原题链接:破解简单密码
解题思路:
注意事项:
参考代码:
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 | dicnum = {} for i in range ( ord ( 'a' ), ord ( 'd' )): dicnum[ chr (i)] = 2 for i in range ( ord ( 'd' ), ord ( 'g' )): dicnum[ chr (i)] = 3 for i in range ( ord ( 'g' ), ord ( 'j' )): dicnum[ chr (i)] = 4 for i in range ( ord ( 'j' ), ord ( 'm' )): dicnum[ chr (i)] = 5 for i in range ( ord ( 'm' ), ord ( 'p' )): dicnum[ chr (i)] = 6 for i in range ( ord ( 'p' ), ord ( 't' )): dicnum[ chr (i)] = 7 for i in range ( ord ( 't' ), ord ( 'w' )): dicnum[ chr (i)] = 8 for i in range ( ord ( 'w' ), ord ( 'z' ) + 1 ): dicnum[ chr (i)] = 9 while True : try : st = input () for i in st: if i.isupper() and i! = 'Z' : print ( chr ( ord (i) + 1 ).lower(),end = '') elif i = = 'Z' : print ( 'a' ,end = '') elif i in dicnum: print (dicnum[i],end = '') else : print (i,end = '') print () except : break |
0 分
0 人评分
C语言网提供由在职研发工程师或ACM蓝桥杯竞赛优秀选手录制的视频教程,并配有习题和答疑,点击了解:
一点编程也不会写的:零基础C语言学练课程
解决困扰你多年的C语言疑难杂症特性的C语言进阶课程
从零到写出一个爬虫的Python编程课程
只会语法写不出代码?手把手带你写100个编程真题的编程百练课程
信息学奥赛或C++选手的 必学C++课程
蓝桥杯ACM、信息学奥赛的必学课程:算法竞赛课入门课程
手把手讲解近五年真题的蓝桥杯辅导课程
发表评论 取消回复