分析完代码后，可以通过第二关得到字符串为 ???KCTF??????????
分析第四关，可以知道后面接的CTF字符串肯定是 1234567890等，因为会排序，那么可能是以下几种
1, 12, 123, 1234, 12345, 123456, 1234567, 12345678, 123456789, 1234567890
通过动态调试可以知道如果是1234[\x0\x0]这种会被第三关干扰从而生成别的字符串，不符合逻辑，那么大胆猜测就是1-9的字符串
通过第一关可以快速遍历得到所有 前三个字符
如下:
然后我就目测了一下，感觉那个 Hi[ 很像, 然后就走入了误区
合在一起是 Hi[KCTF]1-8, 然后就越走越远了，最后才发现可能是其他的字符+KCTF+1-9
写下如下代码，进行尝试性破解
思路就是遍历所有三个字符符合条件的来匹配，因为实在不知道哪三个字符符合条件，包括后面的判断也没有看出来，只能盲猜所有可能，之前误判为Hi[走错了后，改进为所有可能。
进行第五关的校验，就是所有的字符串参与计算，分拆前面7个字符先初始化得到值。
然后就是1-9的排列组合，直接生成所有排列可能，然后参与计算对比得到结果，结果如下：
带入到程序中， 421KCTF381654729， 验证通过。
补一个sbox表格
char first(unsigned char* str, int len)
{
char tmp = 0;
for (int i = 0; i < len; i++)
{
tmp ^= str[i];
for (int j = 0; j < 8; j++)
{
char v9 = 2 * tmp;
tmp = v9 >= 0 ? v9 : (v9 ^ 7);
}
}
return tmp;
}
void second(char v, int out[200])
{
unsigned char v15 = v;
for (int i = 1; i < 200; i++)
{
if ((v15 & 1) != 0)
v15 = 3 * v15 + 1;
else
v15 >>= 1;
out[i] = v15;
printf("%.2x ", v15);
if ((i + 1) % 0x10 == 0)
printf("\r\n");
}
printf("\r\n");
}
unsigned char to_hex(unsigned char a)
{
return (a >= 0x3a) ? (a - 0x37) : (a - 0x30);
}
int sub_4010B7(unsigned char* str, int len)
{
for (int i = 0; i < len; ++i)
{
str[i] = to_hex(str[i]);
}
return 0;
}
int ctf_1(unsigned char* ctf, int len)
{
int tmp = 0;
int tk = 1;
for (int i = 0; i < len; i++)
{
int v23 = ctf[i] + tmp * 10;
int v24 = v23 - 0x37373737;
if (v23 <= 0x4b435445)
v24 = v23;
tmp = v24;
if (v24 % tk++)
return -1;
}
return 0;
}
int __cdecl main(int argc, const char **argv, const char **envp)
{
/// 取字符串
sub_40103A("%s", (char)str);
/// 异或处理
int len = strlen(str);
char a = first(str, len);
/// 初始化sbox, 直接拷贝即可
sub_40106C();
/// 哈希计算
int v = -1;
for (int i = 0; i < len; i++)
{
v = ((unsigned int*)sbox)[(unsigned char)(v ^ s[i])] ^ (v >> 8);
}
/// must be 0xf52e0765
v ~= v;
/// 字符转换hex?
sub_4010B7(str, len);
/// 第一关，其值为0x07
char tbox[200] = { 0 };
second(a, tbox);
/// 观察tbox的结果, tbox[198] | tbox[197] | tbpx[196] 为 0x07
/// 注意：有符号的情况下不能通过第一关，可以修改 0x12A7 edi==0x07, 后面会用到
char k = str[0] ^ str[1] ^ str[2];
if (k != 0x07) failed(-1);
/// 第二关 {3-5} -> 'KCTF'
if (str[3] != 0x14 || str[4] != 0xc || str[5] != 0x1d || str[6] != 0x0f)
failed(-1);
/// 第三关 {未解}
/// 0x13E4 -> nop*6
ctf = &str[7];
if (ctf_1(ctf, strlen(ctf)!=0)
failed(-1);
/// 第四关
/// 1 - 排序 从低到高
sort(ctf, strlen(ctf));
/// 2 - 比较字符串
char cmp[] = '1234567890_A...';
sub_4010B7(cmp, 10);
/// 3 - 比较ctf
if (strncmp(ctf, cmp, strlen(ctf)))
failed(-1);
/// 第五关
if (v!= 0xF52E0765)
failed(-1);
return ok();
}
char first(unsigned char* str, int len)
{
char tmp = 0;
for (int i = 0; i < len; i++)
{
tmp ^= str[i];
for (int j = 0; j < 8; j++)
{
char v9 = 2 * tmp;
tmp = v9 >= 0 ? v9 : (v9 ^ 7);
}
}
return tmp;
}
void second(char v, int out[200])
{
unsigned char v15 = v;
for (int i = 1; i < 200; i++)
{
if ((v15 & 1) != 0)
v15 = 3 * v15 + 1;
else
v15 >>= 1;
out[i] = v15;
printf("%.2x ", v15);
if ((i + 1) % 0x10 == 0)
printf("\r\n");
}
printf("\r\n");
}
unsigned char to_hex(unsigned char a)
