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DES

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Challenge | 2019 | 湖湘杯 | Crypto | DES

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题目考点

  • DES加密

解题过程

题目未知key,采用DES加密，已知Kn，也就是子密钥，已知mes加密的结果，要求求出mes以及key。 主要思路是这样，第一步通过DES子密钥求出密钥key，之后通过key直接进行解密求出mes,当然也可以直接用子密钥求解出mes。 通过子密钥求解key可以参考一道有关密钥编排的DES题目(飘零师傅NB,一搜又是你)，一步一步求解，但是密钥生成子密钥的过程中有8位是没有用的，所以逆向推导子密钥会有8位是不确定的，遍历出来会有256个密钥，这256个密钥都可以用来加密解密，而且结果都是正确的。具体参考过程如下：

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key1 = [1,0,1,0,0,0,0,0,1,0,0,1,0,1,1,0,0,1,0,0,0,1,1,0,0,0,1,1,1,0,1,1,0,0,0,0,0,1,1,1,1,0,0,1,1,0,0,0]
__pc2 = [
    13,16,10,23,0,4,
    2,27,14,5,20,9,
    22,18,11,3,25,7,
    15,6,26,19,12,1,
    40,51,30,36,46,54,
    29,39,50,44,32,47,
    43,48,38,55,33,52,
    45,41,49,35,28,31
]
C1D1 = ['*']*56
for i in range(0,len(key1)):
    C1D1[__pc2[i]] = key1[i]
print C1D1
#[00000001*11111100*110*00*000
# 011001*01*1101*0001011000*01]
# you1
C0 = '000000001*11111100*110*00*00'
D0 = '1011001*01*1101*0001011000*01'
__pc1 = [56,48,40,32,24,16,8,
    0,57,49,41,33,25,17,
    9,1,58,50,42,34,26,
    18,10,2,59,51,43,35,
    62,54,46,38,30,22,14,
    6,61,53,45,37,29,21,
    13,5,60,52,44,36,28,
    20,12,4,27,19,11,3
]
C0D0 = C0+D0
res = ['*']*64
deskey = ""
for i in range(0,len(__pc1)):
    res[__pc1[i]] = C0D0[i]
for i in res:
    deskey += i
print deskey
def zuoyiwei(str,num):
    my = str[num:len(str)]
    my = my+str[0:num]
    return my
def key_change_1(str):
    key1_list = [57,49,41,33,25,17,9,1,58,50,42,34,26,18,10,2,59,51,43,35,27,19,11,3,60,52,44,36,63,55,47,39,31,23,15,7,62,54,46,38,30,22,14,6,61,53,45,37,29,21,13,5,28,20,12,4]
    res = ""
    for i in key1_list:
        res+=str[i-1]
    return res
def key_change_2(str):
    key2_list = [14,17,11,24,1,5,3,28,15,6,21,10,23,19,12,4,26,8,16,7,27,20,13,2,41,52,31,37,47,55,30,40,51,45,33,48,44,49,39,56,34,53,46,42,50,36,29,32]
    res = ""
    for i in key2_list:
        res+=str[i-1]
    return res
def key_gen(str):
    key_list = []
    key_change_res = key_change_1(str)
    key_c = key_change_res[0:28]
    key_d = key_change_res[28:]
    for i in range(1,17):
        if (i==1) or (i==2) or (i==9) or (i==16):
            key_c = zuoyiwei(key_c,1)
            key_d = zuoyiwei(key_d,1)
        else:
            key_c = zuoyiwei(key_c,2)
            key_d = zuoyiwei(key_d,2)
        key_yiwei = key_c+key_d
        key_res = key_change_2(key_yiwei)
        key_list.append(key_res)
    return key_list
