2019 红帽杯

XX

https://en.wikipedia.org/wiki/XXTEA XXTEA加密

输入前四位应为小写字母+数字, 取前四位为key, 后面补\x00, 补足16位

之后对flag进行XXTEA加密, 然后按2 0 3 1 的顺序混淆错位, 再谜之异或, 最后和

0x6B40BCCE, 0xC0953A7C, 0x20209BEF, 0x3502F791, 0xC8021823, 0xFA5656E7进行对比.。

求异或:

# include<stdio.h>

int main() {
    unsigned char cipherX[] = { 0xCE, 0xBC, 0x40, 0x6B, 0x7C, 0x3A, 0x95, 0xC0, 0xEF, 0x9B, 0x20, 0x20, 0x91, 0xF7, 0x02, 0x35, 0x23, 0x18, 0x02, 0xC8, 0xE7, 0x56, 0x56, 0xFA};
    unsigned char *v21 = cipherX + 23;
    int v20 = 23;
    int v22;
    char v23;
    for (; v20 > 0; --v21) {
        v22 = 0;
        if ( v20 / 3 > 0 )
            {
              v23 = *v21;
              do
              {
                    v23 ^= cipherX[v22++];
                    *v21 = v23;
              }
              while ( v22 < v20 / 3 );
        }
        --v20;
    }
    for (int i = 0; i < 24; i++) {
        printf("0x%x, ", cipherX[i]);
    }
}

解题脚本, flag格式猜测key为flag, python3:

import string
from itertools import product
from ctypes import *

def decrypt(cipher, key):
        k = [0] * 4
        for i in range(4):
            tmp = c_uint32(0)
            for j in range(4):
                tmp.value = tmp.value << 8
                tmp.value = tmp.value + key[4 * i + (3 - j)]
            k[i] = tmp
        length = len(cipher)
        n = length // 4
        v = [c_uint32()]*n
        for i in range(n):
            tmp = c_uint32(0)
            for j in range(4):
                tmp.value = tmp.value << 8
                tmp.value = tmp.value + cipher[4 * i + (3 - j)]
            v[i] = tmp
        q = 6 + 52 // n
        z = c_uint32(0)
        y = c_uint32(v[0].value)
        delta = c_uint32(0x9e3779b9)
        sum_delta = c_uint32(q * delta.value)
        for i in range(q)[::-1]:
            e = (sum_delta.value >> 2) & 3
            for p in range(1, n)[::-1]:
                z.value = v[p-1].value
                v[p].value -=  ((z.value >> 5 ^ y.value << 2) + (y.value >> 3 ^ z.value<<4)) ^ ((sum_delta.value ^ y.value) + (k[(p&3)^e].value ^ z.value))
                y.value = v[p].value
            z.value = v[n-1].value
            v[0].value -= ((z.value >> 5 ^ y.value << 2) + (y.value >> 3 ^ z.value<<4)) ^ ((sum_delta.value ^ y.value) + (k[(0&3)^e].value ^ z.value))
            y.value = v[0].value
            sum_delta.value -= delta.value
        plain = b""
        for i in range(n):
            for j in range(4):
                plain+=bytes([v[i].value&0xff])
                v[i].value = v[i].value >> 8
        return plain
# cipherX = [ 0xCE, 0xBC, 0x40, 0x6B, 0x7C, 0x3A, 0x95, 0xC0, 0xEF, 0x9B, 0x20, 0x20, 0x91, 0xF7, 0x02, 0x35, 0x23, 0x18, 0x02, 0xC8, 0xE7, 0x56, 0x56, 0xFA] print(len(cipherX))
cipherX = [0xce, 0xbc, 0x40, 0xa5, 0xb2, 0xf4, 0xe7, 0xb2, 0x9d, 0xa9, 0x12, 0x12, 0xc8, 0xae, 0x5b, 0x10, 0x6, 0x3d, 0x1d, 0xd7, 0xf8, 0xdc, 0xdc, 0x70]
cipher = [0 for i in range(24)]
for i in range(6):
    cipher[i * 4 + 2] = cipherX[i * 4 + 0]
    cipher[i * 4 + 0] = cipherX[i * 4 + 1]
    cipher[i * 4 + 3] = cipherX[i * 4 + 2]
    cipher[i * 4 + 1] = cipherX[i * 4 + 3]

