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[原创]KCTF2021-秋 第二题 迷失丛林 writeup
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2021-11-18 11:20
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KCTF2021-秋 第二题 迷失丛林 writeup
输入32位{0-9，A-F}，每两位组成1byte,处理成16bytes。Serial前8bytes填充到.data:00404000映射表。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 | { 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0xA2, 0x9B, 0xF4, 0xDF, 0xAC, 0x7C, 0xA1, 0xC6, 0x16, 0xD0, 0x0F, 0xDD, 0xDC, 0x73, 0xC5, 0x6B, 0xD1, 0x96, 0x47, 0xC2, 0x26, 0x67, 0x4E, 0x41, 0x82, 0x20, 0x56, 0x9A, 0x6E, 0x33, 0x92, 0x88, 0x29, 0xB5, 0xB4, 0x71, 0xA9, 0xCE, 0xC3, 0x34, 0x50, 0x59, 0xBF, 0x2D, 0x57, 0x22, 0xA6, 0x30, 0x04, 0xB2, 0xCD, 0x36, 0xD5, 0x68, 0x4D, 0x5B, 0x45, 0x9E, 0x85, 0xCF, 0x9D, 0xCC, 0x61, 0x78, 0x32, 0x76, 0x31, 0xE3, 0x80, 0xAD, 0x39, 0x4F, 0xFA, 0x72, 0x83, 0x4C, 0x86, 0x60, 0xB7, 0xD7, 0x63, 0x0C, 0x44, 0x35, 0xB3, 0x7B, 0x19, 0xD4, 0x69, 0x08, 0x0B, 0x1F, 0x3D, 0x11, 0x79, 0xD3, 0xEE, 0x93, 0x42, 0xDE, 0x23, 0x3B, 0x5D, 0x8D, 0xA5, 0x77, 0x5F, 0x58, 0xDB, 0x97, 0xF6, 0x7A, 0x18, 0x52, 0x15, 0x74, 0x25, 0x62, 0x2C, 0x05, 0xE8, 0x0D, 0x98, 0x2A, 0x43, 0xE2, 0xEF, 0x48, 0x87, 0x49, 0x1C, 0xCA, 0x2B, 0xA7, 0x8A, 0x09, 0x81, 0xE7, 0x53, 0xAA, 0xFF, 0x6F, 0x8E, 0x91, 0xF1, 0xF0, 0xA4, 0x46, 0x3A, 0x7D, 0x54, 0xEB, 0x2F, 0xC1, 0xC0, 0x0E, 0xBD, 0xE1, 0x6C, 0x64, 0xBE, 0xE4, 0x02, 0x3C, 0x5A, 0xA8, 0x9F, 0x37, 0xAF, 0xA0, 0x13, 0xED, 0x1B, 0xEC, 0x8B, 0x3E, 0x7E, 0x27, 0x99, 0x75, 0xAB, 0xFE, 0xD9, 0x3F, 0xF3, 0xEA, 0x70, 0xF7, 0x95, 0xBA, 0x1D, 0x40, 0xB0, 0xF9, 0xE5, 0xF8, 0x06, 0xBC, 0xB6, 0x03, 0xC9, 0x10, 0x9C, 0x2E, 0x89, 0x5C, 0x7F, 0xB1, 0x1A, 0xD6, 0x90, 0xAE, 0xDA, 0xE6, 0x5E, 0xB9, 0x84, 0xE9, 0x55, 0xBB, 0xC7, 0x0A, 0xE0, 0x66, 0xF2, 0xD8, 0xCB, 0x00, 0x12, 0xB8, 0x17, 0x94, 0x6A, 0x4A, 0x01, 0x24, 0x14, 0x51, 0x07, 0x65, 0x21, 0xC8, 0x38, 0xFD, 0x8F, 0xC4, 0xF5, 0xFC} |
1 | _404100 dd 2, 4, 8, 10h, 20h, 40h, 80h |
根据映射表，生成二叉树，共生成0x100棵树。_404100作为树每层的长度,通过左右节点生成下一层的节点。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 | // 生成第d棵树（0<=d<0x100）void genTree(int d){ genMp[0] = Mp400[d]; genMp[1] = d + 1; int p = 0; for (int i = 0; i < sizeof(mi) / sizeof(int); i++) { for (int j = 0; j < mi[i]; j++) { uint8_t pm = genMp[p]; p++; genMp[p * 2] = Mp400[pm]; genMp[p * 2 + 1] = pm + 1; } }} |
每棵树最后一层长度为0x100,标记最后一层出现过的值，最后统计0x100棵树中每个值被标记多少次（最多0x100次）。
需要满足sumctMp[0] == 0xA9 && sumctMp[0xE] == 0xAC && sumctMp[0x28] == 0xA7 && sumctMp[0x4F] > 0xC8。
根据这个条件来确定Serial前8bytes，算法不会逆，只能靠猜：.data:00404000映射表初始值都是非0唯一的，先找到缺失的8字节的值：
1 | {0x1e, 0x28, 0x4b, 0x6d, 0x8c, 0xa3, 0xd2, 0xfb} |
验证时，把这8字节排列组合，过了条件的只有这一组
1 | {0x4b, 0x6d, 0x28, 0x8c, 0xfb, 0xd2, 0x1e, 0xa3} |
继续往下走，就是纯暴力了。
先对映射表进行处理，之后Serial后8bytes的解就按照字节单独暴力解就行。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 | uint8_t keyC[]{0x4b, 0x6d, 0x28, 0x8c, 0xfb, 0xd2, 0x1e, 0xa3}; memcpy(Mp400, keyC, 8); isOk(); uint8_t sgs[8]; uint8_t input2[8]; char *out = "GoodJob~"; for (int it = 0; it < 8; it++) { int w = 0; memcpy(sgs, Mp400, 8); while (0x100 != w) { input2[it] = w; int v17 = 0; do { for (int i = 0; i < 8; ++i) { if (v17 >= 8) { if (!i || i == 7) sgs[i]--; } else { uint8_t v19; if (input2[i] & 1) v19 = sgs[i] + 1; else v19 = Mp400[sgs[i]]; sgs[i] = v19; input2[i] >>= 1; } } ++v17; } while (v17 < 9); if ((uint8_t)out[it] == sgs[it]) break; memcpy(sgs, Mp400, 8); w++; } printf("%02x ", w); } printf("\n"); |
输出后8bytes值
1 | 9d 6b ea 2f a4 08 bc 22 |
Serial: B4D682C8BF2DE13AD9B6AEF24A80CB22
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