輸入字串,將該字串當中的每個字母進行加密。加密法則如下:A->Z,B->Y,C->X...Z->A,小寫同樣的映對方式。
數字法則:0->9,1->8,2->7...9->0。寫一個加密函式(f3),
驗證:輸入“I am 15”,輸出“R zn 84”
uj5u.com熱心網友回復:
l = [[48, 57], [65, 90], [97, 122]]
def f3(tmps):
result = ''
for i in tmps:
r = ''
for j in l:
if j[1] >= ord(i) >= j[0]:
r = chr(j[1] - (ord(i) - j[0]))
result = result + (r if r else i)
return result
print(f3('I am 15'))
uj5u.com熱心網友回復:
s="".join([chr(letter) for letter in range(65,91)])s1=s[::-1]
s=s+"".join(x.lower() for x in s)+"9876543210"
s1=s1+"".join(x.lower() for x in s1)[::-1]+"9876543210"[::-1]
print(s)
print(s1)
ss=list(s)
ss1=list(s1)
dicts=dict(zip(ss,ss1))
print(dicts)
tt='aeidk'
tts=map(lambda x: dicts[x] if x in dicts.keys() else x, list(tt))
print("".join(list(tts)))
轉載請註明出處,本文鏈接:https://www.uj5u.com/qita/40194.html
