[Nikas]

Hi, I have a small issue here.

int main()
{
    // Declaring variable
    int n, r, y;
    bool found;
    
    do
    {
        found = false;
        
        cout << "Enter an integer: ";
        cin >> n;
 
        while (n > 0)
        {    
            r = n % 10;
 
            if (r == 0)
            {
                found = true;
                break;
            }
            else if (r == 1)
            {
                found = true;
                break;
            }
            else 
            {
                do
                {
                    for (int k = r; k > 0; k--)
                    {
                        cout << " ";
                    }
        
                    for (int y = r; y > 0; y--)
                    {
                        cout << y;
                    }
                            
                    cout << endl;
                            
                    r--;
                        
                } while (r > 0); 
            }
 
            n /= 10;
 
        }
 
        
    } while (found == true);
    
    return 0;
} 

Basically, a user will input a integer and it will check if it's the integer contains 1 or 0, if it is, then reject the integer and ask the user to input again.

Once it is checked that it does not contain 1 or 0, it will print out the value using the for-loop.

My issue here is only when the the last digit is 1 then it will ask for the new integer straight else if it is something like 5215, it will loop through for the for-loop before asking for the new integer.

Some logic error here but I can't make it out.

[Nikas]

This is what I've come up with. Seem to be working fine, but not sure if I'm doing any extra step.

int main()
{
    // Declaring variable
    int n, r, t, y, count;
    bool found;
    
    
    
    do
    {
        found = false;
        count = 0;
        
        cout << "Enter an integer: ";
        cin >> n;
        t = n;
        
        while (n > 0)
        {    
            r = n % 10;
 
            if (r == 0)
            {
                found = true;
                // break;
                count++;
            }
            else if (r == 1)
            {
                found = true;
                // break;
                count++;
            }
            
            n /= 10;
        }
 
        if (count == 0)
        {
            while (t > 0)
            {
                r = t % 10;
                
                do
                {
                    for (int k = r; k > 0; k--)
                    {
                        cout << " ";
                    }
                    
                    for (int y = r; y > 0; y--)
                    {
                        cout << y;
                    }
                                
                    cout << endl;
                                
                    r--;
                            
                } while (r > 0); 
                
                t /= 10;
            }     
        }
 
        
    } while (found == true);
    
    return 0;
} 

[UnrealEd]

As far as I can tell, I'd say that's the way to do it.

An other option is to convert the number to a string, and do a character based search for 1 and 0. I have no idea though, if this has any advantages over the method you use

[Nikas]

Thanks for confirming that.

[sagar474]

making to many division operations is a hell to the processes.
Hear is your code you can easily convert it to c++

#include<stdlib.h>
#include<string.h>
#include<stdio.h>
int main(){
char *integer=(char *)malloc(10);
int len=0, i=0;
printf("Enter an integer");
while(1)
{
scanf("%s",integer);
len=strlen(integer);
while(i<len && integer[i]!='0' && integer[i]!='1')++i;//check for zeros and ones
if((integer[i]=='0' || integer[i]=='1'))printf("enter again \n"),i=0;//if there is zero or one prints enter again
else break;
}
return 0;
}