2.5现在我们有两种选择,一是根据给出的错误信息,搜索对应的字符串,二是下API断点,栈回溯分析。我们这里先用简单点的字符串搜索试试,如果搜索不到,或者有多处,我们再改用第二种方法
2.19我们先将[EDI+0x1C]的值放一放,我们发现上面要走到这个关键跳转,需要满足几个条件,所以这个函数应该就是我们要找的进行用户名与密码校验的函数,我们从这个函数开始处开始分析。
2.20分析如下,第一个,第二个call分别判断用户名与密码是否为空,第3个call将密码转换为十六进制字节数据,并保存在K[10]字节数组中,后面是对密码的校验,其中K[3]有3种选择,对应三种不同算法,今天我们只分析K[3] = 9C的情况,这种情况其实密码只有16个符号。
2.21,其中有两个关键的函数需要详细的分析,0x00407644处的函数,该函数是对K[0](密码第1个字节)与K[6](第7个字节)进行处理,我们将该函数叫CALL1
2.25这里先将用户名转换ascii版,再传入一个加密函数对用户名进行加密,返回一个4字节的数,再将每一个字节与K[10]数组中的一个元素进行对比,看是否一样。我们将返回的4字节的数看成有4个元素的字节数组R[4]的话,K[4]=R[0],K[5]=R[1],K[6]=R[2],K[7]=R[3],接下来就是我们之前分析的[EDI+0x1C]>=9,现在我们往上可以找到[EDI+0x1C]保存的就是我们前面CALL1的返回值(扩充为4字节之后),所以EAX1>=9,那么现在就只剩下用户名加密函数要分析了。
直接用。这里注意,该加密函数有四个参数,参数1为需要加密的用户名,参数2为1,参数3为0,参数4为上面的EAX2 / 0xB.
#include "pch.h" #include <iostream> #include <windows.h> #include <time.h> DWORD g_EcodeArray[]={ 0x39cb44b8, 0x23754f67, 0x5f017211, 0x3ebb24da, 0x351707c6, 0x63f9774b, 0x17827288, 0x0fe74821, 0x5b5f670f, 0x48315ae8, 0x785b7769, 0x2b7a1547, 0x38d11292, 0x42a11b32, 0x35332244, 0x77437b60, 0x1eab3b10, 0x53810000, 0x1d0212ae, 0x6f0377a8, 0x43c03092, 0x2d3c0a8e, 0x62950cbf, 0x30f06ffa, 0x34f710e0, 0x28f417fb, 0x350d2f95, 0x5a361d5a, 0x15cc060b, 0x0afd13cc, 0x28603bcf, 0x3371066b, 0x30cd14e4, 0x175d3a67, 0x6dd66a13, 0x2d3409f9, 0x581e7b82, 0x76526b99, 0x5c8d5188, 0x2c857971, 0x15f51fc0, 0x68cc0d11, 0x49f55e5c, 0x275e4364, 0x2d1e0dbc, 0x4cee7ce3, 0x32555840, 0x112e2e08, 0x6978065a, 0x72921406, 0x314578e7, 0x175621b7, 0x40771dbf, 0x3fc238d6, 0x4a31128a, 0x2dad036e, 0x41a069d6, 0x25400192, 0x00dd4667, 0x6afc1f4f, 0x571040ce, 0x62fe66df, 0x41db4b3e, 0x3582231f, 0x55f6079a, 0x1ca70644, 0x1b1643d2, 0x3f7228c9, 0x5f141070, 0x3e1474ab, 0x444b256e, 0x537050d9, 0x0f42094b, 0x2fd820e6, 0x778b2e5e, 0x71176d02, 0x7fea7a69, 0x5bb54628, 0x19ba6c71, 0x39763a99, 0x178d54cd, 0x01246e88, 0x3313537e, 0x2b8e2d17, 0x2a3d10be, 0x59d10582, 0x37a163db, 0x30d6489a, 0x6a215c46, 0x0e1c7a76, 0x1fc760e7, 0x79b80c65, 0x27f459b4, 0x799a7326, 0x50ba1782, 0x2a116d5c, 0x63866e1b, 0x3f920e3c, 0x55023490, 0x55b56089, 0x2c391fd1, 0x2f8035c2, 0x64fd2b7a, 0x4ce8759a, 0x518504f0, 0x799501a8, 0x3f5b2cad, 0x38e60160, 