一、题目
1、android 逆向题
小伙伴发来的题，说是某比赛线下题目
2、涉及算法
AES
魔改RC5
二、解题过程
1、java层
2、native层
sub_F08
int __fastcall sub_F08(JNIEnv *a1, int a2, int a3)
{
const char *v3; // r9
size_t v4; // r6
char *v5; // r4
void *v6; // r8
_DWORD *v7; // r5
int v8; // r9
unsigned __int8 *v9; // r6
int v10; // r0
char *v11; // r1
int v12; // r2
int v13; // t1
int v14; // t1
v3 = (*a1)->GetStringUTFChars(a1, a3, 0);
v4 = strlen(v3);
v5 = malloc(v4);
v6 = malloc(4 * v4 + 64);
v7 = malloc(0x20u);
*v7 = 0x1010101;
v7[1] = 0x1010101;
v7[2] = 0x1010101;
v7[3] = 0x1010101;
qmemcpy(v5, v3, v4);
v8 = sub_D68(v6, v5, v4, v7);
free(v5);
memset(v7, 2, 0x20u);
v9 = malloc(4 * v8 + 32);
v10 = sub_E80(v9, v6, v8, v7);
v11 = &c_cipher;
while ( v10 )
{
v13 = *v11++;
v12 = v13;
--v10;
v14 = *v9++;
if ( v14 != v12 )
return 0;
}
return 1;
}
（1）对伪码进行分析，流程如下
第一加密 v8 = sub_D68(v6, v5, v4, v7);
第二次加密 v10 = sub_E80(v9, v6, v8, v7);
解密结果与 &c_cipher进行比较
（2）动态调试
c_cipher
unsigned char c_cipher[56] = {
0xCA, 0x60, 0x55, 0x30, 0xB5, 0xDB, 0xD4, 0xA6, 0x01, 0x15, 0x3F, 0xB8, 0xBC, 0x4C, 0x9C, 0x88,
0xEA, 0xF4, 0x76, 0xDD, 0x8D, 0x7B, 0x1A, 0x26, 0xDA, 0x74, 0x2C, 0x1D, 0x28, 0x63, 0x4B, 0x88,
0x44, 0x22, 0x7E, 0x21, 0x0E, 0x6C, 0xF4, 0xAE, 0xE4, 0x21, 0xC7, 0x67, 0x21, 0x40, 0xC5, 0x3B,
0xB2, 0x55, 0x92, 0x21, 0x9B, 0x29, 0xFA, 0x33
};
第一加密 v8 = sub_D68(v6, v5, v4, v7);
AES的S盒
输入 flag{0123456789}
key 01010101010101010101010101010101
加密结果
7E D8 C7 01 11 B4 88 27 0E 42 0B 31 59 CB 42 63
68 43 4D 37 D7 F6 9F 0E 03 B7 5B B1 5B C9 4B 6C
验证一下
第二次加密 v10 = sub_E80(v9, v6, v8, v7);
输入
7E D8 C7 01 11 B4 88 27 0E 42 0B 31 59 CB 42 63
68 43 4D 37 D7 F6 9F 0E 03 B7 5B B1 5B C9 4B 6C
key
02020202020202020202020202020202
02020202020202020202020202020202
加密结果
8B A8 1D 1F DF E9 09 98 78 B0 0A CA F7 8F DE 1D
56 6F B4 65 E0 32 DC BA BE 7B 8A 81 A1 41 59 BA
B2 55 92 21 9B 29 FA 33
sub_E80
unsigned int __fastcall sub_E80(int a1, int a2, int a3, int a4)
{
unsigned int result; // r0
int i; // r10
int v9; // r9
int v10; // r1
int v11; // r4
int j; // r5
__int64 v13; // kr00_8
sub_DA8(a4);
result = sub_994(a1, a2, a3, 8);
for ( i = 0; i != result >> 3; ++i )
{
v9 = 2 * i + 1;
v10 = *(a1 + 8 * i) + dword_5E50[0];
v11 = *(a1 + 4 * v9) + unk_5E54;
for ( j = 0; j != 12; ++j )
{
v13 = *&dword_5E50[2 * j + 2];
v10 = __ROR4__(v10 ^ v11, 32 - v11) + v13;
v11 = __ROR4__(v10 ^ v11, 32 - v10) + HIDWORD(v13);
}
*(a1 + 8 * i) = v10;
*(a1 + 4 * v9) = v11;
}
return result;
}
我使用trace进行的算法还原
对trace进行分析，算法如下
输入
7E D8 C7 01 11 B4 88 27 0E 42 0B 31 59 CB 42 63
68 43 4D 37 D7 F6 9F 0E 03 B7 5B B1 5B C9 4B 6C
加密结果
8B A8 1D 1F DF E9 09 98 78 B0 0A CA F7 8F DE 1D
56 6F B4 65 E0 32 DC BA BE 7B 8A 81 A1 41 59 BA
