printf problem

A friend ask me to help with his code. the input will be as following:

1st input -> number of loop
2nd input -> given number
3rd input -> number of rotation

example:

1
123045
3

The result is 045123

He told me that he can only get 45123 as the result. the 0 is missing. Is there any way to display it ?? i'm not very familiar with C or C++ though

here is his code. thx in advance

Code:

#include <stdio.h>
#include <stdlib.h>

int main()
{
long int num,n,value, digitNum, ctr, NumX;
int left, right;
long b;

scanf("%ld", &b);

while (b != 0)
{
      
scanf("%d %d", &num, &n);

value = num;
NumX = 1;

     while (value!= 0)
     {
      NumX = NumX * 10;
      value = value / 10;
     }
     
     NumX = NumX / 10;
     
     ctr = 0;
     
     while (ctr != n)
     {
           left = num / 10;
           right = num % 10;
           num = right * NumX + left;
           ctr++;
           
           }
           
           printf("%d", num);
           
printf("\n");
b--;
}

fflush(stdin);getch();
return 0;
}

Maybe it is easyer to convert number to string and work with the string?

As for the original question - yes using format modifiers the number can be printed with zeroes filling it up to desired length

can u explain more about this format modifiers ??? is it something like

Code:

{
    ("%.2f", variable) ;
}

i'm not very sure as well.

and for the code, my friend told me that his so called "Genius" lecture insist of using number instead of string. This because my friend is only in his 2nd month of his new semester.

i tried to help him by using array but still this "genius" lecture reject it. he still insist to use number without array or anything.

yes it is format modifier
http://msdn.microsoft.com/library/de..._functions.asp

read about width and preceiding zeroes

Code:

{
    ("%.2f", variable) ;
}

as far as i knw abt this if the statement is like as follows

Code:

 printf  ("%.2f", variable) ;

the value after the float will be ristricted to two digits

the answer wud be sumthing like
10.22

What about something like ...

Code:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main()
{
    unsigned int i, num, num_digits, num_loops, num_rots;
    int t;

    printf("Enter number of loops : ");

    if ( scanf("%u", &num_loops) != 1 )
    {
        /* Error */
    }

    for (i = 0; i < num_loops; i++)
    {
        printf("Enter number to rotate : ");

        if ( scanf("%u", &num) != 1 )
        {
            /* Error */
        }

        if ( num == 0 ) num_digits = 1;
        else            num_digits = 1 + (int) log10( (double) num );

        printf("Enter number of rotations : ");
        if ( scanf("%u", &num_rots) != 1 )
        {
            /* Error */
        }

        printf("Rotated number is : ");
        for (i = num_digits; i > 0; i--)
        {
            t = i + num_rots - 1;
            if ( t >= num_digits) t -= num_digits;
            printf("%u", (num / (int)pow(10,t)) % 10);
        }
        printf("\n");
    }

    getch();

    return 0;
}