2020网鼎杯 白虎组crypto：bs4

这道密码学题..我...脑洞不出来

下载得到“密文.txt”内容如下，根据提示bs4是base64加密

密文:uLdAuO8duojAFLEKjIgdpfGeZoELjJp9kSieuIsAjJ/LpSXDuCGduouz

泄露的密文:pTjMwJ9WiQHfvC+eFCFKTBpWQtmgjopgqtmPjfKfjSmdFLpeFf/Aj2ud3tN7u2+enC9+nLN8kgdWo29ZnCrOFCDdFCrOFoF=
泄露的明文:ashlkj!@sj1223%^&*Sd4564sd879s5d12f231a46qwjkd12J;DJjl;LjL;KJ8729128713

所以将泄露的明文进行base64加密得到正经密文：YXNobGtqIUBzajEyMjMlXiYqU2Q0NTY0c2Q4NzlzNWQxMmYyMzFhNDZxd2prZDEySjtESmpsO0xqTDtLSjg3MjkxMjg3MTM=

位数和泄露密文一样，想到了凯撒加密，但是凯撒不行，又想到了维基利亚加密，所以这里把泄露密文和正经密文进行比较得到一个字典

a='pTjMwJ9WiQHfvC+eFCFKTBpWQtmgjopgqtmPjfKfjSmdFLpeFf/Aj2ud3tN7u2+enC9+nLN8kgdWo29ZnCrOFCDdFCrOFoF'
b='YXNobGtqIUBzajEyMjMlXiYqU2Q0NTY0c2Q4NzlzNWQxMmYyMzFhNDZxd2prZDEySjtESmpsO0xqTDtLSjg3MjkxMjg3MTM'
s=''
a_list =[]
b_list= []
for i in a:
          a_list.append(i)
for i in b:
          b_list.append(i)
for i in range(0,len(a)):
          print(b_list[i],':',ord(b_list[i])-ord(a_list[i]),end='\t\t')
          #print('\'',a_list[i],'\'',':',ord(a_list[i])-ord(b_list[i]),end=',')

用这个字典去解泄密密文是可以解出泄露明文的，换成需要求的密文：uLdAuO8duojAFLEKjIgdpfGeZoELjJp9kSieuIsAjJ/LpSXDuCGduouzF

悲剧来了，有六个字母不在字典里，想来想去，也没找到规律可以推出这六个字母对应的值，所以暴力破解吧

import base64
d={'p':23,'T':-4,'j':28,'M':-34,'w':21,'J':3,'9':-59,'W':-26,'i':32,'Q':-4,'H':6,'f':-20,'v':21,'C':-39,'+':-26,'e':-20,'F':-7,'C':-39,'F':-7,'K':-33,'T':-4,'B':-39,'p':23,'W':-26,'Q':-4,'t':66,'m':28,'g':55,'j':28,'o':27,'p':23,'g':55,'q':14,'t':66,'m':28,'P':28,'j':28,'f':-20,'K':-33,'f':-20,'j':28,'S':-4,'m':28,'d':-20,'F':-7,'L':-33,'p':23,'e':-20,'F':-7,'f':-20,'/':-23,'A':-39,'j':28,'2':-18,'u':27,'d':-20,'3':-49,'t':66,'N':-34,'7':-59,'u':27,'2':-18,'+':-26,'e':-20,'n':27,'C':-39,'9':-59,'+':-26,'n':27,'L':-33,'N':-34,'8':-59,'k':28,'g':55,'d':-20,'W':-26,'o':27,'2':-18,'9':-59,'Z':14,'n':27,'C':-39,'r':11,'O':28,'F':-7,'C':-39,'D':-39,'d':-20,'F':-7,'C':-39,'r':11,'O':28,'F':-7,'o':27,'F':-7}

miwen="uLdAuO8duojAFLEKjIgdpfGeZoELjJp9kSieuIsAjJ/LpSXDuCGduouzF"

s=''
i=0

for j in miwen:
          if j=='E' or j=='I' or j =='G' or j=='s' or j=='X' or j=='z':
                    s += j
          else:
                    tmp = ord(j)
                    #print(d[j],end='\t')
                    #print(chr(tmp-d[j]),end='')
                    s += chr(tmp-d[j])
                    
