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浙江省2023网络安全信息竞赛(初赛)Reverse 复现

字数统计: 821阅读时长: 4 min
2023/11/07

浙江省2023网络安全信息竞赛(初赛)Reverse 复现

我是fw,赛场上纯纯坐牢,只会复现

小菲哭

luare

主函数

1.1

这里是一段lua的字节码,调试到这一步,提取

1.2

学了一手,有个java的unluac.jar反编译脚本,输入

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java -jar .\unluac.jar lua

1.3

里面的Oo00Oo0在ida里面

1.4

仔细研究一下,是一个查表+异或,关键就在那个ptr的for函数里

给出脚本

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pw=[109,-73,-72,46,-73,-5,99,-100,46,59,32,-76,109,3,59,20,-61,-56,
-119,48,100,118,36,118,82,3,95,106,14,-80,5,-89,89,-85,5,14,46,
-73,7,127]
for i in range(len(pw)):
pw[i]=pw[i]&0xff
table=[ 0x3C, 0x95, 0xC8, 0x28, 0x10, 0x6D, 0x85, 0x60, 0x59, 0x03,0xB3, 0x4C, 0x76, 0x49, 0x48, 0x96,
0xB8, 0x5F, 0xB7, 0x79, 0xC4, 0x64, 0x71, 0x2E, 0x38, 0x8C, 0xAC,0xA7, 0x91, 0x72, 0x63, 0x80,
0xB0, 0x9E, 0x33, 0x4B, 0xAE, 0xF3, 0x8B, 0x7B, 0x4D, 0x5B, 0xB4,0x52, 0xEC, 0x6F, 0xE0, 0xCF,
0xAD, 0xC3, 0x20, 0xAB, 0xEA, 0x67, 0xDC, 0x05, 0x00, 0x9F, 0x40,0x56, 0xD6, 0xFB, 0xFC, 0x24,
0x92, 0xCA, 0x0B, 0x3D, 0x46, 0x0D, 0xF0, 0x4A, 0x5A, 0x55, 0x11,0x1A, 0x3B, 0x8A, 0xBC, 0x7D,
0x6C, 0xE7, 0xA9, 0x13, 0x75, 0xCE, 0x61, 0x30, 0x14, 0xA6, 0x6A,0x27, 0x07, 0xD0, 0x54, 0x9C,
0x5C, 0x8E, 0x89, 0xD8, 0x58, 0x01, 0xC2, 0x34, 0xE8, 0x69, 0x35,0x2F, 0xC0, 0x2A, 0xA0, 0x50,
0x36, 0x88, 0xFF, 0x39, 0x1D, 0x68, 0x0E, 0x0C, 0x93, 0xE6, 0xB1,0xFE, 0x18, 0x7F, 0x6E, 0xB6,
0x78, 0x53, 0x31, 0x2B, 0xE9, 0xD2, 0xF5, 0x29, 0x0F, 0x2C, 0x17,0x84, 0xDE, 0xDB, 0xD9, 0x41,
0x06, 0x19, 0xF7, 0xA1, 0x99, 0xA8, 0x45, 0x7A, 0x3E, 0x23, 0xA5,0x1B, 0xAF, 0x0A, 0xAA, 0xE5,
0xEF, 0xA4, 0xE1, 0xF8, 0xFA, 0x82, 0x3A, 0x9A, 0xDF, 0x8F, 0x1C,0x65, 0xC7, 0x73, 0xD1, 0xC1,
0xC5, 0xD7, 0xA2, 0x5E, 0x87, 0xDD, 0x9D, 0x8D, 0xF9, 0xC9, 0x81,0xCD, 0x90, 0x97, 0xEE, 0x66,
0xDA, 0x4F, 0x42, 0x3F, 0xC6, 0x74, 0x08, 0x37, 0x25, 0xCB, 0x77,0x26, 0xE3, 0x83, 0x32, 0xB9,
0xBD, 0xD3, 0xF2, 0x44, 0xD5, 0x4E, 0x2D, 0xBA, 0x62, 0x98, 0x04,0x1E, 0x12, 0x21, 0xE4, 0xBF,
0x47, 0xF6, 0x86, 0xF4, 0xFD, 0x94, 0x16, 0xA3, 0xEB, 0x1F, 0x70,0x7C, 0xB2, 0x51, 0x02, 0x43,
0x22, 0x15, 0xCC, 0x7E, 0x09, 0x6B, 0xE2, 0x5D, 0xBB, 0x9B, 0xBE,0xB5, 0xD4, 0xED, 0x57, 0xF1]
s=[0]*40
for j in range(40):
for i in range(256):
if table[i]==pw[j]:
s[j]=i
break
print(s[j],end=",")
print()
start=0x44 #"D"
for i in range(40):
print(chr(start),end='')
start=start^s[i]
#DASCTF{e:-aSy|u9aPR0gr~AMfo~$RrE^VeR$3!}

