buu刷题 红帽杯2019-XX wp

首先找到关键函数，

找到我们输入的数据经过一系列操作之后得到的，即最终结果（注意小端序存储）：

把数据dump出来：

int a[] = {0xCE, 0xBC, 0x40, 0x6B, 0x7C, 0x3A, 0x95, 0xC0, 0xEF, 0x9B, 0x20, 0x20, 0x91, 0xF7, 0x02, 0x35, 0x23, 0x18, 0x02, 0xC8, 0xE7, 0x56, 0x56, 0xFA };

然后就来看一下整个程序对我们输入的数据进行了哪些操作：
首先确定字符串长度：

从后往前推：
下面脚本是逆向解出xxtea加密之后的结果，（三个一组进行异或）

#include
#include
int main()
{int count = 0;int b[24];int a[] = { 0xCE, 0xBC, 0x40, 0x6B, 0x7C, 0x3A, 0x95, 0xC0, 0xEF, 0x9B, 0x20, 0x20, 0x91, 0xF7, 0x02, 0x35, 0x23, 0x18, 0x02, 0xC8, 0xE7, 0x56, 0x56, 0xFA };for (int i = 23; i >=3; i--){for (int j = 6-count; j >= 0; j--){a[i]^=a[j];		}if (i % 3 == 0){count++;}}for (int i = 0; i < 24; i++)printf("0x%x,", a[i]);printf("\n");b[2] = a[0];b[0] = a[1];b[3] = a[2];b[1] = a[3];b[6] = a[4];b[4] = a[5];b[7] = a[6];b[5] = a[7];b[10] = a[8];b[8] = a[9];b[11] = a[10];b[9] = a[11];b[14] = a[12];b[12] = a[13];b[15] = a[14];b[13] = a[15];b[18] = a[16];b[16] = a[17];b[19] = a[18];b[17] = a[19];b[22] = a[20];b[20] = a[21];b[23] = a[22];b[21] = a[23];for (int i = 0; i < 24; i++)printf("0x%x,", b[i]);system("pause");
}

运行后可以得到数据0xbc,0xa5,0xce,0x40,0xf4,0xb2,0xb2,0xe7,0xa9,0x12,0x9d,0x12,0xae,0x10,0xc8,0x5b,0x3d,0xd7,0x6,0x1d,0xdc,0x70,0xf8,0xdc

然后就是进行xxtea解密，密钥是输入字符的前四位，猜测是“flag”

#include   
#include   
#include
#define DELTA 0x9e3779b9  
#define MX (((z>>5^y<<2) + (y>>3^z<<4)) ^ ((sum^y) + (key[(p&3)^e] ^ z)))  void btea(uint32_t *v, int n, uint32_t const key[4])
{uint32_t y, z, sum;unsigned p, rounds, e;if (n > 1)            /* Coding Part */{rounds = 6 + 52 / n;sum = 0;z = v[n - 1];do{sum += DELTA;e = (sum >> 2) & 3;for (p = 0; p> 2) & 3;for (p = n - 1; p>0; p--){z = v[p - 1];y = v[p] -= MX;}z = v[n - 1];y = v[0] -= MX;sum -= DELTA;} while (--rounds);}
}int main()
{//03e0164dd30553aa// uint32_t a[2] = { (unsigned int)0xd2cfdad7, 0x9ac8d5d4 };uint32_t v[6] = { (unsigned int)0x40cea5bc, (unsigned int)0xe7b2b2f4,(unsigned int)0x129d12a9,(unsigned int)0x5bc810ae,(unsigned int)0x1d06d73d,(unsigned int)0xdcf870dc };uint32_t const k[4] = { (unsigned int)0x67616c66, (unsigned int)0x0, (unsigned int)0X0, (unsigned int)0x0 };int n = 6; //n的绝对值表示v的长度，取正表示加密，取负表示解密  // v为要加密的数据是两个32位无符号整数  // k为加密解密密钥，为4个32位无符号整数，即密钥长度为128位  //printf("加密前原始数据：%x %x\n", v[0], v[1]);btea(v, -n, k);printf("加密后的数据：%x %x %x %x %x\n", v[0], v[1],v[2],v[3],v[4],v[5]);//btea(v, -n, k);//printf("解密后的数据：%x %x\n", v[0], v[1]);system("pause");		}

然后得到以下数据：67616c66 5858437b 646e615f 742b2b5f 7d6165
按照小端序，再把16进制转成字符就get flag了
以上是用c写的脚本。

