♎♏♋♏♓♏♋♎♋♓♋♏♋♏♏♓♏♋♏♓♋♓♏♋♋♋♎♎♋♎♏♏♏♋♎♋♋♋♏♓♏♏♎♏♋♏♏♎♏♏...

Creado el: 26 de mayo de 2026

Respondido usando GPT-5.5 Thinking por Chat01

Pregunta

♎♏♋♏♓♏♋♎♋♓♋♏♋♏♏♓♏♋♏♓♋♓♏♋♋♋♎♎♋♎♏♏♏♋♎♋♋♋♏♓♏♏♎♏♋♏♏♎♏♏♎♓♋♓♏♋♏♎♓♎♋♎♏♋♏♓♏♋♋♋♏♏♋♋♓♏♋♏♏♓♋♏♓♓♋♓♏♋♏♓♎♎♋♎♏♏♋♋♋♋♋♋♏♓♋♋♏♏♋♏♏♓♋♏♏♓♋♓♏♋♋♎♏♎♋♎♏♏♋♋♋♋♋♋♏♓♋♎♋♏♋♏♏♎♋♓♎♓♋♓♏♋♏♋♋♎♋♎♏♋♋♎♏♋♋♋♏♓♋♏♓♏♋♏♏♎♋♏♓♓♋♓♏♋♋♓♋♎♋♎♏♋♋♎♓♋♋♋♏♓♏♏♓♏♋♏♏♓♋♏♓♓♋♓♏♎♏♎♋♎♋♎♏♏♏♏♓♋♋♋♏♏♓♎♓♏♋♏♏♓♏♋♋♓♋♓♏♎♏♋♓♎♋♎♏♏♋♋♎♋♋♋♏♓♋♎♎♏♋♏♏♓♏♏♎♓♋♓♏♎♎♎♎♎♋♎♏♏♏♋♓♋♋♋♏♏♎♏♋♏♋♏♏♎♏♓♎♓♋♓♏♋♋♏♋♎♋♎♏♋♏♓♋♋♋♋♏♓♏♏♓♏♋♏♏♓♏♎♋♓♋♓♏♋♋♏♏♎♋♎♏♋♓♋♎♋♋♋♏♏♓♏♋♏♋♏♏♎♋♋♏♓♋♓♏♎♎♎♎♎♋♎♏♏♏♏♏♋♋♋♏♏♓♏♋♏♋♏♏♋♋♏♎♓♋♓♏♓♓♏♏♋♓♋♋♓♎♎♎♏♏♏♋♎♎♓♏♓♎♎♏♏♓♓♋♓♋♓♏♎♏♏♏♎♋♎♏♋♏♏♓♋♋♋♏♓♏♓♏♏♋♏♏♎♋♋♋♓♋♓♏♎♏♋♏♎♋♎♏♎♓♓♏♏♓♏♋♎♋♓♋♏♋♏♏♓♏♋♏♓♋♓♏♋♋♋♎♎♋♎♏♏♏♋♎♋♋♋♏♓♏♏♎♏♋♏♏♎♏♏♎♓♋♓♏♋♏♎♓♎♋♎♏♋♏♓♏♋♋♋♏♏♋♋♓♏♋♏♏♓♋♏♓♓♋♓♏♋♏♓♎♎♋♎♏♏♋♋♋♋♋♋♏♓♋♋♏♏♋♏♏♓♋♏♏♓♋♓♏♋♋♎♏♎♋♎♏♏♋♋♋♋♋♋♏♓♋♎♋♏♋♏♏♎♋♓♎♓♋♓♏♋♏♋♋♎♋♎♏♋♋♎♏♋♋♋♏♓♋♏♓♏♋♏♏♎♋♏♓♓♋♓♏♋♋♓♋♎♋♎♏♋♋♎♓♋♋♋♏♓♏♏♓♏♋♏♏♓♋♏♓♓♋♓♏♎♏♎♋♎♋♎♏♋♓♎♓♋♋♋♏♏♏♏♏♏♋♏♏♓♎♎♏♓♋♓♏♎♎♋♓♎♋♎♏♏♋♋♓♋♋♋♏♓♋♏♎♏♋♏♏♎♏♓♋♓♋♓♏♎♎♓♎♎♋♎♏♋♎♋♎♋♋♋♏♏♎♓♓♏♋♏♏♓♓♎♋♓♋♓♏♋♏♎♏♎♋♎♏♏♋♓♎♋♋♋♏♏♓♏♋♏♋♏♏♓♏♓♓♓♋♓♏♋♋♋♎♎♋♎♏♋♏♏♏♋♋♋♏♏♎♎♏♏♋♏♏♓♓♋♏♓♋♓♏♋♏♋♋♎♋♎♏♏♏♓♋♋♋♋♏♏♓♏♋♏♋♏♏♋♋♏♎♓♋♓♏♓♓♏♏♋♓♋♋♓♎♎♎♏♏♏♋♎♎♓♏♓♎♎♏♏♓♓♋♓♋♓♏♎♏♏♏♎♋♎♏♋♏♓♋♋♋♋♏♏♏♓♎♏♋♏♏♎♋♎♓♓♋♓♏♋♋♎♎♓
解码一下上述网盘链接