0x637641d8, 0x33352a42, 0x51a22c19, 0x085c5851, 0x032917ab, 0x2b770ac7, 0x30ac77b3, 0x2bec1907, 0x035202d0, 0x0fa933d3, 0x61255df3, 0x22ad06bf, 0x58b86971, 0x5fca0de5, 0x700d6456, 0x56a973db, 0x5ab759fd, 0x330e0be2, 0x5b3c0ddd, 0x495d3c60, 0x53bd59a6, 0x4c5e6d91, 0x49d9318d, 0x103d5079, 0x61ce42e3, 0x7ed5121d, 0x14e160ed, 0x212d4ef2, 0x270133f0, 0x62435a96, 0x1fa75e8b, 0x6f092fbe, 0x4a000d49, 0x57ae1c70, 0x004e2477, 0x561e7e72, 0x468c0033, 0x5dcc2402, 0x78507ac6, 0x58af24c7, 0x0df62d34, 0x358a4708, 0x3cfb1e11, 0x2b71451c, 0x77a75295, 0x56890721, 0x0fef75f3, 0x120f24f1, 0x01990ae7, 0x339c4452, 0x27a15b8e, 0x0ba7276d, 0x60dc1b7b, 0x4f4b7f82, 0x67db7007, 0x4f4a57d9, 0x621252e8, 0x20532cfc, 0x6a390306, 0x18800423, 0x19f3778a, 0x462316f0, 0x56ae0937, 0x43c2675c, 0x65ca45fd, 0x0d604ff2, 0x0bfd22cb, 0x3afe643b, 0x3bf67fa6, 0x44623579, 0x184031f8, 0x32174f97, 0x4c6a092a, 0x5fb50261, 0x01650174, 0x33634af1, 0x712d18f4, 0x6e997169, 0x5dab7afe, 0x7c2b2ee8, 0x6edb75b4, 0x5f836fb6, 0x3c2a6dd6, 0x292d05c2, 0x052244db, 0x149a5f4f, 0x5d486540, 0x331d15ea, 0x4f456920, 0x483a699f, 0x3b450f05, 0x3b207c6c, 0x749d70fe, 0x417461f6, 0x62b031f1, 0x2750577b, 0x29131533, 0x588c3808, 0x1aef3456, 0x0f3c00ec, 0x7da74742, 0x4b797a6c, 0x5ebb3287, 0x786558b8, 0x00ed4ff2, 0x6269691e, 0x24a2255f, 0x62c11f7e, 0x2f8a7dcd, 0x643b17fe, 0x778318b8, 0x253b60fe, 0x34bb63a3, 0x5b03214f, 0x5f1571f4, 0x1a316e9f, 0x7acf2704, 0x28896838, 0x18614677, 0x1bf569eb, 0x0ba85ec9, 0x6aca6b46, 0x1e43422a, 0x514d5f0e, 0x413e018c, 0x307626e9, 0x01ed1dfa, 0x49f46f5a, 0x461b642b, 0x7d7007f2, 0x13652657, 0x6b160bc5, 0x65e04849, 0x1f526e1c, 0x5a0251b6, 0x2bd73f69, 0x2dbf7acd, 0x51e63e80, 0x5cf2670f, 0x21cd0a03, 0x5cff0261, 0x33ae061e, 0x3bb6345f, 0x5d814a75, 0x257b5df4, 0x0a5c2c5b, 0x16a45527, 0x16f23945}; int __cdecl EncodeUserName(const char *pszUserName, int a2, char a3, unsigned short a4) { const char *v4; // edx@1 signed int v5; // esi@1 signed int v6; // edi@1 unsigned __int8 v7; // bl@2 int v8; // eax@3 int v9; // ecx@3 int v10; // ecx@4 int result; // eax@4 int v12; // ecx@5 unsigned __int8 v13; // [sp+8h] [bp-10h]@2 unsigned __int8 