参考方法：

peple =[]
for i in range(1,31):
    person = 'person%d'%i
    peple.append(person)
c = 0
max =30
t = 0
while max >15:
    god = []
    for pe in peple:
        c += 1
        if c % 9 == 0:
            print(c,pe,'离开')
            max -= 1
            god.append(pe)
    for pe in god:
        peple.remove(pe)
print('留下人员',peple)

参考方法：

people = range(1,31)
tempList = list()
timer = 1
flag = True
while flag:
    for x in people:
        if people.index(x) in tempList:
            continue
        if timer == 9:
            tempList.append(people.index(x))
            timer = 0
            if len(people) - len(tempList) == 15:
                flag = False
                break
        timer += 1
for x in tempList:
    print(x+1,end=" , ")

上面写的有点复杂了，其实可以很简单的几行代码，如下直接定义成函数调用吧：

def yuesefu(nums, step, stay):
    #参数 nums：人数，step: 数到几的步数，stay: 最后留下多少人
    lists = list(range(1, nums+1))
    check = 0
    while len(lists) > stay:
        for i in lists[:]:
            check += 1
            if check == step:
                check = 0
                lists.remove(i)
                print("{}号下船了".format(i))
    return lists
stays = yuesefu(30, 9, 15)
print("最后留下的人：", stays)

这个问题的关键在于每次循环只选出一个人。因为当留在船上的人数大于报数最大值的2倍时，若每次循环选出所有的可能数，则会造成结果中留在船上人少于目标值。将问题定义成函数如下：

def JosephDeathGame(total, onboard, step):
    peoples = list(range(1, 31))
    x = 0 #下船人数计数
    y = 1 #报数计数
    z = 0 #索引计数
    while x < total - onboard:
        if z > len(peoples)-1:
            z = 0
        if y == step:
            print("{0}号下船了！".format(peoples.pop(z)))
            z -= 1
            y = 0
            x += 1
        y += 1
        z +=1
        
JosephDeathGame(30, 15, 9)

people = list(range(1, 31))
i = 0
while len(people) > 15:
    i += (9 - 1)
    if i >= len(people):
        i -= len(people)
    print("{}号下船了".format(people[i]))
    del people[i]

用从字典里移除键值的方法来解决这个问题:

people={}
for x in range(1,31):
    people[x]="p{}".format(x)

c = 0
pn = 30
skip = []

while pn >15:
    
    for i in range(1,31):
      
        if i not in skip: # 跳过在移除人员列表中的键值

            c += 1        # 累计数的次数
            
            if c%9 == 0:  # 当数到9的整数倍时
                print("{:2d}".format(i) + " " + "号下船了")
                skip.append(i) # 创建移除人员列表
                pn -= 1        # 人员名单长度因移除而缩短1
                people.pop(i)  # 移除相应人员

print ('\n留下人员', people)

紧跟上一位同学的思路：

sum = list(range(1,31))
while len(sum) > 15:
    print('{}号下船了'.format(sum.pop(8)))
    sum +=sum[0:8]
    del sum[0:8]

利用切片，每次生成一个新的序列:

s=list(range(1,31))
for i in s:
    if len(s)>15:
            print(s[8])
            s=s[9:]+s[:8]

参考方法：

def check_count(count1, count2, list):
    if count1 == 9:
        count1 = 1
        if count2 == len(list):
            count2 = 0
        else:
            count2 = count2
    
    else:
        count1 += 1
        if count2 == len(list) - 1:    
            count2 = 0
        else:
            count2 += 1
    
    return  count1,  count2

num_ppl = int(input('Key in the number of people: '))

list = []

for i in range(1, num_ppl+1):
    list.append(i)
print(list)     

count1 = 1
count2 = 0

while len(list) > 15:
    if count1 == 9:
        print('No.' + str(list.pop(count2)) + ' jump into the river!')
        count1, count2 = check_count(count1, count2, list)

    else:
        count1, count2 = check_count(count1, count2, list)

print('Survivors:', end = ' ')
print(list)

使用队列（先进先出原理），代码如下：

from queue import Queue

N = 30
q = Queue(N)  #建立一个长度为30的堆栈
for i in range(1, N + 1):  #给堆栈赋值1~30
    q.put(i)
    #print(q.get())
check = 0  #报数检测9
leaveN = 0  #下船人数记录
while leaveN < 15:
    check += 1
    temp = q.get()  #前进先出
    if check == 9:
        print(f'{temp}号下船')
        check = 0
        leaveN += 1
    else:
        q.put(temp)  #进入队列