Hgame-2023-note

WEEK1

RSA

TASK：

from Crypto.Util.number import *

flag = open('flag.txt', 'rb').read()

p = getPrime(512)
q = getPrime(512)
n=p*q
e = 65537
m = bytes_to_long(flag)
c = pow(m, e, n)
print(f"c={c}")
print(f"n={n}")

"""
c=110674792674017748243232351185896019660434718342001686906527789876264976328686134101972125493938434992787002915562500475480693297360867681000092725583284616353543422388489208114545007138606543678040798651836027433383282177081034151589935024292017207209056829250152219183518400364871109559825679273502274955582
n=135127138348299757374196447062640858416920350098320099993115949719051354213545596643216739555453946196078110834726375475981791223069451364024181952818056802089567064926510294124594174478123216516600368334763849206942942824711531334239106807454086389211139153023662266125937481669520771879355089997671125020789
"""

WP：

直接分解N，求解

#HGAME-RSA
from Crypto.Util.number import *
import gmpy2  

n=135127138348299757374196447062640858416920350098320099993115949719051354213545596643216739555453946196078110834726375475981791223069451364024181952818056802089567064926510294124594174478123216516600368334763849206942942824711531334239106807454086389211139153023662266125937481669520771879355089997671125020789
e=65537
c=110674792674017748243232351185896019660434718342001686906527789876264976328686134101972125493938434992787002915562500475480693297360867681000092725583284616353543422388489208114545007138606543678040798651836027433383282177081034151589935024292017207209056829250152219183518400364871109559825679273502274955582
p=11239134987804993586763559028187245057652550219515201768644770733869088185320740938450178816138394844329723311433549899499795775655921261664087997097294813
q=12022912661420941592569751731802639375088427463430162252113082619617837010913002515450223656942836378041122163833359097910935638423464006252814266959128953

phi=(p-1)*(q-1)
d=gmpy2.invert(e,phi)
m=pow(c,d,n)
print(m)
print(long_to_bytes(m))

flag: hgame{factordb.com_is_strong!}

MORSE+

Task:

学校突然放假了，tr0uble正在开开心心的收拾东西准备回家，但是手机铃声突然响起，tr0uble接起电话，但是只听到滴答滴答的声音。努力学习密码学的tr0uble一听就知道这是什么，于是马上记录下来并花了亿点时间成功破译了，但是怎么看这都不像是人能看懂的，还没等tr0uble反应过来，又一通电话打来，依然是滴答滴答的声音。tr0uble想到兔兔也在学习密码学，于是不负责任地把密文都交给了兔兔，兔兔收到密文后随便看了一眼就不屑地说”这么简单都不会？自己解去，别耽误我抢车票”。

Encrypted_message:

5Yeg5Liq5pif5pyf5YmN77yM5oiR5Lus5pS25Yiw5LiA5Liq56We56eY55qE5raI5oGv44CC5L2G5piv6L+Z5Liq5raI5oGv6KKr6YeN6YeN5Yqg5a+G77yM5oiR5Lus5LiN55+l6YGT5a6D55qE55yf5q2j5ZCr5LmJ5piv5LuA5LmI44CC5ZSv5LiA55+l6YGT55qE5L+h5oGv5piv5YWz5LqO5a+G6ZKl55qE77ya4oCc5Y+q5pyJ5YCS552A57+76L+H5Y2B5YWr5bGC55qE56+x56yG5omN6IO95oq16L6+5YyX5qyn56We6K+d55qE57uI54K54oCd44CC

base64解密后得到：

几个星期前，我们收到一个神秘的消息。但是这个消息被重重加密，我们不知道它的真正含义是什么。唯一知道的信息是关于密钥的：“只有倒着翻过十八层的篱笆才能抵达北欧神话的终点”。

使用Audacity解析Morse得到加密信息：

0223e_priibly__honwa_jmgh_fgkcqaoqtmfr

逆序：

rfmtqoaqckgf_hgmj_awnoh__ylbiirp_e3220

W型栅栏密码：

def generate_w(string, n): 
    '''将字符排列成w型'''
    array = [['.']*len(string) for i in range(n)] #生成初始矩阵
    row =0
    upflag = False
    for col in range(len(string)): #在矩阵上按w型画出string
        array[row][col] = string[col]
        if row == n-1:
            upflag = True
        if row == 0:
            upflag = False
        if upflag:
            row -= 1
        else:
            row += 1
    return array

def decode(string, n):
    '''解密'''
    array = generate_w(string, n)
    sub = 0
    for row in range(n): #将w型字符按行的顺序依次替换为string
        for col in range(len(string)):
            if array[row][col] != '.':
                array[row][col] = string[sub]
                sub += 1
    msg = []
    for col in range(len(string)): #以列的顺序依次连接各字符
        for row in range(n):
            if array[row][col] != '.':
                msg.append(array[row][col])
    return array, msg



def crack_cipher(string):
    for n in range(2,len(string)):
        print(str(n)+'栏:'+''.join(decode(string,n)[1]))


if __name__ == "__main__":
    string = "rfmtqoaqckgf_hgmj_awnoh__ylbiirp_e3220"
    crack_cipher(string)
#18栏: rmocfhm_wo_ybipe2023_ril_hnajg_katfqqg

