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本文最后更新于 2023-02-01，文中内容可能已过时。
misc
SimpleFlow
一个流量包 又是熟悉的蚁剑流量
在最后有一个压缩包 提出来发现有密码
在50流找到
Y2QgIi9Vc2Vycy9jaGFuZy9TaXRlcy90ZXN0Ijt6aXAgLVAgUGFTc1ppUFdvckQgZmxhZy56aXAgLi4vZmxhZy50eHQ7ZWNobyBbU107cHdkO2VjaG8gW0Vd
PaSsZiPWorD
crypto
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import random
import hashlib
flag = 'xxxxxxxxxxxxxxxxxxxx'
key = random.randint(1,10)
for i in range(len(flag)):
crypto += chr(ord(flag[i])^key)
m = crypto的ascii十六进制
e = random.randint(1,100)
print(hashlib.md5(e))
p = 64310413306776406422334034047152581900365687374336418863191177338901198608319
q = xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
n = p*q
c = pow(m,e,n)
print(n)
print(c)
#37693cfc748049e45d87b8c7d8b9aacd
#4197356622576696564490569060686240088884187113566430134461945130770906825187894394672841467350797015940721560434743086405821584185286177962353341322088523
#3298176862697175389935722420143867000970906723110625484802850810634814647827572034913391972640399446415991848730984820839735665233943600223288991148186397
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逆推m
然后我被卡在crypto到ASCII的转换了 没想到libnum
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import libnum
m = 2976168736142380455841784134407431434784057911773423743751382131043957
flag = libnum.n2s(m)
flag = 'ndios_;9kgE;WK8e;W?gWn<\\;k|nu'
for i in range(1,11):
for j in range(len(flag)):
print(chr(ord(flag[j]) ^ i),end='')
print('')
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flag{W31coM3_C0m3_7o_f4T3ctf}
shenghuo2