放完整python脚本：

enc = 'CEBC406B7C3A95C0EF9B202091F70235231802C8E75656FA'.decode('hex')dec1 = ''
for i in range(len(enc)/3):j = len(enc)/3 - i - 1res = ''for k in enc[j*3:j*3+3]:tmp = ord(k)for l in range(j):tmp ^= ord(enc[l])res += chr(tmp)dec1 = res + dec1
print dec1.encode('hex')dec2 = [''] * len(dec1)
for i in range(len(dec1)):dec2[i] = dec1[i]box = [1,3,0,2,5,7,4,6,9,11,8,10,13,15,12,14,17,19,16,18,21,23,20,22]
print len(box)for i in range(len(box)):dec2[i] = dec1[box[i]]
dec3 = ''.join(dec2)
print dec3.encode('hex')import struct_DELTA = 0x9E3779B9  def _long2str(v, w):  n = (len(v) - 1) << 2  if w:  m = v[-1]  if (m < n - 3) or (m > n): return ''  n = m  s = struct.pack('<%iL' % len(v), *v)  return s[0:n] if w else s  def _str2long(s, w):  n = len(s)  m = (4 - (n & 3) & 3) + n  s = s.ljust(m, "\0")  v = list(struct.unpack('<%iL' % (m >> 2), s))  if w: v.append(n)  return v  def encrypt(str, key):  if str == '': return str  v = _str2long(str, True)  k = _str2long(key.ljust(16, "\0"), False)  n = len(v) - 1  z = v[n]  y = v[0]  sum = 0  q = 6 + 52 // (n + 1)  while q > 0:  sum = (sum + _DELTA) & 0xffffffff  e = sum >> 2 & 3  for p in xrange(n):  y = v[p + 1]  v[p] = (v[p] + ((z >> 5 ^ y << 2) + (y >> 3 ^ z << 4) ^ (sum ^ y) + (k[p & 3 ^ e] ^ z))) & 0xffffffff  z = v[p]  y = v[0]  v[n] = (v[n] + ((z >> 5 ^ y << 2) + (y >> 3 ^ z << 4) ^ (sum ^ y) + (k[n & 3 ^ e] ^ z))) & 0xffffffff  z = v[n]  q -= 1  return _long2str(v, False)  def decrypt(str, key):  if str == '': return str  v = _str2long(str, False)  k = _str2long(key.ljust(16, "\0"), False)  n = len(v) - 1  z = v[n]  y = v[0]  q = 6 + 52 // (n + 1)  sum = (q * _DELTA) & 0xffffffff  while (sum != 0):  e = sum >> 2 & 3  for p in xrange(n, 0, -1):  z = v[p - 1]  v[p] = (v[p] - ((z >> 5 ^ y << 2) + (y >> 3 ^ z << 4) ^ (sum ^ y) + (k[p & 3 ^ e] ^ z))) & 0xffffffff  y = v[p]  z = v[n]  v[0] = (v[0] - ((z >> 5 ^ y << 2) + (y >> 3 ^ z << 4) ^ (sum ^ y) + (k[0 & 3 ^ e] ^ z))) & 0xffffffff  y = v[0]  sum = (sum - _DELTA) & 0xffffffff  return _long2str(v, True)  key = 'flag'
dec4 = decrypt(dec3, key)
print len(dec4)
print dec4

flag：flag{CXX_and_++tea}

最后总结一下这个程序的整个流程：

1. 判断输入的字符串的每个字符是否包含在"qwertyuiopasdfghjklzxcvbnm1234567890"中
2. 取输入字符串的前4位字符，即"flag"，扩展为16位，作为xxtea加密的秘钥key
3. 将输入的字符串使用key加密，加密后的字符保存在字符数组v18，共24位字符
4. 打乱v18数组，保存到v19数组中
5. 将24位字符，每3位为一组，每一组异或值（具体看代码），得到新的加密字符串
6. 将新的加密字符串与已经存在的字符串比较，相同即获得胜利

难点在于判断是xxtea加密：
这里有另一种方法，装一个插件，FindCrypt ，安装教程自行谷歌。注意，此插件只能简单识别是TEA加密。

参考博客：
https://www.cnblogs.com/playmak3r/p/12063593.html
https://impakho.com/post/redhat-2019-online-writeup
https://www.cnblogs.com/Mayfly-nymph/p/11869959.html
利用TEA算法进行数据加密

本文来自互联网用户投稿，文章观点仅代表作者本人，不代表本站立场，不承担相关法律责任。如若转载，请注明出处。 如若内容造成侵权/违法违规/事实不符，请点击【内容举报】进行投诉反馈！