用python制作剪刀石头布的小游戏

末蓝、 2024-03-23 16:15 75阅读 0赞

## **1** **问题** ##

在python中我们学习了条件语句，那么我们是否可以通过python中条件判断的功能来写出可以判断胜负的剪刀石头布小游戏呢?

## **2** **方法** ##

导入随机函数，保证胜负的随机性  
![13c840bc6a95428b600e20579b615106.png][]

设置对应数值，写好判断输赢的条件语句  
![f51412b8bf44dd142533108efda1e89f.png][]

运行并查看结果

![b3a5035ddc0bdb8250ddc7551330f1c5.png][]

代码清单 1

<table> 
 <tbody> 
 <tr> 
 <td>import random &nbsp; num = 1 yin_num = 0 shu_num = 0 while num &lt;= 3: &nbsp; &nbsp;if shu_num == 2 or yin_num == 2: &nbsp; &nbsp; &nbsp; &nbsp;break &nbsp; &nbsp;user = int(input('请出拳 0（石头） 1（剪刀） 2（布）')) &nbsp; &nbsp;if user &gt; 2: &nbsp; &nbsp; &nbsp; &nbsp;print('不能出大于2的值') &nbsp; &nbsp;else: &nbsp; &nbsp; &nbsp; &nbsp;data = ['石头', '剪刀', '布'] &nbsp; &nbsp; &nbsp; &nbsp;com = random.randint(0, 2) &nbsp; &nbsp; &nbsp; &nbsp;print("您出的是{}，电脑出的是{}".format(data[user], data[com])) &nbsp; &nbsp; &nbsp; &nbsp;if user == com: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;print('平局') &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;continue &nbsp; &nbsp; &nbsp; &nbsp;elif (user == 0 and com == 1) or (user == 1 and com == 2) or (user == 2 and com == 0): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;print('你赢了') &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;yin_num += 1 &nbsp; &nbsp; &nbsp; &nbsp;else: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;print('你输了') &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;shu_num += 1 &nbsp; &nbsp;num += 1</td> 
 </tr> 
 </tbody> 
</table>

## **3** **结语** ##

通过运行结果发现这样的思路是完全可行的，但是我们并没有学过文字直接与文字比较的方法，所以需要用数字来替代文字进行比较才可以得到正确的结果。

[13c840bc6a95428b600e20579b615106.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/23/e4807e29be354dd4b11f24a342d6ba2a.png
[f51412b8bf44dd142533108efda1e89f.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/23/fbb45b27052448e991862b387ddd9b74.png
[b3a5035ddc0bdb8250ddc7551330f1c5.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/23/ef7fe81b465b48c9bfa7abc08a2e74a1.png