2021西湖論劍 Reverse(Re) Writeup(WP)
阿新 • • 發佈:2021-11-22
ROR
二進位制位線性變換,直接上z3。
from z3 import * box = [ 0x65, 0x08, 0xF7, 0x12, 0xBC, 0xC3, 0xCF, 0xB8, 0x83, 0x7B, 0x02, 0xD5, 0x34, 0xBD, 0x9F, 0x33, 0x77, 0x76, 0xD4, 0xD7, 0xEB, 0x90, 0x89, 0x5E, 0x54, 0x01, 0x7D, 0xF4, 0x11, 0xFF, 0x99, 0x49, 0xAD, 0x57, 0x46, 0x67, 0x2A, 0x9D, 0x7F, 0xD2, 0xE1, 0x21, 0x8B, 0x1D, 0x5A, 0x91, 0x38, 0x94, 0xF9, 0x0C, 0x00, 0xCA, 0xE8, 0xCB, 0x5F, 0x19, 0xF6, 0xF0, 0x3C, 0xDE, 0xDA, 0xEA, 0x9C, 0x14, 0x75, 0xA4, 0x0D, 0x25, 0x58, 0xFC, 0x44, 0x86, 0x05, 0x6B, 0x43, 0x9A, 0x6D, 0xD1, 0x63, 0x98, 0x68, 0x2D, 0x52, 0x3D, 0xDD, 0x88, 0xD6, 0xD0, 0xA2, 0xED, 0xA5, 0x3B, 0x45, 0x3E, 0xF2, 0x22, 0x06, 0xF3, 0x1A, 0xA8, 0x09, 0xDC, 0x7C, 0x4B, 0x5C, 0x1E, 0xA1, 0xB0, 0x71, 0x04, 0xE2, 0x9B, 0xB7, 0x10, 0x4E, 0x16, 0x23, 0x82, 0x56, 0xD8, 0x61, 0xB4, 0x24, 0x7E, 0x87, 0xF8, 0x0A, 0x13, 0xE3, 0xE4, 0xE6, 0x1C, 0x35, 0x2C, 0xB1, 0xEC, 0x93, 0x66, 0x03, 0xA9, 0x95, 0xBB, 0xD3, 0x51, 0x39, 0xE7, 0xC9, 0xCE, 0x29, 0x72, 0x47, 0x6C, 0x70, 0x15, 0xDF, 0xD9, 0x17, 0x74, 0x3F, 0x62, 0xCD, 0x41, 0x07, 0x73, 0x53, 0x85, 0x31, 0x8A, 0x30, 0xAA, 0xAC, 0x2E, 0xA3, 0x50, 0x7A, 0xB5, 0x8E, 0x69, 0x1F, 0x6A, 0x97, 0x55, 0x3A, 0xB2, 0x59, 0xAB, 0xE0, 0x28, 0xC0, 0xB3, 0xBE, 0xCC, 0xC6, 0x2B, 0x5B, 0x92, 0xEE, 0x60, 0x20, 0x84, 0x4D, 0x0F, 0x26, 0x4A, 0x48, 0x0B, 0x36, 0x80, 0x5D, 0x6F, 0x4C, 0xB9, 0x81, 0x96, 0x32, 0xFD, 0x40, 0x8D, 0x27, 0xC1, 0x78, 0x4F, 0x79, 0xC8, 0x0E, 0x8C, 0xE5, 0x9E, 0xAE, 0xBF, 0xEF, 0x42, 0xC5, 0xAF, 0xA0, 0xC2, 0xFA, 0xC7, 0xB6, 0xDB, 0x18, 0xC4, 0xA6, 0xFE, 0xE9, 0xF5, 0x6E, 0x64, 0x2F, 0xF1, 0x1B, 0xFB, 0xBA, 0xA7, 0x37, 0x8F ] enc = [ 0x65, 0x55, 0x24, 0x36, 0x9D, 0x71, 0xB8, 0xC8, 0x65, 0xFB, 0x87, 0x7F, 0x9A, 0x9C, 0xB1, 0xDF, 0x65, 0x8F, 0x9D, 0x39, 0x8F, 0x11, 0xF6, 0x8E, 0x65, 0x42, 0xDA, 0xB4, 0x8C, 0x39, 0xFB, 0x99, 0x65, 0x48, 0x6A, 0xCA, 0x63, 0xE7, 0xA4, 0x79 ] v5 = [0]*8 v5[0] = 128 v5[1] = 64 v5[2] = 32 v5[3] = 16 v5[4] = 8 v5[5] = 4 v5[6] = 2 v5[7] = 1 flag = [] for i in range(len(enc)): for n in range(len(box)): if enc[i] == box[n]: flag.append(n) print(flag) Str = [BitVec('instr%d' % i, 8) for i in range(40)] v4 = BitVec('v4', 8) s = Solver() for i in range(0, 40, 8): for j in range(8): v4 = ((v5[j] & Str[i + 3]) << (8 - (3 - j) % 8)) | ((v5[j] & Str[i + 3]) >> ((3 - j) % 8)) | ( (v5[j] & Str[i + 2]) << (8 - (2 - j) % 8)) | ((v5[j] & Str[i + 2]) >> ((2 - j) % 8)) | ( (v5[j] & Str[i + 1]) << (8 - (1 - j) % 8)) | ((v5[j] & Str[i + 1]) >> ((1 - j) % 