Python

python
s = """♎♏♋♏♓♏♋♎♋♓♋♏♋♏♏♓♏♋♏♓♋♓♏♋♋♋♎♎♋♎♏♏♏♋♎♋♋♋♏♓♏♏♎♏♋♏♏♎♏♏♎♓♋♓♏♋♏♎♓♎♋♎♏♋♏♓♏♋♋♋♏♏♋♋♓♏♋♏♏♓♋♏♓♓♋♓♏♋♏♓♎♎♋♎♏♏♋♋♋♋♋♋♏♓♋♋♏♏♋♏♏♓♋♏♏♓♋♓♏♋♋♎♏♎♋♎♏♏♋♋♋♋♋♋♏♓♋♎♋♏♋♏♏♎♋♓♎♓♋♓♏♋♏♋♋♎♋♎♏♋♋♎♏♋♋♋♏♓♋♏♓♏♋♏♏♎♋♏♓♓♋♓♏♋♋♓♋♎♋♎♏♋♋♎♓♋♋♋♏♓♏♏♓♏♋♏♏♓♋♏♓♓♋♓♏♎♏♎♋♎♋♎♏♏♏♏♓♋♋♋♏♏♓♎♓♏♋♏♏♓♏♋♋♓♋♓♏♎♏♋♓♎♋♎♏♏♋♋♎♋♋♋♏♓♋♎♎♏♋♏♏♓♏♏♎♓♋♓♏♎♎♎♎♎♋♎♏♏♏♋♓♋♋♋♏♏♎♏♋♏♋♏♏♎♏♓♎♓♋♓♏♋♋♏♋♎♋♎♏♋♏♓♋♋♋♋♏♓♏♏♓♏♋♏♏♓♏♎♋♓♋♓♏♋♋♏♏♎♋♎♏♋♓♋♎♋♋♋♏♏♓♏♋♏♋♏♏♎♋♋♏♓♋♓♏♎♎♎♎♎♋♎♏♏♏♏♏♋♋♋♏♏♓♏♋♏♋♏♏♋♋♏♎♓♋♓♏♓♓♏♏♋♓♋♋♓♎♎♎♏♏♏♋♎♎♓♏♓♎♎♏♏♓♓♋♓♋♓♏♎♏♏♏♎♋♎♏♋♏♏♓♋♋♋♏♓♏♓♏♏♋♏♏♎♋♋♋♓♋♓♏♎♏♋♏♎♋♎♏♎♓♓♏♏♓♏♋♎♋♓♋♏♋♏♏♓♏♋♏♓♋♓♏♋♋♋♎♎♋♎♏♏♏♋♎♋♋♋♏♓♏♏♎♏♋♏♏♎♏♏♎♓♋♓♏♋♏♎♓♎♋♎♏♋♏♓♏♋♋♋♏♏♋♋♓♏♋♏♏♓♋♏♓♓♋♓♏♋♏♓♎♎♋♎♏♏♋♋♋♋♋♋♏♓♋♋♏♏♋♏♏♓♋♏♏♓♋♓♏♋♋♎♏♎♋♎♏♏♋♋♋♋♋♋♏♓♋♎♋♏♋♏♏♎♋♓♎♓♋♓♏♋♏♋♋♎♋♎♏♋♋♎♏♋♋♋♏♓♋♏♓♏♋♏♏♎♋♏♓♓♋♓♏♋♋♓♋♎♋♎♏♋♋♎♓♋♋♋♏♓♏♏♓♏♋♏♏♓♋♏♓♓♋♓♏♎♏♎♋♎♋♎♏♋♓♎♓♋♋♋♏♏♏♏♏♏♋♏♏♓♎♎♏♓♋♓♏♎♎♋♓♎♋♎♏♏♋♋♓♋♋♋♏♓♋♏♎♏♋♏♏♎♏♓♋♓♋♓♏♎♎♓♎♎♋♎♏♋♎♋♎♋♋♋♏♏♎♓♓♏♋♏♏♓♓♎♋♓♋♓♏♋♏♎♏♎♋♎♏♏♋♓♎♋♋♋♏♏♓♏♋♏♋♏♏♓♏♓♓♓♋♓♏♋♋♋♎♎♋♎♏♋♏♏♏♋♋♋♏♏♎♎♏♏♋♏♏♓♓♋♏♓♋♓♏♋♏♋♋♎♋♎♏♏♏♓♋♋♋♋♏♏♓♏♋♏♋♏♏♋♋♏♎♓♋♓♏♓♓♏♏♋♓♋♋♓♎♎♎♏♏♏♋♎♎♓♏♓♎♎♏♏♓♓♋♓♋♓♏♎♏♏♏♎♋♎♏♋♏♓♋♋♋♋♏♏♏♓♎♏♋♏♏♎♋♎♓♓♋♓♏♋♋♎♎♓"""
len(s), {c:s.count(c) for c in set(s)}


