解密结尾的密文
Creado el: 24 de abril de 2026
Respondido usando GPT-5.5 Thinking por Chat01
Creado el: 24 de abril de 2026
Respondido usando GPT-5.5 Thinking por Chat01
解密结尾的密文
pythoncipher="Xkrsg iu xoh xloo,xlkh ju d xay" shifts=[0,4,1,3] out="" i=0 for ch in cipher: if ch.isalpha(): s=shifts[i%len(shifts)] base='A' if ch.isupper() else 'a' out+=chr((ord(ch)-ord(base)-s)%26+ord(base)) i+=1 else: out+=ch print(out)
pythonimport string cipher="Xkrsg iu xoh xloo,xlkh ju d xay" seqs={"0413":[0,4,1,3],"413":[4,1,3],"20260413":[2,0,2,6,0,4,1,3],"2026413":[2,0,2,6,4,1,3],"4132026":[4,1,3,2,0,2,6],"1304":[1,3,0,4],"3140":[3,1,4,0]} def dec(seq, reset=False, count_non=False): out=""; i=0 for ch in cipher: if ch.isalpha(): s=seq[i%len(seq)] base=65 if ch.isupper() else 97 out+=chr((ord(ch)-base-s)%26+base) i+=1 else: out+=ch if count_non: i+=1 if reset and ch in " ,.;": i=0 return out for name,seq in seqs.items(): for reset in [False, True]: for count in [False, True]: print(name, reset, count, dec(seq,reset,count)) print()
pythondef transform(seq, op=1, reset=False, count_non=False): out=''; i=0 for ch in cipher: if ch.isalpha(): s=seq[i%len(seq)]*op base=65 if ch.isupper() else 97 out+=chr((ord(ch)-base+s)%26+base) i+=1 else: out+=ch if count_non:i+=1 if reset and ch in " ,.;": i=0 return out for name,seq in seqs.items(): print(name, transform(seq,1,False,False), transform(seq,1,True,False))
pythonplain="There is the hill,high is a way" for seq in ([0,4,1,3],[4,1,3]): out='';i=0 for ch in plain: if ch.isalpha(): base=65 if ch.isupper() else 97 out+=chr((ord(ch)-base+seq[i%len(seq)])%26+base); i+=1 else: out+=ch print(seq,out)
按题图给出的规则,日期 2026-4-13 应取移位序列 4,1,3(或按示例补零为 0,4,1,3),解密就是反向减去对应位移。
但我按这两种常见理解解出来都不是通顺明文:
所以,结尾密文 “Xkrsg iu xoh xloo,xlkh ju d xay” 很可能有抄写/OCR错误,或者题目还有未写出的规则。
如果严格按图片里的“日期滑动移位”规则,不能得到有效明文。建议核对一下密文里的几个字母,尤其是 iu / xoh / xloo / xlkh 这些位置。