[SCTF2019]babyre
IDA 打开，没法 F5，发现几处花指令
main 函数里总共有三处这样的花指令，把每处的这三句都 nop 掉，再往下又发现一处花
上面两句 jb，jnb 都 nop 掉，然后看到 in 那里
in 的字节码是 E4，后面紧跟一个 C7，C7 是 mov，所以这一定是花，把 E4 改为 90（对应操作码 nop），然后将 test 按 d 还原为数据
修改完后 798~C21 按 p 声明 main 函数，然后再将 C22~F66 声明函数，F5 得到主函数
要求输入三个 password，都正确就能得到 flag，先看第一部分
strcpy(
v17,
"**************.****.**s..*..******.****.***********..***..**..#*..***..***.********************.**..*******..**...*..*.*.**.*");
v9[0] = 0LL;
v9[1] = 0LL;
v10 = 0;
v7 = &v17[22];
strcpy(v8, "sctf_9102");
puts((const char *)(unsigned int)"plz tell me the shortest password1:");
scanf("%s", v15);
v6 = 1;
while ( v6 )
{
v4 = *((_BYTE *)v15 + v5);
switch ( v4 )
{
case 'w':
v7 -= 5;
break;
case 's':
v7 += 5;
break;
case 'd':
++v7;
break;
case 'a':
--v7;
break;
case 'x':
v7 += 25;
break;
case 'y':
v7 -= 25;
break;
default:
v6 = 0;
break;
}
++v5;
if ( *v7 != 46 && *v7 != 35 )
v6 = 0;
if ( *v7 == 35 )
{
puts("good!you find the right way!\nBut there is another challenge!");
break;
}
}
先声明了一个迷宫，提取出来如下
然后一共有 w，s，d，a，x，y 五个操作，迷宫视为三维迷宫，走法为 w：上，s：下，d：右，a：左，x：下一层，y：上一层，手动走一下得到路径 ddwwxxssxaxwwaasasyywwdd
然后看第二部分
输入 pass2 经过 sub_C22() 加密后与 v8 比较，查看 v8 的值如下
然后看一眼 sub_C22()
输入字符分成四个一组，每次取 8 位二进制索引字符，取字符低 6 位，最终一组得到 3 个字符，猜测是 base64，加密一下 "sctf_9102"，得到 "c2N0Zl85MTAy"，验证正确
往下看第三部分
pass3 传入 sub_FFA() 进行验证，返回一则成功，看一下这个函数
其中，sub_143B() 为
sub_1464() 为
变换赋值主要是异或运算，直接可逆，用它的代码写脚本就好
#include <stdio.h>
#include "defs.h"
unsigned int ROR4(unsigned int x, int y) {
return (x << (32 - y) | x >> y) & 0xffffffff;
}
unsigned int ROL4(unsigned int x, int y) {
return (x >> (32 - y) | x << y) & 0xffffffff;
}
__int64 sub_1464(unsigned int a1) {
int v2;
unsigned int v3[288] = {
0xD6, 0x90, 0xE9, 0xFE, 0xCC, 0xE1, 0x3D, 0xB7,
0x16, 0xB6, 0x14, 0xC2, 0x28, 0xFB, 0x2C, 0x05,
0x2B, 0x67, 0x9A, 0x76, 0x2A, 0xBE, 0x04, 0xC3,
0xAA, 0x44, 0x13, 0x26, 0x49, 0x86, 0x06, 0x99,
0x9C, 0x42, 0x50, 0xF4, 0x91, 0xEF, 0x98, 0x7A,
0x33, 0x54, 0x0B, 0x43, 0xED, 0xCF, 0xAC, 0x62,
0xE4, 0xB3, 0x1C, 0xA9, 0xC9, 0x08, 0xE8, 0x95,
0x80, 0xDF, 0x94, 0xFA, 0x75, 0x8F, 0x3F, 0xA6,
