题目 flowers
解法
先exeinfo看看
32位进去看看
既然是花指令，找找看吧
很明显是这个地方
重新学了一下花指令，以下两篇博客给我很大的帮助
https://blog.csdn.net/demondev/article/details/91978869
https://blog.csdn.net/abel_big_xu/article/details/117927674
过两天再根据花指令专项发一个。
这里的思路就是看代码执行流程
401200后面的程序没法正确的反编译，由于都没有什么用，直接nop掉，指导里面的e8出，这是一个jmp
就自动反汇编了，这时再去反编译
代码就有了
这是一个rc4的魔改版本，魔改在这里有个异或37
上网找到个师傅的标准rc4解密脚本
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]) % 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
a=[0xc6,0x21,0xca,0xbf,0x51,0x43,0x37,0x31,0x75,0xe4,0x8e,0xc0,0x54,0x6f,0x8f,0xee,0xf8,0x5a,0xa2,0xc1,0xeb,0xa5,0x34,0x6d,0x71,0x55,0x8,0x7,0xb2,0xa8,0x2f,0xf4,0x51,0x8e,0xc,0xcc,0x33,0x53,0x31,0x0,0x40,0xd6,0xca,0xec,0xd4]
s=""
for i in a:
s+=chr(i)
s=str(base64.b64encode(s.encode('utf-8')), 'utf-8')
rc4_main("Nu1Lctf233", s)
写的非常完备，我们魔改一下
然后用这道题的数据
完美
