[WP]XCTF- 攻防世界-crypto-新手练习区

1、base64

Y3liZXJwZWFjZXtXZWxjb21lX3RvX25ld19Xb3JsZCF9

 

 2、Caesar

凯撒加密

oknqdbqmoq{kag_tmhq_xqmdzqp_omqemd_qzodkbfuaz}

 

 3、Morse

摩斯密码（摩尔斯密码）

11 111 010 000 0 1010 111 100 0 00 000 000 111 00 10 1 0 010 0 000 1 00 10 110

 

4、幂数加密

8842101220480224404014224202480122

把数字串用 0 分割开后，再将每组密码相加后的和代换为字符

a = [23, 5, 12, 12, 4, 15, 14, 5]
s = ''
for i in range(26):
    s += chr(i+0x41)
flag = ''
for i in a:
    flag += s[i-1]
print(flag)

5、Railfence

ccehgyaefnpeoobe{lcirg}epriec_ora_g

栅栏密码 w型

 

6、不仅仅是Morse

--/.-/-.--/..--.-/-..././..--.-/..../.-/...-/./..--.-/.-/-./---/-/...././.-./..--.-/-.././-.-./---/-.././..../..../..../..../.-/.-/.-/.-/.-/-.../.-/.-/-.../-.../-.../.-/.-/-.../-.../.-/.-/.-/.-/.-/.-/.-/.-/-.../.-/.-/-.../.-/-.../.-/.-/.-/.-/.-/.-/.-/-.../-.../.-/-.../.-/.-/.-/-.../-.../.-/.-/.-/-.../-.../.-/.-/-.../.-/.-/.-/.-/-.../.-/-.../.-/.-/-.../.-/.-/.-/-.../-.../.-/-.../.-/.-/.-/-.../.-/.-/.-/-.../.-/.-/-.../.-/-.../-.../.-/.-/-.../-.../-.../.-/-.../.-/.-/.-/-.../.-/-.../.-/-.../-.../.-/.-/.-/-.../-.../.-/-.../.-/.-/.-/-.../.-/.-/-.../.-/.-/-.../.-/.-/.-/.-/-.../-.../.-/-.../-.../.-/.-/-.../-.../.-/.-/-.../.-/.-/-.../.-/.-/.-/-.../.-/.-/-.../.-/.-/-.../.-/.-/-.../.-/-.../.-/.-/-.../-.../.-/-.../.-/.-/.-/.-/-.../-.../.-/-.../.-/.-/-.../-.../.-

首先将其处理为 01 串

a = '--/.-/-.--/..--.-/-..././..--.-/..../.-/...-/./..--.-/.-/-./---/-/...././.-./..--.-/-.././-.-./---/-.././..../..../..../..../.-/.-/.-/.-/.-/-.../.-/.-/-.../-.../-.../.-/.-/-.../-.../.-/.-/.-/.-/.-/.-/.-/.-/-.../.-/.-/-.../.-/-.../.-/.-/.-/.-/.-/.-/.-/-.../-.../.-/-.../.-/.-/.-/-.../-.../.-/.-/.-/-.../-.../.-/.-/-.../.-/.-/.-/.-/-.../.-/-.../.-/.-/-.../.-/.-/.-/-.../-.../.-/-.../.-/.-/.-/-.../.-/.-/.-/-.../.-/.-/-.../.-/-.../-.../.-/.-/-.../-.../-.../.-/-.../.-/.-/.-/-.../.-/-.../.-/-.../-.../.-/.-/.-/-.../-.../.-/-.../.-/.-/.-/-.../.-/.-/-.../.-/.-/-.../.-/.-/.-/.-/-.../-.../.-/-.../-.../.-/.-/-.../-.../.-/.-/-.../.-/.-/-.../.-/.-/.-/-.../.-/.-/-.../.-/.-/-.../.-/.-/-.../.-/-.../.-/.-/-.../-.../.-/-.../.-/.-/.-/.-/-.../-.../.-/-.../.-/.-/-.../-.../.-'
v = a.split('/')
bin_list = []
for i in v:
    value = ''
    for j in i:
        if j == '-': value += '1'
        else: value += '0'
    bin_list.append(value)
flag = ''
for i in bin_list:
    flag += i + ' '
print(flag)