0x47, 0x07, 0xA7, 0xFC, 0xF3, 0x73, 0x17, 0xBA,
0x83, 0x59, 0x3C, 0x19, 0xE6, 0x85, 0x4F, 0xA8,
0x68, 0x6B, 0x81, 0xB2, 0x71, 0x64, 0xDA, 0x8B,
0xF8, 0xEB, 0x0F, 0x4B, 0x70, 0x56, 0x9D, 0x35,
0x1E, 0x24, 0x0E, 0x5E, 0x63, 0x58, 0xD1, 0xA2,
0x25, 0x22, 0x7C, 0x3B, 0x01, 0x21, 0x78, 0x87,
0xD4, 0x00, 0x46, 0x57, 0x9F, 0xD3, 0x27, 0x52,
0x4C, 0x36, 0x02, 0xE7, 0xA0, 0xC4, 0xC8, 0x9E,
0xEA, 0xBF, 0x8A, 0xD2, 0x40, 0xC7, 0x38, 0xB5,
0xA3, 0xF7, 0xF2, 0xCE, 0xF9, 0x61, 0x15, 0xA1,
0xE0, 0xAE, 0x5D, 0xA4, 0x9B, 0x34, 0x1A, 0x55,
0xAD, 0x93, 0x32, 0x30, 0xF5, 0x8C, 0xB1, 0xE3,
0x1D, 0xF6, 0xE2, 0x2E, 0x82, 0x66, 0xCA, 0x60,
0xC0, 0x29, 0x23, 0xAB, 0x0D, 0x53, 0x4E, 0x6F,
0xD5, 0xDB, 0x37, 0x45, 0xDE, 0xFD, 0x8E, 0x2F,
0x03, 0xFF, 0x6A, 0x72, 0x6D, 0x6C, 0x5B, 0x51,
0x8D, 0x1B, 0xAF, 0x92, 0xBB, 0xDD, 0xBC, 0x7F,
0x11, 0xD9, 0x5C, 0x41, 0x1F, 0x10, 0x5A, 0xD8,
0x0A, 0xC1, 0x31, 0x88, 0xA5, 0xCD, 0x7B, 0xBD,
0x2D, 0x74, 0xD0, 0x12, 0xB8, 0xE5, 0xB4, 0xB0,
0x89, 0x69, 0x97, 0x4A, 0x0C, 0x96, 0x77, 0x7E,
0x65, 0xB9, 0xF1, 0x09, 0xC5, 0x6E, 0xC6, 0x84,
0x18, 0xF0, 0x7D, 0xEC, 0x3A, 0xDC, 0x4D, 0x20,
0x79, 0xEE, 0x5F, 0x3E, 0xD7, 0xCB, 0x39, 0x48,
0xC6, 0xBA, 0xB1, 0xA3, 0x50, 0x33, 0xAA, 0x56,
0x97, 0x91, 0x7D, 0x67, 0xDC, 0x22, 0x70, 0xB2,
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00,
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00
};
v2 = (v3[BYTE2(a1)] << 16) | v3[(unsigned __int8)a1] | (v3[BYTE1(a1)] << 8) | (v3[HIBYTE(a1) ] << 24);
return ROL4(v2, 12) ^ (unsigned int)(ROL4(v2, 8) ^ ROR4(v2, 2)) ^ ROR4(v2, 6);
}
int main() {
int n = 25;
unsigned int v10[30] = {0};
v10[26] = 0xBE040680;
v10[27] = 0xC5AF7647;
v10[28] = 0x9FCC401F;
v10[29] = 0xD8BF92EF;
do {
v10[n] =v10[n + 4] ^ sub_1464(v10[n + 1] ^ v10[n + 2] ^ v10[n + 3]);
--n;
} while (n >= 0);
for (int i = 0; i < 4; i++)
printf("%c%c%c%c", ((char*)&v10[i])[0], ((char*)&v10[i])[1], ((char*)&v10[i])[2], ((char*)&v10[i])[3]);
return 0;
}
得到：fl4g_is_s0_ug1y!
然后输入得到 flag
sctf{ddwwxxssxaxwwaasasyywwdd-c2N0Zl85MTAy(fl4g_is_s0_ug1y!)}
