2023 0xGame Week1

2023 0xGame
match
发布日期:   2023-10-11
更新日期:   2023-11-27
文章字数:   1.7k
阅读时长:   9 分
阅读次数:  

Crypto

Vigenere

题目

[TOC]

0dGmqk{79ap4i0522g0a67m6i196he52357q60f}

密钥game

flag: 0xGame{79ad4e0522a0a67a6e196be52357e60b}

babyrsa

题目

from Crypto.Util.number import *
from random import getrandbits
from secret import flag

def getN():
	N = 1
	for i in range(16):
		tmp = getPrime(32)
		N *= tmp
	return N

mask = getrandbits(256)
e = 65537
n = getN()
m = bytes_to_long(flag)
c = pow(m*mask,e,n)
print(f'n = {n}')
print(f'e = {e}')
print(f'c = {c}')
print(f'mask = {mask}')


'''
n = 93099494899964317992000886585964221136368777219322402558083737546844067074234332564205970300159140111778084916162471993849233358306940868232157447540597
e = 65537
c = 54352122428332145724828674757308827564883974087400720449151348825082737474080849774814293027988784740602148317713402758353653028988960687525211635107801
mask = 54257528450885974256117108479579183871895740052660152544049844968621224899247
'''

sage脚本

from Crypto.Util.number import *
n = 93099494899964317992000886585964221136368777219322402558083737546844067074234332564205970300159140111778084916162471993849233358306940868232157447540597
e = 65537
c = 54352122428332145724828674757308827564883974087400720449151348825082737474080849774814293027988784740602148317713402758353653028988960687525211635107801
mask = 54257528450885974256117108479579183871895740052660152544049844968621224899247
phi = euler_phi(n)
d = inverse(e,phi)
m = pow(c,d,n)
flag =  int(m) // int(mask)
print(long_to_bytes(flag))
# 0xGame{Magic_M@th_Make_Crypt0}

密码，觅码，先有*再密

题目

from secret import flag #从中导入秘密的flag，这是我们要破解的信息
from Crypto.Util.number import bytes_to_long #从函数库导入一些编码函数
from base64 import b64encode

#hint:也许下列函数库会对你有些帮助，但是要怎么用呢……
from base64 import b64decode
from gmpy2 import iroot
from Crypto.Util.number import long_to_bytes

flag = flag.encode()
lent = len(flag)
flag = [flag[i*(lent//4):(i+1)*(lent//4)] for i in range(4)]#将flag切割成四份

c1 = bytes_to_long(flag[0])
c2 = ''.join([str(bin(i))[2:] for i in flag[1]])
c3 = b64encode(flag[2])
c4 = flag[3].hex()
print(f'c1?= {pow(c1,5)}\nc2 = {c2}\nc3 = {c3}\nc4 = {c4}')

'''
c1?= 2607076237872456265701394408859286660368327415582106508683648834772020887801353062171214554351749058553609022833985773083200356284531601339221590756213276590896143894954053902973407638214851164171968630602313844022016135428560081844499356672695981757804756591891049233334352061975924028218309004551
c2 = 10010000100001101110100010100111101000111110010010111010100001101110010010111111101000011110011010000001101011111110011010011000101011111110010110100110100000101110010010111101100101011110011110111100
c3 = b'lueggeeahO+8jOmCo+S5iOW8gOWni+aIkQ=='
c4 = e4bbace79a8443727970746fe68c91e68898e590a72121217d
'''
#全是乱码，那咋办嘛？

