python2.7练习小例子(十)

    10):古典问题:有一对兔子,从出生后第3个月起每个月都生一对兔子,小兔子长到第三个月后每个月又生一对兔子,假如兔子都不死,问每个月的兔子总数为多少?

    程序分析:兔子的规律为数列1,1,2,3,5,8,13,21....

    程序源代码:

 

#!/usr/bin/python
# -*- coding: UTF-8 -*-

f1 = 1
f2 = 1
for i in range(1,22):
    print '%12ld %12ld' % (f1,f2),
    if (i % 3) == 0:
        print ''
    f1 = f1 + f2
    f2 = f1 + f2

    以上实例输出结果为:

 

           1           1            2           3            5           8 
          13          21           34          55           89         144 
         233         377          610         987         1597        2584 
        4181        6765        10946       17711        28657       46368 
       75025      121393       196418      317811       514229      832040 
     1346269     2178309      3524578     5702887      9227465    14930352 
    24157817    39088169     63245986   102334155    165580141   267914296

    

#!/usr/bin/python
# -*- coding: UTF-8 -*-#
#递归做,非常慢。计算n=36就要大概七八秒吧
def fib(n):
    if n==1 or n==2:
        return 1
    else:
        return fib(n-1)+fib(n-2)
print fib(36)

    

#!/usr/bin/python
# -*- coding: UTF-8 -*-

# time 为第几个月,n 为 3
def rabbit(time,n):
    if time<1:
        return 0
    elif time==1:
        num=1
    elif time<n:
        num=1
    else:
        num=rabbit(time-1,n)+rabbit(time-(n-1),n)
    return num
print rabbit(25,3)

    

#!/usr/bin/python
# -*- coding: UTF-8 -*-

def rabbit(num):
    f1 = 1
    #第一个月为1
    f2 = 1
    #第二个月为1
    if num == 1 or num == 2:
        return 1
    else:
        for i in xrange(num-1):
            f1,f2 = f2,f1+f2
    return f1

# 第三十六月兔子数量
print rabbit(36)

    使用斐波那契数列:

 

# !/usr/bin/python
# coding=UTF-8

n = int(raw_input("第几个月: "))
# 斐波那契数列的通项公式
f =(1/(5**0.5))*(((1+(5**0.5))/2)**n - ((1-(5**0.5))/2)**n)
print "第%d个月:共%d只"  % ( n,f)

    Python3 参考方法:

 

#!/usr/bin/python3

def rabbit(n):
    count = [1,0,0]  #将兔子成长期分为三个月
    for i in range(1,n): #每个月更新一次不同成长期的兔子对数
        count[2] = count[2] + count[1]
        count[1] = count[0]
        count[0] = count[2]
    return count[0]+count[1]+count[2] #返回兔子对数总数

n = int(input("查看第几个月的兔子对数:"))
rabbit_sum = rabbit(n)
print("第%d个月的兔子对数为%d"%(n,rabbit_sum))

    

#!/usr/bin/env python
# -*- coding: UTF-8 -*-

def Rabbit(num):
     i = 1
     a,b = 1,1
     while i <= num:
         yield a
         i += 1
         a,b = b,a+b


list = [x for x in Rabbit(20)]
print(list)

    

# -*- coding: utf-8 -*-

Rabbits={'rabbits':0}    # 新生兔子
home=[{'rabbits':1}]     # 所有兔子对都在这个列表中,初始按过完一月计算
month=int(raw_input('请输入月份:'))
time=1                   # time 表示已经过完的月数,到了该月月末。

while time <month:
    for j in home:
        if j['rabbits']>=2:
            home.append(Rabbits.copy())
        else:
            j['rabbits']+=1
    time+=1

print '兔子数量为 %u。'% len(home)

    

#!/usr/bin/python
# -*- coding: UTF-8 -*-

a = 1
b = 1
for i in range(1,21,2):
    print '%d %d'%(a,b),
    a += b
    b += a

    

# -*- coding: UTF-8 -*-
def rabbit(n):
    if n == 1:
        return [1]
    if n == 2:
        return  [1,1]
    rabbits = [1,1]
    for i in range(2,n):
        rabbits.append( rabbits [-1] + rabbits [-2])  #取List倒数第一个和倒数第二个数值相加
    return rabbits
print rabbit(18)  #第十八个月的数量

    

all_rabbit = []    
    
class Rabbit():
    def __init__(self, birthday):
        self.birthday = birthday
        all_rabbit.append(self)
    def makechild(self, month):
        if month-self.birthday>=2:
            Rabbit(month)

Rabbit(1)
for i in range(1, 22):
    [j.makechild(i) for j in all_rabbit[:]]
    print(len(all_rabbit))

    简单的练习小例子,有兴趣的可以来试试。如果感觉不错的话,请多多点赞支持哦。。。

  原文链接:https://blog.csdn.net/luyaran/article/details/80046801

posted @ 2018-04-25 11:44  luyaran  阅读(316)  评论(0编辑  收藏  举报