网鼎杯朱雀组逆向 tree

找到main函数

进入到chkflag函数，发现其是将输入的flag中的xxx每一个x都换成2进制的形式，每一个4位，然后存在glockflag中

parse函数是将glockflag中的二进制拿出来，0代表左，1代表右，开始遍历叶子节点，如果找到的叶子节点是zvzjyvosgnzkbjjjypjbjdvmsjjyvsjx，就正确

我们先将 每个节点及其路径打印出来

def traverse_leaf(pnode):
    if pnode != 0:
        if Dword(pnode + 12) == 0 and Dword(pnode + 16) == 0:
            print(chr(Byte(pnode)))
            print("".join(a))
            lujing.append([chr(Byte(pnode)), "".join(a)])
        a.append('0')
        traverse_leaf(Dword(pnode + 12))
        a.append('1')
        traverse_leaf(Dword(pnode + 16))
    if pnode != 0X0406530:
        a.pop()


traverse_leaf(0X0406530)
print(lujing)

[[‘y’, ‘0000’], [‘b’, ‘00010’], [‘q’, ‘00011’], [‘g’, ‘0010’], [‘f’, ‘0011’], [‘j’, ‘010’], [‘w’, ‘01100’], [‘p’, ‘01101’], [‘x’, ‘011100’], [‘d’, ‘0111010’], [‘i’, ‘0111011’], [‘k’, ‘01111’], [‘s’, ‘100’], [‘z’, ‘1010’], [‘n’, ‘1011’], [‘c’, ‘11000’], [‘t’, ‘110010’], [‘e’, ‘110011’], [‘h’, ‘1101’], [‘o’, ‘11100’], [‘l’, ‘1110100’], [‘u’, ‘11101010’], [‘r’, ‘111010110’], [‘a’, ‘111010111’], [‘m’, ‘111011’], [‘v’, ‘1111’]]

然后在开始写脚本，将 zvzjyvosgnzkbjjjypjbjdvmsjjyvsjx 转成路径，然后拼起来，4位4位的分开就是flag中的xxxx

lujing = [['y', '0000'], ['b', '00010'], ['q', '00011'], ['g', '0010'], ['f', '0011'], ['j', '010'], ['w', '01100'], ['p', '01101'], ['x', '011100'], ['d', '0111010'], ['i', '0111011'], ['k', '01111'], ['s', '100'], ['z', '1010'], ['n', '1011'], ['c', '11000'], ['t', '110010'], ['e', '110011'], ['h', '1101'], ['o', '11100'], ['l', '1110100'], ['u', '11101010'], ['r', '111010110'], ['a', '111010111'], ['m', '111011'], ['v', '1111']]
res = "zvzjyvosgnzkbjjjypjbjdvmsjjyvsjx"
flag01 = ""
flagx = ""
for i in res:
    for j in lujing:
        if i in j[0]:
            flag01 += j[1]
print(flag01)
for i in range(0, len(flag01), 4):
    tmp = "%x" % int(flag01[i:i+4], 2)
    flagx += tmp
print(flagx)

最终打印出flagx是afa41fc8574f12481a849d7f7120f89c
将flag{xxxxxxxx-xxxx-xxxx-xxxx-xxxxxxxxxxxx}替换掉，即
flag为flag{afa41fc8-574f-1248-1a84-9d7f7120f89c}

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