脚本

from base64 import b64decode,b16decode
from gmpy2 import iroot
from Crypto.Util.number import long_to_bytes

c1= 2607076237872456265701394408859286660368327415582106508683648834772020887801353062171214554351749058553609022833985773083200356284531601339221590756213276590896143894954053902973407638214851164171968630602313844022016135428560081844499356672695981757804756591891049233334352061975924028218309004551
c2 = 10010000100001101110100010100111101000111110010010111010100001101110010010111111101000011110011010000001101011111110011010011000101011111110010110100110100000101110010010111101100101011110011110111100
c3 = 'lueggeeahO+8jOmCo+S5iOW8gOWni+aIkQ=='
c4 = 'e4bbace79a8443727970746fe68c91e68898e590a72121217d'
flag1 = long_to_bytes(iroot(c1,5)[0])
print(flag1)
for i in range(25):
    print(str(long_to_bytes(int(str(c2)[i*8:(i+1)*8],2)))[2:-1],end='')
print('\n')
print(b64decode(c3))
print(bytes.fromhex(c4))
# b'0xGame{ \xe6\x81\xad\xe5\x96\x9c\xe4\xbd\xa0,\xe5\xb7\xb2\xe7\xbb\x8f\xe7'
# \x90\x86\xe8\xa7\xa3\xe4\xba\x86\xe4\xbf\xa1\xe6\x81\xaf\xe6\x98\xaf\xe5\xa6\x82\xe4\xbd\x95\xe7\xbc
#
# b'\x96\xe7\xa0\x81\xe7\x9a\x84\xef\xbc\x8c\xe9\x82\xa3\xe4\xb9\x88\xe5\xbc\x80\xe5\xa7\x8b\xe6\x88\x91'
# b'\xe4\xbb\xac\xe7\x9a\x84Crypto\xe6\x8c\x91\xe6\x88\x98\xe5\x90\xa7!!!}'
# 把他们拼一块儿decode
flag = b'0xGame{ \xe6\x81\xad\xe5\x96\x9c\xe4\xbd\xa0,\xe5\xb7\xb2\xe7\xbb\x8f\xe7\x90\x86\xe8\xa7\xa3\xe4\xba\x86\xe4\xbf\xa1\xe6\x81\xaf\xe6\x98\xaf\xe5\xa6\x82\xe4\xbd\x95\xe7\xbc\x96\xe7\xa0\x81\xe7\x9a\x84\xef\xbc\x8c\xe9\x82\xa3\xe4\xb9\x88\xe5\xbc\x80\xe5\xa7\x8b\xe6\x88\x91\xe4\xbb\xac\xe7\x9a\x84Crypto\xe6\x8c\x91\xe6\x88\x98\xe5\x90\xa7!!!}'
print(flag.decode())
# 0xGame{ 恭喜你,已经理解了信息是如何编码的，那么开始我们的Crypto挑战吧!!!}

Take me bag!

题目

from Crypto.Util.number import *
from secret import flag

def encrypt(m):
	m = str(bin(m))[2:][::-1]
	enc = 0
	for i in range(len(m)):
		enc += init[i] * int(m[i]) % n
	return enc
#
w = getPrime(64)
n = getPrime(512)
init = [w*pow(3, i) % n for i in range(512)]

c = encrypt(bytes_to_long(flag))

print(f'w={w}')
print(f'n={n}')
print(f'c={c}')

'''
w=16221818045491479713
n=9702074289348763131102174377899883904548584105641045150269763589431293826913348632496775173099776917930517270317586740686008539085898910110442820776001061
c=4795969289572314590787467990865205548430190921556722879891721107719262822789483863742356553249935437004378475661668768893462652103739250038700528111
'''

脚本

import random
from collections import namedtuple
import gmpy2
from Crypto.Util.number import isPrime, bytes_to_long, inverse, long_to_bytes


w=16221818045491479713
c=4795969289572314590787467990865205548430190921556722879891721107719262822789483863742356553249935437004378475661668768893462652103739250038700528111

nbit=512
N=9702074289348763131102174377899883904548584105641045150269763589431293826913348632496775173099776917930517270317586740686008539085898910110442820776001061
pubkey=[w*pow(3, i) % N for i in range(512)]
L=Matrix(QQ,nbit + 1, nbit + 1)
#构造矩阵L
for i in range(nbit):
    L[i,i]=1

for i in range(nbit):
    L[i,nbit]=pubkey[i]*N

for i in range(nbit):
    L[nbit,i]=1/2

L[nbit,nbit]=c*N


print("LLL start")
res=L.LLL()

for i in range(0, nbit + 1):
    # print solution
    M = res.row(i).list()[:-1]#最后面密文恢复后变成0
    flag = True
    for m in M:
        if m != 1/2 and m != -1/2:#根据破解原理，恢复的明文应只包含-1/2和1/2
            flag = False
            break
    if flag:
        mm=""
        print (i, M)
        for j in M:
            if j==-1/2:#不确定-1/2和1/2哪个代表二进制1
                mm+="1"
            else:
                mm+="0"
flag=mm[::-1]
print(long_to_bytes(int(flag,2)))

