攻防世界密码学新手区解题思路

1.base64

直接使用工具解密base64

2.Caesar

使用工具进行解密，发现偏移为12

3.Morse

将1看成-，0看成.

转化成小写

4.混合编码

首先base64解码：

LzExOS8xMDEvMTA4Lzk5LzExMS8xMDkvMTAxLzExNi8xMTEvOTcvMTE2LzExNi85Ny85OS8xMDcvOTcvMTEwLzEwMC8xMDAvMTAxLzEwMi8xMDEvMTEwLzk5LzEwMS8xMTkvMTExLzExNC8xMDgvMTAw

然后将解码后的Unicode转化成ASCII,或者直接将其放到html文件中打开

LzExOS8xMDEvMTA4Lzk5LzExMS8xMDkvMTAxLzExNi8xMTEvOTcvMTE2LzExNi85Ny85OS8xMDcvOTcvMTEwLzEwMC8xMDAvMTAxLzEwMi8xMDEvMTEwLzk5LzEwMS8xMTkvMTExLzExNC8xMDgvMTAw

然后在进行一次base64解密

/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

最后ASCII转化成字符串

5.幂数加密

说是幂数加密其实不然，这里我们直接把0看成分隔符，将每一组的数字相加，得到的数字按照字母表映射

s=['88421','122','48','2244','4','142242','248','122']
flag=''
x=0
a={
    
    1:"A",2:'B',3:'C',4:'D',5:'E',6:'F',7:'G',8:'H',9:'I',10:'J',11:'K',12:'L',13:'M',14:'N',15:'O',16:'P',17:'Q',18:'R',19:'S',20:'T',21:'U',22:'V',23:'W',24:'X',25:'Y',26:'Z'}
for i in range(8):
    for j in range(len(s[i])):
        x+=int(s[i][j])
    flag+=a[x]
    x=0
print(flag)

#WELLDONE

得到结果：WELLDONE

6.Railfence

栅栏密码：
number为5

7.easy_RSA

已知p，q，e求d

 
def exgcd(a, n):
    if n == 0:
        return 1, 0
    else:
        x, y = exgcd(n, a % n)
        x, y = y, x - (a // n) * y
        return x, y
 
 
def getReverse(a, n):
    re, y = exgcd(a, n)
    while(re < 0):
        re += n
    return re
 
 
if __name__ == "__main__":
    p = int(input("请输入p:"))
    q = int(input("请输入q:"))
    e = int(input("请输入e:"))
    d = getReverse(e, (p - 1)*(q - 1))
    print('d = ' + str(d))

8.不仅仅是Morse

莫斯解密之后进行bacon解密

may_be_have_another_decodehhhhaaaaabaabbbaabbaaaaaaaabaababaaaaaaabbabaaabbaaabbaabaaaababaabaaabbabaaabaaabaababbaabbbabaaabababbaaabbabaaabaabaabaaaabbabbaabbaabaabaaabaabaabaababaabbabaaaabbabaabba

取decode后面的

#!/usr/bin/python
# -*- coding=utf -*-
import re

table ={
    
    'a': 'aaaaa', 'b': 'aaaab', 'c': 'aaaba', 'd': 'aaabb', 'e': 'aabaa', 'f': 'aabab', 'g': 'aabba',
        'h': 'aabbb', 'i': 'abaaa', 'j': 'abaab', 'k': 'ababa', 'l': 'ababb', 'm': 'abbaa', 'n': 'abbab',
        'o': 'abbba', 'p': 'abbbb', 'q': 'baaaa', 'r': 'baaab', 's': 'baaba', 't': 'baabb', 'u': 'babaa',
        'v': 'babab', 'w': 'babba', 'x': 'babbb', 'y': 'bbaaa', 'z': 'bbaab'}
 
 
def bacon(cipher):
    msg = ''
    codes = re.findall(r'.{5}', cipher)
    for code in codes:
        if code == '':
            msg += ' '
        else:
            UNCODE = dict(map(lambda t: (t[1], t[0]), table.items()))
            msg += UNCODE[code]
    return msg
 
 
if __name__ == '__main__':
    cipher = 'aaaaabaabbbaabbaaaaaaaabaababaaaaaaabbabaaabbaaabbaabaaaababaabaaabbabaaabaaabaababbaabbbabaaabababbaaabbabaaabaabaabaaaabbabbaabbaabaabaaabaabaabaababaabbabaaaabbabaabba'
    plaintext = bacon(cipher)
    print(plaintext)

9.easychallenge

pyc转化成py代码

https://tool.lu/pyc/

解密：

import base64


def decode1(ans):
    s = ''
    for i in ans:
        x=(ord(i)-25)^36
        s += chr(x)

    return s


def decode2(ans):
    s = ''
    for i in ans:
        i=i^36
        x=i-36
        s += chr(x)

    return s


def decode3(ans):
    return base64.b32decode(ans)
final= 'UC7KOWVXWVNKNIC2XCXKHKK2W5NLBKNOUOSK3LNNVWW3E==='
flag=decode1(decode2(decode3(final)))
print(flag)