 莫斯密码在线转换

 

 根据提示，还有另一种加密——培根密码

 

7、混合编码

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

base64 解密

HTML 解密

在进行一次 base64 解密

 

 之后对得到的数字串根据 ASCII 码转换为字符串

a = '119/101/108/99/111/109/101/116/111/97/116/116/97/99/107/97/110/100/100/101/102/101/110/99/101/119/111/114/108/100'
b = a.split('/')
flag = ''
for i in b:
    value = int(i)
    flag += chr(value)
print(flag)

 

8、easy_RSA

在一次RSA密钥对生成中，假设p=473398607161，q=4511491，e=17
求解出d

由所给的 p, q 我们可以求得 m = (p-1)*(q-1)

之后再由 d * e = 1 mod m 可以知道 d 即是 e 模 m 的逆元，运用扩展欧几里得算法就能得到逆元（我这里直接从网上找到一段扩展欧几里得算法的代码）

p = 473398607161
q = 4511491
e = 17
n = p * q
m = (p-1) * (q-1)
print('m=' + str(m))


#欧几里得算法求最大公约数
def get_gcd(a, b):
    k = a // b
    remainder = a % b
    while remainder != 0:
        a = b 
        b = remainder
        k = a // b
        remainder = a % b
    return b
    
#改进欧几里得算法求线性方程的x与y
def get_(a, b):
    if b == 0:
        return 1, 0
    else:
        k = a // b
        remainder = a % b        
        x1, y1 = get_(b, remainder)
        x, y = y1, x1 - k * y1            
    return x, y
    
a, b = 17, m
 
#将初始b的绝对值进行保存
if b < 0:
    m = abs(b)
else:
    m = b
flag = get_gcd(a, b)
 
#判断最大公约数是否为1，若不是则没有逆元
if flag == 1:    
    x, y = get_(a, b)    
    x0 = x % m #对于Python '%'就是求模运算，因此不需要'+m'
    print(x0) #x0就是所求的逆元
else:
    print("Do not have!")

 

9、easychallenge

题目所给的 pyc 文件，先在线反编译为 python 代码

import base64

def encode1(ans):
    s = ''
    for i in ans:
        x = ord(i) ^ 36
        x = x + 25
        s += chr(x)
    
    return s

def encode2(ans):
    s = ''
    for i in ans:
        x = ord(i) + 36
        x = x ^ 36
        s += chr(x)
    
    return s

def encode3(ans):
    return base64.b32encode(ans)

flag = ' '
print 'Please Input your flag:'
flag = raw_input()
final = 'UC7KOWVXWVNKNIC2XCXKHKK2W5NLBKNOUOSK3LNNVWW3E==='
if encode3(encode2(encode1(flag))) == final:
    print 'correct'
else:
    print 'wrong'

先后对 flag 进行 encode1, encode2, encode3 进行加密得到 final

那么将顺序反过来解密 final 就能得到 flag

import base64

def main():
    decode()

def decode():
    final = 'UC7KOWVXWVNKNIC2XCXKHKK2W5NLBKNOUOSK3LNNVWW3E==='
    result = base64.b32decode(final)
    result1 = []
    for i in result:
        result1.append(i)
    print(result1)

    result2 = []
    for i in result1:
        value = i ^ 36
        value -= 36
        result2.append(value)
    print(result2)

    result3 = []
    for i in result2:
        value = i - 25
        value ^= 36
        result3.append(value)
    print(result3)

    flag = ''
    for i in result3:
        flag += chr(i)
    print(flag)

main()

 

