有m个人玩萝卜蹲游戏，每个人说出一个不重复的词汇，组成序列s，大家依次报s里面的单词，每一轮最长有b个单词，如下图所示，

b=5，m=3，s=[‘a’,’b’,’c’],按照下列顺序进行游戏

[‘a’]

[‘a’, ‘b’]

[‘a’, ‘b’, ‘c’]

[‘a’, ‘b’, ‘c’, ‘a’]

[‘a’, ‘b’, ‘c’, ‘a’, ‘b’]

[‘a’, ‘b’, ‘c’, ‘a’]

[‘a’, ‘b’, ‘c’]

[‘a’, ‘b’]

求已经知道某个单词出现的次数，求这是第几轮

想法：求出一个循环该单词出现的最大次数counter，然后n=n%counter，然后计算n出现在第几轮，加上之前的轮数

b=5
n=4000 #出现次数
word=’b’#已知
s=[‘a’,’b’,’c’]
ss=s*b
ss=ss[:b]
ss_max=[]
for i in range(len(ss)):
ss_max+=ss[:i+1]
print(ss[:i+1])
for i in range(len(ss)):

if len(ss[:len(ss)-i-1])>1:
    ss_max+=ss[:len(ss)-i-1]
    print(ss[:len(ss)-i-1])

counter=0
m=0
for i in ss_max:
if i==word:
counter+=1
if n>counter:
m=int(n/counter)
n=n%counter
lunshu=(2b-1)m
print(m)
for i in range(len(ss)):
lunshu += 1
for j in ss[:i+1]:
if j==word:
n-=1

    if n==0:
        break
if n==0:
    break

if n>0:
for i in range(len(ss)):

    if len(ss[:len(ss) - i - 1]) >1:
        lunshu += 1
        for j in ss[:len(ss) - i - 1]:
            if j==word:
                n-=1

            if n==0:
                break
        if n==0:
            break

print(lunshu)