(956, {'♏': 318, '♋': 316, '♎': 154, '♓': 168})

Python

python
import itertools, string, codecs, re, math, collections, textwrap, binascii
symbols = sorted(set(s))
patterns = ['00','01','10','11']
def decode_map(mapping, bit_order='normal'):
    bits = ''.join(mapping[ch] for ch in s)
    if bit_order == 'revpair':
        bits = ''.join(mapping[ch][::-1] for ch in s)
    out = bytes(int(bits[i:i+8],2) for i in range(0,len(bits),8))
    return out

candidates=[]
for perm in itertools.permutations(patterns):
    mp=dict(zip(symbols,perm))
    b=decode_map(mp)
    try:
        txt=b.decode('utf-8')
    except:
        txt=b.decode('latin1', errors='replace')
    printable=sum(ch in string.printable for ch in txt)/len(txt)
    ascii_print=sum(32<=x<127 or x in (9,10,13) for x in b)/len(b)
    # url features
    score=ascii_print + 2*('http' in txt or 'pan' in txt or 'baidu' in txt)
    candidates.append((score,ascii_print,printable,mp,txt[:200],b[:20]))
for item in sorted(candidates, reverse=True)[:10]:
    print(item[0], item[1], item[3], repr(item[4]), item[5])

Python

python
patterns=['00','01','10','11']
symbols=list(set(s))
def bytes_from_map(mp, mode):
    chunks=[mp[ch] for ch in s]
    if mode=='normal':
        bits=''.join(chunks)
    elif mode=='rev_pairs_each_byte':
        # reverse order of 4 2-bit pairs per byte
        bits=''.join(''.join(chunks[i:i+4][::-1]) for i in range(0,len(chunks),4))
    elif mode=='rev_bits_each_pair':
        bits=''.join(c[::-1] for c in chunks)
    elif mode=='rev_bits_byte':
        bits=''.join((''.join(chunks[i:i+4]))[::-1] for i in range(0,len(chunks),4))
    elif mode=='pair_rev_and_order_rev':
        bits=''.join(''.join(c[::-1] for c in chunks[i:i+4][::-1]) for i in range(0,len(chunks),4))
    elif mode=='bytes_reversed':
        bits=''.join(chunks)
        # reverse bytes order
        bs=[bits[i:i+8] for i in range(0,len(bits),8)]
        bits=''.join(bs[::-1])
    return bytes(int(bits[i:i+8],2) for i in range(0,len(bits),8))
def score_text(b):
    txt=b.decode('latin1','replace')
    ascii_print=sum(32<=x<127 or x in (9,10,13) for x in b)/len(b)
    common=sum(chr(x) in (string.ascii_letters+string.digits+":/.?&_=#-% \n\r\t") for x in b)/len(b)
    keywords=sum(k in txt.lower() for k in ['http','pan','baidu','lanzou','123pan','aliyun','quark','提取','密码','code','pwd','s/'])
