't mind the mechanism if it is confusing to understand, the problem is with the last if statement, I cannot seem to get the right output.

Code:

#include <stdio.h>
int main()
{
    int number;
    int remainder, remainder2, remainder3, sum=0, sum2=0, sum3=0;
    
    scanf("%d", &number);
    
    while (number!=0)
        {
        remainder=number%10;
        sum=sum+remainder;
            number=number/10;
        }
    
    if(sum>9)
    {
        remainder2=sum%10;
            sum2=sum2+remainder2;
                sum=sum/10;
                    sum2=remainder2+sum;
    }
    if(sum2>9){
        remainder3=sum2%10;
            sum3=sum3+remainder3;    
                sum2=sum2/10;
                    sum3=remainder3+sum2;
    }

    if(sum<=9)
    printf("%d \n", sum);
    if else(sum2>9)
    printf("%d \n", sum2);
    else
    printf("%d \n", sum3);
        
}

Specifically the wrong part of the code is

Code:

 if else(sum2>9)

Printing out all the different sums works as I want it to, the If statement logic does not work at all in the part mentioned above.

I really appreciate your help guys, this problem has been a pain in the ass for the past 3 days.

Hey there, thanks for your reply!

I tried switching it to if else and when I enter 399 into the output the result should be a 3, but it still displays a 2. Any ideas?

Can you try to show me your fixed version of the code, if you don't mind?

I got my code to work without your help, who's laughing now mr. Code factory?!

Originally Posted by Ashley95

Hey guys, I wanted to create a program that will read an integer and display the sum of the digits of the integer, if the sum of the digits of the integer is more than 9 then the sum of the digits of that number will be calculated as well. This is done until the answer becomes an integer that is between 1 to 9. Don't mind the mechanism if it is confusing to understand, the problem is with the last if statement, I cannot seem to get the right output.

Code:

#include <stdio.h>
int main()
{
    int number;
    int remainder, remainder2, remainder3, sum=0, sum2=0, sum3=0;
    
    scanf("%d", &number);
    
    while (number!=0)
        {
        remainder=number%10;
        sum=sum+remainder;
            number=number/10;
        }
    
    if(sum>9)
    {
        remainder2=sum%10;
            sum2=sum2+remainder2;
                sum=sum/10;
                    sum2=remainder2+sum;
    }
    if(sum2>9){
        remainder3=sum2%10;
            sum3=sum3+remainder3;    
                sum2=sum2/10;
                    sum3=remainder3+sum2;
    }

    if(sum<=9)
    printf("%d \n", sum);
    if else(sum2>9)
    printf("%d \n", sum2);
    else
    printf("%d \n", sum3);
        
}

Specifically the wrong part of the code is

Code:

 if else(sum2>9)

Printing out all the different sums works as I want it to, the If statement logic does not work at all in the part mentioned above.

I really appreciate your help guys, this problem has been a pain in the ass for the past 3 days.

this type of problems are better to solve using dynamic programming technique, it'll also optimize your code in terms of time, try it if you can.