8)) | ( (v5[j] & Str[i]) << (8 - -j % 8)) | ((v5[j] & Str[i]) >> (-j % 8)) s.add(((((v5[j] & Str[i + 7]) << (8 - (7 - j) % 8)) | ((v5[j] & Str[i + 7]) >> ((7 - j) % 8)) | ((v5[j] & Str[i + 6]) << (8 - (6 - j) % 8)) | ((v5[j] & Str[i + 6]) >> ((6 - j) % 8)) | ((v5[j] & Str[i + 5]) << (8 - (5 - j) % 8)) | ((v5[j] & Str[i + 5]) >> ((5 - j) % 8)) | ((v5[j] & Str[i + 4]) << (8 - (4 - j) % 8)) | ((v5[j] & Str[i + 4]) >> ((4 - j) % 8)) | v4)) == flag[i+j]) if s.check() == sat: m = s.model() s = [] for i in range(len(m)): s.append(m[Str[i]].as_long()) print(bytes(s).decode()) # Q5la5_3KChtem6_HYHk_NlHhNZz73aCZeK05II96
TacticalArmed
單指令SMC解密執行。程式將一個一個完整的40位元組TEA加密流程的指令放入了資料區,在執行時動態的解密單指令並執行,下一次執行的單指令將覆蓋上一次執行的單指令。
可以利用OD或者x64dbg的指令碼,將所有指令dump下來。od指令碼如下:
VAR opcode_size VAR dump_address BP 00401504 Dump: BP 0040140 RUN MOV opcode_size, [ebp-10] MOV dump_address, [405640] DMA dump_address, opcode_size, "my_code.bin" CMP eip, 00401504 JNE Dump MSG "執行完畢" RET
由於原程式沒有基質隨機,不用修復重定位偏移,所以可以直接將dump下來的指令在作為一個新增的節新增進題目程式中,並稍微修復使IDA可以反彙編。
很顯然是一個tea加密,只不過輪數變為33論,並且key在tls中會檢測偵錯程式,只有偵錯程式不存在的時候才會給予正確的key。
需要注意的是,在整個40位元組的加密過程中,sum只初始化了一次,所以與正常呼叫tea是有所不同的,那麼解密函式如下。
#include <stdio.h> #include <stdint.h> void TeaDecrypt(uint32_t* v, uint32_t* k) { uint32_t v0 = v[0], v1 = v[1], sum = 0, i; // B6528EEE uint32_t delta = 0x81A5692E; uint32_t k0 = k[0], k1 = k[1], k2 = k[2], k3 = k[3]; uint32_t sums[] = { 0xb6528eee ,0x6ca51ddc ,0x22f7acca ,0xd94a3bb8,0x8f9ccaa6 }; for (int n = 0; n < 10; n += 2) { sum = sums[n / 2]; v0 = v[0 + n], v1 = v[1 + n]; for (i = 0; i < 33; i++) { v1 -= ((v0 << 4) + k2) ^ (v0 + sum) ^ ((v0 >> 5) + k3); v0 -= ((v1 << 4) + k0) ^ (v1 + sum) ^ ((v1 >> 5) + k1); sum -= delta; } v[0 + n] = v0; v[1 + n] = v1; } } int main() { uint8_t enc[] = { 0xED, 0x1D, 0x2F, 0x42, 0x72, 0xE4, 0x85, 0x14, 0xD5, 0x78, 0x55, 0x03, 0xA2, 0x80, 0x6B, 0xBF, 0x45, 0x72, 0xD7, 0x97, 0xD1, 0x75, 0xAE, 0x2D, 0x63, 0xA9, 0x5F, 0x66, 0x74, 0x6D, 0x2E, 0x29, 0xC1, 0xFC, 0x95, 0x97, 0xE9, 0xC8, 0xB5, 0x0B, 0 }; uint32_t key[] = { 0x7CE45630,0X58334908,0X66398867,0XC35195B1 }; TeaDecrypt((uint32_t*)enc, key); puts((char*)enc); // kgD1ogB2yGa2roiAeXiG8_aqnLzCJ_rFHSPrn55K return 0; }
虛假的粉絲
一個腦洞題。
開始需要輸入3個部分通過check,第一個check是有U和S字元的檔案,第二個check是這些字元在檔案的偏移,第三個是讀取多少個字元,grep一下所以檔案內容,可以獲得三個資訊。
分別是:
4457 1118 40
輸入後,會列印一個UzNDcmU3X0szeSUyMCUzRCUyMEFsNE5fd0FsSzNS,base64一下是最後一個key
S3Cre7_K3y = Al4N_wAlK3R
輸入最後一個keyAl4N_wAlK3R,程式會解密5135號檔案,該檔案的字元圖裡面的字母就是flag。
flag: A_TrUe_AW_f4ns
to be or not to be, is a question.