猫史档案馆


【Python作品分享】新的作品【求助帖】

用户:无敌的隔壁老王无敌的隔壁老王查看:0 回复:4 评论:0 创建时间:2024-08-16T19:00:33


【作品展示】

center_image

 

【作品介绍】

这样做对吗?

 

【作品源代码】

n = input()
s1 = ""
for i in n:
    if i == "a":
        s1 += "01"
    if i == "b":
        s1 += "02"
    if i == "c":
        s1 += "03"
    if i == "d":
        s1 += "04"
    if i == "e":
        s1 += "05"
    if i == "f":
        s1 += "06"
    if i == "g":
        s1 += "07"
    if i == "h":
        s1 += "08"
    if i == "i":
        s1 += "09"
    if i == "j":
        s1 += "10"
    if i == "k":
        s1 += "11"
    if i == "l":
        s1 += "12"
    if i == "m":
        s1 += "13"
    if i == "n":
        s1 += "14"
    if i == "o":
        s1 += "15"
    if i == "p":
        s1 += "16"
    if i == "q":
        s1 += "17"
    if i == "r":
        s1 += "18"
    if i == "s":
        s1 += "19"
    if i == "t":
        s1 += "20"
    if i == "u":
        s1 += "21"
    if i == "v":
        s1 += "22"
    if i == "w":
        s1 += "23"
    if i == "x":
        s1 += "24"
    if i == "y":
        s1 += "25"
    if i == "z":
        s1 += "26"
    if i == " ":
        s1 += " "
print(s1)
s2 = ""
for x in s1:
    if x == "0":
        s2 += "27"
    if x == "1":
        s2 += "28"
    if x == "2":
        s2 += "29"
    if x == "3":
        s2 += "30"
    if x == "4":
        s2 += "31"
    if x == "5":
        s2 += "32"
    if x == "6":
        s2 += "33"
    if x == "7":
        s2 += "34"
    if x == "8":
        s2 += "35"
    if x == "9":
        s2 += "36"
    if x == " ":
        s2 += "00"
print(s2)

 

【提示】

部分含有Python第三方库相关内容的作品,在海龟编辑器网页端无法运行哦!如遇到这种情况,可以打开下面的链接,下载海龟编辑器客户端:

https://python.codemao.cn


回复

上一页1 页 / 共 1下一页
code猫的code猫的

你提供的代码片段是将字母 a-z 和空格转换为特定的数字编码。然后,再进一步将这些数字编码转换为另一组数字编码。代码可以通过使用字典来简化,减少冗余,同时提高可读性和效率。以下是修改后的版本:

n = input("请输入字符串: ")

# 首先,我们定义两个字典来进行编码
char_to_num = {chr(i): str(i - 96).zfill(2) for i in range(97, 123)}  # a-z
char_to_num[' '] = ' '

# 编码第一步
s1 = ''.join(char_to_num[i] for i in n if i in char_to_num)

print("第一步编码结果:", s1)

# 第二步编码的字典
num_to_num = {str(i): str(i + 27) for i in range(10)}  # 0-9
num_to_num[' '] = '00'  # 处理空格

# 编码第二步
s2 = ''.join(num_to_num[i] for i in s1 if i in num_to_num)

print("第二步编码结果:", s2)

点赞0


评论


九爪金龙九爪金龙

 &nbsp你好

点赞0


评论


九爪金龙九爪金龙

 &nbsp你好

点赞0


评论


九爪金龙九爪金龙

 &nbsp请开放图书馆评论区

点赞0


评论