The Josephus Problem

I am having a bit of trouble. The Josephus problem is where you have a circle of people, body_count people big (like, if body_count were 5, that means the circle is made up of 5 people). And another term, excuse_counter, is set to a number. Using this number we go around the circle elinimating people until one is left.

For instance, let's say the circle is made of 9 people, and we choose our excuse_counter to be 5:

5, 1, 7, 4, 3, 6, 9, 2, 8

We reloop around this circle nailing off the person we come to when we count to excuse_counter. Using the remaining people we count again, etc, until we are left with one man left.

However I am having a bit of trouble with the coding portion. It finds the first person alright, but from here it's all downhill:

Code:

person* pick_person(person *group, int excuse_counter, int body_count){
	person *temp = new person;
	temp -> position = 12345;
	temp -> next = group;
	
	person *previous = temp;
	person *current;
	
	for (int i = 1; i < body_count; i++){
		
		for (int x = 1; x <= excuse_counter; x++){

			if (temp -> next -> next == NULL)
				temp -> next = previous -> next;
			
			
			else
				temp = temp -> next;
					
		}
		
		cout << "\nPerson " << temp -> position << " is leaving...";
		
		current = temp;
		
			
		while (previous -> next != current){
			previous = previous -> next;
		}
			
		previous -> next = current -> next;
		delete current;
				   	   	   	   
	}	 

return group;
} //end person *pick_person

The struct:

Code:

struct person{
	int position;
	person *next;
};

I know this is very close to working but it needs a push in the right direction. I am using a linklist

I would use a queue. I had to do this problem in college.

Also, would it be better to use a random number to choose which person who would be killed??

No the user decides the values

That and the problem states that we must use a linklist

well my approach to this would be a circular list. This can be rotated and nodes removed. I would think that a circular list will model a circle of people better

Try to find the trick to solving this problem fast without a
list.