    return ascii_print+common+5*keywords
res=[]
for perm in itertools.permutations(patterns):
    mp=dict(zip(sorted(set(s)),perm))
    for mode in ['normal','rev_pairs_each_byte','rev_bits_each_pair','rev_bits_byte','pair_rev_and_order_rev','bytes_reversed']:
        b=bytes_from_map(mp, mode)
        txt=b.decode('latin1','replace')
        res.append((score_text(b), sum(32<=x<127 or x in (9,10,13) for x in b)/len(b), mp, mode, repr(txt[:300]), b[:10]))
for r in sorted(res, reverse=True)[:20]:
    print('score',r[0],'print',r[1],'mode',r[3], 'mp',r[2], 'txt',r[4])

Python

python
import itertools, string, re, math, collections, binascii, textwrap
patterns=['00','01','10','11']
modes=['normal','rev_pairs_each_byte','rev_bits_each_pair','rev_bits_byte','pair_rev_and_order_rev','bytes_reversed']
def bytes_from_map(mp, mode):
    chunks=[mp[ch] for ch in s]
    if mode=='normal':
        bits=''.join(chunks)
    elif mode=='rev_pairs_each_byte':
        bits=''.join(''.join(chunks[i:i+4][::-1]) for i in range(0,len(chunks),4))
    elif mode=='rev_bits_each_pair':
        bits=''.join(c[::-1] for c in chunks)
    elif mode=='rev_bits_byte':
        bits=''.join((''.join(chunks[i:i+4]))[::-1] for i in range(0,len(chunks),4))
    elif

Python

python
# choose likely mapping by symbol frequencies: ♏=00, ♋=01, ♎=10/11, ♓=11/10
mp={'♏':'00','♋':'01','♎':'11','♓':'10'}  # candidate top
b=bytes_from_map(mp,'normal')
for k in range(256):
    x=bytes([c^k for c in b])
    txt=x.decode('latin1','ignore')
    if any(w in txt.lower() for w in ['http','pan','baidu','lanzou','123pan','aliyun','quark','www']):
        print('xor',k,repr(txt[:500]))
# score print for xor
best=[]
for k in range(256):
    x=bytes([c^k for c in b])
    pr=sum(32<=c<127 or c in (9,10,13) for c in x)/len(x)
    best.append((pr,k,x[:80]))
print(sorted(best, reverse=True)[:10])
# Try add/sub constant
for op in ['add','sub']:
    hits=[]
    for k in range(256):
        if op=='add': x=bytes((c+k)%256 for c in b)
        else: x=bytes((c-k)%256 for c in b)
        txt=x.decode('latin1','ignore')
        if any(w in txt.lower() for w in ['http','pan','baidu','lanzou','123pan','aliyun','quark','www']):
            print(op,k,repr(txt[:500]))
            break
        pr=sum(32<=c<127 or c in (9,10,13) for c in x)/len(x)
        hits.append((pr,k))
    print(op, sorted(hits, reverse=True)[:5])