10.Normal_RSA

使用kali的OpenSSL提取pem文件信息

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

这里的Modulus即为RSA的n的十六进制表示

十六进制转10进制，然后大数分解：
http://www.factordb.com/

275127860351348928173285174381581152299
319576316814478949870590164193048041239

使用rsatool.py生成私钥,e取65537

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

然后再使用OpenSSL解密flag.enc

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

11.轮转机加密

直接暴力轮转，取出每一个轮转的值，然后一个个试flag

import re
sss='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<'
m='NFQKSEVOQOFNP'
#将sss转化成列表,正则表达式，re.S在此模式下"."的匹配不受限制，可匹配所有字符
content=re.findall(r'<(.*?)<',sss,re.S)
iv=[2,3,7,5,13,12,9,1,8,10,4,11,6]
print(content)
lll=[]
for i in range(13):
    index=content[iv[i]-1].index(m[i])
    lll.append(index)
print(lll)
for i in range(0,26):
    flag=""
    for j in range(13):
        flag+=content[iv[j]-1][(lll[j]+i)%26]
    print(flag.lower())

12.easy_ECC

很难的数学问题，没有深入学习，这里值提供代码，需要了解更多请看 https://bbs.pediy.com/thread-253672.htm

import collections
 
 
def inverse_mod(k, p):
    """Returns the inverse of k modulo p.
   This function returns the only integer x such that (x * k) % p == 1.
   k must be non-zero and p must be a prime.
   """
    if k == 0:
        raise ZeroDivisionError('division by zero')
 
    if k < 0:
        # k ** -1 = p - (-k) ** -1 (mod p)
        return p - inverse_mod(-k, p)
 
    # Extended Euclidean algorithm.
    s, old_s = 0, 1
    t, old_t = 1, 0
    r, old_r = p, k
 
    while r != 0:
        quotient = old_r // r
        old_r, r = r, old_r - quotient * r
        old_s, s = s, old_s - quotient * s
        old_t, t = t, old_t - quotient * t
 
    gcd, x, y = old_r, old_s, old_t
 
    assert gcd == 1
    assert (k * x) % p == 1
 
    return x % p
 
 
# Functions that work on curve points #########################################
 
def is_on_curve(point):
    """Returns True if the given point lies on the elliptic curve."""
    if point is None:
        # None represents the point at infinity.
        return True
 
    x, y = point
 
    return (y * y - x * x * x - curve.a * x - curve.b) % curve.p == 0
 
 
def point_neg(point):
    """Returns -point."""
    assert is_on_curve(point)
 
    if point is None:
        # -0 = 0
        return None
 
    x, y = point
    result = (x, -y % curve.p)
 
    assert is_on_curve(result)
 
    return result
 
 
def point_add(point1, point2):
    """Returns the result of point1 + point2 according to the group law."""
    assert is_on_curve(point1)
    assert is_on_curve(point2)
 
    if point1 is None:
        # 0 + point2 = point2
        return point2
    if point2 is None:
        # point1 + 0 = point1
        return point1
 
    x1, y1 = point1
    x2, y2 = point2
 
    if x1 == x2 and y1 != y2:
        # point1 + (-point1) = 0
        return None
 
    if x1 == x2:
        # This is the case point1 == point2.
        m = (3 * x1 * x1 + curve.a) * inverse_mod(2 * y1, curve.p)
    else:
        # This is the case point1 != point2.
        m = (y1 - y2) * inverse_mod(x1 - x2, curve.p)
 
    x3 = m * m - x1 - x2
    y3 = y1 + m * (x3 - x1)
    result = (x3 % curve.p,
              -y3 % curve.p)
 
    assert is_on_curve(result)
 
    return result
 
 
def scalar_mult(k, point):
    """Returns k * point computed using the double and point_add algorithm."""
    assert is_on_curve(point)
 
    if k < 0:
        # k * point = -k * (-point)
        return scalar_mult(-k, point_neg(point))
 
    result = None
    addend = point
 
    while k:
        if k & 1:
            # Add.
            result = point_add(result, addend)
 
        # Double.
        addend = point_add(addend, addend)
 
        k >>= 1
 
    assert is_on_curve(result)
 
    return result
 
 
# Keypair generation and ECDHE ################################################
 
def make_keypair():
    """Generates a random private-public key pair."""
    private_key = curve.n
    public_key = scalar_mult(private_key, curve.g)
 
    return private_key, public_key
 
 
EllipticCurve = collections.namedtuple('EllipticCurve', 'name p a b g n h')
curve = EllipticCurve(
    'secp256k1',
    # Field characteristic.
    p=15424654874903,
    # Curve coefficients.
    a=16546484,
    b=4548674875,
    # Base point.
    g=(6478678675,5636379357093),
    # Subgroup order.
    n=546768,
    # Subgroup cofactor.
    h=1,
)
private_key, public_key = make_keypair()
print("private key:", hex(private_key))
print("public key: (0x{:x}, 0x{:x})".format(*public_key))
print("x + y = " + str(public_key[0] + public_key[1]))

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