约瑟夫环问题是数学和计算机科学中的一个经典问题，详见约瑟夫问题百度词条。之前在计算机二级考试中遇到过，当时用C语言写的，感觉有点复杂。现在学习python，发现可以用列表的pop方法完美模拟把人移除的操作，因此重新用python写了一下：
问题的具体描述：有34个人围成一圈，顺序排号。从第一个人开始报数（从1到3报数），凡报到3的人退出圈子，问最后留下的是原来第几号的那位。
python代码：
ring_list = [ i+1 for i in range(34)]
bias = 0
print(ring_list)
while(len(ring_list)>1):
pop_time = (len(ring_list)+bias)//3
residue = (len(ring_list)+bias)%3
pop_index = [ i*3-bias+2 for i in range(pop_time)]
pop_index.sort(reverse =True)
for index in pop_index:
ring_list.pop(index)
print(ring_list)
bias = residue
输出结果：
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34]
[1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 25, 26, 28, 29, 31, 32, 34]
[1, 4, 5, 8, 10, 13, 14, 17, 19, 22, 23, 26, 28, 31, 32]
[1, 4, 8, 10, 14, 17, 22, 23, 28, 31]
[1, 4, 10, 14, 22, 23, 31]
[1, 10, 14, 23, 31]
[10, 14, 31]
[10, 31]
[10]
[Finished in 0.7s]
我们可以看到，最终留下的人为10号。
我们也可以把这段代码封装成函数，把34和3作为参数传入，也可得到同样的结果：
def joseph(total_num,pop_num):
ring_list = [ i+1 for i in range(total_num)]
bias = 0
print(ring_list)
while(len(ring_list)>1):
pop_time = (len(ring_list)+bias)//pop_num
residue = (len(ring_list)+bias)%pop_num
pop_index = [ i*3-bias+2 for i in range(pop_time)]
pop_index.sort(reverse =True)
for index in pop_index:
ring_list.pop(index)
print(ring_list)
bias = residue
joseph(34,3)