B2 55 92 21 9B 29 FA 33
输入由32补全40位，08补全
输入
7E D8 C7 01 11 B4 88 27 0E 42 0B 31 59 CB 42 63
68 43 4D 37 D7 F6 9F 0E 03 B7 5B B1 5B C9 4B 6C
08 08 08 08 08 08 08 08
输入长度 40 >> 3 = 5
进行了5次大循环，每次循环进行了12+1次
每次小循环使用的数组：
uint32_t dword_C30DDE50[] = {
0x751EF375, 0x9A777FE8,
0xCF327F57, 0x378B39DC,
0xB0CF04C8, 0xA0AA7A08,
0xCE01CEA5, 0xF400183,
0x58FBD0BB, 0x8D44D42A,
0x59F0BB54, 0xB632AA3A,
0x40030399, 0xA9C58CAC,
0xB1F1F560, 0x29796A99,
0x513FE4A6, 0x8204DA32,
0xC435BFCA, 0x465706D,
0xECB351E9, 0x9A83C02C,
0x4615414F, 0x8728979D,
0x8CED0CFA, 0x15647C9C};
逆推
还原算法
uint32_t dword_C30DDE50[] = {
0x751EF375, 0x9A777FE8,
0xCF327F57, 0x378B39DC,
0xB0CF04C8, 0xA0AA7A08,
0xCE01CEA5, 0xF400183,
0x58FBD0BB, 0x8D44D42A,
0x59F0BB54, 0xB632AA3A,
0x40030399, 0xA9C58CAC,
0xB1F1F560, 0x29796A99,
0x513FE4A6, 0x8204DA32,
0xC435BFCA, 0x465706D,
0xECB351E9, 0x9A83C02C,
0x4615414F, 0x8728979D,
0x8CED0CFA, 0x15647C9C};
uint32_t plaintext[] = {0x01C7D87E, 0x2788B411, 0x310B420E, 0x6342CB59, 0x374D4368, 0x0E9FF6D7, 0xB15BB703, 0x6C4BC95B, 0x08080808, 0x08080808};
int len = sizeof(plaintext) / sizeof(uint32_t);
for(int i = 0; i < len/2; i++){
int R1 = plaintext[i*2];
int R4 = plaintext[i*2+1];
int R2,R5,R6,R9;
int R3=0xC30DDE50;
for(int j = 0; j < 13; j++){
if(j == 0){
R1 = R1 + dword_C30DDE50[j*2];
R4 = R4 + dword_C30DDE50[j*2+1];
}else{
R1 = R1 ^ R4;
R6 = 0x20 - R4;
R1 = ror(R1, R6);
R2 = dword_C30DDE50[j*2];
R6 = dword_C30DDE50[j*2+1];
R1 = R1 + R2;
R2 = R1 ^ R4;
R4 = 0x20 - R1;
R2 = ror(R2, R4);
R4 = R2 + R6;
}
}
printf("0x%x,0x%x, \n", R1, R4);
}
对加密算法进行逆运算
uint32_t dword_C30DDE50[] = {
0x751EF375, 0x9A777FE8,
0xCF327F57, 0x378B39DC,
0xB0CF04C8, 0xA0AA7A08,
0xCE01CEA5, 0xF400183,
0x58FBD0BB, 0x8D44D42A,
0x59F0BB54, 0xB632AA3A,
0x40030399, 0xA9C58CAC,
0xB1F1F560, 0x29796A99,
0x513FE4A6, 0x8204DA32,
0xC435BFCA, 0x465706D,
0xECB351E9, 0x9A83C02C,
0x4615414F, 0x8728979D,
0x8CED0CFA, 0x15647C9C};
uint32_t cipher[] = {0x305560CA, 0xA6D4DBB5, 0xB83F1501, 0x889C4CBC, 0xDD76F4EA, 0x261A7B8D,
0x1D2C74DA, 0x884B6328, 0x217E2244, 0xAEF46C0E, 0x67C721E4, 0x3BC54021,
0x219255B2, 0x33FA299B};
// 解密
int len2 = sizeof(cipher) / sizeof(uint32_t);
for(int i = 0; i< len2/2; i++){
int R1_cipher = cipher[i*2];
int R4_cipher = cipher[i*2+1];
int R1, R4;
for(int j = 12; j >=0; j--){
if(j == 0){
R1_cipher = R1_cipher - dword_C30DDE50[j*2];
R4_cipher = R4_cipher - dword_C30DDE50[j*2+1];
}else{
R4_cipher = rol(R4_cipher - dword_C30DDE50[j*2+1], 0x20 - R1_cipher) ^ R1_cipher;
R1_cipher = rol(R1_cipher - dword_C30DDE50[j*2], 0x20 - R4_cipher) ^ R4_cipher;
}
}
printf("%x %x \n", R1_cipher, R4_cipher);
}
使用AES Decrypt
[复制链接]
发表于 2021-11-2 11:40
