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JZzCcd8P3cpFS13A6Px6FCE3ZRbk8TmSidCcG 这是CTF解密题目 他...

创建于：2026年1月9日

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问题

JZzCcd8P3cpFS13A6Px6FCE3ZRbk8TmSidCcG

这是CTF解密题目 他可能是什么加密方法 帮我解密一下

Python

python
import base64, binascii, textwrap, math, re, string
s="JZzCcd8P3cpFS13A6Px6FCE3ZRbk8TmSidCcG"
len(s)


37

Python

python
s="JZzCcd8P3cpFS13A6Px6FCE3ZRbk8TmSidCcG"
def b64_decode(s):
    pad = '=' * ((4 - len(s)%4)%4)
    try:
        return base64.urlsafe_b64decode(s+pad)
    except Exception as e:
        return None, e
b64_decode(s)[:10]


(None,
 binascii.Error('Invalid base64-encoded string: number of data characters (37) cannot be 1 more than a multiple of 4'))

Python

python
alphabet = "123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz"
def b58decode(s):
    n=0
    for ch in s:
        n*=58
        if ch not in alphabet:
            raise ValueError(f"invalid char {ch}")
        n+=alphabet.index(ch)
    # convert to bytes
    h = n.to_bytes((n.bit_length()+7)//8, 'big') if n>0 else b''
    # add leading zeros for leading '1's
    pad = 0
    for ch in s:
        if ch=='1': pad+=1
        else: break
    return b'\x00'*pad + h
b=b58decode(s)
len(b), b[:50]


(27,
 b'\x81\xfd]\xcf7\xa2AY\x06H\x1b\xa5h\x9eh\xd2\xb1h \xc7\x0b@\xad\xe9X\x0bY')

Python

python
b


b'\x81\xfd]\xcf7\xa2AY\x06H\x1b\xa5h\x9eh\xd2\xb1h \xc7\x0b@\xad\xe9X\x0bY'

Python

python
import hashlib, binascii
def b58check_decode(s):
    raw=b58decode(s)
    if len(raw)<4: return None
    data, cks = raw[:-4], raw[-4:]
    chk = hashlib.sha256(hashlib.sha256(data).digest()).digest()[:4]
    return data, cks, chk, cks==chk
data, cks, chk, ok = b58check_decode(s)
ok, len(data), data, cks, chk


(False,
 23,
 b'\x81\xfd]\xcf7\xa2AY\x06H\x1b\xa5h\x9eh\xd2\xb1h \xc7\x0b@\xad',
 b'\xe9X\x0bY',
 b'\xb4_,\x94')

Python

python
b=b58decode("JZzCcd8P3cpFS13A6Px6FCE3ZRbk8TmSidCcG")
def score_english(bs):
    # simple scoring: count printable and common letters
    printable=sum(32<=c<127 for c in bs)
    freq=sum(bs.lower().count(ch) for ch in b'etaoinshrdlu')
    return printable*2+freq
best=[]
for k in range(256):
    x=bytes([c^k for c in b])
    if b'flag' in x.lower() or b'ctf' in x.lower() or b'{' in x:
        best.append((k,x))
len(best), best[:5]


(23,
 [(19,
   b'\x92\xeeN\xdc$\xb1RJ\x15[\x08\xb6{\x8d{\xc1\xa2{3\xd4\x18S\xbe\xfaK\x18J'),
  (34,
   b'\xa3\xdf\x7f\xed\x15\x80c{$j9\x87J\xbcJ\xf0\x93J\x02\xe5)b\x8f\xcbz){'),
  (35, b'\xa2\xde~\xec\x14\x81bz%k8\x86K\xbdK\xf1\x92K\x03\xe4(c\x8e\xca{(z'),
  (38,
   b'\xa7\xdb{\xe9\x11\x84g\x7f n=\x83N\xb8N\xf4\x97N\x06\xe1-f\x8b\xcf~-\x7f'),
  (51, b'\xb2\xcen\xfc\x04\x91rj5{(\x96[\xad[\xe1\x82[\x13\xf48s\x9e\xdak8j')])

Python

python
scores=[]
for k in range(256):
    x=bytes([c^k for c in b])
    scores.append((score_english(x),k,x))
scores.sort(reverse=True)
scores[:10]