v14; // [sp+Ch] [bp-Ch]@2 unsigned __int8 v15; // [sp+10h] [bp-8h]@2 int v16; // [sp+14h] [bp-4h]@1 v4 = pszUserName; v16 = 0; v5 = strlen(pszUserName); v6 = 0; if (v5 <= 0) { result = 0; } else { v13 = 0; v14 = 0; v7 = 15 * a4; v15 = 17 * a3; do { v8 = toupper(v4[v6]); v9 = v16 + g_EcodeArray[v8]; if (a2) { v10 = g_EcodeArray[v7] + g_EcodeArray[v15] + g_EcodeArray[(unsigned __int8)(v8 + 47)] * (g_EcodeArray[(unsigned __int8)(v8 + 13)] ^ v9); result = g_EcodeArray[v14] + v10; v16 = g_EcodeArray[v14] + v10; } else { v12 = g_EcodeArray[v7] + g_EcodeArray[v15] + g_EcodeArray[(unsigned __int8)(v8 + 23)] * (g_EcodeArray[(unsigned __int8)(v8 + 63)] ^ v9); result = g_EcodeArray[v13] + v12; v16 = g_EcodeArray[v13] + v12; } v14 += 19; ++v6; v15 += 9; v7 += 13; v13 += 7; v4 = pszUserName; } while (v6 < v5); } return result; } int main() { srand(time(NULL)); //设置随机种子 byte k[10] = { 0x11,0x22,0x33,0x9C,0x55,0x66,0x77,0x88,0x99,0xAA }; int dwRet =rand()% 0x3E8; char szName[100] = { 0 }; printf("请输入用户名:"); scanf_s("%s", szName, 100); //用户名,加密 DWORD dwKey = EncodeUserName(szName, 1, 0, dwRet); //cmp k[4] ,retValue&0xFF //cmp k[5] ,retValue>>8&0xFF //cmp k[6] ,retValue>>16&0xFF //cmp k[7] ,retValue>>24&0xFF k[4] = dwKey & 0xFF; k[5] = dwKey >> 8 & 0xFF; k[6] = dwKey >> 16 & 0xFF; k[7] = dwKey >> 24 & 0xFF; //AL = (K[0] ^ K[6] ^ 0X18 + 0X3D) ^ 0XA7 while (true) { byte k0 = rand()%0xFF; byte k6 = k[6]; DWORD EAX1 = ((k0^k6 ^ 0x18 + 0x3D) ^ 0xA7) & 0xFF; if (EAX1 >= 9) { k[0] = k0; k[6] = k6; break; } } //ESI = (0X100*(K[1]^K[7] & 0XFF) + K[2]^K[5] & 0XFF)&0XFFFF //EAX = (((ESI^0X7892)+0X4D30)^0X3421)&0XFFFF / 0XB //判断余数 是否为0,为0返回商,不为0返回0 while (true) { byte k1 = rand() % 0xFF; byte k7 = k[7]; byte k2 = rand() % 0xFF; byte k5 = k[5]; DWORD ESI = (0X100 * (k1 ^ k7 & 0xFF) + k2 ^ k5 & 0xFF) & 0xFFFF; DWORD EAX2 = (((ESI ^ 0x7892) + 0x4D30) ^ 0x3421) & 0xFFFF; if (EAX2 % 0xB ==0 && EAX2 / 0xB == dwRet) { k[1] = k1; k[7] = k7; k[2] = k2; k[5] = k5; break; } } printf("%02X%02X-%02X%02X-%02X%02X-%02X%02X-%02X%02X\n", k[0], k[1], k[2], k[3], k[4], k[5], k[6], k[7], k[8], k[9]); system("pause"); return 0; }