deskey = "01000abc01de111f0100100h0110010i0110111j01k00L1m0n0o010p0100001q"
print key_gen(deskey)
deskey = "0100000c0110111f0100100h0110010i0110111j0110011m0100010p0100001q"
# ['101000001001011001000110001110110000011110011000','1k1o000000d101100n010010100l011000110110be1a0110','01100100kn010d10011100000011110ae00010111110010b','1d0001101101000n01010ok0000100b011l01e0011010011','00k01110d10000110101001n01l0e1111010010b00010a01','0010111n0101000100o010d1101010110e100101a10010l0','o0d0101100000001n10k10010b101l00110100110000011a','000n1k01010010o010011001e10101a0010001001l10b110','00o111010100n00k1d0010000l00000e11111b01010101a0','000100100dn010011000110k0110ba01101001001011l000','0001100k00101n0d000o0101111010010b0l11000a001e11','0n000o0100101100k010110d00001110a1010010e0l11110','110k00odn010010010100100b00101010101la011110010e','1101o00010001110d01000n01000a0e0100010b0111l0001','11nd0000101100100010ok10110b001l1a10111e00010101','10100000101ndk10o010011010100e011001a0l000001b11']
# [101000001001011001000110001110110000011110011000], [111000000011011001010010100101100011011000100110],[011001001101011001110000001111000000101111100100],[110001101101000101010010000100001110100011010011], [001011101100001101010011011001111010010000010001],[001011110101000100001011101010110010010101001010],[001010110000000111011001001011001101001100000110],[000111010100100010011001010101000100010011100110],[000111010100100111001000010000001111100101010100],[000100100110100110001101011000011010010010111000],[000110010010110100000101111010010001110000001011],[010000010010110010101101000011100101001000111110],[110100011010010010100100000101010101100111100100],[110100001000111010100010100000001000100011110001],[111100001011001000100110110000111010111000010101],[101000001011111000100110101000011001001000001011]
print key_gen(deskey)
# deskey = "0100000"+c+"0110111"+f+"0100100"+h+"0110010"+i+"0110111"+j+"0110011"+m+"0100010"+p+"0100001"+q
def bintostr(str):
    res = ""
    for i in range(0, len(str), 8):
        res += chr(int(str[i:i+8],2))
    return res
for c in "01":
    for f in "01":
        for h in "01":
            for i in "01":
                for j in "01":
                    for m in "01":
                        for p in "01":
                            for q in "01":
                                str = "0100000" + c + "0110111" + f + "0100100" + h + "0110010" + i + "0110111" + j + "0110011" + m + "0100010" + p + "0100001" + q
                                str = bintostr(str)
                                print str

