Given a number N, the task is to find the required numbers consist of only 0 and 1 digit whose sum is equal to N.
Input: 9 Output: 1 1 1 1 1 1 1 1 1 Only numbers smaller than or equal to 9 with digits 0 and 1 only are 0 and 1 itself. So to get 9, we have to add 1 - 9 times. Input: 31 Output: 11 10 10
- Initialize product p to 1 and m to zero.
- Create the vector that stores the resultant integer counts of 0s and 1s.
- Loop for N and check if N is multiple of 10 if yes get the decimal and update p by multiplying 10 and store this value in a vector and decrease N by m do this for each decimal and print the total size of vector.
- Finally traverse the vector and print the elements.
Below is the implementation of the above approach.
11 10 10
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