Reference:L1B0

Re

RSA256

春秋欢乐赛原题。。flag都不变的
给了三个加密文件和公钥证书public.key，可以使用openssl进行处理
$openssl rsa -pubin -text -modulus -in ./public.key
Public-Key: (256 bit)
Modulus:
    00:d9:9e:95:22:96:a6:d9:60:df:c2:50:4a:ba:54:
    5b:94:42:d6:0a:7b:9e:93:0a:ff:45:1c:78:ec:55:
    d5:55:eb
Exponent: 65537 (0x10001)
Modulus=D99E952296A6D960DFC2504ABA545B9442D60A7B9E930AFF451C78EC55D555EB
writing RSA key
-----BEGIN PUBLIC KEY-----
MDwwDQYJKoZIhvcNAQEBBQADKwAwKAIhANmelSKWptlg38JQSrpUW5RC1gp7npMK
/0UceOxV1VXrAgMBAAE=
-----END PUBLIC KEY-----
rsa参数中， Exponent=65537 即为 e 值，Modulus即为n
使用python解密即可
#!/usr/bin/env python
#coding:utf-8   
import gmpy2   
import rsa   
p = 302825536744096741518546212761194311477   
q = 325045504186436346209877301320131277983   
n = 98432079271513130981267919056149161631892822707167177858831841699521774310891   
e = 65537   
d = int(gmpy2.invert(e , (p-1) * (q-1)))   
privatekey = rsa.PrivateKey(n , e , d , p , q)        
with open("encrypted.message1" , "rb") as f:   
    print(rsa.decrypt(f.read(), privatekey).decode())         
    with open("encrypted.message2" , "rb") as f:   
        print(rsa.decrypt(f.read(), privatekey).decode())          
        with open("encrypted.message3" , "rb") as f:   
            print(rsa.decrypt(f.read(), privatekey).decode())
结果

leftleftrightright

一个upx加壳的exe程序
脱壳后，能够在ida中发现疑似经过换位的flag:s_imsaplw_e_siishtnt{g_ialt}F
按照英文单词猜测，还是有一定几率能猜出正确的flag的。
程序去壳之后就不能运行了，根据学长指示，可以在winedbg中进行调试。但是系统自带的老版本wine会遇到很多错误，所以编译安装wine3.8

编译安装wine3.8

下载源码到用户目录
$tar Jxf wine-3.8.tar.xz
$cd wine-3.8/
$./configure --enable-win64
遇到一个错误error: no suitable bison found. Please install the 'bison' package.
$sudo apt-get install bison
又一个错误error: FreeType 64-bit development files not found. Fonts will not be built.
根据提示判断是和字体相关的包，试了几次一直处错误，索性暂时不安了影响应该不大
$./configure --enable-win64 --without-freetype
$make $sudo make install

wine好像还有点问题，日后再说。。。
只要将断点下在比较函数部分，输入和flag等长的字符串，对比换位前后的变化，即可得到flag变化规则，将之前的字符串逆向变换即可得到flag 借用M4x大佬的图和代码说明
#!/usr/bin/env python
# -*- coding: utf-8 -*-
__Auther__ = 'M4x'    

encrypt = "s_imsaplw_e_siishtnt{g_ialt}F"    
before = "abcdefghijklmnopqrstuvwxyzABC"
after = "onpqmlrskjtuihvwgfxyedzAcbBCa"    
flag = [encrypt[after.find(c)] for c in before]

print "".join(flag)

#Flag{this_was_simple_isnt_it}
My math is bad

解方程类的题目，使用z3比较容易解决
详见z3学习档案
#!/usr/bin/env python
# -*- coding: utf-8 -*-
__Auther__ = 'M4x'    

encrypt = "s_imsaplw_e_siishtnt{g_ialt}F"    
before = "abcdefghijklmnopqrstuvwxyzABC"
after = "onpqmlrskjtuihvwgfxyedzAcbBCa"    
flag = [encrypt[after.find(c)] for c in before]

print "".join(flag)

