Arbitrary number of nested-loops? [duplicate]

Answer 1

You can use fundamental principal of counting, like incrementing the last digit till it reaches its max value, then increment the second last one and so on, like a countdown does Here is a sample code, assuming there might be diff length of diff lists.

#include 
using namespace std;
int main() {
    int n;
    cin>>n;
    int a[n], len[n],i,j;
    for(i = 0 ; i < n ; i++)
    {
        cin>>len[i];
        a[i]=0;
    }
    while(1)
    {
        for(i = 0 ; i< n;i++)
            cout<=0 ; j--)
        {
            if(++a[j]<=len[j])
                break;
            else
                a[j]=0;
        }
        if(j<0)
            break;
    }    
    return 0;
}

Try to run the code with 4 1 1 1 1 and it will give all 4 digit permutations of 0 and 1.

0 0 0 0 
0 0 0 1 
0 0 1 0 
0 0 1 1 
0 1 0 0 
0 1 0 1 
0 1 1 0 
0 1 1 1 
1 0 0 0 
1 0 0 1 
1 0 1 0 
1 0 1 1 
1 1 0 0 
1 1 0 1 
1 1 1 0 
1 1 1 1 

You can use 2d arrays for getting combinations of nos.