What’s CBC

题目

from Crypto.Util.number import *
from secret import flag,key

def bytes_xor(a,b):
	a,b=bytes_to_long(a),bytes_to_long(b)
	return long_to_bytes(a^b)

def pad(text):
	if len(text)%8:
		return text
	else:
		pad = 8-(len(text)%8)
		text += pad.to_bytes(1,'big')*pad
		return text

def Encrypt_CBC(text,iv,key):
	result = b''
	text = pad(text)
	block=[text[_*8:(_+1)*8] for _ in range(len(text)//8)]
	for i in block:
		tmp = bytes_xor(iv,i)
		iv = encrypt(tmp,key)
		result += iv
	return result

def encrypt(text,key):
	result = b''
	for i in text:
		result += ((i^key)).to_bytes(1,'big')
	return result

iv = b'11111111'
enc = (Encrypt_CBC(flag,iv,key))
print(f'enc = {enc}')

enc = b"\x8e\xc6\xf9\xdf\xd3\xdb\xc5\x8e8q\x10f>7.5\x81\xcc\xae\x8d\x82\x8f\x92\xd9o'D6h8.d\xd6\x9a\xfc\xdb\xd3\xd1\x97\x96Q\x1d{\\TV\x10\x11"

脚本

from Crypto.Util.number import *


def bytes_xor(a, b):
    a, b = bytes_to_long(a), bytes_to_long(b)
    return long_to_bytes(a ^ b)

def pad(text):
    if len(text) % 8:
        return text
    else:
        pad = 8 - (len(text) % 8)
        text += pad.to_bytes(1, 'big') * pad
        return text

def decrypt_CBC(ciphertext, iv, key):
    result = b''
    block = [ciphertext[_ * 8:(_ + 1) * 8] for _ in range(len(ciphertext) // 8)]
    for i in block:
        tmp = decrypt(i, key)
        result += bytes_xor(tmp, iv)
        iv = i
    return result

def decrypt(ciphertext, key):
    result = b''
    for i in ciphertext:
        result += ((i ^ key)).to_bytes(1, 'big')
    return result

expected_prefix = b'0xGame'
iv = b'11111111'
ciphertext = b"\x8e\xc6\xf9\xdf\xd3\xdb\xc5\x8e8q\x10f>7.5\x81\xcc\xae\x8d\x82\x8f\x92\xd9o'D6h8.d\xd6\x9a\xfc\xdb\xd3\xd1\x97\x96Q\x1d{\\TV\x10\x11"

for key_candidate in range(256):
    decrypted = decrypt_CBC(ciphertext, iv, key_candidate)
    if decrypted.startswith(expected_prefix):
        print(f"Found key: {key_candidate}")
        print(f"Decrypted flag: {decrypted}")

猜谜

题目

from secret import flag,key
from Crypto.Util.number import *

def dec(text):
	text = text.decode()
	code = 'AP3IXYxn4DmwqOlT0Q/JbKFecN8isvE6gWrto+yf7M5d2pjBuk1Hh9aCRZGUVzLS'
	unpad = 0
	tmp = ''
	if (text[-1] == '=') & (text[-2:] != '=='):
		text = text[:-1]
		unpad = -1
	if text[-2:] == '==':
		text = text[:-2]
		unpad = -2
	for i in text:
		tmp += str(bin(code.index(i)))[2:].zfill(3)
	tmp = tmp[:unpad]
	result = long_to_bytes(int(tmp,2))
	return result

