xD MAZE

迷宫题，不过不是上下左右，而是只有前进
0-512，1-64，2-8，3-1
每一次v14加上前进的单位后，需要保证map[v14]==32

地图map的大小为4096，即64x64
将map转成64x64的形式，并将map中所有的32化为1，所有的非32化为0
于是0-512-向下8行，1-64-向下1行，2-8-向右8个单位，3-1-向右1个单位，手动走完迷宫即可

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for i in range(64):tmp=""for j in range(64):if map[i*64+j]==32:tmp+="1,"else:tmp+="0,"print(tmp)
# hgame{3120113031203203222231003011}

upx magic 0

输入中的每个字符先左移8位，然后经过一个CRC16变换，结果与数组v14比较

直接写爆破脚本即可得到flag

arr=[0x8d68,0x9d49,0x2a12,0xab1a,0xcbdc,0xb92b,0x2e32,0x9f59,0xddcd,0x9d49,0xa90a,0xe70,0xf5cf,0xa50,0x5af5,0xff9f,0x9f59,0xbd0b,0x58e5,0x3823,0xbf1b,0x78a7,0xab1a,0x48c4,0xa90a,0x2c22,0x9f59,0x5cc5,0x5ed5,0x78a7,0x2672,0x5695]
flag=""
for i in range(len(arr)):for j in range(32,128):tmp=(j<<8)&0xfffffor k in range(8):if tmp&0x8000!=0:tmp=((2*tmp)^0x1021)&0xffffelse:tmp=(2*tmp)&0xffffif tmp==arr[i]:flag+=chr(j)
print(flag)
# hgame{noW_YOu~koNw-UPx~mAG|C_@Nd~crC16}

fake shell

对输入的验证在sub_176C函数，是一个常规的RC4加密过程

直接用密钥"happyhg4me!“解密发现不对，对aHappyhg4me查找交叉引用发现密钥被修改为"w0wy0ugot1t”

用"w0wy0ugot1t"作为密钥解密RC4即可得到flag

#include<stdio.h>
void rc4_init(unsigned char* s, unsigned char* key, unsigned long Len_k)
{int i = 0, j = 0;char k[256] = { 0 };unsigned char tmp = 0;for (i = 0; i < 256; i++) {s[i] = i;k[i] = key[i % Len_k];}for (i = 0; i < 256; i++) {j = (j + s[i] + k[i]) % 256;tmp = s[i];s[i] = s[j];s[j] = tmp;}
}void rc4_crypt(unsigned char* Data, unsigned long Len_D, unsigned char* key, unsigned long Len_k) //加解密
{unsigned char s[256];rc4_init(s, key, Len_k);int i = 0, j = 0, t = 0;unsigned long k = 0;unsigned char tmp;for (k = 0; k < Len_D; k++) {i = (i + 1) % 256;j = (j + s[i]) % 256;tmp = s[i];s[i] = s[j];s[j] = tmp;t = (s[i] + s[j]) % 256;Data[k] = Data[k] ^ s[t];}
}
void main()
{//密钥unsigned char key[] = "w0wy0ugot1t";unsigned long key_len = sizeof(key) - 1;//密文unsigned char data[] = {182, 148, 250, 143, 61, 95, 178, 224, 234, 15, 210, 102, 152, 108, 157, 231, 27, 8, 64, 113, 197, 190, 111, 109, 124, 123, 9, 141, 168, 189, 243, 246};//解密rc4_crypt(data, sizeof(data), key, key_len);for (int i = 0; i < sizeof(data); i++){printf("%c", data[i]);}printf("\n");return;
}
//hgame{s0meth1ng_run_bef0r_m4in?}

creakme2

main函数中的逻辑很清晰
输入经过XTEA加密，密钥key为[1,2,3,4]，密文与已知的res比较

进入XTEA加密函数，ida反编译出来就是常规XTEA加密算法

其实不然，查看汇编代码，0x140001139地址处有条idiv ecx指令，ecx是sum左移31位后的结果
即如果sum的二进制最高位为0，idiv ecx会触发除0异常，跳到0x140001141地址处进行处理，使sum^=0x1234567