{
return (a >= 0x3a) ? (a - 0x37) : (a - 0x30);
}
int sub_4010B7(unsigned char* str, int len)
{
for (int i = 0; i < len; ++i)
{
str[i] = to_hex(str[i]);
}
return 0;
}
int ctf_1(unsigned char* ctf, int len)
{
int tmp = 0;
int tk = 1;
for (int i = 0; i < len; i++)
{
int v23 = ctf[i] + tmp * 10;
int v24 = v23 - 0x37373737;
if (v23 <= 0x4b435445)
v24 = v23;
tmp = v24;
if (v24 % tk++)
return -1;
}
return 0;
}
int __cdecl main(int argc, const char **argv, const char **envp)
{
/// 取字符串
sub_40103A("%s", (char)str);
/// 异或处理
int len = strlen(str);
char a = first(str, len);
/// 初始化sbox, 直接拷贝即可
sub_40106C();
/// 哈希计算
int v = -1;
for (int i = 0; i < len; i++)
{
v = ((unsigned int*)sbox)[(unsigned char)(v ^ s[i])] ^ (v >> 8);
}
/// must be 0xf52e0765
v ~= v;
/// 字符转换hex?
sub_4010B7(str, len);
/// 第一关，其值为0x07
char tbox[200] = { 0 };
second(a, tbox);
/// 观察tbox的结果, tbox[198] | tbox[197] | tbpx[196] 为 0x07
/// 注意：有符号的情况下不能通过第一关，可以修改 0x12A7 edi==0x07, 后面会用到
char k = str[0] ^ str[1] ^ str[2];
if (k != 0x07) failed(-1);
/// 第二关 {3-5} -> 'KCTF'
if (str[3] != 0x14 || str[4] != 0xc || str[5] != 0x1d || str[6] != 0x0f)
failed(-1);
/// 第三关 {未解}
/// 0x13E4 -> nop*6
ctf = &str[7];
if (ctf_1(ctf, strlen(ctf)!=0)
failed(-1);
/// 第四关
/// 1 - 排序 从低到高
sort(ctf, strlen(ctf));
/// 2 - 比较字符串
char cmp[] = '1234567890_A...';
sub_4010B7(cmp, 10);
/// 3 - 比较ctf
if (strncmp(ctf, cmp, strlen(ctf)))
failed(-1);
/// 第五关
if (v!= 0xF52E0765)
failed(-1);
return ok();
}
unsigned char st = 0x20;
unsigned char se = 0x7f;
for (unsigned char x = st; x <= se; x++)
{
for (unsigned char y = st; y <= se; y++)
{
for (unsigned char z = st; z <= se; z++)
{
if ((to_hex(x) ^ to_hex(y) ^ to_hex(z)) == 0x07)
{
printf("%c%c%c\n", x, y, z);
}
}
}
}
unsigned char st = 0x20;
unsigned char se = 0x7f;
for (unsigned char x = st; x <= se; x++)
{
for (unsigned char y = st; y <= se; y++)
{
for (unsigned char z = st; z <= se; z++)
{
if ((to_hex(x) ^ to_hex(y) ^ to_hex(z)) == 0x07)
{
printf("%c%c%c\n", x, y, z);
}
}
}
}
unsigned char st = 0x20;
unsigned char se = 0x7f;
unsigned char buf[9] = { 0 };
memcpy(&buf[3], "KCTF", 4);
for (unsigned char x = st; x <= se; x++)
{
for (unsigned char y = st; y <= se; y++)
{
for (unsigned char z = st; z <= se; z++)
{
if ((to_hex(x) ^ to_hex(y) ^ to_hex(z)) == 0x07)
{
buf[0] = x;
buf[1] = y;
buf[2] = z;
int v = get_ctf((char*)buf, 7);
check(v, (char *)buf);
}
}
}
}
int get_ctf(char* s, int len)
{
int v = -1;
for (int i = 0; i < len; i++)
{
v = ((unsigned int*)sbox)[(unsigned char)(v ^ s[i])] ^ (v >> 8);
}
return v;
}
void check(int v, char *s)
{
char arr[] = { '1', '2', '3', '4', '5', '6', '7', '8', '9', '0'};
char arr2[sizeof(arr)] = { 0 };
for (int i = 1; i < sizeof(arr); i++)
{
arr2[i] = 0;
memcpy(arr2, arr, i);
do {
if (check_ctf2(arr2, i, v)) {
printf("found: \n");
printf("%s\n", s);
print_hex((unsigned char*)arr2, i);
printf("%s\n", arr2);
}
} while (std::next_permutation(arr2, arr2 + i));
}
}
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