[Week 1] Caesar Cipher

• 简单凯撒，偏移量为1
• 0xGame{The_Beginning_Of_Crypto}

[Week 1] RSA-Easy

from Crypto.Util.number import *
from hashlib import md5
def MD5(m):return md5(str(m).encode()).hexdigest()
Pub_Key = (689802261604270193, 620245111658678815)
e = Pub_Key[1]
n = Pub_Key[0]
p = 823642439
q = 837502087
c = 289281498571087475
d = inverse(e,(p-1)*(q-1))
m = pow(c,d,n)
flag = flag = '0xGame{'+ MD5(m) +'}'
print(flag)

• 0xGame{5aa4603855d01ffdc5dcf92e0e604f31}

[Week 1] Code

• 逆向，编写脚本

from Crypto.Util.number import *
from base64 import b64decode
m0 = b'0xGame{73d7'
m1 = 60928972245886112747629873
m2 = '3165662d393339332d3034'
m3 = b'N2YwZTdjNGRlMX0='
m1 = long_to_bytes(m1)
m2 = bytes.fromhex(m2)
m3 = b64decode(m3)
print(m0+m1+m2+m3)

• 0xGame{73d72f64-7656-11ef-9393-047f0e7c4de1}

[Week 1] Code-Vigenere

• 由脚本可以知道key长度为5，恰巧0xGame正好五个字母
• 编写脚本可以得到key是oWccl

alpha1 = 'abcdefghijklmnopqrstuvwxyz'
alpha2 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
m = '0lCcop{oyd94092-g8mq-4963-88b6-4helrxdhm6q7}'
clue = '0xGame'
key = ''
for i in range(len(clue)):
    if clue[i] in alpha1:
        key += alpha1[(alpha1.find(m[i])-alpha1.find(clue[i]))%26]
    elif clue[i] in alpha2:
        key += alpha2[(alpha2.find(m[i])-alpha2.find(clue[i]))%26]
print(key)

• 后续可以使用工具，或者使用脚本

alpha1 = 'abcdefghijklmnopqrstuvwxyz'
alpha2 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
m = '0lCcop{oyd94092-g8mq-4963-88b6-4helrxdhm6q7}'
key = 'oWccl'
key_nums = []
pointer = 0
ans = ''
for i in key:
    if i in alpha1:
        key_nums += [alpha1.find(i)]
    elif i in alpha2:
        key_nums += [alpha2.find(i)]
for i in m:
    if i in alpha1:
        new_index = (alpha1.find(i) - key_nums[pointer]) % 26
        ans += alpha1[new_index]
        pointer = (pointer + 1) % len(key_nums)
    elif i in alpha2:
        new_index = (alpha2.find(i) - key_nums[pointer]) % 26
        ans += alpha2[new_index]
        pointer = (pointer + 1) % len(key_nums)
    else:
        ans += i
print(ans)

• 得到0xGame{acb94092-e8bc-4963-88f6-4fcadbbfb6c7}

[Week 1] RSA-Baby

• 写脚本

from Crypto.Util.number import *from hashlib import md5
def MD5(m):return md5(str(m).encode()).hexdigest()
Pub_Key = (547938466798424179, 80644065229241095)
Prv_Key = (547938466798424179, 488474228706714247)
Encrypt_msg = 344136655393256706
m = pow(Encrypt_msg,Prv_Key[1],Prv_Key[0])
flag = '0xGame{'+ MD5(m) +'}'
print(flag)

[Week 1] Number-Theory-CRT

• hint提示CRT中国剩余定理和二次剩余

• 注意到print(gmpy2.gcd(e,phi))的输出是2，也就是说令$e’=\frac{e}{2}$有$(m^\left. e’\right. )^2\equiv c\ mod\ n$
• 要求解任意模数二次剩余先要求解奇素数模数二次剩余–具体理论查看此链
• 据此推导过程编写脚本

from Crypto.Util.number import *
from hashlib import md5
import gmpy2
import sympy
from sympy.ntheory.modular import crt
def MD5(m):return md5(str(m).encode()).hexdigest()
Pub_Key = (1022053332886345327, 294200073186305890)
c = 107033510346108389
e = Pub_Key[1]
n = Pub_Key[0]
p = 970868179
q = 1052721013
phi = (p-1) * (q-1)
print(gmpy2.gcd(e,phi)) # 发现公约数2,分析得到二次剩余

def find_quadratic_residues(a, p):
    # 首先检查 a 是否是模 p 下的二次剩余
    if not sympy.is_quad_residue(a, p):
        return None  # 如果 a 不是二次剩余，返回 None
    
    # 使用 sympy 的 nthroot_mod 找到一个解
    x = sympy.nthroot_mod(a, 2, p, all_roots=False)
    
    # 计算另一个解
    second_solution = p - x
    
    return (x, second_solution)

x1 = find_quadratic_residues(c,p)       # 求解模p下的二次剩余
x2 = find_quadratic_residues(c,q)       # 求解模q下的二次剩余

for i in x1:
    for j in x2:
        remainders = [i,j]
        mods = [p,q]
        m_ = crt(mods, remainders)[0]   # CRT合并得到模n的二次剩余解
        c_ = m_%n

        e_ = e//2
        d = inverse(e_,phi)
        m = pow(c_,d,n)

        flag = '0xGame{'+ MD5(m) +'}'
        print(flag)

• 注意的是，对于求解二次剩余有两解，所以在合并的时候都可以作为合并用的解
• 所以遍历所有组合，得到正确flag
• 0xGame{127016d0be858ef48a99723710ad4d49}