AndriodELF

2.1

有个UPX壳,还是魔改壳

2.2

把最后两个upx改UPX才能用脚本出

主函数

2.3

然后进sub_218FB4函数

2.4

四个加密逐一逆向,给出脚本

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#include<stdio.h>
int main()
{
unsigned char pw[64] = {
0x3D, 0x45, 0x38, 0x7E, 0x78, 0x4B, 0x6A, 0x5C, 0x5B, 0x52, 0x4C, 0x73, 0x4E, 0x39, 0x49, 0x5F,
0x49, 0x40, 0x38, 0x5E, 0x74, 0x40, 0x66, 0x44, 0x46, 0x7A, 0x39, 0x3B, 0x67, 0x39, 0x70, 0x6C,
0x71, 0x5E, 0x6D, 0x4D, 0x5A, 0x4C, 0x7F, 0x3B, 0x4D, 0x63, 0x5E, 0x4E, 0x44, 0x5A, 0x7B, 0x51,
0x38, 0x61, 0x29, 0x63, 0x75, 0x5B, 0x67, 0x46, 0x4E, 0x5D, 0x79, 0x29, 0x4D, 0x29, 0x6D, 0x71
};
unsigned int table[16] = {
0x0000000D, 0x00000004, 0x00000000, 0x00000005, 0x00000002, 0x0000000C, 0x0000000B, 0x00000008,
0x0000000A, 0x00000006, 0x00000001, 0x00000009, 0x00000003, 0x0000000F, 0x00000007, 0x0000000E,
};
unsigned char flag[64];
unsigned char v[64];
unsigned char pw1[16];
for(int o=0;o<4;o++)
{
for(int j=0;j<16;j++)
{
pw1[j]=pw[o*16+j];
}
for(int j=15;j>=0;j--)
{
for(int i=0;i<16;i++)
{
v[i]=pw1[i]^(120*j);//加密4 异或120*j
v[i]=(v[i]<<5)|(v[i]>>3);//加密3 移位
}
for(int i=0;i<16;i++)
{
flag[i]=v[table[i]];
}//加密2 换表
for(int i=0;i<16;i++)
{
int v1=0;
for(int l=0;l<8;l++)
{
v1|=((flag[i]>>l)&1)<<(7-l);
}
flag[i]=v1;
}//加密1 循环移位
for(int x=0;x<16;x++)
{
pw1[x]=flag[x];
}
}
for(int i=0;i<16;i++)
{
printf("%c",pw1[i]);
}
}
}
//DASCTF{51bWZvM0p1xNHLo3A1ndrVH0|VsED3LFyRwYkEVeRqeFSNE!0!oyUki!}
//flag给这么长我也醉了
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  1. 1. 浙江省2023网络安全信息竞赛(初赛)Reverse 复现
    1. 1.1. luare
    2. 1.2. AndriodELF