最后维吉尼亚密码解密：

Key: vidar

x def phase5(n1, c1, n2, c2, n3, c3):    r = CRT([c1, c2, c3], [n1, n2, n3])    m = int(r)^(1/3)    print(hex(m)[2:])    n1 = 78642188663937191491235684351005990853149481644703243255021321296087539054265733392095095639539412823093600710316645130404423641473150336492175402885270861906530337207734106926328737198871118125840680572148601743121884788919989184318198417654263598170932154428514561079675550090698019678767738203477097731989c1 = 23419685303892339080979695469481275906709035609088426118328601771163101123641599051556995351678670765521269546319724616458499631461037359417701720430452076029312714313804716888119910334476982840024696320503747736428099717113471541651211596481005191146454458591558743268791485623924245960696651150688621664860n2 = 98174485544103863705821086588292917749386955237408645745685476234349659452606822650329076955303471252833860010724515777826660887118742978051231030080666542833950748806944312437614585352818344599399156268450521239843157288915059003487783576003027303399985723834248634230998110618288843582573006048070816520647c2 = 72080679612442543693944655041130370753964497034378634203383617624269927191363529233872659451561571441107920350406295389613006330637565645758727103723546610079332161151567096389071050158035757745766399510575237344950873632114050632573903701015749830874081198250578516967517980592506626547273178363503100507676n3 = 91638855323231795590642755267985988356764327384001022396221901964430032527111968159623063760057482761918901490239790230176524505469897183382928646349163030620342744192731246392941227433195249399795012672172947919435254998997253131826888070173526892674308708289629739522194864912899817994807268945141349669311c3 = 22149989692509889061584875630258740744292355239822482581889060656197919681655781672277545701325284646570773490123892626601106871432216449814891757715588851851459306683123591338089745675044763551335899599807235257516935037356212345033087798267959242561085752109746935300735969972249665700075907145744305255616​phase5(n1,c1,n2,c2,n3,c3)# 464c41477b325e3872736138633566336366663462633039353334396665633635666332323633653837387dpython

Stream Cipher

Task:

很喜欢李小龙先生的一句话”Be water my friend”，但是这条小溪的水好像太多了。

from flag import flag
assert type(flag) == bytes

key = [int.from_bytes(b"Be water", 'big'), int.from_bytes(b"my friend", 'big')]

def stream(i):
    if i==0:
        return key[0]
    elif i==1:
        return key[1]
    else:
        return (stream(i-2)*7 + stream(i-1)*4)

enc = b""
for i in range(len(flag)):
    water = stream((i//2)**6) % 256
    enc += bytes([water ^ flag[i]])

print(enc)
# b'\x1a\x15\x05\t\x17\tu"-\x06lm\x01-\xc7\xcc2\x1eXA\x1c\x15\xb7\xdb\x06\x13\xaf\xa1-\x0b\xd4\x91-\x06\x8b\xd4-\x1e\xab\xaa\x15-\xf0\xed\x1f\x17\x1bY'

wp:

参考官方wp: 这道题给了⼤家⼀个跑不出来的程序，需要优化程序得到flag。优化⽅法还挺多的，预期是⽤矩阵快速幂的⽅法。

回到本题，根据递归函数可以得到stream的递推式 S[ i ] = 4 ∗ S [ i − 1] + 7 ∗ S [ i − 2]

类似于斐波那契数列，只不过增加了系数。 考虑⽤斐波那契矩阵和矩阵快速幂优化

enc = b'\x1a\x15\x05\t\x17\t\xf5\xa2-\x06\xec\xed\x01-\xc7\xcc2\x1eXA\x1c\x157[\x06\x13/!-\x0b\xd4\x91-\x06\x8b\xd4-\x1e+*\x15-pm\x1f\x17\x1bY'
def mul(a, b):
   c = [[0, 0], [0, 0]]
   for i in range(2):
      for j in range(2):
         for k in range(2):
            c[i][j] += (a[i][k] * b[k][j]) % 256
            c[i][j] %= 256
   return c
 
def power(n):
   if n==1: return key[1] % 256
   if n==0: return key[0] % 256
   res = [[1, 0], [0, 1]]
   A = [[4, 7], [1, 0]]
   while n:
      if n & 1: res = mul(A, res)
      A = mul(A, A)
      n >>= 1
   return (res[1][0] * key[1] + res[1][1] * key[0]) % 256
flag = b''
for i in range(0, len(enc)):
   water = power((i//2)**6)
   flag += bytes([water ^ enc[i]])
print(flag)

from flag import flag        #flag未知

assert type(flag) == bytes   #assert要求表达式的布尔值必须为真，说明flag的数据形式为bytes

key = [int.from_bytes(b"Be water", 'big'), int.from_bytes(b"my friend", 'big')]
#key=[4784265876259235186,2019423192753765707364]
def stream(i):
    if i==0:
        return key[0]   #key[0]=4784265876259235186
    elif i==1:
        return key[1]   #key[1]=2019423192753765707364
    else:
        return (stream(i-2)*7 + stream(i-1)*4)

enc = b""
for i in range(len(flag)):   
    water = stream((i//2)**6) % 256     #地板除向下取整 114 110
    enc += bytes([water ^ flag[i]])

print(enc)
# b'\x1a\x15\x05\t\x17\tu"-\x06lm\x01-\xc7\xcc2\x1eXA\x1c\x15\xb7\xdb\x06\x13\xaf\xa1-\x0b\xd4\x91-\x06\x8b\xd4-\x1e\xab\xaa\x15-\xf0\xed\x1f\x17\x1bY'