Survivor Solution in Java using Recursion

Survivor.java

//There are 100 people sat at a circular table. 
//You go around the table asking every second person to leave (starting by asking the guy at seat 1 to leave) until there is one person left. 
//Which seat is left?
//usage: java Survivor 100
class Survivor {
  public static void main(String[] args){
    int n = Integer.parseInt(args[0]);
    int result = eliminate(1, 1, n, false);
    System.out.println(result);
  }
  
  public static int eliminate(int distance, int first, int n, boolean safe){
    if(n == 1)
      return first;
      
    int new_first = (safe ? first : first + distance); //if safe is not true then the person at first is eliminated and the new first becomes first + distance.
    int new_n = (n%2==1 && safe ? n/2 + 1 : n/2); //if n is odd and the first person is safe, then n/2 + 1 people remain.
    boolean new_safe = (n%2==1 ? !safe : safe); //if n is odd then the safety of the first person next round will flip
 
    return eliminate(distance*2, new_first, new_n, new_safe);
  }
}