Displaying Numbers divisible by user's desired number

This is a discussion on Displaying Numbers divisible by user's desired number within the C Programming forums, part of the General Programming Boards category; well...I was doing a program to ask the user from the range he wants the program to generate numbers from ...

  1. #1
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    Exclamation Displaying Numbers divisible by user's desired number

    well...I was doing a program to ask the user from the range he wants the program to generate numbers from and also ask the user for a number and then generates all the numbers divisible by this number I did this so far but not working well...its always in a never-ending loop...Any Help?? Suggestion??

    Code:
    Code:
    #include <stdio.h>
    int main()
    {
    int x,ul,ll,j=0;
    
    printf("Please enter your desired number you want its divisible numbers\n");
    scanf("%i",&x);
    
    
    printf("Please enter the lower limit you desire\n");
    scanf("%i",&ll); //ALERT: getchar(); --> its only with characters!
    
    printf("Please enter the upper limit you desire\n");
    scanf("%i",&ul);
    
    
    	for (j <= ll; j <= ul; j++)
    	{
    		if (x < j)
    		{
    	if (j % x == 0)
    			{
    				printf("\t%i\t", x);
    				j++;
    
    			}
    		}
    		else
    		{       if ( j == x)
    			{
    				j = 0;
    				printf("\n");
    			}
    		}
    	}
    
    	return (0);
    }

  2. #2
    and the hat of wrongness Salem's Avatar
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    Shouldn't you be using your previous program (the one where things are divisible by 6), and replacing all those 6's with your user input variable?
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
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  3. #3
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    well..i have a problem in the lower and upper limit thing

  4. #4
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    Well..I got this so far

    Code:
    #include <stdio.h>
    int main()
    {
    int x,ul,ll,j=1;
    
    printf("Please enter your desired number you want its divisible numbers\n");
    scanf("&#37;i",&x);
    
    
    printf("Please enter the lower limit you desire\n");
    scanf("%i",&ll); //ALERT: getchar(); --> its only with characters!
    
    printf("Please enter the upper limit you desire\n");
    scanf("%i",&ul);
    
    
    	for (j >= ll && j <= ul; j--)
    	{
    		if (j < x)
    		{
    	if (j % x  == 0)
    			{
    				printf("\n%i", x);
    				j++;
    
    			}
    		}
    		else
    		{       if ( j == x)
    			{
    				j = 0;
    				printf("\t\t");
    			}
    		}
    	}
    
    	return (0);
    }

  5. #5
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    I think the error is in here

    Code:
    for (j >= ll && j <= ul; j--)
    or what do you think? I truely dunno ?

  6. #6
    C++ Witch laserlight's Avatar
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    Okay, I see that you have been trying hard, so I am going to give you an example of what you have been trying to do. However, I have included a bug, namely, the program does not print out the multiples of the number in the given range, unless the lower limit of the range is itself a multiple of the number. Your task is to fix this bug.
    Code:
    #include <stdio.h>
    
    int main(void)
    {
        int num, lower, upper;
        int i;
        printf("Enter the integer: ");
        scanf("&#37;i", &num);
        printf("Enter the lower limit: ");
        scanf("%i", &lower);
        printf("Enter the upper limit: ");
        scanf("%i", &upper);
    
        printf("The multiples of %i in the given range are:\n", num);
        /* Insert bug fixing code here. */
        for (i = lower; i <= upper; i += num)
        {
            printf("%i ", i);
        }
        printf("\n");
    
        return 0;
    }
    Note how well formatted my example code is. I have used meaningful variable names (or at least tried to, I am not sure what to call the number that the multiples are based on. Divisor?) instead of abbreviations that may be difficult to understand at a glance.
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  7. #7
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    Well..That's too advanced for my tiny little brain..but,I will try

  8. #8
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    Code:
    #include <stdio.h>
    
    int main(void)
    {
    	 int num, lower, upper;
        int i;
    	 printf("Enter the integer: ");
        scanf("&#37;i", &num);
    	 printf("Enter the lower limit: ");
        scanf("%i", &lower);
        printf("Enter the upper limit: ");
        scanf("%i", &upper);
    
        printf("The multiples of %i in the given range are:\n", num);
    	 /* multiples won't be all in one variable
    	 and it can't be outside the loop */
        for (i = lower; i <= upper; i += num)
    	 {
    		  printf("%i ", i);
        }
    	 printf("\n");
    
    	 return 0;
    }

    Here's the fixed one:

    Code:
    #include <stdio.h>
    
    int main(void)
    {
    	 int num, lower, upper;
    	 int i;
    	 printf("Enter the integer: ");
    	 scanf("%i", &num);
    	 printf("Enter the lower limit: ");
    	 scanf("%i", &lower);
        printf("Enter the upper limit: ");
    	 scanf("%i", &upper);
    
    
    	 for (i = lower; i <= upper; i += num)
    	 {
    if (lower < num)
    		{
    	if (lower % num  == 0)
    			{
    				printf("\n%i", num);
    				lower++;
    
    			}
    		}
    		else
    		{       if ( lower == num)
    			{
    				lower = 0;
    				printf("\t\t");
    			}
    		}
    
    	 printf("\n");
    
    	 return 0;
    }
    }
    I tried...will post if i did any changes
    Thanks for your help lasernight its soooooooooo apreciated

  9. #9
    C++ Witch laserlight's Avatar
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    My hint is what I gave in my example: insert the bug fixing code before the loop, and do not modify the loop. The trick is to find the correct value of lower such that it is the smallest multiple of num greater than or equal to the lower limit provided by the user.

    To find this new lower limit, you can either use a loop with the idea of checking if the number is a multiple, or you can use a little bit of mathematics.
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  10. #10
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    I will try again

    Thanks a million!!

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