No repeating digits and reconstructing int - c++

Question

The program i am designing is for an assignment, but as a do distant learning it is not easy finding a solution.
The program that I have to create must first ask user for an unsigned long int and then break that number down to each digit without repeating number (for example 3344 the program should list 3 and 4), my program just lists all digits. After they have been listed the position of that digits needs to be dispayed with the position (digit at the right is position 0). Then the program should be "reconstruct" to make the original unsigned long int.
An example of what it should look like :
7377683
3 : 0 5
6 : 2
7 : 3 4 6
8 : 1
7377683
The code that i am using currently :
#include <iostream>
using namespace std;
int main()
{
unsigned long int number;
cout << "Enter an integer " << endl;
cin >> number;
for(int i=0; i<10 ; i++)
{
if (number > 0)
{
cout << number%10 << " : " << i; //output digit and position
cout << "\n";
number /= 10;
}
}
return 0;
}
I cannot use arrays or strings to complete this task and that is what i am finding challenging.

Answer 1

You could store digit positions in a decimal bitmask type thing.
unsigned long n, digits[10]{};
// Input
std::cin >> n;
// Break down
for (int i = 1; n; i *= 10, n /= 10)
digits[n % 10] += i;
// Reconstruct and print digit positions
for (int i = 0; i < 10; i++) {
if (!digits[i])
continue;
n += digits[i] * i;
std::cout << i << ":";
for (int j = 0; digits[i]; j++, digits[i] /= 10)
if (digits[i] % 10)
std::cout << " " << j;
std::cout << std::endl;
}
// Output
std::cout << n;
It's kinda neat because you don't need to know how many digits your number has. Also, you could construct the new number and output the positions of all digits in the same loop which you are breaking it down, thus removing the need to store the digits anywhere, but that feels like cheating.

Answer 2

Since you can't use arrays or strings you can probably get away with using an integral type as a bitmap. Any time you output a number in your loop set a bit in the bitmap that corresponds to that number. Then when you need to output that number you check to see if that bit is set and if it is you skip printing it out. Something like the following maybe.
for (int mask = 0, i = 0; i<10; i++)
{
if (number > 0)
{
int value = number % 10;
if ((mask & (1 << value)) == 0)
{
cout << value << " : " << i << endl; //output digit and position
mask |= 1 << value;
}
number /= 10;
}
}

Answer 3

Taking a number down into individual digits works like this:
int number = 4711;
vector<int> v;
while(number > 0)
{
int digit = number % 10;
number /= 10;
v.push_back(digit);
}
Putting it back together again into an integer (we need to go "backwards", as the digits come out "back to front" in the above code)
int number = 0;
for(int i = v.size()-1; i >= 0; i--)
{
number *= 10;
number += v[i];
}
I'm intentionally not showing a complete program to solve your problem, since part of learning programming is to learn how to solve problems. But you sometimes need a few "steps" on the way.
Something like this would solve it with arrays:
int array[10][10] = { { 0 } }; // Position of each digit.
int count[10] = { 0 }; // Number of each digit
int number = 4711;
int pos = 0;
while(number > 0)
{
int digit = number % 10;
number /= 10;
count[digit]++;
array[digit][count[digit]] = pos;
pos++;
}
I'm leaving it to you to fill in the rest of the code (to print and reassemble the number). [The above code doesn't cope with the number zero].

Answer 4

This is the working solution which address to the most crucial problem in your question:
int number = 7377683;
int temp = number;
int pos = 0;
int counter = 0;
int currNum;
int uniqueCount = 0;
Added: Codes to check number of unique digits in number:
for (int x=0; x<9; x++)
for (int y=temp; y>0; y/=10)
if (y%10 == x)
{
uniqueCount ++;
break;
}
Codes to generate the output of every unique elements and positions:
for (int y=0; y<uniqueCount; y++)
{
pos = counter;
currNum = number%10;
cout << temp%10 << " : ";
for (int x=temp; x>0; x/=10)
{
if (temp%10 == currNum)
cout << pos << " ";
pos++;
temp /= 10;
}
counter++;
number /=10;
temp = number;
cout << endl << endl;
}
Program Output:
3 : 0 5
8 : 1
6 : 2
7 : 3 4 6
This solution is using the most basic construct without array (according to your requirements).