cipher_b = b""
for i in cipher:
    cipher_b += bytes([i])

key = b'flag'
for i in range(12):
    key += b'\x00'
plain = decrypt(cipher_b, key)
flag = "".join(list(map(chr, plain)))
print(flag)

easyRE

出题挖了坑, 实际逻辑在fini里, 参考 https://luomuxiaoxiao.com/?p=516

主逻辑就是个异或加密, 根据提示知道异或key是flag

解密脚本:

s = [0x49 , 0x6F , 0x64 , 0x6C , 0x3E , 0x51 , 0x6E , 0x62 , 0x28 , 0x6F , 0x63 , 0x79 , 0x7F , 0x79 , 0x2E , 0x69 , 0x7F , 0x64 , 0x60 , 0x33 , 0x77 , 0x7D , 0x77 , 0x65 , 0x6B , 0x39 , 0x7B , 0x69 , 0x79 , 0x3D , 0x7E , 0x79 , 0x4C , 0x40 , 0x45 , 0x43]

input = ""

for i in range(len(s)):
    s[i] ^= i
    input += chr(s[i])

print(input)

# input = "Info:The first four chars are `flag`"

cipher = [  0x40, 0x35, 0x20, 0x56, 0x5D, 0x18, 0x22, 0x45, 0x17, 0x2F,  0x24, 0x6E, 0x62, 0x3C, 0x27, 0x54, 0x48, 0x6C, 0x24, 0x6E,  0x72, 0x3C, 0x32, 0x45, 0x5B]

key = []

xor = list(map(ord, "flag"))
for i in range(4):
    key.append(cipher[i] ^ xor[i])

flag = ""
for i in range(len(cipher)):
    flag += chr(cipher[i] ^ key[i % 4])
print(flag)

childRE

先把输入当成满二叉树然后后序遍历再输出进行打乱, 然后UnDecorate, 然后置换加密进行对比

爆破UnDecorate之后的内容, 然后进行还原得到

private: char * __thiscall R0Pxx::My_Aut0_PWN(unsigned char *)

根据维基百科 https://en.wikipedia.org/wiki/Name_mangling 进行

Decorate, 然后还原打乱再进行md5得flag

解密脚本:

# -*- coding: utf-8 -*-
# cmp1 = "(_@4620!08!6_0*0442!@186%%0@3=66!!974*3234=&0^3&1@=&0908!6_0*&"
# cmp2 = "55565653255552225565565555243466334653663544426565555525555222"
# table = [  0x31, 0x32, 0x33, 0x34, 0x35, 0x36, 0x37, 0x38, 0x39, 0x30,#   0x2D, 0x3D, 0x21, 0x40, 0x23, 0x24, 0x25, 0x5E, 0x26, 0x2A,#   0x28, 0x29, 0x5F, 0x2B, 0x71, 0x77, 0x65, 0x72, 0x74, 0x79,#   0x75, 0x69, 0x6F, 0x70, 0x5B, 0x5D, 0x51, 0x57, 0x45, 0x52,#   0x54, 0x59, 0x55, 0x49, 0x4F, 0x50, 0x7B, 0x7D, 0x61, 0x73,#   0x64, 0x66, 0x67, 0x68, 0x6A, 0x6B, 0x6C, 0x3B, 0x27, 0x41,#   0x53, 0x44, 0x46, 0x47, 0x48, 0x4A, 0x4B, 0x4C, 0x3A, 0x22,#   0x5A, 0x58, 0x43, 0x56, 0x42, 0x4E, 0x4D, 0x3C, 0x3E, 0x3F,#   0x7A, 0x78, 0x63, 0x76, 0x62, 0x6E, 0x6D, 0x2C, 0x2E, 0x2F,#   0x00]