#Flag{this_was_simple_isnt_it}
obfuscation and encode

程序逻辑可以说乱透了。。
输入的flag经过fencode和encode两个函数进行加密之后与lUFBuT7hADvItXEGn7KgTEjqw8U5VQUq进行比较。
经过分析，encode部分是进行了三位变四位的操作
Result为最终比较结果，trans为flag经过fencode处理得到的串
Reault[0] = alpha[trans[0] >> 2 & 0x3f]  == 'l'
v11 = 1
v10 = 2
Reault[1] = alpha[((trans[1]>>4)|16*trans[0]) & 0x3f] == 'U'
Reault[2] = alpha[((trans[2]>>6)|4*trans[1]) & 0x3f] == 'F'
v12 = 3
v24 = 4
Reault[3] = alpha[trans[2]&0x3f] == 'B'
可以根据这一规律进行每四位进行爆破，trans
这是类base64，一般是更改替换表和偏移位数进行编码，所以可以通过改写base64标准解码代码，实现一步到位

这是改写的爱测试国赛的代码。
# -*- coding: UTF-8 -*-
table = 'FeVYKw6a0lDIOsnZQ5EAf2MvjS1GUiLWPTtH4JqRgu3dbC8hrcNo9/mxzpXBky7+'

def decodeBase64(src):
    delPaddingTail = {0: 0, 2: 4, 1: 2}
    value = ''
    n = src.count('=')
    sin = src[:len(src) - n]
    for c in sin:
        value += bin(table.find(c))[2:].zfill(6).replace('0b', '')
    value = value[:len(value) - delPaddingTail[n]]
    print value
    middle = []
    for i in range(8, len(value) + 1, 8):
        middle.append(int(value[i-8:i], 2))
    output = middle
    print output
    return ''.join(map(chr, output)) 

res = decodeBase64("lUFBuT7hADvItXEGn7KgTEjqw8U5VQUq")
print res
得到trans
[37, 192, 59, 166, 31, 175, 76, 165, 203, 139, 164, 155, 59, 225, 40, 133, 38, 38, 22, 231, 17, 9, 7, 38]  
然后，我是通过暴力运行性加代码分析解决的fencode
写gdb脚本的方法倒是不错，程序汇编的0x4008c6和0x400906两行和生成trans有关

在调试的时候只要查看这两句就能看到操作数，其余的地方不用管
也能看到对24位flag，分成6组，每四位和m数组的对应位相乘求和再%127得到trans，同样可以使用z3进行求解。
#!/usr/bin/env python
# -*-coding=utf-8-*-
from z3 import *

trans = [37, 192, 59, 166, 31, 175, 76, 165, 203, 139, 164, 155, 59, 225, 40, 133, 38, 38, 22, 231, 17, 9, 7, 38]
print len(trans)
m = [2,2,4,-5,1,1,3,-3, -1, -2, -3, 4, -1, 0, -2,2]
a = BitVec('a',64)
b = BitVec('b',64)
c = BitVec('c',64)
d = BitVec('d',64)

for i in range(6):
    s = Solver()
    s.add((2 * a + 2 * b + 4 * c - 5 * d) & 0xff== trans[4 * i])
s.add((a + b + 3 * c - 3 * d)& 0xff== trans[4*i+1])
s.add((-1 * a - 2 * b -3 * c + 4 * d) & 0xff == trans[4 * i + 2])
s.add(( -1 * a  - 2 * c + 2 * d) & 0xff == trans[4 * i + 3])
s.add(a<256)
s.add(b<256)
s.add(c<256)
s.add(d<256)
if s.check() == sat:
    print s.model()
else :
    print s.check()

#[b = 108, a = 102, c = 97, d = 103]
#[b = 100, a = 123, c = 79, d = 95]
#[b = 48, a = 121, c = 85, d = 95]
#[b = 78, a = 75, c = 111, d = 87]
#[b = 48, a = 95, c = 73, d = 108]
#[b = 109, a = 86, c = 63, d = 125]

f = [102,108,97,103,123,100,79,95,121,48,85,95,75,78,111,87,95,48,73,108,86,109,63,125]
print map(chr,f)
flag = ''
for i in f:
    flag += chr(i)
print flag
可以得到六组解，排好顺序转字符即可

Pwn

作者：辣鸡小谱尼
出处：http://www.cnblogs.com/ZHijack/
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ISCC2018 Reverse & Pwn writeup

Reference:L1B0

Re

RSA256

leftleftrightright

编译安装wine3.8

My math is bad

obfuscation and encode

Pwn

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