求出256个密钥后用任一解密可以得到mes。然后就是让人秃头的地方，讲道理这里一般都有个啥暗示可以筛选或者看出来一个明显特定的密钥，开始我盲猜要是可见字符，然后全部都是，然后又根据其它flag盲猜是uuid,然后全部都不是。 然后比赛快结束的时候，木的办法的我只好遍历了256个flag，开始一个一个试，在经历了卡顿掉线重新登陆提交的20分钟后我试出来了，具体结果就不说了，这个key确实是一个特定的词，但是混在256个里面很难看出来。密码表如下

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@nHdnfDB, @nHdnfDC, @nHdnfEB, @nHdnfEC, @nHdngDB, @nHdngDC, @nHdngEB, @nHdngEC,
@nHdofDB, @nHdofDC, @nHdofEB, @nHdofEC, @nHdogDB, @nHdogDC, @nHdogEB, @nHdogEC,
@nHenfDB, @nHenfDC, @nHenfEB, @nHenfEC, @nHengDB, @nHengDC, @nHengEB, @nHengEC,
@nHeofDB, @nHeofDC, @nHeofEB, @nHeofEC, @nHeogDB, @nHeogDC, @nHeogEB, @nHeogEC,
@nIdnfDB, @nIdnfDC, @nIdnfEB, @nIdnfEC, @nIdngDB, @nIdngDC, @nIdngEB, @nIdngEC,
@nIdofDB, @nIdofDC, @nIdofEB, @nIdofEC, @nIdogDB, @nIdogDC, @nIdogEB, @nIdogEC,
@nIenfDB, @nIenfDC, @nIenfEB, @nIenfEC, @nIengDB, @nIengDC, @nIengEB, @nIengEC,
@nIeofDB, @nIeofDC, @nIeofEB, @nIeofEC, @nIeogDB, @nIeogDC, @nIeogEB, @nIeogEC,
@oHdnfDB, @oHdnfDC, @oHdnfEB, @oHdnfEC, @oHdngDB, @oHdngDC, @oHdngEB, @oHdngEC,
@oHdofDB, @oHdofDC, @oHdofEB, @oHdofEC, @oHdogDB, @oHdogDC, @oHdogEB, @oHdogEC,
@oHenfDB, @oHenfDC, @oHenfEB, @oHenfEC, @oHengDB, @oHengDC, @oHengEB, @oHengEC,
@oHeofDB, @oHeofDC, @oHeofEB, @oHeofEC, @oHeogDB, @oHeogDC, @oHeogEB, @oHeogEC,
@oIdnfDB, @oIdnfDC, @oIdnfEB, @oIdnfEC, @oIdngDB, @oIdngDC, @oIdngEB, @oIdngEC,
@oIdofDB, @oIdofDC, @oIdofEB, @oIdofEC, @oIdogDB, @oIdogDC, @oIdogEB, @oIdogEC,
@oIenfDB, @oIenfDC, @oIenfEB, @oIenfEC, @oIengDB, @oIengDC, @oIengEB, @oIengEC,
@oIeofDB, @oIeofDC, @oIeofEB, @oIeofEC, @oIeogDB, @oIeogDC, @oIeogEB, @oIeogEC,
AnHdnfDB, AnHdnfDC, AnHdnfEB, AnHdnfEC, AnHdngDB, AnHdngDC, AnHdngEB, AnHdngEC,
AnHdofDB, AnHdofDC, AnHdofEB, AnHdofEC, AnHdogDB, AnHdogDC, AnHdogEB, AnHdogEC,
AnHenfDB, AnHenfDC, AnHenfEB, AnHenfEC, AnHengDB, AnHengDC, AnHengEB, AnHengEC,
AnHeofDB, AnHeofDC, AnHeofEB, AnHeofEC, AnHeogDB, AnHeogDC, AnHeogEB, AnHeogEC,
AnIdnfDB, AnIdnfDC, AnIdnfEB, AnIdnfEC, AnIdngDB, AnIdngDC, AnIdngEB, AnIdngEC,
AnIdofDB, AnIdofDC, AnIdofEB, AnIdofEC, AnIdogDB, AnIdogDC, AnIdogEB, AnIdogEC,
AnIenfDB, AnIenfDC, AnIenfEB, AnIenfEC, AnIengDB, AnIengDC, AnIengEB, AnIengEC,
AnIeofDB, AnIeofDC, AnIeofEB, AnIeofEC, AnIeogDB, AnIeogDC, AnIeogEB, AnIeogEC,
AoHdnfDB, AoHdnfDC, AoHdnfEB, AoHdnfEC, AoHdngDB, AoHdngDC, AoHdngEB, AoHdngEC,
AoHdofDB, AoHdofDC, AoHdofEB, AoHdofEC, AoHdogDB, AoHdogDC, AoHdogEB, AoHdogEC,
AoHenfDB, AoHenfDC, AoHenfEB, AoHenfEC, AoHengDB, AoHengDC, AoHengEB, AoHengEC,
AoHeofDB, AoHeofDC, AoHeofEB, AoHeofEC, AoHeogDB, AoHeogDC, AoHeogEB, AoHeogEC,
AoIdnfDB, AoIdnfDC, AoIdnfEB, AoIdnfEC, AoIdngDB, AoIdngDC, AoIdngEB, AoIdngEC,
AoIdofDB, AoIdofDC, AoIdofEB, AoIdofEC, AoIdogDB, AoIdogDC, AoIdogEB, AoIdogEC,
AoIenfDB, AoIenfDC, AoIenfEB, AoIenfEC, AoIengDB, AoIengDC, AoIengEB, AoIengEC,
AoIeofDB, AoIeofDC, AoIeofEB, AoIeofEC, AoIeogDB, AoIeogDC, AoIeogEB, AoIeogEC,

FLAG

1
flag{AnHengDB}