于是加密过程实际如下

void encipher(unsigned int num_rounds, uint32_t v[2], uint32_t const key[4]) {unsigned int i;uint32_t v0 = v[0], v1 = v[1], sum = 0, delta = 0x9E3779B1;for (i = 0; i < num_rounds; i++) {v0 += (((v1 << 4) ^ (v1 >> 5)) + v1) ^ (sum + key[sum & 3]);sum += delta;if (sum >> 31 == 0)sum ^= 0x1234567;v1 += (((v0 << 4) ^ (v0 >> 5)) + v0) ^ (sum + key[(sum >> 11) & 3]);}v[0] = v0; v[1] = v1;
}

对解密过程做相应的变化，解密XTEA即可得到flag

#include <stdio.h>
#include <stdint.h>void encipher(unsigned int num_rounds, uint32_t v[2], uint32_t const key[4]) {unsigned int i;uint32_t v0 = v[0], v1 = v[1], sum = 0, delta = 0x9E3779B1;for (i = 0; i < num_rounds; i++) {v0 += (((v1 << 4) ^ (v1 >> 5)) + v1) ^ (sum + key[sum & 3]);sum += delta;if (sum >> 31 == 0)sum ^= 0x1234567;v1 += (((v0 << 4) ^ (v0 >> 5)) + v0) ^ (sum + key[(sum >> 11) & 3]);}v[0] = v0; v[1] = v1;
}void decipher(unsigned int num_rounds, uint32_t v[2], uint32_t const key[4]) {unsigned int i;uint32_t v0 = v[0], v1 = v[1], delta = 0x9E3779B1, sum = -946975483;for (i = 0; i < num_rounds; i++) {v1 -= (((v0 << 4) ^ (v0 >> 5)) + v0) ^ (sum + key[(sum >> 11) & 3]);if (sum >> 31 == 0)sum ^= 0x1234567;sum -= delta;v0 -= (((v1 << 4) ^ (v1 >> 5)) + v1) ^ (sum + key[sum & 3]);}v[0] = v0; v[1] = v1;
}int main()
{uint32_t v[2] = { 0x457E62CF,0x9537896C };//hgame{SE//0x1F7E7F72,0xF7A073D8->H_s0und5//0x8E996868,0x40AFAF99->_50_1nte//0x0F990E34,0x196F4086->re5ting}uint32_t const k[4] = { 1,2,3,4 };int n = sizeof(v) / sizeof(uint32_t);unsigned int r = 32;decipher(r, v, k);for (int i = 0; i < n; i++){for (int j = 0; j < 4; j++){printf("%c", (v[i] >> (j * 8)) & 0xFF);}}printf("\n");return 0;
}
//hgame{SEH_s0und5_50_1ntere5ting}

upx magic 1

elf文件加了upx壳，脱壳工具去不掉壳，于是手动调试去壳
参考：ELF64手脱UPX壳实战
这里没有dump内存文件，直接在ida中分析
在0x401417地址的call rax指令处下断点，此时rax寄存器中保存的即为main函数的地址

F7步入sub_400B8D函数，F5反编译

实际上就是和upx magic 0一样的逻辑，爆破即可得到flag

arr=[0x8d68,0x9d49,0x2a12,0xab1a,0xcbdc,0xb92b,0x2e32,0x9f59,0xddcd,0x9d49,0xa90a,0xe70,0xf5cf,0x5ed5,0x3c03,0x7c87,0x2672,0xab1a,0xa50,0x5af5,0xff9f,0x9f59,0xbd0b,0x58e5,0x3823,0xbf1b,0x78a7,0xab1a,0x48c4,0xa90a,0x2c22,0x9f59,0x5cc5,0x5ed5,0x78a7,0x2672,0x5695]
flag=""
for i in range(len(arr)):for j in range(32,128):tmp=(j<<8)&0xfffffor k in range(8):if tmp&0x8000!=0:tmp=((2*tmp)^0x1021)&0xffffelse:tmp=(2*tmp)&0xffffif tmp==arr[i]:flag+=chr(j)
print(flag)
# hgame{noW_YOu~koNw-rea1_UPx~mAG|C_@Nd~crC16}

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