Python

python
hexmap={'♋':'B','♎':'E','♏':'F','♓':'3'}
hexstr=''.join(hexmap[ch] for ch in s)
len(hexstr), hexstr[:100]


(956,
 'EFBF3FBEB3BFBFF3FBF3B3FBBBEEBEFFFBEBBBF3FFEFBFFEFFE3B3FBFE3EBEFBF3FBBBFFBB3FBFF3BF33B3FBF3EEBEFFBBBB')

Python

python
hb=bytes.fromhex(hexstr)
print(len(hb), hb[:50])
for enc in ['utf-8','gbk','latin1']:
    try:
        print(enc, hb.decode(enc)[:300])
    except Exception as e:
        print(enc, e)
# try see printable ratio and search
print(sum(32<=x<127 or x in (9,10,13) for x in hb)/len(hb))
print(hb[:100].hex())
print(set(hb))

Pensamiento

undefined

Python

python
# Analyze repeats at symbol level
n=len(s)
# find longest repeated substring positions using brute force suffix? n small
best=[]
for L in range(200,20,-1):
    seen={}
    found=[]
    for i in range(n-L+1):
        sub=s[i:i+L]
        if sub in seen:
            found.append((seen[sub],i,L))
            break
        seen[sub]=i
    if found:
        print("longest >=", L, found[0])
        print(s[found[0][0]:found[0][0]+50], s[found[0][1]:found[0][1]+50])
        break
# maybe first half repeats:
for offset in range(1,n):
    # LCP first vs offset
    L=0
    while offset+L<n and s[L]==s[offset+L]:
        L+=1
    if L>50:
        print('prefix repeat at',offset,'len',L)
# occurrences

Python

python
for off in [480,483]:
    L=0
    while off+L<n and s[3+L]==s[off+L]:
        L+=1
    print('from3 vs',off,L)
print(s[:10], s[480:490], s[483:493])
# check blocks:
A=s[3:483]
B=s[483:963]
print(len(A),len(B),A==B, sum(a==b for a,b in zip(A,B)), len(B))
# display difference positions
diff=[i for i,(a,b) in enumerate(zip(A,B)) if a!=b]
print(len(diff), diff[:20])
print(A[:20], B[:20])

Python

python
def decode_segment(seq, mp, offset=0, mode='normal'):
    seq=seq[offset:]
    L=(len(seq)//4)*4
    seq=seq[:L]
    chunks=[mp[ch] for ch in seq]
    if mode=='normal':
        bits=''.join(chunks)
    elif mode=='rev_pairs_each_byte':
        bits=''.join(''.join(chunks[i:i+4][::-1]) for i in range(0,len(chunks),4))
    elif mode=='rev_bits_byte':
        bits=''.join((''.join(chunks[i:i+4]))[::-1] for i in range(0,len(chunks),4))
    elif mode=='rev_bits_each_pair':
        bits=''.join(c[::-1] for c in chunks)
    elif mode=='pair_rev_and_order_rev':
        bits=''.join(''.join(c[::-1] for c in chunks[i:i+4][::-1]) for i in range(0,len(chunks),4))
    return bytes(int(bits[i:i+8],2) for i in range(0,len(bits),8))
def candidates_for_seq(seq, top=10):
    ress=[]
    for perm in itertools.permutations(patterns):
        mp=dict(zip(sorted(set(s)),perm))
        for offset in range(4):
            for mode in modes[:-1]:
                b=decode_segment(seq, mp, offset, mode)
                if not b: continue
                text=b.decode('latin1','replace')
                # try raw and xor constants
                for trans in ['raw'] + [('xor', k) for k in range(256)]:
                    if trans=='raw':
                        x=b; k=None
                    else:
                        x=bytes(c^trans[1] for c in b); k=trans[1]
                    tt=x.decode('latin1','replace')
                    score=0
                    score += 10*sum(w in tt.lower() for w in ['http','pan','baidu','lanzou','123pan','aliyun','quark','www','pwd','提取'])
                    score += sum(32<=c<127 or c in (9,10,13) for c in x)/len(x)
                    if 'http' in tt.lower(): score +=5
                    if score>5:
                        ress.append((score, mp, offset, mode, trans, repr(tt[:200])))
    return sorted(ress, key=lambda x:x[0], reverse=True)[:top]
for name,seq in [('whole',s),('first480',s[:480]),('second476',s[480:]),('A3_483',s[3:483]),('B483',s[483:])]:
    res=candidates_for_seq(seq,5)
    print(name, len(seq), res[:2])