# # for i in range(len(cmp1)):
# #     for j in range(len(table)):
# #         if ord(cmp1[i]) == table[j]:
# #             print hex(j)+', ',
# # for i in range(len(cmp2)):
# #     for j in range(len(table)):
# #         if ord(cmp2[i]) == table[j]:
# #             print hex(j)+', ',
# index1 = [0x14, 0x16, 0xd, 0x3, 0x5, 0x1, 0x9, 0xc, 0x9, 0x7, 0xc, 0x5, 0x16, 0x9, 0x13, 0x9, 0x3, 0x3, 0x1, 0xc, 0xd, 0x0, 0x7, 0x5, 0x10, 0x10, 0x9, 0xd, 0x2, 0xb, 0x5, 0x5, 0xc, 0xc, 0x8, 0x6, 0x3, 0x13, 0x2, 0x1, 0x2, 0x3, 0xb, 0x12, 0x9, 0x11, 0x2, 0x12, 0x0, 0xd, 0xb, 0x12, 0x9, 0x8, 0x9, 0x7, 0xc, 0x5, 0x16, 0x9, 0x13, 0x12]
# index2 = [0x4, 0x4, 0x4, 0x5, 0x4, 0x5, 0x4, 0x2, 0x1, 0x4, 0x4, 0x4, 0x4, 0x1, 0x1, 0x1, 0x4, 0x4, 0x5, 0x4, 0x4, 0x5, 0x4, 0x4, 0x4, 0x4, 0x1, 0x3, 0x2, 0x3, 0x5, 0x5, 0x2, 0x2, 0x3, 0x5, 0x4, 0x2, 0x5, 0x5, 0x2, 0x4, 0x3, 0x3, 0x3, 0x1, 0x5, 0x4, 0x5, 0x4, 0x4, 0x4, 0x4, 0x4, 0x1, 0x4, 0x4, 0x4, 0x4, 0x1, 0x1, 0x1]
# # print(len(index2))
# name = ""

# for j in range(len(index1)):
#     for i in range(0x20, 0x7e):
#         if i % 23 == index1[j] and i // 23 == index2[j]:
#             name += chr(i)
#             break
# print name
# private: char * __thiscall R0Pxx::My_Aut0_PWN(unsigned char *)
# ?表示模板
# * P
# char D; unsigned char E
# private near A
# 64位编程时, 唯一可用的调用协议的编码是A
# @: 形参表结束标志
# Z: 缺省的异常规范
# PAE unsiged char *
# PAD 返回类型
# A表示private的成员函数;A表示非只读成员函数;E表示thiscall
# ?My_Aut0_PWN@R0Pxx@@AAEPADPAE@Z

s1 = "1234567890abcdefghijklmnopqrstu"
s2 = "fg8hi94jk0lma52nobpqc6rsdtue731"
dic = []
for i in range(len(s1)):
    for j in range(len(s2)):
        if s1[i] == s2[j]:
            dic.append(j)

# print dic

name = "?My_Aut0_PWN@R0Pxx@@AAEPADPAE@Z"
input = ""
for i in range(len(name)):
    input += name[dic[i]]
# print input


def md5(str):
    import hashlib
    m = hashlib.md5()
    m.update(str)
    return m.hexdigest()

print md5(input)

calc

C++写的 STL+VS 又开了Release

自己实现了大数算法, 输入三个数, 然后进行一波计算之后检查

三个数的大小排序为 p2 < p1 < p3

对算法进行化简, 发现就是42的三立方和问题

p1**3 + p2 ** 3 - p3 ** 3 == 42

直接百度答案进行排序

p1 == 80435758145817515
p2 == 12602123297335631
p3 == 80538738812075974

snake

先玩了一波hhhh

在.\Snake_Data\Plugins\找到Interface.dll, 分析导出函数GameObject

发现就是个RSA

def f(v):
    s = ""
    while v != 0:
        v, r = divmod(v, 255)
        s = chr(r) + s
    return s

n = 139907262641720884635250105449327463531131227516500497307311002094885245322386805049406878643982216326493527702414689439930090794753345844178528356178539094825247389836142928474607108262267087850211322640806135698076207986818086837911361480181444157057782599277473843153161174504240064610043962720953514451563
c = 79981856490856999850671700360733120831999995589421207460490185876531860518527597767905168099182891345123878966403548022646956365158864209467614850251731806682037300712511185681164865174187586907707195428804234739667769742078793162639867922056194688917569369338005327309973680573581158754297630654105882382426

for i in range(2, 100):
    r = pow(c, i, n)
    s = f(r)
    if s.startswith("flag"):
        print(s)