 10、转轮机加密

1:  < ZWAXJGDLUBVIQHKYPNTCRMOSFE <
2:  < KPBELNACZDTRXMJQOYHGVSFUWI <
3:  < BDMAIZVRNSJUWFHTEQGYXPLOCK <
4:  < RPLNDVHGFCUKTEBSXQYIZMJWAO <
5:  < IHFRLABEUOTSGJVDKCPMNZQWXY <
6:  < AMKGHIWPNYCJBFZDRUSLOQXVET <
7:  < GWTHSPYBXIZULVKMRAFDCEONJQ <
8:  < NOZUTWDCVRJLXKISEFAPMYGHBQ <
9:  < XPLTDSRFHENYVUBMCQWAOIKZGJ <
10: < UDNAJFBOWTGVRSCZQKELMXYIHP <
11： < MNBVCXZQWERTPOIUYALSKDJFHG <
12： < LVNCMXZPQOWEIURYTASBKJDFHG <
13： < JZQAWSXCDERFVBGTYHNUMKILOP <

密钥为：2,3,7,5,13,12,9,1,8,10,4,11,6
密文为：NFQKSEVOQOFNP

根据密钥重新排序十三组字符串

将密文的十三个字符分别排到每组的起始位置

之后更改横纵方向方便查看，当然仅仅为了是方便查看

s = ['ZWAXJGDLUBVIQHKYPNTCRMOSFE ',\
 'KPBELNACZDTRXMJQOYHGVSFUWI ',\
 'BDMAIZVRNSJUWFHTEQGYXPLOCK ',\
 'RPLNDVHGFCUKTEBSXQYIZMJWAO ',
 'IHFRLABEUOTSGJVDKCPMNZQWXY ',\
 'AMKGHIWPNYCJBFZDRUSLOQXVET ',\
 'GWTHSPYBXIZULVKMRAFDCEONJQ ',\
 'NOZUTWDCVRJLXKISEFAPMYGHBQ ',\
 'XPLTDSRFHENYVUBMCQWAOIKZGJ ',\
 'UDNAJFBOWTGVRSCZQKELMXYIHP ',\
 'MNBVCXZQWERTPOIUYALSKDJFHG ',\
 'LVNCMXZPQOWEIURYTASBKJDFHG ',\
 'JZQAWSXCDERFVBGTYHNUMKILOP ']

for i in range(len(s)):
    s[i] = s[i].strip()

key = [2,3,7,5,13,12,9,1,8,10,4,11,6]
v = 'NFQKSEVOQOFNP'

t = []
#更换字符串的顺序
for i in range(13):
    t.append(s[key[i]-1])
print(t)

u = []
#改变每个字符串的排列顺序
for i in range(13):
    new_sting = ''
    index = t[i].index(v[i])
    new_sting += t[i][index:] + t[i][0:index]
    u.append(new_sting)
print(u)

w = []
#改变横纵方向方便查看，并修改为小写
for i in range(26):
    new_sting = ''
    for j in range(13):
        new_sting += u[j][i].lower()
    w.append(new_sting)

print(w)

之后在得到的结果中找到一个可以看得懂的明文串即是 flag

 

 11、Normal_RSA

下载得到两个文件

需要运用到工具 openssl, rsatool, yafu

openssl rsa -pubin -text -modulus -in warmup -in pubkey.pem

 得到 e, n ，之后由 yafu 分解得到 p, q

python rsatool.py -o prikey.pem -e 65537 -p 319576316814478949870590164193048041239 -q 275127860351348928173285174381581152299

 得到 prikey.pem （自己命名的文件）后和 flag.enc 放在同一目录下，解密得到 flag

openssl rsautl -decrypt -in flag.enc -inkey prikey.pem

 

12、easy_ECC

已知椭圆曲线加密Ep(a,b)参数为

p = 15424654874903

a = 16546484

b = 4548674875

G(6478678675,5636379357093)

私钥为

k = 546768

求公钥K(x,y)

借助 ECCTOOL  运算

print(13957031351290+5520194834100)