catchme

安卓逆向，java层传递输入，调用native层的check方法
ida打开.so文件，没有直接找到check方法，JNI_OnLoad也看不出什么
Findcrypt查到AES的S盒，通过对S盒查找交叉引用，来到sub_B2A4函数

输入经过AES加密，再进行变表base64，最后与true_cipher进行比较
在AES解密前，AES_key和true_cipher都需要进行异或变换

from Crypto.Cipher import AES
import base64key=[0x24, 0x3C, 0x3D, 0x37, 0x36, 0x21, 0x35, 0x26, 0x3F, 0x37, 0x32, 0x2A, 0x72, 0x72, 0x72, 0x72]
for i in range(len(key)):key[i]^=0x53
print(bytes(key))
# wonderfulday!!!!enc=[0x4F, 0x1C, 0x36, 0x49, 0x09, 0x3A, 0x3F, 0x07, 0x4D, 0x3D, 0x22, 0x39, 0x00, 0x0A, 0x22, 0x25, 0x06, 0x09, 0x01, 0x20, 0x4A, 0x1B, 0x51, 0x51]
for i in range(len(enc)):enc[i]^=0x6C
print(bytes(enc))
# #pZ%eVSk!QNUlfNIjemL&w==ori_table="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"
cur_table="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz!@#$%^&*()+/"cipher=base64.b64decode(bytes(enc).decode().translate(str.maketrans(cur_table,ori_table)))aes=AES.new(bytes(key),AES.MODE_ECB)
print(aes.decrypt(cipher))
# flag{weu/.,iopl}

FFunction

ida打开my_plugin.dll，函数窗口第一个函数f，最后调用了memcmp函数，猜测这个f就是校验函数
在f函数的第一行代码处下断点，通过尝试，当输入的长度为30时，程序会断下来

观察此时的input，发现程序在调用f函数前，对原本的输入进行了位置变换和标准base64变换

s="flag{abcdefghijklmnopqrstuvwx}"
# 1
# f}lxawgv{uatbscrdqepfognhmiljk
# Zn1seGF3Z3Z7dWF0YnNjcmRxZXBmb2duaG1pbGpr