[(36,
  60,
  b"\xbd\xc1a\xf3\x0b\x9e}e:t'\x99T\xa2T\xee\x8dT\x1c\xfb7|\x91\xd5d7e"),
 (36, 45, b'\xac\xd0p\xe2\x1a\x8flt+e6\x88E\xb3E\xff\x9cE\r\xea&m\x80\xc4u&t'),
 (36,
  44,
  b"\xad\xd1q\xe3\x1b\x8emu*d7\x89D\xb2D\xfe\x9dD\x0c\xeb'l\x81\xc5t'u"),
 (35, 61, b'\xbc\xc0`\xf2\n\x9f|d;u&\x98U\xa3U\xef\x8cU\x1d\xfa6}\x90\xd4e6d'),
 (35, 41, b'\xa8\xd4t\xe6\x1e\x8bhp/a2\x8cA\xb7A\xfb\x98A\t\xee"i\x84\xc0q"p'),
 (34,
  36,
  b'\xa5\xd9y\xeb\x13\x86e}"l?\x81L\xbaL\xf6\x95L\x04\xe3/d\x89\xcd|/}'),
 (33,
  53,
  b'\xb4\xc8h\xfa\x02\x97tl3}.\x90]\xab]\xe7\x84]\x15\xf2>u\x98\xdcm>l'),
 (33,
  33,
  b"\xa0\xdc|\xee\x16\x83`x'i:\x84I\xbfI\xf3\x90I\x01\xe6*a\x8c\xc8y*x"),
 (33,
  32,
  b'\xa1\xdd}\xef\x17\x82ay&h;\x85H\xbeH\xf2\x91H\x00\xe7+`\x8d\xc9x+y'),
 (32,
  59,
  b'\xba\xc6f\xf4\x0c\x99zb=s \x9eS\xa5S\xe9\x8aS\x1b\xfc0{\x96\xd2c0b')]

Python

python
import zlib, gzip, io
b=b58decode("JZzCcd8P3cpFS13A6Px6FCE3ZRbk8TmSidCcG")
for name, func in [("zlib", lambda: zlib.decompress(b)), ("gzip", lambda: gzip.decompress(b))]:
    try:
        out=func()
        print(name, out[:100])
    except Exception as e:
        print(name, "fail", e)

Python

python
alphabets = {
"bitcoin":"123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz",
"flickr":"123456789abcdefghijkmnopqrstuvwxyzABCDEFGHJKLMNPQRSTUVWXYZ",
"ripple":"rpshnaf39wBUDNEGHJKLM4PQRST7VWXYZ2bcdeCg65jkm8oFqi1tuvAxyz"
}
def b58decode_alpha(s,alphabet):
    n=0
    for ch in s:
        n*=58
        if ch not in alphabet:
            return None
        n+=alphabet.index(ch)
    h = n.to_bytes((n.bit_length()+7)//8,'big') if n>0 else b''
    pad=0
    for ch in s:
        if ch==alphabet[0]: pad+=1
        else: break
    return b'\x00'*pad + h
def is_mostly_printable(bs):
    return sum(32<=c<127 for c in bs)/len(bs) if bs else 0
for name,alpha in alphabets.items():
    raw=b58decode_alpha(s,alpha)
    print(name, len(raw), raw[:10], is_mostly_printable(raw))

Python

python
alpha=alphabets["bitcoin"]
# Precompute index map
idx={ch:i for i,ch in enumerate(alpha)}
def shift_str(s,shift):
    return ''.join(alpha[(idx[ch]+shift)%58] if ch in idx else ch for ch in s)
best=[]
for sh in range(58):
    ss=shift_str(s,sh)
    raw=b58decode_alpha(ss,alpha)
    if raw is None: 
        continue
    pr=is_mostly_printable(raw)
    if pr>0.8:
        best.append((sh,pr,raw))
best[:10], len(best)


([], 0)

Python

python
alpha=alphabets["bitcoin"]
idx={ch:i for i,ch in enumerate(alpha)}
indexes=[idx[ch] for ch in s]
indexes[:20], len(indexes), min(indexes), max(indexes)


([17, 32, 57, 11, 35, 36, 7, 22, 2, 35, 47, 14, 25, 0, 2, 9, 5, 22, 55, 5],
 37,
 0,
 57)

Python

python
import itertools, numpy as np, math
b=b58decode_alpha(s,alphabets["bitcoin"])
def best_key_for_len(L):
    key=[]
    for i in range(L):
        col=b[i::L]
        best=None
        for k in range(256):
            x=bytes([c^k for c in col])
            pr=sum(32<=c<127 for c in x)
            # penalize weird
            score=pr
            if best is None or score>best[0]:
                best=(score,k)
        key.append(best[1])
    return bytes(key)
def decrypt_with_key(key):
    out=bytes([c^key[i%len(key)] for i,c in enumerate(b)])
    return out
cands=[]
for L in range(1,13):
    key=best_key_for_len(L)
    pt=decrypt_with_key(key)
    pr=is_mostly_printable(pt)
    cands.append((pr,L,key,pt))
cands.sort(reverse=True)
cands[:5]


