RE_crypto

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ida

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upload successful

程序逻辑不难,输入的字符作为unk_4021A0数组的下标,定位该下标所对应的值,与((v11 + v12) ^ 0x19)比较,这个值是可以自己算出来的

exp

byte1=[0x32,0x61,0x34,0x39,0x66,0x36,0x39,0x63,0x33,0x38,0x33,0x39,0x35,0x63,0x64,0x65,0x39,0x36,0x64,0x36,0x64,0x65,0x39,0x36,0x64,0x36,0x66,0x34,0x65,0x30,0x32,0x35,0x34,0x38,0x34,0x39,0x35,0x34,0x64,0x36,0x31,0x39,0x35,0x34,0x34,0x38,0x64,0x65,0x66,0x36,0x65,0x32,0x64,0x61,0x64,0x36,0x37,0x37,0x38,0x36,0x65,0x32,0x31,0x64,0x35,0x61,0x64,0x61,0x65,0x36]

byte = [ 0x63, 0x7C, 0x77, 0x7B, 0xF2, 0x6B, 0x6F, 0xC5, 0x30, 0x01, 0x67, 0x2B, 0xFE, 0xD7, 0xAB, 0x76, 0xCA, 0x82, 0xC9, 0x7D, 0xFA, 0x59, 0x47, 0xF0, 0xAD, 0xD4, 0xA2, 0xAF, 0x9C, 0xA4, 0x72, 0xC0, 0xB7, 0xFD, 0x93, 0x26, 0x36, 0x3F, 0xF7, 0xCC, 0x34, 0xA5, 0xE5, 0xF1, 0x71, 0xD8, 0x31, 0x15, 0x04, 0xC7, 0x23, 0xC3, 0x18, 0x96, 0x05, 0x9A, 0x07, 0x12, 0x80, 0xE2, 0xEB, 0x27, 0xB2, 0x75, 0x09, 0x83, 0x2C, 0x1A, 0x1B, 0x6E, 0x5A, 0xA0, 0x52, 0x3B, 0xD6, 0xB3, 0x29, 0xE3, 0x2F, 0x84, 0x53, 0xD1, 0x00, 0xED, 0x20, 0xFC, 0xB1, 0x5B, 0x6A, 0xCB, 0xBE, 0x39, 0x4A, 0x4C, 0x58, 0xCF, 0xD0, 0xEF, 0xAA, 0xFB, 0x43, 0x4D, 0x33, 0x85, 0x45, 0xF9, 0x02, 0x7F, 0x50, 0x3C, 0x9F, 0xA8, 0x51, 0xA3, 0x40, 0x8F, 0x92, 0x9D, 0x38, 0xF5, 0xBC, 0xB6, 0xDA, 0x21, 0x10, 0xFF, 0xF3, 0xD2, 0xCD, 0x0C, 0x13, 0xEC, 0x5F, 0x97, 0x44, 0x17, 0xC4, 0xA7, 0x7E, 0x3D, 0x64, 0x5D, 0x19, 0x73, 0x60, 0x81, 0x4F, 0xDC, 0x22, 0x2A, 0x90, 0x88, 0x46, 0xEE, 0xB8, 0x14, 0xDE, 0x5E, 0x0B, 0xDB, 0xE0, 0x32, 0x3A, 0x0A, 0x49, 0x06, 0x24, 0x5C, 0xC2, 0xD3, 0xAC, 0x62, 0x91, 0x95, 0xE4, 0x79, 0xE7, 0xC8, 0x37, 0x6D, 0x8D, 0xD5, 0x4E, 0xA9, 0x6C, 0x56, 0xF4, 0xEA, 0x65, 0x7A, 0xAE, 0x08, 0xBA, 0x78, 0x25, 0x2E, 0x1C, 0xA6, 0xB4, 0xC6, 0xE8, 0xDD, 0x74, 0x1F, 0x4B, 0xBD, 0x8B, 0x8A, 0x70, 0x3E, 0xB5, 0x66, 0x48, 0x03, 0xF6, 0x0E, 0x61, 0x35, 0x57, 0xB9, 0x86, 0xC1, 0x1D, 0x9E, 0xE1, 0xF8, 0x98, 0x11, 0x69, 0xD9, 0x8E, 0x94, 0x9B, 0x1E, 0x87, 0xE9, 0xCE, 0x55, 0x28, 0xDF, 0x8C, 0xA1, 0x89, 0x0D, 0xBF, 0xE6, 0x42, 0x68, 0x41, 0x99, 0x2D, 0x0F, 0xB0, 0x54, 0xBB, 0x16]

ans = ""
for i in range(0,len(byte1),2):
    v9 = 0
    v10 = 0
    if byte1[i] < 48 or byte1[i] > 57:
        v9 = byte1[i] - 87
    else:
        v9 = byte1[i] - 48

    if byte1[i+1] < 48 or byte1[i+1] > 57:
        v10 = byte1[i+1] - 87
    else:
        v10 = byte1[i+1] - 48
    flag = (16*v9+v10) ^ 0x19
    for j in range(0,len(byte)):
        if flag == byte[j]:
            ans += chr(j)
print ans

总结

看了encrypt,只理解了一小部分,后面的部分实在不太明白到底是干嘛,还是太菜了,近几天做的逆向题目,让我深刻认识到密码学、算法(acm)在逆向中的重要地位,一步一个jio印吧……



reverse      writeup jactf

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