第32到第45行代码，是将input倒置

# 2
# rpGbp1Gaud2bmBXZxRmcjNnY0FWd7Z3Z3FGes1nZ

然后是将倒置后的input，每个字符的高四位和低四位拆分成两个字节

# 3
# ord("r")==0x72 -> 0x70 0x02
# ord("p")==0x70 -> 0x70 0x00

最后就是TEA加密，delta已知，key已知，密文(a2)已知

于是，先解密TEA

#include <stdio.h>
#include <stdint.h>//加密函数
void encrypt(unsigned int num_rounds, uint32_t* v, uint32_t* k) {uint32_t v0 = v[0], v1 = v[1], sum = 0, i;uint32_t delta = 0x79B99E37;uint32_t k0 = k[0], k1 = k[1], k2 = k[2], k3 = k[3];for (i = 0; i < num_rounds; i++) {sum += delta;v0 -= ((v1 << 4) + k0) ^ (v1 + sum) ^ ((v1 >> 5) + k1);v1 -= ((v0 << 4) + k2) ^ (v0 + sum) ^ ((v0 >> 5) + k3);}v[0] = v0; v[1] = v1;
}//解密函数
void decrypt(unsigned int num_rounds, uint32_t* v, uint32_t* k) {uint32_t v0 = v[0], v1 = v[1], i;uint32_t delta = 0x79B99E37,sum = delta*num_rounds;uint32_t k0 = k[0], k1 = k[1], k2 = k[2], k3 = k[3];for (i = 0; i<num_rounds; i++) {v1 += ((v0 << 4) + k2) ^ (v0 + sum) ^ ((v0 >> 5) + k3);v0 += ((v1 << 4) + k0) ^ (v1 + sum) ^ ((v1 >> 5) + k1);sum -= delta;}v[0] = v0; v[1] = v1;
}//打印数据 hex_or_chr: 1-hex 0-chr
void dump_data(uint32_t * v,int n,bool hex_or_chr)
{if(hex_or_chr){for(int i=0;i<n;i++){printf("0x%x,",v[i]);}}else{for (int i = 0; i < n; i++){for (int j = 0; j < sizeof(uint32_t)/sizeof(uint8_t); j++){printf("0x%02x,", (v[i] >> (j * 8)) & 0xFF);}}}printf("\n");return;
}int main()
{// v为要加解密的数据uint32_t v[] = { 0x5c15754c,0xd1d781e7,0x501bf173,0xcb4db222,0x215d61f5,0x3fca9ee7,0x7c76b5c7,0xc7dd8cb9,0x990d23fa,0xbab1ad3,0x8e12c932,0xd307baf2,0xe52dd123,0xfbb68f2c,0xbdd853e3,0x892e1e4e,0x39dd66fa,0x87feec65,0x307c5e60,0x340c6c00 };// k为加解密密钥，4个32位无符号整数，密钥长度为128位uint32_t k[4] = { 0x0BABEC0FE,0x0DEADBEEF,0x0FACEB00C,0x0DEADC0DE };// num_rounds，建议取值为32unsigned int r = 32;int n = sizeof(v) / sizeof(uint32_t);/*printf("加密前明文数据：");dump_data(v,n,1);for(int i=0;i<n/2;i++){encrypt(r,&v[i*2], k);}printf("加密后密文数据：");dump_data(v,n,1);*/for(int i=0;i<n/2;i++){decrypt(r,&v[i*2], k);}printf("解密后明文数据：");dump_data(v,n,1);printf("解密后明文字符：");dump_data(v,n,0);return 0;
}
// 解密后明文数据：0x6400130,0x4600440,0x2500740,0x8500330,0xa400c60,0x3600e60,0x9300c60,0x4600130,0x2400330,0xd400450,0x2500770,0x2600850,0x6500f60,0x8500030,0xe400730,0xa500030,0x5400860,0x9400750,0x1300370,0xa500e60,
// 解密后明文字符：0x30,0x01,0x40,0x06,0x40,0x04,0x60,0x04,0x40,0x07,0x50,0x02,0x30,0x03,0x50,0x08,0x60,0x0c,0x40,0x0a,0x60,0x0e,0x60,0x03,0x60,0x0c,0x30,0x09,0x30,0x01,0x60,0x04,0x30,0x03,0x40,0x02,0x50,0x04,0x40,0x0d,0x70,0x07,0x50,0x02,0x50,0x08,0x60,0x02,0x60,0x0f,0x50,0x06,0x30,0x00,0x50,0x08,0x30,0x07,0x40,0x0e,0x30,0x00,0x50,0x0a,0x60,0x08,0x40,0x05,0x50,0x07,0x40,0x09,0x70,0x03,0x30,0x01,0x60,0x0e,0x50,0x0a,

再按照3、2、1的顺序将解TEA得到的明文转换成flag即可

import base64
s="flag{abcdefghijklmnopqrstuvwx}"
# 1
# f}lxawgv{uatbscrdqepfognhmiljk
# Zn1seGF3Z3Z7dWF0YnNjcmRxZXBmb2duaG1pbGpr# 2
# rpGbp1Gaud2bmBXZxRmcjNnY0FWd7Z3Z3FGes1nZ# 3
# ord("r")==0x72 -> 0x70 0x02
# ord("p")==0x70 -> 0x70 0x00# 4
# TEAtea_plain=[0x30,0x01,0x40,0x06,0x40,0x04,0x60,0x04,0x40,0x07,0x50,0x02,0x30,0x03,0x50,0x08,0x60,0x0c,0x40,0x0a,0x60,0x0e,0x60,0x03,0x60,0x0c,0x30,0x09,0x30,0x01,0x60,0x04,0x30,0x03,0x40,0x02,0x50,0x04,0x40,0x0d,0x70,0x07,0x50,0x02,0x50,0x08,0x60,0x02,0x60,0x0f,0x50,0x06,0x30,0x00,0x50,0x08,0x30,0x07,0x40,0x0e,0x30,0x00,0x50,0x0a,0x60,0x08,0x40,0x05,0x50,0x07,0x40,0x09,0x70,0x03,0x30,0x01,0x60,0x0e,0x50,0x0a]
cipher=""
for i in range(0,len(tea_plain),2):cipher+=chr(tea_plain[i]+tea_plain[i+1])
cipher=cipher[::-1]
cipher=base64.b64decode(cipher)flag=""
for i in range(0,len(cipher),2):flag+=chr(cipher[i])
for i in range(len(cipher)-1,0,-2):flag+=chr(cipher[i])
print(flag)
# flag{Emp0wer_F1utter_w1th_C!!}