[(0.8148148148148148,
  11,
  b'\xc0\x80\x00 \x00\x80\x00\x00   ',
  b'A}]\xef7"AY&h;e\xe8\x9eH\xd21h \xe7+`miX+Y'),
 (0.7777777777777778,
  12,
  b'\x00\xa0\x00\x80\x00\x00\x00\x00 \x00 \x80',
  b'\x81]]O7\xa2AY&H;%h>hR\xb1h \xc7+@\x8diX\xabY'),
 (0.7777777777777778,
  10,
  b' \x80\x00\xa0\x00\x80\x00\x00@\x00',
  b'\xa1}]o7"AYFH;%h>hR\xb1h`\xc7+\xc0\xadIX\x8bY'),
 (0.7777777777777778,
  8,
  b'  `\x80@\xc0\x00\x80',
  b'\xa1\xdd=OwbA\xd9&h{%(^hR\x91H@GK\x80\xadix+9'),
 (0.7037037037037037,
  9,
  b'\x00\x80 \x00\xc0\x80\x00  ',
  b'\x81}}\xcf\xf7"Ay&H\x9b\x85h^\xe8\xd2\x91H G+@miX+y')]

Python

python
pt=bytes([c^0x80 for c in b])
pt, is_mostly_printable(pt)


(b'\x01}\xddO\xb7"\xc1\xd9\x86\xc8\x9b%\xe8\x1e\xe8R1\xe8\xa0G\x8b\xc0-i\xd8\x8b\xd9',
 0.3333333333333333)

Python

python
s="JZzCcd8P3cpFS13A6Px6FCE3ZRbk8TmSidCcG"
rev=s[::-1]
for name in ["orig","rev"]:
    raw=b58decode_alpha(rev if name=="rev" else s, alphabets["bitcoin"])
    pr=is_mostly_printable(raw)
    print(name, len(raw), pr, raw)

Python

python
import codecs
pt=b58decode_alpha(s[::-1], alphabets["bitcoin"]).decode()
pt, codecs.decode(pt,'rot_13')


('synt{flfgrz_u4px3q_fhpp3ff}', 'flag{system_h4ck3d_succ3ss}')

思考

import codecs pt=b58decode_alpha(s[::-1], alphabets["bitcoin"]).decode() pt, codecs.decode(pt,'rot_13')

('synt{flfgrz_u4px3q_fhpp3ff}', 'flag{system_h4ck3d_succ3ss}')

Thought for 3m 0s

回答

这个串很像 Base58（Bitcoin 字母表）：它只用到了 1-9A-HJ-NP-Za-km-z 这些字符，刻意避开了 0/O/I/l。

但它还做了两步“套娃”：

解密过程

  1. 先把字符串反转
text
G c C d i S m T 8 k b R Z 3 E C F 6 x P 6 A 3 1 S F p c 3 P 8 d c C z Z J
=> "GcCd...ZJ"  (完整反转见下面代码)
  1. 对反转后的字符串做 Base58 解码
    得到可读文本：
synt{flfgrz_u4px3q_fhpp3ff}
  1. 再做 ROT13
flag{system_h4ck3d_succ3ss}

✅ 最终答案 / flag： flag{system_h4ck3d_succ3ss}

复现用 Python（可直接跑）

python
s = "JZzCcd8P3cpFS13A6Px6FCE3ZRbk8TmSidCcG"

alphabet = "123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz"
idx = {c:i for i,c in enumerate(alphabet)}

def b58decode(x: str) -> bytes:
    n = 0
    for ch in x:
        n = n * 58 + idx[ch]
    b = n.to_bytes((n.bit_length()+7)//8, "big") if n else b""
    # leading '1' => leading zero bytes
    pad = 0
    for ch in x:
        if ch == "1": pad += 1
        else: break
    return b"\x00"*pad + b

rev = s[::-1]
mid = b58decode(rev).decode()
print("base58(reversed) =", mid)

# rot13
import codecs
print("rot13 =", codecs.decode(mid, "rot_13"))

如果你还有同题的其它样本（或者题目提示/描述），也可以贴出来，我可以帮你判断它还可能混了哪些常见套路（倒序、ROT、异或、压缩等）。

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