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[原创]第2题 变形金刚 Writeup
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2019-3-11 16:17 2739
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这是一个Android逆向
有lib,里面做了字符串解密
默认用户名是13909876543
手动解密,发现了android/support/v7/app/AppCompiatActivity
这个Activity很可疑,去查看这个Activity
发现用到了https://ivonhoe.github.io/2017/02/09/%E7%BE%8E%E5%9B%A2%E5%A6%82%E4%BD%95%E9%98%B2dex2jar/
的反反编译手法
手动修改一下Dex2Jar的源码,再反编译,得到class
查看这个Activity的源代码,发现是使用了AES加密,不过是输入Check之后的操作了,暂时不用看
Native层实现了eq函数,函数内先对650f909c-7217-3647-9331-c82df8b98e98进行了翻转,替换,然后作为RC4算法的key,对传入的参数进行加密,然后进行自定义base64编码,再比较是否正确。
变换之后的Key是36f36b3c-a03e-4996-8759-8408e626c215
RC4 Base64算法都是修改过的
Base64逆向解码
>>> charset = "!:#$%&()+-*/`~_[]{}?<>,.@^abcdefghijklmnopqrstuvwxyz0123456789\\';" >>> origin = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=" >>> import string >>> t = string.maketrans(charset, origin) >>> cipher = ''.join(map(chr, [0x20, 0x7B, 0x39, 0x2A, 0x38, 0x67, 0x61, 0x2A, 0x6C, 0x21, 0x54, 0x6E, 0x3F, 0x40, 0x23, 0x66, 0x6A, 0x27, 0x6A, 0x24, 0x5C, 0x67, 0x3B, 0x3B])) >>> for i in xrange(len(cipher)): ... if (i % 4 == 0): ... cc += chr(ord(cipher[i]) ^ 7) ... elif (i % 4 == 2): ... cc += chr(ord(cipher[i]) ^ 0xF) ... else: ... cc += cipher(i) ... >>> map(ord, cc.translate(t).decode('base64')) [253, 30, 138, 78, 9, 202, 144, 3, 231, 241, 133, 159, 155, 247, 131, 62, 14]
RC4复现
#include <cstdio> #include <cstring> unsigned char sbox[] = { 0xD7, 0xDF, 0x02, 0xD4, 0xFE, 0x6F, 0x53, 0x3C, 0x25, 0x6C, 0x99, 0x97, 0x06, 0x56, 0x8F, 0xDE, 0x40, 0x11, 0x64, 0x07, 0x36, 0x15, 0x70, 0xCA, 0x18, 0x17, 0x7D, 0x6A, 0xDB, 0x13, 0x30, 0x37, 0x29, 0x60, 0xE1, 0x23, 0x28, 0x8A, 0x50, 0x8C, 0xAC, 0x2F, 0x88, 0x20, 0x27, 0x0F, 0x7C, 0x52, 0xA2, 0xAB, 0xFC, 0xA1, 0xCC, 0x21, 0x14, 0x1F, 0xC2, 0xB2, 0x8B, 0x2C, 0xB0, 0x3A, 0x66, 0x46, 0x3D, 0xBB, 0x42, 0xA5, 0x0C, 0x75, 0x22, 0xD8, 0xC3, 0x76, 0x1E, 0x83, 0x74, 0xF0, 0xF6, 0x1C, 0x26, 0xD1, 0x4F, 0x0B, 0xFF, 0x4C, 0x4D, 0xC1, 0x87, 0x03, 0x5A, 0xEE, 0xA4, 0x5D, 0x9E, 0xF4, 0xC8, 0x0D, 0x62, 0x63, 0x3E, 0x44, 0x7B, 0xA3, 0x68, 0x32, 0x1B, 0xAA, 0x2D, 0x05, 0xF3, 0xF7, 0x16, 0x61, 0x94, 0xE0, 0xD0, 0xD3, 0x98, 0x69, 0x78, 0xE9, 0x0A, 0x65, 0x91, 0x8E, 0x35, 0x85, 0x7A, 0x51, 0x86, 0x10, 0x3F, 0x7F, 0x82, 0xDD, 0xB5, 0x1A, 0x95, 0xE7, 0x43, 0xFD, 0x9B, 0x24, 0x45, 0xEF, 0x92, 0x5C, 0xE4, 0x96, 0xA9, 0x9C, 0x55, 0x89, 0x9A, 0xEA, 0xF9, 0x90, 0x5F, 0xB8, 0x04, 0x84, 0xCF, 0x67, 0x93, 0x00, 0xA6, 0x39, 0xA8, 0x4E, 0x59, 0x31, 0x6B, 0xAD, 0x5E, 0x5B, 0x77, 0xB1, 0x54, 0xDC, 0x38, 0x41, 0xB6, 0x47, 0x9F, 0x73, 0xBA, 0xF8, 0xAE, 0xC4, 0xBE, 0x34, 0x01, 0x4B, 0x2A, 0x8D, 0xBD, 0xC5, 0xC6, 0xE8, 0xAF, 0xC9, 0xF5, 0xCB, 0xFB, 0xCD, 0x79, 0xCE, 0x12, 0x71, 0xD2, 0xFA, 0x09, 0xD5, 0xBC, 0x58, 0x19, 0x80, 0xDA, 0x49, 0x1D, 0xE6, 0x2E, 0xE3, 0x7E, 0xB7, 0x3B, 0xB3, 0xA0, 0xB9, 0xE5, 0x57, 0x6E, 0xD9, 0x08, 0xEB, 0xC7, 0xED, 0x81, 0xF1, 0xF2, 0xBF, 0xC0, 0xA7, 0x4A, 0xD6, 0x2B, 0xB4, 0x72, 0x9D, 0x0E, 0x6D, 0xEC, 0x48, 0xE2, 0x33 }; int main() { unsigned char key[256] = {0}; char k[] = "36f36b3c-a03e-4996-8759-8408e626c215" ; unsigned char plain[] = {253, 30, 138, 78, 9, 202, 144, 3, 231, 241, 133, 159, 155, 247, 131, 62, 14}; // unsigned char plain[] = {253, 31, 74, 242, 6, 138, 148, 4, 231, 77, 128, 159, 143, 248, 195, 250}; for (int i = 0; i < 256; i++) { key[i] = k[i % strlen(k)]; } int j = key[0] - 0x29; sbox[0] = sbox[j]; sbox[j] = 0xd7u; // int j; unsigned char temp; for(int i=1;i<256;i++) { j=(j+sbox[i]+key[i])%256; temp=sbox[i]; sbox[i]=sbox[j]; sbox[j]=temp; } int i = 0; j = 0; int t = 0, kk = 0; for(kk=0;kk<strlen((char *)plain);kk++) { i=(i+1)%256; j=(j+sbox[i])%256; temp=sbox[i]; sbox[i]=sbox[j]; sbox[j]=temp; t=(sbox[i]+sbox[j])%256; plain[kk]^=sbox[t]; } puts((char *)plain); }
解得输入
fu0kzHp2aqtZAuY6
// 一开始算的时候多了一个6,怎么搞都不对。。。
输入正确结果后拿到一个Flag,提交的时候发现是提交输入的结果 = =
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