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program to get/display digits which are odd positioned in a number.

//program to get/display digits which are odd positioned in a number.
#include<stdio.h>
#include<conio.h>
void main()
{
   int n,rem,count=0;
   printf("enter a number for 'n'\n");
  scanf("%d",&n);
  while(n!=0)
    {
      rem=n%10;
       count++;
       if(count%2!=0)
        {
          printf("%d",rem);
          }
       n=n/10;    
   }
getch();
}
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logics in mind:-
>first we enter a number(for 'n').
->If we have number 123 then its odd positioned digits are 1 and 3(from both side). It means, first we have to get 3 then 2 and then 1 to check all digits.        
                  ->For this, 
                 ->we have to divide by 10 to get last digit. 
                 ->we also use 'count' with initial value '0' to to know position  of digits.
                 ->we divide that count by 2 to get remainder.If count is not equal to 2 then we understand that the digit is in odd position and if not then that is not.
                 ->Then we display that.
                -> to get second digit, we get first 12 and for this , we use 123/10.It is done to get integer                           part  only.
->We repeat this until the value reaches 0. We use loop for this as shown above in the code.        

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