Python

python
import itertools, string, math, re
pair_orders=list(itertools.permutations(range(4)))
swap_masks=range(16)
def decode_custom(seq, mp, offset=0, order=(0,1,2,3), mask=0):
    seq=seq[offset:]
    L=(len(seq)//4)*4
    seq=seq[:L]
    outs=[]
    for i in range(0,L,4):
        pairs=[mp[ch] for ch in seq[i:i+4]]
        bits=''
        for j,idx in enumerate(order):
            p=pairs[idx]
            if (mask>>j)&1:
                p=p[::-1]
            bits += p
        outs.append(int(bits,2))
    return bytes(outs)

def text_score(x):
    txt=x.decode('latin1','replace')
    lower=txt.lower()
    printable=sum(32<=c<127 or c in (9,10,13) for c in x)/len(x)
    # penalize high

Python

python
def codons(seq, offset=0):
    seq=seq[offset:]
    return [seq[i:i+3] for i in range(0, len(seq)-len(seq)%3,3)]
def compatible(pattern_chars, codons):
    # for equal chars codon equal, for codon equal chars equal
    if len(pattern_chars)>len(codons): return False
    f={}
    g={}
    for ch,co in zip(pattern_chars,codons):
        if ch in f and f[ch]!=co: return False
        if co in g and g[co]!=ch: return False
        f[ch]=co; g[co]=ch
    return True

import base64
prefixes=[
    "https://pan.baidu.com/s/",
    "https://pan.baidu.com/share/init?surl=",
    "https://www.aliyundrive.com/s/",
    "https://pan.quark.cn/s/",
    "https://www.123pan.com/s/",
    "https://www.lanzoui.com/",
    "链接：https://pan.baidu.com/s/",
    "链接: https://pan.baidu.com/s/",
    "pan.baidu.com/s/",
    "https://"
]
for p in prefixes:
    b64=base64.b64encode(p.encode()).decode().rstrip('=')
    print("prefix",p, b64, len(b64))
    for off in range(3):
        cs=codons(s,off)
        # test every position perhaps prefix could start at codon offset
        hits=[]
        for pos in range(0, min(30, len(cs)-len(b64)+1)):
            if compatible(b64, cs[pos:pos+len(b64)]):
                hits.append(pos)
        if hits:
            print(" off",off,"hits",hits[:10])

Python

python
seqs={'orig':s,'rev':s[::-1]}
for name,seq in seqs.items():
    res=[]
    for perm in itertools.permutations(patterns):
        mp=dict(zip(sorted(set(s)),perm))
        for offset in range(4):
            for mode in modes[:-1]:
                b=decode_segment(seq,mp,offset,mode)
                for trans in ['raw','not']:
                    x=b if trans=='raw' else bytes(255-c for c in b)
                    txt=x.decode('latin1','replace')
                    sc=text_score(x)
                    if sc>2 or any(w in txt.lower() for w in ['http','baidu','pan']):
                        res.append((sc,name,offset,mode,trans,mp,txt[:200]))
    print('SEQ',name,'res',len(res))
    for r in sorted(res,key=lambda

