• Jarvis OJ 一些简单的re刷题记录和脚本



    [61dctf] androideasy

    164求解器

    50

    相反

    1

    脚本如下:

    s=''

    a=113, 123, 118, 112, 108, 94, 99, 72, 38, 68, 72, 87, 89, 72, 36, 118, 100, 78, 72, 87, 121, 83, 101, 39, 62, 94, 62, 38, 107, 115, 106

    for i in range(len(a)):

    s+=chr(a[i]^(0x17))

    print(s)

    [61dctf] stheasy

    192求解器

    100

    相反

    2

    3

    脚本如下:

    s=""

    a=[0x48,0x5D,0x8D,0x24,0x84,0x27,0x99,0x9F,0x54,0x18,0x1E,0x69,0x7E,0x33,0x15,0x72,0x8D,0x33,0x24,0x63,0x21,0x54,0x0C,0x78,0x78,0x78,0x78,0x78,0x1B]

    b=[]

    c='lk2j9Gh}AgfY4ds-a6QW1#k5ER_T[cvLbV7nOm3ZeX{CMt8SZo]U'

    for i in range(len(a)):

    b.append(int(a[i]/3-2))

    for j in range(len(b)):

    s+=c[b[j]]

    print (s)

    Flag 是下一关的邮箱地址(以 DD 开头)。

    1.Hello.12b9bde7c0c8558a9da42aa1798cafc8


    DD - Hello

    129 SOLVERS

    100

    REVERSE


    You have solved this Challenge

    4

    5

    脚本如下:

    cip=[0x41, 0x10, 0x11, 0x11, 0x1B, 0x0A, 0x64, 0x67, 0x6A, 0x68, 0x62, 0x68, 0x6E, 0x67, 0x68, 0x6B, 0x62, 0x3D, 0x65, 0x6A, 0x6A, 0x3D, 0x68, 0x4, 0x5, 0x8, 0x3, 0x2, 0x2, 0x55, 0x8, 0x5D, 0x61, 0x55, 0x0A, 0x5F, 0x0D, 0x5D, 0x61, 0x32, 0x17, 0x1D, 0x19, 0x1F, 0x18, 0x20, 0x4, 0x2, 0x12, 0x16, 0x1E, 0x54, 0x20, 0x13, 0x14, 0x0, 0x0]

    start=0x100000CB0

    end=0x100000C90

    v2=((start-end)>>2)^cip[0]

    for i in range(55):

    cip[i]-=2

    cip[i]^=v2

    v2+=1

    s=''

    for j in cip:

    s+=chr(j)

    print(s)

    软件密码破解-1

    118 SOLVERS

    100

    REVERSE

    请对压缩包中的程序进行分析并获取flag。flag形式为xxx-xxxxx_xxxx。

    CTF_100_0.rar.b5abee530fee7cdae2f5cdc33bb849e8

    s=''

    a=[0x28,0x57,0x64,0x6B,0x93,0x8F,0x65,0x51,0xE3,0x53,0xE4,0x4E,0x1A,0xFF]

    b=[0x1B,0x1C,0x17,0x46,0xF4,0xFD,0x20,0x30,0xB7,0x0C,0x8E,0x7E,0x78,0xDE]

    for i in range(len(a)):

    s+=chr(a[i]^b[i])

    print(s)

    Classical Crackme

    293 SOLVERS

    100

    REVERSE

    经典Crackme题目,FLAG就是注册码。

    CrackMe.rar.4b81595bfc90d446ba30f9c9bb03fb49

    简单base64加密

    FindKey

    284 SOLVERS

    150

    REVERSE

    FLAG就是你输入的key

    findkey.31a509f4006ba41368dcf963762388bb

    脚本如下

    key=''

    s=[]

    pwda = [188,155,11,58,251,208,204,202,150,120,206,237,114,92,126,6,42]

    pwdb = [53,222,230,35,67,248,226,216,17,209,32,2,181,200,171,60,108]

    lookup = [196,153,149,206,17,221,10,217,167,18,36,135,103,61,111,31,92,152,21,228,105,191,173,41,2,245,23,144,1,246,89,178,182,119,38,85,48,226,165,241,166,214,71,90,151,3,109,169,150,224,69,156,158,57,181,29,200,37,51,252,227,93,65,82,66,80,170,77,49,177,81,94,202,107,25,73,148,98,129,231,212,14,84,121,174,171,64,180,233,74,140,242,75,104,253,44,39,87,86,27,68,22,55,76,35,248,96,5,56,20,161,213,238,220,72,100,247,8,63,249,145,243,155,222,122,32,43,186,0,102,216,126,15,42,115,138,240,147,229,204,117,223,141,159,131,232,124,254,60,116,46,113,79,16,128,6,251,40,205,137,199,83,54,188,19,184,201,110,255,26,91,211,132,160,168,154,185,183,244,78,33,123,28,59,12,210,218,47,163,215,209,108,235,237,118,101,24,234,106,143,88,9,136,95,30,193,176,225,198,197,194,239,134,162,192,11,70,58,187,50,67,236,230,13,99,190,208,207,7,53,219,203,62,114,127,125,164,179,175,112,172,250,133,130,52,189,97,146,34,157,120,195,45,4,142,139]

    for j in range(17):

    for i in range(255):

    if (i+pwda[j]&255)==lookup[j + pwdb[j]]:

    s.append(i)

    print(s)

    a=17

    while a>0:

    key+=chr(s[a-1])

    a-=1

    print(key)

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  • 原文地址:https://www.cnblogs.com/kk328/p/10266276.html
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