nothing

出题人，richar师傅的wp：ISC-Nothing出题思路及WriteUp
FallW1nd师傅的wp：nothing

bad_apple

附件是1个.elf文件和1张电路图，电路图里是树莓派连接ST7789
根据字符串"frame: %d\n"的交叉引用，来到sub_100003A4函数，猜测是画图函数
frame_data是每一帧填充的像素，总共2192帧，每帧的大小为宽240*高180

dump出byte_100068CC数据

data=get_bytes(0x100068CC,1343128)
with open("D:\\CTFFiles\\ZMatches\\202208\\20220801-360DSCTF-Final\\bad_apple\\data","wb") as f:f.write(data)

直接复制ida反编译出来的伪代码，把每帧的frame_data输出到文件里

#include<stdio.h>
#include"defs.h"int32 frame_id=0;int32 dword_20000FC8=0;char byte_100068CC[1343128];int __fastcall sub_1000035C(int *a1)
{int v1; // r1_BYTE *i; // r3int v3; // r4int v4; // r6char *v5; // r7while ( 1 ){v1 = a1[2];if ( v1 )break;v4 = *a1;v5 = (char *)&byte_100068CC + *a1;v3 = 0;for ( i = (_BYTE *)v5; ; ++i ){v3 |= (*i & 0x7F) << v1;if ( (char)*i >= 0 )break;LOBYTE(v1) = v1 + 7;}a1[2] = v3;*a1 = v4 + i + 1 - (_BYTE *)v5;*((_WORD *)a1 + 2) = ~*((_WORD *)a1 + 2);}a1[2] = v1 - 1;return *((unsigned __int16 *)a1 + 2);
}void sub_100003A4(FILE *fp)
{int i; // r4_WORD frame_data[43204]; // [sp+0h] [bp-15188h] BYREFprintf("frame: %d\n", frame_id);for ( i = 0; i <= 43199; ++i )frame_data[i] = sub_1000035C(&dword_20000FC8);for(int y=0;y<180;y++){for(int x=0;x<240;x++){if(frame_data[y*240+x])fwrite("*",1,1,fp);elsefwrite("0",1,1,fp);}fwrite("\n",1,1,fp);}frame_id++;return;
}int main()
{FILE *fp = NULL;fp = fopen("D:\\CTFFiles\\ZMatches\\202208\\20220801-360DSCTF-Final\\bad_apple\\data", "rb");fseek(fp, 0, SEEK_SET);fread(byte_100068CC, 1, 1343128, (FILE*)fp);fclose(fp);fp = fopen("D:\\CTFFiles\\ZMatches\\202208\\20220801-360DSCTF-Final\\bad_apple\\flag", "wb");for(int i = 0; i < 2192; i++) {sub_100003A4(fp);}fclose(fp);return 0;
}

在最后的地方看到flag

fantastic_cpu

verilog写的vm，flag的4个int，4选2，选4组做加法，和data比较
通过运行.v，发现data是已知的8个int的第0、2、4、6个

用claripy求解即可得到flag

import claripyflag=[claripy.BVS(f"flag_{i}",64) for i in range(4)]data=[0x90771cb9,0xe3520b8,0x65dfd405,0x74c00bcc,0x64b1ca23,0x74942df9,0xb7385ab6,0xec36f96c]cons=[[0,1],[1,3],[2,3],[0,3]]s=claripy.Solver()
for i in range(len(cons)):s.add(flag[cons[i][0]]+flag[cons[i][1]]==data[i*2])for i in s.batch_eval(flag,1):for c in i:print(hex(c)[2:].zfill(8),end="")
# 70e7d1b51f8f4b041e61412246508901