I have some problems with assignment: Remove all digits that can be found on the left from most right digit 1

Question

I started C++ in Uni. Now we are using loops. I have practical assignment: Task: Input: integer number. Remove all digits that can be found on the left from most right digit 1.

For example:

• input: 314195
• result: 95

Or:

• input 45131005
• result: 005

For this task we can not use arrays or strings. However we still can use math functions. Also there must obligatory be while/do while condition.

I have already written the code that finds values before the first digit 1 using % and /. However, as a result it doesn't work properly and when I enter something like 431004 - it will just show 4, not the 004. Here is my code:

#include <iostream>
using namespace std;
int main()
{
int number; 
cout << endl << "Enter integer number: ";
cin >> number;

int result;
int buffer = 0;
int divider = 10;
int check = 1;

do {
    result = buffer;
    buffer = number % divider;
    divider = divider * 10;
    check *= 10;
} while ((buffer / (check / 10)) != 1);

cout << endl << "Result = " << result; 

return 0;
}

Answer 1

004 and 4 are ths ame number 4.

If you need for example to output 4 as 004 then use manupulators std::setfill and std::setw declared in header <iomanip>.

Also you should declare the variable number as having an unsigned integer type instead of the type int. Otherwise you need to take into account the sign of a number.

Here is a demonstration program that shows how to get the expected result using the manipulators.

#include <iostream>
#include <iomanip>

int main()
{
    const unsigned int Base = 10;

    unsigned long long int number;
    
    std::cout << "Enter a non-negative integer number: ";

    if (std::cin >> number)
    {
        unsigned long long int result = 0;
        unsigned long long int multiplier = 1;

        int n = 0;

        while ( number && number % Base != 1 )
        {
            result += ( number % Base ) * multiplier;
            ++n;
            multiplier *= Base;
            number /= Base;
        }

        std::cout << "Result = " 
            << std::setw( n ) << std::setfill( '0' ) << result << std::endl;
    }
}

The program output might look like

Enter a non-negative integer number: 431004
Result = 004

Answer 2

While it's possible to use a combination of std::setw and std::setfill to do the job, that strikes me as pretty much an array in disguise--you're basically just using the stream's buffer as the array to hold the intermediate result.

If you wanted to avoid that, you could do something a bit different. For one example:

#include <iostream>

void cvt(int input) { 
    if ((input != 0) && ((input % 10) != 1)) {
        cvt(input / 10);
        std::cout.put('0' + input % 10);
    }
}

int main() { 
    cvt(45131005);
    std::cout << '\n';
}

As expected, the result is: 005--and (although lack of buffering would make it somewhat slow) this doesn't rely on the rest of the system being able to do any more than write a character to standard output. As a bonus, it's actually shorter and simpler than (most) code that uses things like std::setw and std::setfill.