def enc(text):
	code = 'AP3IXYxn4DmwqOlT0Q/JbKFecN8isvE6gWrto+yf7M5d2pjBuk1Hh9aCRZGUVzLS'
	text = ''.join([str(bin(i))[2:].zfill(8) for i in text])
	length = len(text)
	pad = b''
	if length%3 == 1:
		text += '00'
		pad = b'=='
	elif length%3 == 2:
		text += '0'
		pad = b'='
	result = [code[int(text[3*i:3*(i+1)],2)] for i in range(0,len(text)//3)]
	return ''.join(result).encode()+pad

def encrypt(flag):
	result = b''
	for i in range(len(flag)):
		result += (key[i%7]^(flag[i]+i)).to_bytes(1,'big')
	return result

c = enc(encrypt(flag))
print(f'c = {c}')

'''
c = b'IPxYIYPYXPAn3nXX3IXA3YIAPn3xAYnYnPIIPAYYIA3nxxInXAYnIPAIxnXYYYIXIIPAXn3XYXIYAA3AXnx='
'''

脚本

from Crypto.Util.number import *
c = b'IPxYIYPYXPAn3nXX3IXA3YIAPn3xAYnYnPIIPAYYIA3nxxInXAYnIPAIxnXYYYIXIIPAXn3XYXIYAA3AXnx='
def dec(text):
	text = text.decode()
	code = 'AP3IXYxn4DmwqOlT0Q/JbKFecN8isvE6gWrto+yf7M5d2pjBuk1Hh9aCRZGUVzLS'
	unpad = 0
	tmp = ''
	if (text[-1] == '=') & (text[-2:] != '=='):
		text = text[:-1]
		unpad = -1
	if text[-2:] == '==':
		text = text[:-2]
		unpad = -2
	for i in text:
		tmp += str(bin(code.index(i)))[2:].zfill(3)
	tmp = tmp[:unpad]
	result = long_to_bytes(int(tmp,2))
	return result

m = dec(c)
print(m)
flag0 = b'0xGame{'
m0 = m[:7]
def decrypt(flag):
	result = b''
	for i in range(7):
		result += (m0[i%7]^(flag0[i]+i)).to_bytes(1,'big')
	return result
key = decrypt(flag0)
print(key)

def decrypt2(flag):
	result = b''
	for i in range(len(flag)):
		result += ((key[i%7]^(m[i]))-i).to_bytes(1,'big')
	return result
flag = decrypt2(m)
print(flag)
0xGame{Kn0wn_pl@intext_Att@ck!}

WEB

Signin

image-20231005022747302

baby_php

image-20231005153026778

base64解码

hello_http

image-20231005140155552

repo_leak

image-20231005154141340

image-20231005154109355

image-20231005154045555

MISC

SIGNIN

base64 在 base32

0xGame{Welc0me_To_0xGame2023!!}

least and most

stegsolve

image-20231006174446625

image-20231006174515056

拼接一起

another signin

首先我们进行nc

image-20231007162110850

我们需要发送0.001ether到给我们的账户

成功发送

image-20231007162359634

继续nc输入2

image-20231007162458786

给了一个contact address

继续nc输入4查看合约代码

image-20231007162608311

把合约代码复制进Remix

image-20231007162704725

只需要覆盖Love0xGme就行

可以用简单exp或者直接在Remix上操做也行

exp

contract exp{
    address transcation=0x5F709E66Ef718c005cA1b0976372850E45391114;
    Greeter target=Greeter(transcation);
    constructor()payable {}
    function hack() public returns(bool){
        bool ans=false;
        string memory greeting="Love0xGame";
        target.setGreeting(greeting);
        ans=target.isSolved();
        return ans;
    }
}

image-20231007162912150

攻击合约成功之后就nc输入3就行

image-20231007162952849

RE

数字筑基

image-20231005200017950

代码金丹

image-20231005200232473

文章作者: f14g
文章链接: http://example.com/2023/10/11/2023-0xGame-Week1/
版权声明: 本博客所有文章除特別声明外，均采用 CC BY 4.0 许可协议。转载请注明来源 f14g !
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