使用python从三个角度解决josephus问题的方法

def josephus_A(n, k, m):
  people = list(range(1, (n+1)))
  i = k-1
  for num in range(n):
    count = 0
    while count < m: 
      if people[i] > 0:
        count += 1
      if count == m:
        print(people[i], end=" ")
        people[i] = 0
      i = (i+1) % n # count只是flag，真正记的数是i
    if num < n-1:
      print(end=",", )
    else:
      print(" ")

def josephus_L(n, k, m):
  people = list(range(1, (n+1)))
  i=k-1
  for num in range(n,0,-1):
    i=(i+m-1)%num
    print(people.pop(i),end=", " if num>1 else "\n")

class LNode: # 建立链表结点
  def __init__(self,elem,next_=None):
    self.elem=elem
    self.next=next_
class LCList: # 建立循环链接表
  def __init__(self):
    self._rear=None
  def is_empty(self):
    return self._rear is None
  def prepend(self,elem): # 前端插入
    p=LNode(elem)
    if self._rear is None:
      p.next=p # 建立一个结点的环
      self._rear=p
    else:
      p.next=self._rear.next
      self._rear.next=p
  def append(self,elem): # 尾端插入
    self.prepend(elem)
    self._rear = self._rear.next
  def pop(self): # 前端弹出
    if self._rear is None:
      raise LinkedListUnderflow("in pop of CLList")
    p = self._rear.next
    if self._rear is p:
      self._rear =None
    else:
      self._rear.next=p.next
    return p.elem
  def printall(self): # 输出表元素
    if self.is_empty():
      return
    p = self._rear.next
    while True:
      print(p.elem)
      if p is self._rear:
        break
      p=p.next
class LinkedListUnderflow(ValueError): # 自定义异常
  pass
class Josephus(LCList):
  def __init__(self,n,k,m):
    LCList.__init__(self)
    for i in range(n):
      self.append(i+1)
    self.turn(k-1)
    while not self.is_empty():
      self.turn(m-1)
      print(self.pop(),end=("\n" if self.is_empty() else ", "))
  def turn(self,m):
    for i in range(m):
      self._rear = self._rear.next