hogwarts
(霍格沃兹测试学院官方)
April 27, 2022, 3:09am
1
我们今天的任务是实现一个简单的ascii加密算法,具体的要求是:给定一个纯英文的字符串,将其中的每一个字母 加上各自的索引,最终得到ascii编码表中的最新字符。
【示例】
输入:“AAAA”
输出:“AAAA”
解释:AAAA的索引分别是’0,1,2,3’,而A的ascii编号是65,所以增量后是’65,66,67,68’,对应字符为’ABCD’
题目难度:简单
题目来源:CodeWars-ASCII Shift Encryption/Decryption
def solution(src: str)-> str:
# your code here
assert solution("AAAA") == "ABCD"
assert solution("PASSWORD") == "PBUV[TXK"
assert solution("0123456789") == "02468:<>@B"
joyoko
(徐子宇)
April 27, 2022, 3:56am
2
def solution(src: str)-> str:
# your code here
return "".join(chr(ord(e)+i) for i, e in enumerate(src))
assert solution("AAAA") == "ABCD"
assert solution("PASSWORD") == "PBUV[TXK"
assert solution("0123456789") == "02468:<>@B"
def solution(src: str) -> str:
# your code here
ls = "".join([chr(ord(src[i]) + i) for i in range(len(src))])
return ls
if __name__ == '__main__':
assert solution("AAAA") == "ABCD"
assert solution("PASSWORD") == "PBUV[TXK"
assert solution("0123456789") == "02468:<>@B"
def solution(src: str)-> str:
return "".join([chr(ord(src[i])+i) for i in range(len(src))])
assert solution("AAAA") == "ABCD"
assert solution("PASSWORD") == "PBUV[TXK"
assert solution("0123456789") == "02468:<>@B"
def solution(src: str) -> str:
return ''.join([chr(ord(src[i]) + i) for i in range(len(src))])
assert solution("AAAA") == "ABCD"
assert solution("PASSWORD") == "PBUV[TXK"
assert solution("0123456789") == "02468:<>@B"
def solution(src: str) -> str:
return ''.join(chr(i+ord(src[i])) for i in range(len(src)))
def solution(src: str)-> str:
# your code here
# l = list(src)
# for i in range(len(l)):
# l[i] = chr(i + ord(l[i]))
# return ''.join(l)
return ''.join([chr(i + ord(list(src)[i])) for i in range(len(list(src)))])
assert solution("AAAA") == "ABCD"
assert solution("PASSWORD") == "PBUV[TXK"
assert solution("0123456789") == "02468:<>@B"
def solution(str_1):
print("".join(chr(ord(e)+i) for i,e in enumerate(str_1)))
#enumerate(str_1)元组的形式返回(0,A),(1,A),(i,e)
solution(“PASSWORD”)
solution(“0123456789”)
qunzi
(群子)
March 8, 2023, 1:55am
9
def solution(src: str)-> str:
dic=list(src)
res=[chr(i+ord(dic[i])) for i in range(len(dic))]
result="".join(res)
return result
assert solution("AAAA") == "ABCD"
assert solution("PASSWORD") == "PBUV[TXK"
assert solution("0123456789") == "02468:<>@B"