          #print('ss',j)

out=[]

flag='flag'
s_tmp = s
for a in range(65,123):
          s_tmp.replace('E',chr(a))
          out.append(s_tmp)
          s_tmp = s
          for b in range(65,123):
                    s_tmp.replace('I',chr(b))
                    out.append(s_tmp)
                    s_tmp = s
                    for c in range(65,123):
                              s_tmp.replace('G',chr(c))
                              out.append(s_tmp)
                              s_tmp = s
                              for d in range(65,123):
                                        s_tmp.replace('s',chr(d))
                                        out.append(s_tmp)
                                        s_tmp = s
                                        
                                        for e in range(65,123):
                                                  s_tmp.replace('X',chr(e))
                                                  out.append(s_tmp)
                                                  s_tmp = s
                                                  for f in range(65,123):
                                                            s_tmp.replace('z',chr(f))
                                                            out.append(s_tmp)
                                                            s_tmp = s
                                        
          
                              
for i in out:
          i +='='
          try:
                    code = base64.b64decode(i)
                    if flag in coed:
                              print(code)
          except:
                    continue
                    
                              
                              

out[i]=s
flag='flag'
for i in out:
          if( flag in base64.b64decode(i)):
                    print(i)

好的，没有flag因为内存错误啦~~数字太多超出内存，没办法，等writeup吧

wp出了，说是靠猜，不知道他们怎么猜的，我是这么猜的：

原密文经过字典翻译，有五个字母翻译不出来以?代替

ZmxhZ3sxZTNhMm?lN?0xYz?yLT?mNGYtOWIyZ??hNGFmYW?kZj?xZTZ?M

对flag字符串进行base64加密，发现和密文开头一样

 

尝试以？作为分隔符把每一段密文解出来

第一段：flag{1e3a2

第二段：\x96

第三段：\xd3\x163

第四段：\x98\xd1\x98\xb4\xe5\x88\xc9

第五段：\x84\xd1\x85\x99\x85

第六段：...

从第一段开始爆破下一个flag字符

s='flag{1e3a2'
tmp=s
for i in range(97,128):
          s += chr(i)
          out=base64.b64encode(s.encode('utf-8'))
          str1=''
          for j in range(0,len(out)-1):
                    str1 += chr(out[j])
          str1 += 'lN=='
          try:
                    print(base64.b64decode(str1))
          except:
                    continue
          s=tmp

得到如下结果，当第一段后两个字符为de时，符合密文？之后是l

 假设第一段后面是de，继续往后爆破，没有密文l之后是N的结果

s='flag{1e3a2de'
tmp=s
for i in range(97,128):
          s += chr(i)
          out=base64.b64encode(s.encode('utf-8'))
          print(out)
          str1=''
          for j in range(0,len(out)-1):
                    str1 += chr(out[j])
          str1 += '0xYz=='
          try:
                    print(base64.b64decode(str1))
          except:
                    continue
          s=tmp

换爆破范围为数字，4，5，6，7符合l之后是N

 因为flag的形式是uuid：8-4-4-12，数字之后就是短横线-，所以从第一个数字带入flag查看base64加密后的值和题中给的那个数字一样

s='flag{1e3a2de'
tmp=s
for i in range(52,56):
          s +=chr(i)
          s += '-'
          print(s)
          print(base64.b64encode(s.encode('utf-8')))
          s=tmp

都是0，没有更往后的数字来判断

 那只有在4，5，6，7的base64加密上接上题目中的密文

s='flag{1e3a2de'
tmp=s
for i in range(52,56):
          s += chr(i)
          s += '-'
          out=base64.b64encode(s.encode('utf-8'))
          print(out)
          str1=''
          for j in range(0,len(out)-1):
                    str1 += chr(out[j])
          str1 += 'xYz=='
          try:
                    print(base64.b64decode(str1))
          except:
                    continue
          s=tmp

都能解出来，那么说明目前这四个数字都有可能，排除不了，继续往后猜，套路和前面一样

 最后猜出来的结果是flag{1e3a2de5-1c02-4f4f-9b2e-a4afabdf01e6}