Python

python
# DNA codon table RNA? use DNA codons
codon_table = {
'TTT':'F','TTC':'F','TTA':'L','TTG':'L','TCT':'S','TCC':'S','TCA':'S','TCG':'S',
'TAT':'Y','TAC':'Y','TAA':'*','TAG':'*','TGT':'C','TGC':'C','TGA':'*','TGG':'W',
'CTT':'L','CTC':'L','CTA':'L','CTG':'L','CCT':'P','CCC':'P','CCA':'P','CCG':'P',
'CAT':'H','CAC':'H','CAA':'Q','CAG':'Q','CGT':'R','CGC':'R','CGA':'R','CGG':'R',
'ATT':'I','ATC':'I','ATA':'I','ATG':'M','ACT':'T','ACC':'T','ACA':'T','ACG':'T',
'AAT':'N','AAC':'N','AAA':'K','AAG':'K','AGT':'S','AGC':'S','AGA':'R','AGG':'R',
'GTT':'V','GTC':'V','GTA':'V','GTG':'V','GCT':'A','GCC':'A','GCA':'A','GCG':'A',
'GAT':'D','GAC':'D','GAA':'E','GAG':'E','GGT':'G','GGC':'G','G

Pensamiento

undefined

Python

python
def beast_decode(text, beast):
    if not (text.startswith(beast[3]+beast[1]+beast[0]) and text.endswith(beast[2])):
        return None, 'format'
    body=text[3:-1]
    if len(body)%2: return None, 'odd'
    code=''
    n=0
    for i in range(0,len(body),2):
        try:
            pos1=beast.index(body[i]); pos2=beast.index(body[i+1])
        except ValueError:
            return None, 'badchar'
        k=(pos1*4+pos2 - n) % 16
        code += f"{k:x}"
        n += 1
    # hex chunks 4
    if len(code)%4: print("hex len not mult4",len(code))
    out=''
    vals=[]
    for i in range(0,len(code),4):
        hx=code[i:i+4]
        if len(hx)<4: break
        vals.append(int(hx,16))
        out+=chr(int(hx,16))
    return out, code

be

(119,
 '接:https://pan.baidu.com/s/1wK97dl0TsQti9s5jDAjTvA\xa0\n提取码:7vi6\n接:https://pan.baidu.com/s/1O2VWcdqPTWErlA3h6ZJuyA\xa0\n提取码:98gd',
 '63a5003a00680074007400700073003a002f002f00700061006e002e00620061006900640075002e0063006f006d002f0073',
 476)

Python

python
for perm in itertools.permutations(sorted(set(s))):
    out,h=beast_decode(s, list(perm))
    if out and any(substr in out for substr in ['http','pan','提取','百度','链接','接']):
        print(perm, repr(out[:200]))

Pensamiento

Decoding final answer with citation considerations

For the final answer in Chinese, I need to cite the algorithm properly, especially since I found information from the web about the "Beast translator." I’ll remember to cite sources like CSDN or 博客园 where the algorithm was discussed. I will also avoid raw URLs unless coding-related, as per guidelines.

Thought for 7m 37s

Respuesta

解出来是 两组百度网盘链接。明文前缀显示为“接:”，我按语义补成“链接:”展示：

text
链接: https://pan.baidu.com/s/1wK97dl0TsQti9s5jDAjTvA
提取码: 7vi6

链接: https://pan.baidu.com/s/1O2VWcdqPTWErlA3h6ZJuyA
提取码: 98gd

这是“兽音译者/兽语”一类的四字符表编码；你这段用的字符表可推为 [♋, ♏, ♓, ♎]，头部是 ♎♏♋，尾部是 ♓。该类算法通常把 Unicode 十六进制位加递增偏移后，再用 4 个自定义字符映射成两位密文字符。(blog.csdn.net)

Pregunta

Python

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Python

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Python

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